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•, , Head Office : B-32, Shivalik Main Road, Malviya Nagar, New Delhi-110017, , •, , Sales Office : B-48, Shivalik Main Road, Malviya Nagar, New Delhi-110017, Tel. : 011-26691021 / 26691713, , Page Layout : Prakash Chandra Sahoo, , Typeset by Disha DTP Team, , DISHA PUBLICATION, ALL RIGHTS RESERVED, © Copyright Author, No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the, publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have, tried and made our best efforts to provide accurate up-to-date information in this book., , For further information about the books from DISHA,, Log on to www.dishapublication.com or email to [email protected]

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STUDY PACKAGE IN PHYSICS FOR JEE MAIN & ADVANCED, Booklet No., , Title, , 1, , Units, Measurements &, Motion, , 2, , Laws of Motion and, Circular Motion, , 3, , Work Energy, Power &, Gravitation, , 4, , Rotational Motion, , 5, , Properties of Matter &, SHM, , 6, , Heat & Thermodynamics, , 7, , Waves, , 8, , Electrostatics, , 9, , Current Electricity, , 10, , Magnetism, EMI & AC, , 11, , Ray & Wave Optics, , 12, , Modern Physics, , Chapter Nos., Ch 0. Mathematics Used in Physics, Ch 1. Units and Measurements, Ch 2. Vectors, Ch 3. Motion in a Straight Line, Ch 4. Motion in a Plane, Ch 5. Laws of Motion and Equilibrium, Ch 6. Circular Motion, Ch 7. Work, Energy and Power, Ch 8. Collisions and Centre of Mass, Ch 9. Gravitation, Ch 1. Rotational Mechanics, Ch 2. Properties of Matter, Ch 3. Fluid Mechanics, Ch 4. Simple Harmonic Motion, Ch 5. Thermometry, Expansion &, Calorimetry, Ch 6. Kinetic Theory of Gases, Ch 7. Laws of Thermodynamics, Ch 8. Heat Transfer, Ch 9. Wave – I, Ch 10. Wave –II, Ch 0. Mathematics Used in Physics, Ch 1. Electrostatics, Ch 2. Capacitance & Capacitors, Ch 3. DC and DC circuits, Ch 4. Thermal and Chemical effects of, Current", Ch 5. Magnetic Force on Moving, Charges & Conductor, Ch 6. Magnetic Effects of Current, Ch 7. Permanent Magnet & Magnetic, Properties of Substance, Ch 8. Electromagnetic Induction, Ch 9. AC and EM Waves, Ch 1. Reflection of Light, Ch 2. Refraction and Dispersion, Ch 3. Refraction at Spherical Surface,, Lenses and Photometry, Ch 4. Wave optics, Ch 5. Electron, Photon, Atoms,, Photoelectric Effect and X-rays, Ch 6. Nuclear Physics, Ch 7. Electronics & Communication, , Page Nos., , 1-202, , 203-318, 319-480, 1-120, 121-364, , 365-570, , 571-698, 1-216, 217-338, , 339-618, , 1-244, , 245-384

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Contents, , Contents, , Study Package Booklet 10 - Magnetism, EMI & AC, 5. Magnetic Force on Moving Charges &, Conductor, 339-394, 5.1, 5.2 , 5.3, 5.4 , , Magnetism : an introduction, 340, The magnetic field, 340, Force on a moving charge, 340, Motion of charged particle in uniform, magnetic field, 343, 5.5 Hall effect : 1879, 350, 5.6 Mass spectrograph, 351, 5.7 Cyclotron , 352, 5.8 Magnetic force on current carrying, , conductor, 355, 5.9 Torque on a current loop in, magnetic field, 359, 5.10 Potential energy of magnetic dipole, 359, Review of formulae and important points, 364, Exercise 5.1 - 5.6, 365-382, Hints & solutions, 383-394, , 6. Magnetic Effects of Current, , 395-448, , 6.1 Magnetic field of moving charge, 6.2 The biot savarts law, 6.3 Solenoid , 6.4 Toroid or anchor ring, 6.5 Ampere’s law, Review of formulae and important points, Exercise 6.1 -6.6, , 396, 397, 403, 404, 407, 419, 420-436, , Hints & solutions, , 437-448, , 7. Permanent Magnet & Magnetic, Properties of Substance, 7.1, , Permanent magnet : an introduction, , 449-484, 450, , 7.2 Coulomb’s law in magnetism, , 450, , 7.3 Earth’s magnetism , , 455, , 7.4 Moving coil galvanometer, , 459, , 7.5 Tangent law , , 460, , 7.6 Tangent galvanometer , , 460, , 7.7 Vibration magnetometer, , 460, , 7.8 Deflection magnetometer , , 461, , 7.9 Gauss’s law in magnetism, , 462, , 7.10 Magnetism and electron , , 464, , 7.11 Magnetic properties of the materials, 465, Review of formulae and important points, 471, Exercise 7.1 -7.6, 473-480, Hints & solutions, 481-484, , 8. Electromagnetic Induction, Electromagnetic induction :, an introduction, 8.2 Magnetic flux, 8.3 Faraday’s law of emi, 8.4 Lenz’s law, 8.5 Mechanism of emi, 8.6 Rotating conductor, 8.7 Induced electric field, 8.8 Self inductance, 8.9 Self induction, 8.10 Rl - dc circuit, 8.11 The lc-oscillations, 8.12 Damped oscillations in an rlc circuit, 8.13 Mutual induction, 8.14 Combinations of inductors , 8.15 Eddys current, 8.16 AC generator or dynamo, 8.17 DC generator, 8.18 DC motor, 8.19 Transformer, Review of formulae and important points, Exercise 8.1 - 8.6, Hints & solutions, , 485-576, , 8.1, , 9. AC and EM-Waves, 9.1 Alternating current (ac), 9.2 Average and root mean square, (rms) values, 9.3 Three simple circuits, 9.4 RC-circuit, 9.5 LR-circuit, 9.6 LC-circuit, 9.7 Series RLC-circuit, 9.8 RLC, resonance circuit, 9.9 Power in ac circuits, 9.10 Electromagnetic waves, 9.11 Poynting vector , 9.12 Electromagnetic spectrum, Review of formulae and important points, Exercise 9.1-9.6, Hints & solutions, , 486, 486, 486, 487, 492, 495, 497, 508, 508, 512, 515, 516, 517, 519, 526, 526, 527, 527, 528, 533, 535-560, 561-576, , 577-618, 578, 578, 582, 584, 584, 585, 585, 585, 587, 596, 599, 600, 602, 603-612, 613-618

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340, , ELECTRICITY & MAGNETISM, , 5.1 MAGNETISM :, , AN INTRODUCTION, , The first magnetic phenomenon observed were those associated with naturally occurring, magnets; pieces of iron found near Magnesia (hence the term "magnet"). The study of, magnetic phenomenon remained confined for thousand of years to magnets made in this, way. In 1820 Danish scientist Hans Christian Oersted observed that a compass was, deflected when it placed near a current carrying wire. This made the connection between, electrical and magnetic phenomenon. On this basis Oersted thus demostrated that, magnetic effect could be produced by moving charge. It is now known that all magnetic, phenomena result from forces between electric charges in motion. That is, charges in, motion relative to an observer produces a magnetic field as well as an electric field., , Fig. 5.1, , 5.2 THE, , MAGNETIC FIELD, , In finding the force between the charges, it is very useful to introduce the concept of, electric field and to describe the interaction in two stages :, (i), , Charge creates an electric field E in the space surrounding it;, , (ii) The electric field E exerts a force F qE on a charge placed in the field., We shall follow the same pattern in describing the forces on a moving charge., (i) A moving charge or a current creates a magnetic field in the space surrounding it;, (ii) The magnetic field exerts force on a moving charge or a current in the field., Like electric field, magnetic field is a vector quantity. We shall use the symbol, , B for magnetic field., , 5.3 FORCE, , ON A MOVING CHARGE, , The electric force on a charge does not depend on its velocity; it is same whether the, charge is moving or not. The magnetic force, conversely, is found to have a magnitude, that increases with speed. Further more, the direction of the force depends on the direction, of the field B and the velocity v . The force F always perpendicular to both B and v ., The magnitude of force F is found proportional to the component of v perpendicular, to the field. The force is zero when v and B are parallel or antiparallel. Therefore the, magnetic force, qvB sin, F = qv B qvB, , Direction of force, The direction of magnetic force can be obtained by either of the two rules., 1., , Fig. 5.2, , Right hand rule : If we wrap the fingers of the right hand around the line, perpendicular to the plane of v and B so that they curl around with the sense of, rotation, then thumb points in the direction of force F .

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Fig. 5.3, , Fig. 5.4, In the diagram shown the force on the charge, , 2., , F, In vector notation, we can write, , =, , qvB sin k ., , F, , =, , q v, , B, , Fleming's left hand rule (FLHR) : If we hold fore finger, middle finger and thumb, of the left hand mutually perpendicular to each other; the fore finger shows the, direction of B, the middle finger shows the direction of the velocity of positive, charge, then thumb indicates the direction of force., , More about magnetic field, 1., , 2., , Other name used for magnetic field are :, (i) Magnetic induction (old name)., (ii) Magnetic flux density., Magnetic field can be represented by lines, called lines of field, just as the electric, field. B is related to its line of induction in the following ways :, (i), , The tangent drawn at any point to a line of field gives the direction of B at, that point., (ii) The number of lines per unit area gives an idea about the magnitude of the, magnetic field vector B . This field is large if lines are closer and small if they, are apart., , Fig. 5.5, , 341

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342, , ELECTRICITY & MAGNETISM, The direction of B is represented as follows :, , Fig. 5.6, 3., , F, ., The unit of B : The unit of B can be obtained from, qv, Thus SI unit of B is N/A-m., 1 N/A-m, Also, 1T, , = 1 tesla., = 104 gauss,, = 1 Wb/m2., , Lorentz force, When a charged particle moves through a region of space where both electric and, magnetic fields are acting, the net force on the charged particle is equal to the vector sum, of the forces exert by both the fields. Thus net force, , F, , Eq q v, , =, , B, , This net force is called Lorentz force., , More about magnetic force, 1., , The magnetic force on a charged particle is zero, if either v = 0 or, , 2., , The magnetic force will be maximum, for = 90o, which is F = qvB sin90o = qvB., , 3., , The magnetic force is always perpendicular to the direction of motion of charged, particle, and therefore magnetic force does no work on it. Hence the kinetic energy, of moving charged particle in magnetic field remains constant., , 4., , In using the formula F = qvB sin , put the value of the charge q without sign., , 5., , In using the formula F, , 0 or180 ., , q(v B) , put the value of the charge with the proper, , sign., , Ex. 1, , An electron is moving with a velocity 2 i + 3 j m/s in an, , electric field of intensity 3 i + 6 j + 2k V/m and a magnetic field, of 2 j + 3 k tesla. Find the magnitude and direction of the Lorentz, force acting on the electron., , Sol., , = q E, , v, , = – 1.6 10, , 3i 6 j, , = – 1.6 10, , 19, , 12i, , = 9.6 10, , = cos, , B, 19, , 19, , 3i 6 j, , 19, , 22 12, , It is in xz-plane, making an angle, , Lorentz force is given by, F, , F, , = – 1.6 10, , 2k, , 2i 3 j, , 2 j 3k, , 1, , 2, ., 5, , 6k, , 2k, , 9i, , 6j, , 9.6 10, 2.15 10, , 18, , 4k, , 19, , 2i, , kˆ, , N, , with the x-axis, where, Ans.

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Ex. 2, , 343, , A proton beam moves through a region of space where, there is a uniform magnetic field of magnitude 2.0 T, with direction, along the positive z-axis as shown in fig.5.7. The protons have, , The direction of force is along negative y-axis, thus, , velocity of magnitude 3 ×105 m/s in the xz plane, at an angle 30o to, , Ex. 3, , the positive z-axis. Find force on proton. q = 1.6 × 10-19 C, , -3, particle is projected in a magnetic field of 7.0i - 3.0 j ×10 T . The, , Sol., , acceleration of a particle is found to be, , F, , =, , 4.8 10, , 14, , j N, , Ans., , An experimenter s diary reads as follows : "a charged, , i + 7.0 j × 10-6, , The magnitude of force, F, , m/s2". The number to be left of i in the last expression was not, readable. What can this number be ?, , = qv B, = q v sin 30, , Sol., , B, , As magnetic force is perpendicular to the magnetic field, so, =, , 1.6 10, , 19, , B F = 0, , 1, 2, , 3 105, , 2 Fig. 5.7, , or, , B a = 0., , Let value in box provided is k., =, , 4.8 10, , 14, , N, , 7k, , 3 7 = 0, , k, , 3, , Ans., , 5.4 MOTION OF CHARGED PARTICLE IN UNIFORM MAGNETIC FIELD, The path of charged particle in magnetic field depends on the angle between v and B ., Depending on different values of , the possible cases are :, Case 1 : When is 0° or 180°:, For = 0° or 180°, the force on the moving charge F, , qvB sin 0 or 180, , 0,, , and therefore particle goes undeviated along a straight path., Case 2 : When = 90°:, (i) When particle is projected from inside the field, it experiences a force which, always perpendicular to the velocity and so its path will be circular. The, necessary centripetal force is provided by the magnetic force. If r be the, radius of the path, then, , mv 2, r, or, , =, , qvB sin 90, , r, , =, , mv, qB, , r, , =, , mv, qB, , r, , =, , Fig. 5.8, , or we can write, , or, , Let, , v, q, B, m, , q, , is called specific charge,, m, The equation (1) can be written in the form :, , The K.E. of the particle, , r, , =, , v, B, , K, , =, , P2, 2m, , ...(1), , Fig. 5.9

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344, , ELECTRICITY & MAGNETISM, or, P =, 2mK, If charged particle is accelerated by potential V, then, K = qV, r, , =, , P, qB, , 2mK, qB, , T, , =, , Length of path, speed, , 2mqV, qB, , ...(2), , Time period :, , or, , T, , =, , 2 r, v, , =, , 2 m, qB, , mv, qB, , 2, v, , Also linear frequency of rotation, f, and angular frequency, , =, , 1, T, , =, , 2 f, , qB, ,, 2 m, Bq, ., m, , Note: 1.Time period T, f and are independent of v., 2.The velocity at any instant can be written as, , vx i v y j, (ii) When particle is projected out side the field: If the length of the magnetic, field is enough, then the angle with which the charged particle emerges out, will equal to the angle with which it enters into the field. Thus we have two, cases:, v, , =, , (a)Time spend in magnetic field, t, , =, , T, 2, , m, qB, , PQ = 2r, (b)The time spend in magnetic field, , 2 m, 2, T, T , where T =, qB, 2, PQ = 2r sin, Case 3 : When particle is projected at an angle , which is not equal to 0°, 90° or, 180°:, t, , =, , The velocity of the particle can be resolved into two components; one along B ,, Fig. 5.10, , that is vcos and other perpendicular to B that is vsin . The velocity vsin will, provide circular path and the velocity vcos will provide a straight line. The resulting, path is a helical path., (i), , r, , =, , mv, qB, , mv sin, qB

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, (ii) Since time period of rotation does not depend on velocity, therefore, T, , 2 m, qB, , =, , (iii) Pitch (p) : The centre to centre distance between two consecutive circular, paths is called pitch. Thus, p, , =, , v cos, , T, , =, , v cos, , 2 m, qB, , Fig. 5.11, , 2 mv cos, qB, , Comparison of paths in E - field and B - field, 1., , = 0°, , 1., , = 0°, , 2., , = 90°, , 2., , = 90°, , 3., , 0° <, , < 90°, , 3., , 0° <, , < 90°, , Deviation of charged particle in magnetic field, Suppose a charged particle q enters normally in a uniform magnetic field B . The magnetic, field extends to a distance x, which is less than or equal to radius of the path, that is, x r., (i), , The radius of path, and, , r, sin, , =, , mv, qB, , =, , x, r, , Above relation can be used when x, (ii), , For x, , r, , and deviation,, (iii), , r, , =, , r, , mv, qB, , 180 as clear from the diagram., , If particle moves for time t inside the field, then, =, t, =, , Bq, t, m, , Fig. 5.12, , 345

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346, , ELECTRICITY & MAGNETISM, As, , = q/m, =, , B t, , (iv) Velocity of particle :, We have,, , B0 t ,, , v0, B0, , r, , Velocity of particle at any time t, v =, =, , vx i v y j, v0 cos i v0 sin j, , On substituting the value of , we have, Fig. 5.13, (v), , or, , v, , =, , v0 cos B0 t i v0 sin B0 t j, , Position of particle :, , r, , =, , xi, , =, , r sin i, , r r cos, , =, , r sin i, , 1 cos, , =, , v0, B0, , yj, j, , j, , sin B0 t i, , 1 cos B0 t, , j, , Motion of charged particle in both electric and magnetic, field, Case 1 : When E || B and particle velocity is perpendicular to both the fields, Consider a particle of charge q and mass m projected from the origin with velocity, , v v0 i into a region having electric and magnetic field i.e., E E0 j and B B0 j. The, electric field exerts force only along y-direction and therefore particle accelerates in this, , Fig. 5.17, , Fy, , qE0, . The velocity component in y-direction, m, m, goes on increasing with time. The magnetic field rotates the particle in a circle in xzplane. The resultant path of the particle is a helical path with increasing pitch., The velocity of a particle at any time t would be, direction with the acceleration a y, , v, , =, , vx i v y j v z k, , vy, , =, , ayt, , vz 2, , =, , v02 = constant, , where, , vx, , =, , v0 cos, , v0 cos, , Bqt, m, , and, , vz, , =, , v0 sin, , v0 sin, , Bqt, m, , =, , t, , Here, and, , vx 2, , v = v0 cos, , Bqt, i, m, , qE0 t, m, , Bq, t, m, qE0 t, Bqt, j v0 sin, k, m, m

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, Similarly position vector of the particle, r, , or, , r, , =, , xi, , =, , r sin, , =, , y j zk, 1, a yt 2 j, 2, , i, , r r cos, , mv0, Bqt, i, sin, Bq, m, , k, , 1 qE 2, t j, 2 m, mv0, Bq, , 1 cos, , Bqt, m, , k, , Note:, While moving on a helical path the particle touches the y-axis after every T second,, where T, , 2 m, ., qB, , Case 2: When E B and the particle is released at rest : Consider a particle of, charge q and mass m, is placed at the origin with zero initial velocity into a region of, uniform electric and magnetic field. Let field E is acting along x-axis and field B is, along y-axis, that is E, , B0 j ., , E0 i and B, , The electric force accelerates the particle along x-axis and so, the particle starts gaining, velocity along x-axis. As soon as particle starts moving, the magnetic force starts acting, and bends the particle. The resulting motion of particle is in xz- plane., At any instant its velocity, v, , = vx i vz k, , F, , =, , q E v, , B, , =, , q E0 i, , v x i vz k, , =, , qE0 i qvx B0 k, , F, , =, , q E0, , a, , =, , F, m, , ax, , =, , dvx, dt, , q, E0, m, , az, , =, , dvz, dt, , q, v x B0, m, , The resultant force is thus given by, , or, , B0 j, , qvz B0, , i, , vz B0 i qv x B0 k, , ...(i), , and acceleration, q, E0, m, , vz B0 i, , q, v x B0 k, m, , ...(ii), , This acceleration has two components, , and, , Differentiating equation (iii) w.r.t. time, we get, d 2 vx, dt, , 2, , =, , dvz, q, B0, m, dt, , v z B0, , ...(iii), ...(iv), , Fig. 5.18, , 347

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , or, , vz, , =, , qE0, 1 cos t, m, , =, , qE0, qB0, m, m, , =, , E0, 1 cos t, B0, , =, , E0, sin t, B0, , =, , E0, B0, , 349, , 1 cos t, , ...(ix), , From equation (viii), we have, , dx, dt, or, , x, , or, , x, , t, , sin t dt, 0, , =, , E0, B0, , =, , E0, 1 cos t, B0, , =, , E0, 1 cos t, B0, , =, , E0, B0, , t, , cos t 0, ...(1), , Also from equation (ix), we have, , dz, dt, or, , z, , Fig. 5.19, , t, , 1 cos t dt, 0, , E0, ...(2), t sin t, B0, Equations (1) and (2), together represents a cycloid which is defined as the path generated, by the point on the circumference of a wheel rolling on the ground., or, , z, , =, , Ex. 4, , When a proton is released from rest in a room, it starts, with an initial acceleration a0 towards west. When it is projected, towards north with a speed of v0, it moves with an initial acceleration, 3a0 towards west. Find the electric field and the minimum possible, magnetic field in the room., , Sol., When proton released, it experiences only electric force, so, Fe = ma0 = Eq, E, , =, , ma0, q, , ma0, , towards west, e, , Ans., When proton is projected, it experiences force due to both the fields, and, so, Fe + Fm = m (3a0), As, Fe = ma0,, Fm = 2ma0., The magnetic force on the proton will be ev0B, therefore, ev0B = 2ma0,, , B, , =, , 2ma0, downward, ev0, , Ex. 5, , Ans., , A particle of mass m and charge q is projected into a, region having a perpendicular magnetic field B. Find the angle of, deviation (fig. 5.14) of the particle as it comes out of the magnetic, field if the width d of the region is very slightly smaller than, , (a), , mv, qB, , (b), , mv, 2qB, , (c), , Fig. 5.14, , 2mv, qB

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350, , ELECTRICITY & MAGNETISM, , Sol. The radius of path, For d, , (a), , r , we have, , d=, , mv, qB, , r, , sin, , r,, , sin, , or, (b), , d=, , mv, 2qB, , d=, , r , sin, , d, r, , Fig. 5.15, Show that these paraxial electrons are refocused on the x-axis at a, , mv, qB, , =, , 2, , =, , mv, 2qB, mv, qB, , 6, , distance x =, , Sol., , radian, , d, r, , 2, , 1, , mv, qB, , =, , =, , 2mv, qB, , mv, qB, , =, , =, , or, (c), , =, , mV, , eB 2, , ., , The velocity of electrons as they emerge out,, 1 2, mv, 2, , Ans., or, , =, , 2eV, m, , The velocity can be resolved into two perpendicular components, vcos, along x-axis and vsin, to it. The time taken to complete a circle, , 1, 2, , radian, , v, , = eV, , 2 m, eB, In this time the distance travelled by electrons along x-axis, x = v cos, T, , T, , Ans., , =, , 2eV, m, , =, , r , the deviation of particle is therefore = radian., , 2 m, cos, eB, , for small , cos = 1, , Ans., , Ex. 6, , Electrons emitted with negligible speed from an electron, gun are accelerated through a potential difference V along the xaxis. These electrons emerge from a narrow hole into a uniform, magnetic field B directed along this axis. However, some of the, electrons emerging from the hole make slightly divergent angles, as shown in fig. 5.15., , 5.5 HALL, , EFFECT, , Fig. 5.16, x, , =, , 8, , 2, , mV, , eB 2, , Proved, , : 1879, , According to Hall, if a current carrying conductor is placed in a transverse magnetic, field, an emf is setup across the conductor perpendicular to both current and magnetic, field. This was first observed by Hall, and therefore is called Hall effect., , Fig. 5.20, Suppose we have a steady current i flowing in a uniform conducting strip along x-axis. If, the strip is placed in a uniform magnetic field B, acting along the y-axis, the field B, will

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, exert a deflecting force on electrons along z-axis and on positive ions along negative zaxis. Because of this force, the electrons displace towards front face, ABCD and positive, ions on back face PQRS. The displacement of charge carriers gives rise to a transverse, field, known as Hall electric field which acts along positive z-axis, and opposes the side, ways drift of the carriers. When equilibrium is reached in which the magnetic deflecting, force on the charge carriers is just balanced by the electric force caused by Hall electric, field. Thus, , or, , evdB, , = E He, , EH, , = vd B, , VH, , =, , ...(1), , The Hall potential, EH, , b, , = vd B b, Also, , vd, , As, , or, , 5.6 MASS, , A, , =, , i, neA, , = b ,, , VH, , =, , i, Bb, neb, , VH, , =, , iB, ., ne, , ...(2), , SPECTROGRAPH, , It is used to find mass number, to determine the relative abundance of isotopes or to, produce separated isotopes., An instrument in which we use a photographic plate and obtain a series of lines on it are, known as mass spectrographs. Another instrument in which an electric or magnetic field, is adjusted to make each part of the spectrum in turn to fall on a fixed detecting slit and, measured electrically are called mass spectrometers., All mass spectroscopes start with an ion source where the ions are produced by electron, bombardment of gases. The ions are first set in motion with the help of an accelerating, potential., In the spectrograph designed by Bainbridge, a velocity selector is used to obtain ions, of particular velocity v. The velocity selector allows a beam of a positive ions having the, same velocity v to pass undeviated through crossed electric and magnetic fields E and B., In this case forces due to these fields are equal and opposite. i.e.,, , or, , Fe, , = Fm, , qE, , = qvB, , v, , =, , E, B, , … (i), , All the ions having the same velocity, v = E/B enter the analysing chamber through slit., In this chamber another magnetic field B is applied perpendicular to the plane of paper, and in outward direction. Due to this field, ions of different masses move in circles of, different radii such that, , Fig. 5.21 Mass Spectrograph., , 351

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352, , ELECTRICITY & MAGNETISM, mv 2, r, or, , = qvB, , m, , =, , qrB ', v, , ...(ii), , Assuming equal charges on each ion, the mass of each ion is proportional to the radius, of its path. Ions of different isotopes converse at different points on the photographic, plate. The relative abundance of the isotopes is measured from the densities of the, photographic images they produce. For two isotopes of masses m1 and m2, we have, , m1, m2, , =, , r1, ., r2, , 5.7 CYCLOTRON, Cyclotron is a device which is used to accelerate positive particles like -particle, deuteron, etc., It is based on the fact that the electric field accelerates a charged particle and the magnetic, field keeps it revolving in circular orbits of increasing radius., It consists of two hollow D-shaped metallic chambers D1 and D2 called dees. The dees, are connected to the source of high frequency electric field. The whole apparatus is, placed between the two poles of a strong electromagnet N–S as shown in Fig. 5.22. The, magnetic field acts perpendicular to the plane of the dees., (i), , Cyclotron frequency : Time taken by charged particle to describe a semicircular, path,, t, , Fig. 5.22, , =, , r, v, , m, qB, , 2t, , 2 m, qB, , The period of oscillating electric field, T, , =, , 1, Bq, ., T 2 m, Maximum kinetic energy of the particle :, and cyclotron frequency f, , (ii), , (a) We have, , r, , =, , mv, qB, , for, , r0, , =, , mv0, qB, , v0, , r0 qB, m, , (r0, maximum radius of circular path), K.E., , =, , 1, mv0 2, 2, , r qB, 1, m 0, 2, m, , (b) Also, K.E. = work done by electric source, K.E., , = f qV, = 2fqV., , 2, , 2, , q2 B2, r02, 2m

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Ex. 7, , Two positive ions having the same charge q but different, masses, m1 and m2 are accelerated horizontally from rest through, a potential difference V. They then enter a region where there is, uniform field B normal to the plane of the trajectory.., (a) Show that, if the beam entered the magnetic field along xaxis, the value of the y-coordinate for each ion at any time t is, approximately, y = Bx, (b), , q, 8mV, , 2, , Now, , (b), , From equation (iii) m, , 1, 2, , provided x remain much smaller than y., Can this arrangement be used for isotope separation ?, , Sol., (a), , 1, y =, 2, , The path of ion is shown in fig. 5.23. The magnetic force provides, an acceleration which is towards the centre of path C. At any, instant the acceleration, , qBx 2, 2qB, m, m, , Bx, , 2, , q, 8mV, , 353, , 1, 2, , 1, y, , y2, m1, ., =, y1, m2, , Ex. 8, , A particle of charge +q and mass m moving under the, , influence of a uniform field E i and a uniform magnetic field Bk, follows a trajectory from P to Q. The velocities at P and Q are v i and, –2v j . Find the rate of doing work by the fields., , Sol. Magnetic field does no work on the moving charged particle. It, bends the path of the particle. Electric field does work on the particle., Due to which its speed changes., , Fig. 5.23, ac =, and, , F, m, , qvB, m, , ay = accos, , 0, cos, , for small x,, ay, For, , ac =, , Fig. 5.24, Workdone by electric field =, , 1, , or, , qvB, m, , x = vt,, y =, , ...(i), , 1, a yt 2, 2, , ...(ii), , From above equations, we get, y =, , =, , 1, x, ay, 2, v, 1 qBx 2, 2 mv, , or, , or, , Which gives, , v =, , 2qV, ., m, , Eq, , 2a =, , (Eq) =, , K.E., 1, m 2v, 2, , 2, , v2, , 3mv 2, 4a, , Now rate of doing work at P, 2, , x2, , 1 qvB, 2 m, , = Fe v, , v2, , =, , 3mv 2, 4a, , =, , 3 mv3, ., 4 a, , ...(iii), , Since ions are accelerated through potential V,, 1 2, mv = qV, 2, , Fe s =, , K.E, , Ex. 9, , Eq, , v cos0, , v, , Ans., , A circular wire loop of radius r can with stand a radial, force T before breaking. A particle of mass m and charge q (q > 0) is, sliding over the wire. A magnetic field B is applied normal to the, plane of the wire. What maximum speed vmax the particle can have, before the loop breaks ?

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354, , ELECTRICITY & MAGNETISM, , Sol., The centripetal force available is, = (T + qvB), , or, , 4 10 5 i =, , 4 10, , 9, , 2 104 k, , or, , 4 10 5 i =, , 4 10, , 9, , 2 10 4, , or, , 4 10 5 i = 8 10, , 5, , Bx j, , Bx i, Bx, , By j, j, , By, , i, , By i, , On comparing both sides of above expression, we get, Bx = 0 and By = – 0.5, Fig. 5.25, , B = 0, , Thus, , From Newton s second law, , or, , mv 2, r, , (T + qvB) =, , mv 2, r, , qvB T, , 0, , =, , =, , 0.5 j T, , Ans., , Ex. 11 A proton accelerated by a potential difference V = 500, kV flies through a uniform transverse magnetic field with induction, B = 0.51 T. The field occupies a region of space d = 10 cm thickness, (see fig. 5.27). Find the angle through which the proton deviates, from the initial direction of motion., , Solving above equation for v, we get, , r, v max = 2m qB, , Ex. 6, , 0.5 j T, , q2B2, , 4Tm, r, , .Ans., , A particle carries a charge of 4 × 10-9 C. When it moves, , with a velocity v1 of 3 ×104 m/s at 45o above the x-axis in the xyplane, a uniform field exerts a force F1 along negative z-axis. When, the particle moves with a velocity v2 of 2 × 104 m/s along the z-axis,, , Fig. 5.27, , Sol. If v is the speed of the proton, then, , there is a force of 4 × 10-5 N exerted on it along the x-axis. What, , 1 2, mv = qV, 2, , are the magnitude and direction of the magnetic field ?, , Sol., Given,, , 2qV, ., m, Here q and m are the charge and mass of the proton respectively., The proton traverses circular path of radius R in the magnetic field,, where, or, , v1 =, , 3 104 cos 45 i sin 45 j, , = 3 104, , i, , j, , 2, , 2, , v =, , R =, , mv, qB, , From the figure, sin, , =, , d, R, d, , =, , m, , 2 qV, m, qB, , q, Fig. 5.28, 2mV, On substituting the values, we get = 30o, = Bd, , Fig. 5.26, In first case;, The force acts along z-axis, so field may be, B = Bx i, , B y j., , F2 = q v, , B, , In the second case,, , Ex. 12, , Ans., , A particle with specific charge q/m moves rectilinearly, due to an electric field E = E0–ax, where a is a positive constant, x, is the distance from the point where the particle was initially at, rest. Find, (a) the distance covered by the particle till the moment it come, to a stand still;, (b) the acceleration of the particle at that moment.

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Sol., , The force exerted by the electric field on the particle, F = Eq = q (E0 – ax)., Acceleration of the particle, , v, , dv, q, E0, =, dx, m, , ax, , Sol., , ...(i), Fig. 5.29, , Upon integrating, we have, q, 1 2, E0 x, v =, m, 2, Given at x = 0, v = 0;, , (a), , C, , 1, mv02 = qV, 2, , 2q, E0 x, m, , 1 2, ax, 2, , ...(ii), , 2q, E0 x, m, , 1 2, ax, 2, , x = xm, , q, = m E0, , =, , a, , v0 =, , Ans., , ...(i), 0, , v0, , n 1, , Ans., , particles accelerated by a potential difference V propagates from a, point A along the axis of a straight solenoid. The beam is brought, into focus at a distance from the point A at two successive values, of magnetic induction B1 and B2. Find the specific charge q/m of the, , . In this duration, let, , 2 m, = n qB, 1, =, , qE0, ., m, , v0, , particle completes n circles for B1 and (n + 1) for B2, then, , 2E0, a, , Ex. 13 A slightly divergent beam of non-relativistic charged, , On solving, , n = B, 2, , 2 m, qB2, , ...(ii), , B1, B1, , Substituting this values of v0 and n in equation (ii) and solving, we get, 2qV, m, , particles., or, , 5.8 MAGNETIC, , 2qV, m, , If t is the time to travel a distance , then t, , 2 E0, ., a, The corresponding acceleration, from equation (i), , or, , or, , For slightly divergent particle,, v0cos, v0., , The velocity of the particle again will be zero., 0 =, , (b), , 1 2, ax, 2, , Suppose v0 is the velocity of the particle, then, , C = 0., v2 =, , Thus, , 355, , q, m, , =, , =, , B1, B2, , B1, , 2 m, qB1, , 8 2V, 2, , B2, , B1, , 2, , ., , FORCE ON CURRENT CARRYING CONDUCTOR, , When a current carrying conductor is placed in a magnetic field, magnetic force is acted, upon the moving charges (electrons) within the conductor. This force is transmitted to, the material of the conductor, and the conductor as a whole experiences a force distributed, along its length., Suppose a conducting wire of length carrying current i, lies in a magnetic field B (see, fig. 5.30)., Consider a small element of length d of the wire. The free electrons drift with a speed, vd opposite to the direction of the current. We know that i = jA = neAvd. Here A is the area, of cross section of the wire and n is the number of free electrons per unit volume. Each, electron experiences an average magnetic force, , fm, , =, , e vd, , B, , The number of free electrons in the element is nAd . Thus the magnetic force on the wire, is, , dFm, , =, , nAd, , fm, , Fig. 5.30, , Ans.

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356, , ELECTRICITY & MAGNETISM, nAd, , =, , e vd, , If we take d along the direction of the current, then vd d, , B, = vd d, , and the above, , expression becomes, , or, , Quantity id, , dFm, , =, , nAevd d, , dFm, , =, , i d, , =, , id, , =, , i, , B, , B, B, , is called current element., , The force on the wire of length, , Fm, , B., , The above equation can be used in two stages, To find the magnitude of force by F = Bi sin and the direction of force can be obtained, by Fleming left hand rule or by right hand palm rule., , Force on curved conductor, Let us consider a conducting wire of arbitrary shape and is placed in uniform magnitude, field B ., The force on d, , length of the conductor dFm = id, B . To get force on the whole wire,, integrate above equation over the length of the wire. Thus, Q, , Fm, , id, , =, , B, , P, Q, , = i, , Fig. 5.31, , =, , d, , B, , P, , i PQ B., , The other simpler way to get the force on current carrying wire is : draw straight line, joining the ends of the conductor (here PQ), and then find its component perpendicular, to B , here it is PQ sin . Therefore, F, , = Bi(PQ sin )., , Ex. 14 Find the force on the conductors placed in uniform, magnetic field as shown in the figure., Sol., (c), , (a), , (b), , (d), , F = Bi × PQ = Bi, direction of force along + y-axis, , F = Bi(PQsin ),, direction of force along + z-axis, , PQ = 0, F = Bi × 0 = 0, , F = Bi (PQ),, direction of force along + y -axis, (perpendicular to line PQ)

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , (e), , (f), , = 2 BiR, , cos, , = 2 BiR, , cos, , = Bi, F=0, , or, , F = Bi(PQ), direction of force perpendicular, to line PQ., , 357, , /2, 0, , 2, , cos 0, , 2R, , F = Bi (PQ), , Ans., , The direction of force is perpendicular to line PQ. i.e., along y-axis., , Ex. 16, , Find in brief the force on the conductor shown in fig., , 5.33., , Sol., (g), , (h), (a), , F, , Note:, , 0, , F, , Bi kˆ, , Magnetic force on any close loop in a uniform field acting, , perpendicular to plane of loop always be zero., , Ex. 15, , In the fig. 5.32 shown a semi-circular wire loop of, , (b), , radius R is placed in a uniform magnetic field B. The plane of the, loop is perpendicular to the magnetic field. Find magnetic force on, the loop., Fig. 5.33, , Sol., Consider an element of length d l = Rd in segment PS. The force on it dF, = Bid l = Bi(Rdq) the direction of force is normal to d . Similar element, is taken in segment QS, and will have the same force., , Ex. 17 A circular loop of radius a, carrying a current i, is placed, in a two dimensional magnetic field. The centre of the loop concides, with the centre of the field as shown in Fig. 5.34. The strength of, the magnetic field at the periphery of the loop is B. Find the magnetic, force on the wire., Sol., , Consider an element of length d , the force on it dF = Bi(d ), , sin90o = Bi(d ) and perpendicular to d and B, that is along z-axis. For, any of the element of the loop, the direction is along z-axis. Therefore, force on entire loop, 2 a, , F, , Bi (d ) k, , =, 0, , Fig. 5.32, 2 a, , The resultant force on these elements, = Bi, , = 2dF sinq, = 2 Bi(Rdq)sinq, The resultant force on the entire loop, , or, , F, , Ex. 18, , 2, , dF, , F =, , d k, 0, , = Bi 2 a k, , Fig. 5.34, , Ans., , A hypothetical magnetic field existing in a region is, , given by B = B0er , where er denotes the unit vector along the, , 0, , 2, , = 2 BiR sin d, 0, , radial direction. A circular loop of radius a, carrying a current i, is, placed with its plane parallel to the xy- plane and the centre at, (0, 0, d). Find the magnitude of the magnetic force acting on the, loop.

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358, , ELECTRICITY & MAGNETISM, , Sol. The situation is shown in the fig. 5.35., sin, , =, , a, a, , 2, , d2, , The given field can be resolved into two components: B0sin , which acts, all round the periphery of loop and lie in the plane of the loop and other, component B0cos acts all round the periphery and perpendicular to the, plane of the loop., , Fig. 5.38, Since net force on the entire loop is zero, so the net force on its, element also be zero. Therefore the equilibrium of element gives, , 2T sin, for small, , , sin, , 2T, which gives, , 2, , = Bi(d ), , 2, , 2, , = Bi a, 2, T = Bia, , [d = a ], Ans., , Ex. 20 The magnetic field existing in a region is given by, B = B0 1 +, , Fig. 5.35, , x, , k., , A square loop of edge and carrying a current i, is placed with its, edges parallel to x-y axis. Find the magnitude of the net magnetic, force experienced by the loop., , Sol., Magnetic field at x = 0,, , B1 = B0 k, and at x = ,, Fig. 5.36, Force on the loop due to B0cos is zero. The force is only due to B0sin ,, which is, F =, , B0 sin, , i 2 a, , or, , F =, , a, , 2 Bi a 2, a2, , d2, , 2, , d, , The situation is shown in the fig. 5.39., The forces on the sides BC and DA are equal and opposite and get, cancelled., The force on side AB,, , F1 = B0i j, And the force on side CD,, , a, , = 2Bi a, , B2 = 2B0 k, , 2, , F2 =, ., , 2B0i j, , Ans., , Ex. 19 Fig. 5.37, , shows a circular wire-loop of radius a, carrying, a current i, placed in a perpendicular magnetic field B., (a) Take a small part d of the wire. Find force on this part of the, wire exerted by the magnetic field., (b) Find the force of compression in the wire., , Sol., Fig. 5.39, Therefore net force on the loop, F, , (a), (b), , Fig. 5.37, The force on the element by magnetic field dF = Bi(d ), towards, the centre of loop (By FLHR)., Let us now consider an element, cutout from the loop. The force, exerted by rest part of loop on the element is shown in fig. 5.38., Let this force is T., , = F1, =, , or, , F2, , B0i j, , F = B0i ., , Ans.

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , 5.9 TORQUE, , ON A CURRENT LOOP IN MAGNETIC FIELD, , Let us consider a rectangular loop a b is placed in a magnetic field B. The normal of, loop makes an angle with the direction of B ., The force exerted by the magnetic field on four sides of the loop is shown in fig. 5.41. On, sides AB and CD, the force Biacos is equal and opposite and their line of action is same., While on sides BC and AD, the force on each is Bib and opposite in directions. Their line, of actions are at a separation of a sin . Therefore the net force on entire loop, Fnet = 0., But there is a net torque which is, =, , Bib a sin, , =, , Bi a b sin, , Fig. 5.40, , = BiAsin, If coil consists of N turns, the torque on entire coil is, = (NiA) B sin, or, = MB sin, where M = NiA, is known as magnetic moment of the loop of N-turns., In vector notation, it can be written as, =, , 5.10 POTENTIAL, , M, , ...(1), , Fig. 5.41, , ...(2), , B, , ENERGY OF MAGNETIC DIPOLE, , When the current loop or a magnetic dipole is placed in magnetic field B, work must be, done by an external agent to change the orientation of the dipole. This work done by, agent becomes the potential energy of the magnetic dipole. Magnetic energy is assumed, to be zero when M is, , to B. If the dipole is rotated through an angle, , from zero, , energy position, then, U, , = Wagent, , d, 90, , =, , MB, , sin d, 90, , or, , U, , = –MB cos, , or, , U, , =, , Note:, , M B, , Above results are derived for rectangular loop. But they can be used for a, , plane loop of any shape., 1., , 2., , = iA = i R2, = MB sin = (i R2) Bsin, F net = 0, For solenoid : If windings are closely spaced, the solenoid can be approximated by, a number of circular loops., = NiAB sin ., For circular loop :, and, , M, , Fig. 5.42, , 359

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360, , ELECTRICITY & MAGNETISM, Analogy between electric dipole and magnetic dipole, Electric dipole, , Magnetic dipole, , 1., , P = q, direction from –q to +q, , 1., , M = iA, direction from S to N, , 2., , Fnet = 0, , 2., , Fnet = 0, , 3., , =M, , 3., , = P E, , 4., , U = –PE cos, , B, , U = –MB cos, , More about magnetic moment, , M, , (a), , The direction of M is the direction of thumb of the right hand if the fingers of this, curl around the loop are in the direction of the current., , (b), , Direction of M can also be determined as in the case of electric dipole, the dipole, moment P has a direction from negative to positive charge. In the similar way, direction of M is from south magnetic pole to north magnetic pole. The south and, north poles can be determined by the sence of flow of current. The side from where, the current seems to be clockwise becomes south pole [S] and the opposite side, from it seems anticlockwise becomes north pole [N]., , Fig. 5.44(a), , (c), , M of a rectangular square loop can also the determined by this method, M = iA, = i AB BC, = i BC CD, , Fig. 5.44(b), , = i CD DA, = i DA AB, Gyro-magnetic Ratio, , Fig. 5.45, , M, : The ratio of angular momentum to magnetic moment is called, C, , Gyro-magnetic ratio., , Ex. 21 The fig. 5.43 shows four orientations, at an angle, magnetic dipole moment M in a magnetic field. Find :, (a) the magnitude of the torque on the dipole and, , 3,, 4,, , , of a, , (a), , 3, 4, , = (180° + ), = (360° – ), , Torque on the dipole is given by = MB sin ., = MB sin, , Sol., , 1, 2, , = MB sin (180° – ) = MB sin, , For the dipole, , 3, , = MB sin (180° + ) = –MB sin, , 4, , = MB sin (360° – ) = –MB sin, , (b), , the potential energy of the dipole., , (b), , Potential energy of a dipole in magnetic field is given by, U = –MB cos ., U1 = –MB cos, U2 = –MB cos (180° – ) = MB cos, , Fig. 5.43, 1,, 2,, , 1, 2, , =, = (180° – ), , U3 = –MB cos (180° + ) = MB cos, U4 = –MB cos(360° – ) = –MB cos

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361, , MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Ex. 22, , A square loop of side carries a current i. It is placed, , as shown in Fig. 5.46. Find the magnetic moment of the loop., , Sol., , Method I :, , The magnetic moment of each loop, M = iA, = i 2, Their magnetic moments are perpendicular to each other. Hence, , M net =, , 2M, , 2i, , 2, , ., , Ans., , Ex. 24, , A thin uniform ring of radius R carrying uniform charge, q and mass m rotates about its axis with angular velocity . Find, the ratio of its magnetic moment and angular momentum., , Sol., , The equivalent current in the ring, , Fig. 5.46, where, and, , M = iAn, M = i 2, , n = sin 30 i cos 30 j, =, , Method II :, Here, , i, 2, , 3j, 2, 3i, , =, , M, , = i OC CB, , 2, , CB =, M, , Magnetic moment, , q, 2, , q, T, , q, 2, , i, 2, , 3, j, 2, , Angular momentum, , k, , = i, , = i, , 2, , i, , 2, , 2, , M = iA, =, , q, 2, , =, , q R2, 2, , sin 60 j, , 3 j, 2, , 2, , =, , j, , cos 60 i, , i, = 2, and, , i =, , 2, , i, , M, , OC =, , Fig. 5.48, , i, 2, , 3, j, 2, , j, 2, , 3, i, 2, , 3i, , j ., , carrying loop ABCDEFA. Each side of the loop is, the loop is i., , =, , Ans., , A positive charge q is distributed over a circular ring of, radius a. It is placed in a horizontal plane and is rotated about its, axis at a uniform angular speed . A horizontal magnetic field B, exists in the space . Find the torque acting on the ring due to the, magnetic force., Ans., , Sol., , and current in, , Sol. The given loop can break into two loops to get magnetic moment, , We know that, Here, , Fig. 5.49, = MB sin, = 90o, = MB = i AB, =, , =, Fig. 5.47, , q, ., 2m, , Ex. 25, , Ex. 23 Find the magnitude of magnetic moment of the current, , that is in ABEFA and BCDEB., , L = I, = (mR 2), , M, L, , k, , R2, , q, 2, , a2 B, , 1, q a2B ., 2, , Ans.

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362, , ELECTRICITY & MAGNETISM, , Ex. 26, , A thin insulated wire forms a plane spiral of N light, turns carrying a current i. The radii of inside and outside turns are, equal to a and b. Find the magnetic moment of the spiral with a, given current., , Sol. Consider dr width of the spiral, at a distance r from the centre., Number of turns in this, , (b), , Fig. 5.51, Work done by the agent is given by, 2, , W =, , 2, , MB sin d, , d, 1, , = MB cos, , 1, , 1, , cos, , 2, , Given 1= 0°, 2 = 90°, W = MB [cos0°–cos90°], = MB, = N i AB, , Fig. 5.50, dN =, , =, , N, b, , a, , = 5.4 10, , dN iA, , =, , =, , Ni, 3b a, , 6, , J, , 5.4, , J, , Suppose the disc is placed in xy-plane and is rotated about the z-axis., Consider an annular ring of radius r and of thickness dr, the charge on this, ring, , b, , dr i, , r, , 2, , a, b, , r 2 dr, a, , b3, , a3 ., , Ans., , Ex. 27 Fig. 5.51 shows a circular coil with 250 turns, an area A, of 2.52 × 10 -4 m 2 , and a current of 100 A. The coil is at rest in a, uniform magnetic field of magnitude B = 0.85 T with its magnetic, dipole moment M initially aligned with B., (a) What is the direction of the current in the coil ?, (b) How much work would the torque applied by an external, agent to do on the coil to rotate it 90 o from its initial, orientation, so that M is perpendicular to B and the coil is, again at rest ?, , Sol., (a), , Ans., , Sol., , b a, Ni, b a, , 0.85, , with angular speed, rad/s. If a magnetic field B is directed, perpendicular to the rotation axis, then find the torque acts on the, disc., , a, , =, , 4, , The flat insulating disc of radius a carries an excess, charge on its surface is of C/m2. Consider disc to rotate around, the axis passing through its centre and perpendicular to its plane, , b, , N, , 2.52 10, , Ex. 28, , dM, , =, , 6, , dr, , The magnetic moment of current loop, having dN turns is,, dM = (dN) i A, Total magnetic moment, M =, , 250 100 10, , According to the right-hand rule, the direction of the current through, the wires from the right side of the coil is from top to bottom (see, figure)., , Fig. 5.52, dq = (2 r dr), As the ring rotates with angular velocity , so the current, i =, , dq, dt, , =, r dr, The torque on the current loop, = i A B., Hence the torque on this annular ring, , 2 rdr, 2

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, d, , Ex. 30, , = i dA B, =, , r dr, , r 3 Bdr, , =, , B r 3dr, , a, , and, , Two metal strips, each of length , are clamped parallel, , to each other on a horizontal floor with a separation b between, them. A wire of mass m lies on them perpendicular as shown in Fig., 5.53. A vertically upward magnetic field of strength B exists in the, space. The metal strips are smooth but the coefficient of friction, between wire and the floor is . A current i is established when the, switch S is closed at the instant t = 0. Discuss the motion of the wire, after the switch is closed. How far away from the strips will the wire, reach ?, , r 2 B sin 90, , =, , 363, , 0, , =, , Ba 4 ., 4, , Ans., , Ex. 29 Consider a non conducting plate of radius a and mass m, which has a charge q distributed uniformly over it. The plate is, rotated about its axis with an angular speed . Show that the, magnetic moment M and the angular momentum L of the plate are, , Sol. If, , is the surface charge density, then, q =, , a2., , i =, , rdr (calculated in previous example), , Current, , Fig. 5.53, , Sol., , q, M, related as, =, ., 2, m, L, , When current starts flowing in the wire, it experiences a force, F = Bib of, constant magnitude. Due to which it accelerates on the metal strips., Thereafter when wire falls on to the floor, it retards due to friction and, finally stops. Thus, , The magnetic moment of the element ring, , a =, , dM = i (dA), =, =, , rdr ( r2), r3, , v2 = 0 + 2a, , dr, , = 2, , r 3 dr, , M =, 0, , =, , =, , 0 = 2v2 – a x, , 4, , a2, 4, , q a2, 4, , or, , The angular momentum of the disc about its axis, L =, , Here, , x =, , a =, , ma 2, ., 2, x =, 2, , The ratio, , M, L, , =, , q a, 4, ma 2, 2, , Bib, m, , Let x be the distance moved by the wire on the floor, its final velocity, becomes zero, and so, , a4, , a2, , Bib, ., m, , The velocity gained by the wire on the strips, , a, , and, , F, m, , q, ., 2m, , Ans., , v2, 2a ', , mg, m, , 2 Bib, m, 2 g, , g, , Bib, ., mg, , Ans.

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364, , ELECTRICITY & MAGNETISM, , Review of formulae & Important Points, 1., , Force on a moving charge : The force on a charge q moving with, , 6., , velocity v in a magnetic field is given by, , B. The magnetic field extends to a distance x, which is less than or, equal to the radius of the path, then deviation angle is given by, , F = qvBsin ,, and the direction of force is given by Fleming left hand rule., , sin, , In vector notation, we can write, =, , F, , 2., , q v, , 7., , B, , Lorentz force : When a charged particle moves through a region, of space having both E and B field, the net force on the particle, called Lorentz force and is given by, =, , F, , q E, , v, , In magnetic field speed and hence kinetic energy of the charged, particle remain constant., , 4., , When charged particle is projected perpendicular to the magnetic, field, its path will be circular., The radius of path, , 8., , 9., , iB, ., ne, , Mass spectrograph : For two isotopes of mass numbers m1 and, m2 ,, =, , r1, ., r2, , Cylotron : If V is the potential and f is the frequency of the AC, source used in cylotron, then K.E. of the particle q will be, K = 2fqV., Magnetic force on current carrying conductor :, F = Bi sin ,, , mv, qB, , and the direction of force can be obtained by Flemings left hand, rule. In vector notation F = i, , The time to complete the circle, , 5., , x, ., r, , Hall effect : Hall potential is given by, , m1, m2, , 10., , T =, , =, , VH =, , B ., , 3., , r =, , A charged particle q enters normally in a uniform magnetic field, , 2 m, ., qB, , 11., , B., , Force on curved conductor :, Q, , When charged particle is projected at an angle with the magnetic, field, its path will be helical. The radius of path, , F, , 12., , id, , =, , B, , P, , Torque on a current loop :, =, , M, , B,, , where M = Ni A, is called magnetic moment of the loop., 13., r =, , and pitch ,, , p, , =, , mv sin, ,, qB, 2 mv cos, qB, , Magnetic moment of an electron moving in a circle of radius r with, speed v, M =, , evr, ., 2

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , E xercise 5. 1, , MCQ Type 1, , Only one option correct, 1., The figure shows four directions for the velocity vector v of a, positively charged particle moving through a uniform electric field, , (a) I, (b) II, III, (c) III, (d) I, II, III, A particle is projected in a plane perpendicular to a uniform magnetic, field. The area bounded by the path described by the particle is, proportional to, (a) the velocity, (b) the momentum, (c) the kinetic energy, (d) none of these, A positively charged particle projected towards east is deflected, towards north by a magnetic field. The field may be, (a) downward, (b) upward, (c) towards west, (d) towards south, Given, B = magnetic induction and R = radius of the path, the, energy of a charged particle coming out of a cyclotron is given by, , 4., , E (directed out of the page and represented with an encircled, , dot) and a uniform magnetic field B. Of all four directions, which, might result in a net force of zero ?, 5., , 6., , 2., , (a) 1, (b) 2, (c) 3, (d) 4, The figure shows a metallic, rectangular solid that is to move at a, certain speed v through the uniform magnetic field B. You have, three choices for the direction of the velocity of the solid : vx, vy, and vz parallel to x, y and z directions respectively. For which, choice is the front face at lower potential ?, , 7., , (a), , qB 2 R 2, 2m, , (b), , q 2 BR 2, 2m, , (c), , q 2 BR, 2m, , (d), , q2B2R2, 2m, , A uniform electric field and a uniform magnetic field are produced,, pointed in the same direction. An electron is projected with its, velocity pointing in the same direction, (a) the electron will turn to its right, (b) the electron will turn to its left, (c) the electron velocity will increase in magnitude, (d) the electron velocity will decrease in magnitude, Two particles X and Y having equal charges, after being accelerated, through the same potential difference, enter a region of uniform, magnetic field and describes circular path of radius R1 and R2, respectively. The ratio of mass of X to that of Y is, , 8., , 3., , 365, , (a) v x, (b) v y, (c) v z, (d) none, Figure shows three situations in which a positive particle of, , (a), , R1, R2, , (c), , R1, R2, , velocity v moves through a uniform magnetic field B and, experiences a magnetic force FB . In which situation(s) the, orientations of the vectors are physically reasonable, 9., , 1/ 2, , (b), , R2, R1, , (d), , R1, R2, , 2, , The radius of curvature of the path of the charged particle in a, uniform magnetic field is directly proportional to, (a) the charge on the particle, (b) the momentum of the particle, (c) the energy of the particle, (d) the intensity of the field, , Answer Key, , 1, , (d), , 2, , (b), , 3, , (d), , 4, , (c), , Sol. from page 383, , 6, , (d), , 7, , (d), , 8, , (c), , 9, , (b), , 5, , (a)

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366, 10., , ELECTRICITY & MAGNETISM, , If a particle of charge 10–12 coulomb moving along the x̂ direction, with a velocity 105 m/s experiences a force of 10–10 newton in, , 16., , ŷ direction due to magnetic field, then the minimum magnetic, field is, , 11., , 12., , 13., , 14., , 15., , (a), , 6.25 103 tesla in ẑ, , (b), , 10–15 tesla in ẑ, , (c), , 6.25 10 3 tesla in ẑ, , (d), , 10–3 tesla in ẑ, , direction, , A particle of mass M and charge Q moving with velocity v, describes a circular path of radius R when subjected to a uniform, transverse magnetic field of induction B. The work done by the, field when the particle completes one full circle is, , (a), , (b), , BQv 2 R, , direction, direction, , 17., , direction, , An electron and a proton enter region of uniform magnetic field in, a direction at right angles to the field with the same kinetic energy., They describe circular paths of radius re and rp respectively. Then, (a), , re = rp, , (b), , re < rp, , (c), , re > rp, , (d), , re may be less than or greater than rP depending on the, direction of the magnetic field., , 18., , Mv 2, 2 R, R, , (c) zero, (d) BQ2 R, A particle of charge –16×10–18 coulomb moving with velocity 10, ms–1 along the x-axis enters a region where a magnetic field of, induction B is along the y-axis, and an electric field of magnitude, 104 V/m is along the negative z-axis. If the charged particle continues, moving along the x-axis, the magnitude of B is, (a) 10–3 Wb/m2, (b) 103 Wb/m2, 5, 2, (c) 10 Wb/m, (d) 1016 Wb/m2, A very long straight wire carries a current I. At the instant when a, charge +Q at point P has velocity v , as shown, the force on the, charge is, , A charge moves in a circle perpendicular to a magnetic field. The, time period of revolution is independent of, (a), , magnetic field, , (b), , charge, , (c), , mass of the particle, , (d), , velocity of the particle, , An electron and a proton with equal momentum enter, perpendicularly into a uniform magnetic field, then, (a), , the path of proton shall be more curved than that of electron, , (b), , the path of proton shall be less curved than that of electron, , (c), , both are equally curved, , (d), , path of both will be straight line, , 19., , (a) opposite to OX, (b) along OX, (c) opposite to OY, (d) along OY, A rectangular loop carrying a current i is situated near a long straight, wire such that the wire is parallel to one of the sides of the loop, and is in the plane of the loop. If a steady current I is established, in wire as shown in figure, the loop will, , Two particles A and B of masses mA and mB respectively and, having the same charge are moving in a plane. A uniform magnetic, field exists perpendicular to this plane. The speeds of the particles, are vA and vB respectively, and the trajectories are as shown in the, figure. Then, , 20., , (a), , mAvA < mBvB, , (b), , mAvA > mBvB, , (c), , mA < mB and vA < vB, , (d), , mA = mB and vA = vB, , (a) rotate about an axis parallel to the wire, (b) move away from the wire or towards right, (c) move towards the wire, (d) remain stationary, A conducting circular loop of radius r carries a constant current i., It is placed in a uniform magnetic field B , such that B is, perpendicular to the plane of the loop. The magnetic force acting, on the loop is, , A charged particle is released from rest in a region of steady uniform, electric and magnetic fields which are parallel to each other the, particle will move in a, (a), , straight line, , (b), , circle, , (c), , helix, , (d), , cycloid, , 21., , (a), , irB, , (b), , 2 riB, , (c), , Zero, , (d), , riB, , If a current is passed in a spring, it, (a), , gets compressed, , (b), , gets expanded, , (c), , oscillates, , (d), , remains unchanged, , Answer Key, , 10, , (d), , 11, , (b), , 12, , (d), , 13, , (c), , 14, , (b), , 15, , (a), , Sol. from page 383, , 16, , (c), , 17, , (b), , 18, , (d), , 19, , (c ), , 20, , (c), , 21, , (a)

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, 22., , 23., , 24., , 25., , 26., , A stream of electrons is projected horizontally to the right. A, straight conductor carrying a current is supported parallel to electron, stream and above it. If the current in the conductor is from left to, right then what will be the effect on electron stream?, (a) The electron stream will be pulled upward, (b) The electron stream will be pulled downwards, (c) The electron stream will be retarted, (d) The electron beam will be speeded up towards the right, Two long conductors, separated by a distance d carry current I1, and I2 in the same direction. They exert a force F on each other., Now the current in one of them is increased to two times and its, directions is reversed. The distance is also increased to 3d. The, new value of the force between them is, (a) –2 F, (b) F / 3, (c) 2F / 3, (d) –F / 3, Two thin, long, parallel wires, separated by a distance d carry a, current i A in the same direction. They will, (a), , attract each other with a force of, , (b), , repel each other with a force of, , (c), , attract each other with a force of, , (d), , repel each other with a force of, , 0i, , 2, , 2, , / 2 d2, , 0i, , 0i, 0i, , 2, , 2, , 367, , (a) I > III > II > IV, (b) I > II > III > IV, (c) I > IV > II > III, (c) III > IV > I > II, Two very long, straight and parallel wires carry steady currents I, and I respectively. The distance between the wires is d. At a, certain instant of time, a point charge q is at a point equidistant, from the two wires in the plane of the wires. Its instantaneous, velocity v is perpendicular to this plane. The magnitude of the, force due to the magnetic field acting on the charge at this instant, is, , 28., , (a), (c), 29., , 0 Iqv, , 2 d, 2, , 0 Iqv, , (b), , 0 Iqv, , (d), , d, , d, , 0, , An electron moving with a speed u along the positive x-axis at y =, 0 enters a region of uniform magnetic field B – Bokˆ which exists, to the right of y-axis. The electron exists from the region after some, time with the speed v at co-ordinate y, then, , / 2 d2, , / 2 d, , / 2 d, , A particle of charge q and mass m moves in a circular orbit of, radius r with angular speed . The ratio of the magnitude of its, magnetic moment to that of its angular momentum depends on, (a), and q, (b), q and m, (c) q and m, (d), and m, A conducting loop carrying a current I is placed in a uniform, magnetic field pointing into the plane of the paper as shown. The, loop will have a tendency to, , (a) v > u, y < 0, (b) v = u, y > 0, (c) v > u, y > 0, (d) v = u, y < 0, A proton beam moves through a region of space where there exists, a uniform magnetic field of magnitude 4.0 T along z-axis. The, protons have a velocity of 4 × 105 m/s in the x-z plane at an angle, of 30º to the positive z-axis. The force on the proton is :, , 30., , y, , 27., , B, , (a) contract, (b) expand, (c) move towards + ve x-axis (d) move towards–ve x-axis, A current carrying loop is placed in a uniform magnetic field in, four different orientations, I, II, III and IV arrange them in the, decreasing order of potential energy., , I., , x, 0, , 30, , v, , z, (a), (c), , 25.6 x 10-14 j N, – 12.8 x 10-14 j N, , B, , II., , i, , e, , III., , – 25.6 x 10-14 j N, 12.8 x 10-14 j N, , v, , i, , 31., , (b), (d), , Work done on an electron moving in a solenoid along its axis is, equal to, (a) zero, (b) –e B, (c) 1 / B, (d) None of the above, , IV., , Answer Key, , 22, , (b), , 23, , (c), , 24, , (c), , 25, , (c), , 26, , (b), , Sol. from page 383, , 27, , (c), , 28, , (d), , 29, , (d), , 30, , (c), , 31, , (a)

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368, 32., , ELECTRICITY & MAGNETISM, , The figure shows three situations when an electron moves with, velocity v travels through a uniform magnetic field B . In each, case, what is the direction of magnetic force on the electron., , v, , B, , x, , 38., , B, x, , x, v, , v, , B, , z, , z, , z, , 3, , 2, , 1, , 33., , y, , y, , y, , (a) +ve z-axis, –ve x-axis, + ve y-axis, (b) –ve z-axis, –ve x-axis and zero, (c) +ve z-axis, +ve y-axis and zero, (d) –ve z-axis, +ve x–axis and zero, A closed loop PQRS carrying a current is placed in a uniform, magnetic field. If the magnetic forces on segment PS, SR and RQ, are F1, F2 and F3 respectively and are in the plane of the paper and, along the directions shown, the force on the segment QP is, (a), , F3, , F1, , 2, , (b), , F3, , F1, , F2, , (c), , F3, , F1, , F2, , F3, , (d), , F1, , 2, , Q, , 39., , 40., , F22, P, F3, F1, , F2, , S, , R, , (a), , 2, , F2, , 34., , 35., , A charged particle moves along a circle under the action of constant, electric field and magnetic field. Which of the following are possible, (a), , E, , 0, B, , 0, , (b), , E, , 0, B, , 0, , (c), , E, , 0, B, , 0, , (d), , E, , 0, B, , 0., , A proton of mass 1.67 10, , 27, , 41., , kg and charge 1.6 10, , 19, , a force F given by ......which results in the lowering of the, potential of the face.........Assume the speed of the carriers to be v, , 10 7 s, , (a), , a circle of radius = 0.2 m and time period, , (b), , a circle of radius = 0.1 m and time period 2, , 10 7 s, , a helix of radius = 0.1 m and time period 2, , 7, , a helix of radius = 0.2 m and time period 4 10 7 s, A proton (or charged particle) moving with velocity v is acted, upon by electric field E and magnetic field B. The proton will, move undeflected if, (a) E is perpendicular to B, (b) E is parallel to v and perpendicular to B, , 37., , (c), , E, B and v are mutually perpendicular and v, , (d), , E and B both are parallel to v, , (a), , evBkˆ, ABCD, , (b), , evBkˆ, EFGH, , (c), , evBkˆ, ABCD, , (d), , evBkˆ, EFGH, , 10 s, , (d), , 36., , There will be repulsion between A and B attraction between, x and y, (b) There will be attraction between A and B, repulsion between, x and y, (c) There will be repulsion between A and B and also x and y, (d) There will be attraction between A and B and also x and y, A metallic block carrying current I is subjected to a uniform magnetic, induction B as shown in the figure. The moving charges experience, , C is, , projected with a speed of 2 106 m / s at an angle of 60° to the, x-axis. If a uniform magnetic field of 0.104 tesla is applied along, y-axis, the path of proton is, , (c), , (a) go on moving in the same direction with increasing velocity, (b) go on moving in the same direction with constant velocity., (c) turn to its right, (d) turn to its left, An ionized gas contains both positive and negative ions. If it is, subjected simultaneously to an electric field along the + x direction, and a magnetic field along the + z direction, then, (a) positive ions deflect towards + y direction and negative ions, towards – y direction, (b) all ions deflect towards + y direction, (c) all ions deflect towards – y direction, (d) positive ions deflect towards – y direction and negative ions, towards + y direction, A particle of mass m and charge q moves with a constant velocity, v along the positive x direction. It enters a region containing a, uniform magnetic field B directed along the negative z direction,, extending from x = a to x = b. The minimum value of v required so, that the particle can just enter the region x > b is, (a) qb B / m, (b) q(b – a) B / m, (c) qa B/m, (d) q(b + a) B / 2m, A and B are two conductors carrying a current i in the same direction., x and y are two electron beams moving in the same direction, , E, B, , A homogeneous electric field E and a uniform magnetic field B are, pointing in the same direction. A proton is projected with its, , 42., , Two insulated rings, one of slightly smaller, diameter than the other are suspended, along their common diameter as shown., Initially the planes of the rings are, mutually perpendicular. When a steady, current is set up in each of them, (a), (b), (c), (d), , velocity parallel to E . It will, , the two rings rotate into a common plane, the inner ring oscillates about its initial position, the inner ring stays stationary while the outer one moves, into the plane of the inner ring, the outer ring stays stationary while the inner one moves, into the plane of the outer ring, , Answer Key, , 32, , (b), , 33, , (d), , 34, , (a), , 35, , (c), , 36, , (c), , Sol. from page 383, , 38, , (c), , 39, , (b), , 40, , (b), , 41, , (a ), , 42, , (a), , 37, , (a)

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369, , MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Level -2, Only one option correct, 1., , For four situations, here is the velocity v of a proton at a certain, instant as it moves through a uniform magnetic field B :, (i), , v, , (iii) v, , 2i 3 j and B, , 4kˆ, , (ii), , 2kˆ and B, , 4i, , (iv) v, , 3j, , v, , 3i, , 4kˆ, , 2 j and B, , 20i and B, , – 4i, , If F1, F2, F3 and F4 are the magnitude of forces in the four situations, respectively, then, , 2., , (a), , F1, , F2, , F3, , (c), , F1, , F2, , F3 , F4, , A magnetic field B, , B, , F4, 0, , (b), , F1, , F2 ; F3, , (d), , F1, , F2, , F3, , B0 j exists in the region a, , – B0 j in the region 2 a, , x, , F4, 0, F4, , 0, , 2a and, , x, , 3a , where B0 is a positive, , constant. A positive point charge moving with a velocity v, , v0 i, , 5., , where v0 is a positive constant, enters the magnetic field at x =a., The trajectory of the charge in this region can be like :, , (a) couple on loop P will be the highest, (b) couple on loop Q will be the highest, (c) couple on loop R will be the highest, (d) couple on loop S will be the highest, For a positively charged particle moving in a x-y plane initially, along the x-axis, there is a sudden change in its path due to the, presence of electric and/or magnetic fields beyond P. The curved, path is shown in the x-y plane and is found to be non-circular., Which one of the following combinations is possible., , (a), , (b), , 6., , (c), , (a), , E, , 0; B, , biˆ ckˆ, , (b), , E, , ai; B, , ckˆ aiˆ, , (c), , E, , 0; B, , cjˆ bkˆ, , (d), , E, , ai; B, , ckˆ bjˆ, , A wire ABCD is bent in the form shown here in the figure. Segments, AB and CD are of length 1 m each while the semicircular loop is of, radius 1 m. A current of 5 A flows from A towards the end D and, the whole wire is placed in a magnetic field of 0.5 T directed out of, the page. The force acting on the wire is :, , (d), , B, 3., , 4., , A proton, a deuteron and an –particle having the same kinetic, energy are moving in circular trajectories in a constant magnetic, field. If rp, rd and r denote respectively the radii of the trajectories, of these particles, then, (a) r = rp < rd, (b) r > rd > rp, (c) r = rd > rp, (d) rp = rd = r, Four wires each of length 2.0 metres are bent into four loops P, Q,, R and S and then suspended into uniform magnetic field. Same, current is passed in each loop. Which statement is correct?, , A, , (a), (c), , Answer Key, , 1, , (c), , 2, , (a), , Sol. from page 384, , 5, , (b), , 6, , (c), , B, , 40 N, 10 N, , (b), (d), , 3, , D, , C, , (a), , 5N, 20 N, , 4, , (d)

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370, 7., , ELECTRICITY & MAGNETISM, , A charged particle moves along the x-axis in crossed magnetic and, , 9., , electric fields given as B Bzˆ and E Eyˆ . The motion of the, particle as observed in the laboratory frame will look like :, , z, , A mass spectrograph is a device for separating charged particles, having different masses. Consider two particles of the same charge, q but different masses m1 and m2 injected into the region of a, uniform magnetic field B with a known velocity v normal to the, magnetic field as shown in the figure. The particles are separated, by a distance d, given by :, , (a), , o, , y, , x, z, d, v, , (b), y, , o, x, z, , 10., , d, , | (m2 m1 )v |, | qB |, , (b), , d, , | 2(m2 m1 )v |, | qB |, , (c), , d, , | (m2 m1 )v |, | 2qB |, , (d), , d, , | (m2 m1 )v |, | 4qB |, , The electric field and the magnetic field in a region are given by, E, , (c), , y, , o, , z, , (d), , 11., , y, , o, x, , 1iˆ m/s at a point in a, , An electron moving with a velocity v1, , E0, B0, , (b), , (c), , E0, 2 B0, , (d), , v2, , (1iˆ 1 ˆj ) m/s at the same point. It experiences a force, , F2, , e(iˆ, , ˆj ) . The force the electron would experience if it were, , moving with a velocity v3, , (c), , zero, , ˆj ), , (c), 12., , v 1 v2 , at the same point is :, (b), , e(iˆ, , (d), , ejˆ, , ˆj ), , 7, , (b), , 8, , (c), , Sol. from page 384, , 11, , (b), , 12, , (d), , none of these, , q, 1, , a2 B, a2 B, , q, , (b), , 1, q, 2, , (d), , 1, q, 2, , a2 B, a2 B, , An insulating rod of length l carries a charge q distributed uniformly, on it. The rod is pivoted at an end and is rotated at a frequency f, about a fixed perpendicular axis. The magnetic moment of the, system is :, (a) zero, (b), q f l2, (c), , Answer Key, , 2E0, B0, , A positive electric charge q is distributed over a circular ring of, radius a. It is placed in a horizontal plane and is rotated about its, axis at a uniform angular speed . A horizontal magnetic field B, exists in the space. The torque acting on the ring due to the magnetic, force is :, , of the electron. If the electron is moving with a velocity, , e(iˆ, , ˆjB . Consider a frame of reference moving with, 0, , (a), , (a), , ejˆ where e is the charge, , magnetic field experiences a force F1, , (a), , ˆ and B, iE, 0, , a velocity v0 k̂ . The electric field in this frame will be zero if v0 is, equal to :, , x, , 8., , (a), , 1, 2, , q f l2, , 9, , (d), , (b), , 1, 3, , q f l2, , 10, , (a)

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, 13., , A particle of charge q and mass m starts moving from the origin, E0 iˆ and B, , under the action of an electric field E, , 16., , B0 iˆ with, , A conducting wire bent in the form of a parabola y2 = 2x carries a, current i = 2 A as shown in figure. This wire is placed in a uniform, , after a time, , 14., , (a), , t, , 2mv0, qE, , (c), , t, , 3Bq, mv0, , (b), , t, , (d), , t, , 4kˆ tesla. The magnetic force on the wire (in, , magnetic field B, newton), , ˆ, 0 j . The speed of the particle will become 2 0, , velocity v, , y (m), , 2Bq, mv0, , A, , 3mv0, qE, , 2 x (m), , A mass spectrometer is a device which select particle of equal, mass. An in with electric charge q > 0 starts at rest from a source, S and is accelerated through potential difference V. It passes through, a hole into a region of constant magnetic field B perpendicular to, the plane of the paper as shown in the figure. The particle is, deflected by the magnetic field and emerges through the bottom, hole at a distance d from the top hole. The mass of the particle is, , B, 16iˆ, , (a), 17., , (b), , 32iˆ, , (c), (d) 16iˆ, 32iˆ, A circular current loop of radius a is placed in a radial field B as, shown. Then net force acting on the loop is, , B, , I, , S, , 371, , B, , a, , V, , (a), , 18., , qB 2d 2, 4V, , (b), , and current i = 10A. The magnetic moment M of the loop is, , qB 2d 2, , qBd, (d), 2V, 8V, OABC is current carrying square loop an electron is projected, from the centre of loop along its diagonal AC as shown. Unit, vector in the direction of initial acceleration will be, , (c), , 15., , qBd, V, , (a) zero, (b) 2 BaIcos, (c) 2 aIBsin, (d) none, Figure shows a square current carrying loop ABCD of side 10 cm, , y, B, A, , C, i = 10, , x, , 30°, , y, , D, z, , B, , A, , v, O, , x, , C, , 19., , (a), , 0.05 iˆ, , (c), , 0.05, , 3iˆ, , 3kˆ A-m 2, , (b), , 0.05, , kˆ A-m 2, , (d), , iˆ, , ˆj, , kˆ A-m 2, , kˆ A-m 2, , A charged particle with charge q enters a region of constant uniform, and mutually orthogonal fields E and B with a velocity v, , (a), , (c), , (b), , k̂, , (d), , k̂, , –, ˆi, , ˆi, , perpendicular to both E and B , and comes out without any, change in magnitude or direction of v . Then, , ˆj, 2, , ˆj, 2, , (a), , v, , E, , B / B2, , (b), , v, , B, , E / B2, , (c), , v, , E, , B / E2, , (d), , v, , B, , E / E2, , Answer Key, , 13, , (d), , 14, , (c), , 15, , (b), , Sol. from page 384, , 17, , (c), , 18, , (a), , 19, , (a), , 16, , (b)

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372, 20., , ELECTRICITY & MAGNETISM, , A current carrying loop is placed in a uniform magnetic field, pointing in negative z direction. Branch PQRS is a three quarter, circle, while branch PS is straight. If force on branch PS is F, force, on branch PQR is, , y, B, , P, x, , S, , Q, , 21., , 2F, , (b), , (c), , F, 2, , (d), , 24., , 2 F, , (c), , Bi 2, x, , (b), , Bix 2, , (d), , 25., , Two particles A and B of same mass and having charges of same, magnitude but of opposite nature are thrown into a region of, magnetic field (as shown) with speed v1 and v2 (v1 > v2). At the, time particle A escapes out of the magnetic field, angular momentum, of particle B w.r.t. particle A is proportional to (Assume both the, particles escape the region after traversing half circle), , A, , v2, , Wall, , m, 2qB, , 4, , (b), , m, qB, , (c), , m, 4qB, , 2, , (d), , m, 2, 4qB, , 2, 3, , A jumper of mass m and length is placed on two parabolic rails, in x-y plane. Shape of the rails can be described by, Rail 1 : y = x2 (and z = 0), Rail 2 : y = x2 (and z = ), If x is horizontal and y is vertical direction and magnetic field in the, , (a), , iB0, 2mg, , (c), , iB0, mg, , (b), 2, , (d), , iB0, mg, , iB0, 2mg, , 2, , A conducting loop is placed in a magnetic field of strength B, perpendicular to its plane. Radius of the loop is r, current in the, loop is i and linear mass density of the wire of loop is m. Speed of, any transverse wave in the loop will be, (a), , Bir, m, , (b), , Bir, 2m, , 2Bir, Bir, (d) 2, m, m, A charged sphere of mass m and charge – q starts sliding along the, surface of a smooth hemispherical bowl, at position P. The region, has a transverse uniform magnetic field B. Normal force by the, surface of bowl on the sphere at position Q is, (c), , 26., v1, , 2 qB, , coordinate of its ends is (i = current in jumper), , Bi 2, 4x, Bix 2, 4, , 2 qB, , space is B0 ˆj , the jumper can remain in equilibrium when y, , A vertical wire of length has its both ends fixed. Space around the, wire has a horizontal magnetic field of strength B. If current i is, passed through the wire such that its mid-point is deflected by, amount x, tension in the wire is, (a), , 22., , F, 2, , mv, , (a), , R, , (a), , mv, , B, , P, , 23., , (a), , v1 + v2, , (b), , v1 – v2, , (c), , v12, , (d), , v12, , v22, , R, , v22, , B, , Q, , A particle of mass m and charge q enters a region of magnetic field, (as shown) with speed v. There is a region in which the magnetic, field is absent, as shown. The particle after entering the region, collides elastically with a rigid wall. Time after which the velocity, of particle becomes antiparallel to its initial velocity is, , (a), , mg sin + qB, , 2 gR sin, , (b), , 3 mg sin + qB, , 2 gR sin, , (c), , mg sin – qB, , 2 gR sin, , (d), , 3 mg sin – qB, , 2 gR sin, , Answer Key, , 20, , (a), , 21, , (b), , 22, , (c), , Sol. from page 384, , 24, , (d), , 25, , (a), , 26, , (b), , 23, , (d)

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373, , MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, 27., , The figure shows two infinite semi-cylindrical shells: shell-1 and, shell-2. Shell-1 carries current i1, in inward direction normal to the, plane of paper, while shell-2 carries same current i1, in opposite, direction. A long straight conductor lying along the common axis of, the shells is carrying current i2 in direction same as that of current, in shell-1. Force per unit length on the wire is, , (a), , 30., , r, , (b), , 31., , 0i1i2, , 2 r, , 2 0i1i2, 2 0i1i2, (d), 2, r, r, A square loop of side carrying current i is part of the shown, arrangement. Minimum current i to just move the block up the, inclined plane is, (c), , 28., , 1, (d) 2, 2, A particle of mass m and charge q is thrown from origin at t = 0, with velocity 2iˆ 3 ˆj 4 kˆ units in a region with uniform magnetic, , E may be, (a) 5iˆ 10 ˆj units, , Shell-2, , zero, , 2, , m, field 2iˆ units. After time t = qB , an electric field E is switched, on, such that particle moves on a straight line with constant speed., , Insulation, , (a), , (b), , (c), , Shell-1, , Insulation, , 1, , 6iˆ 9kˆ units, (d) 6iˆ 8kˆ units, (c), 6kˆ 8 ˆj units, Shown in the figure are two horizontal parallel conducting rails, separated by distance . A uniform magnetic field B exists in vertical, upward direction. A wire of mass m can slide on the rails. The rails, are connected to a source, which drives current i in the circuit given, (b), , a, , where a and b are positive, b, constants. Coefficient of friction between the rails and the wire is, . Minimum so that wire does not slide is, , by i = at – bt2 between t = 0 and t =, , Rail 1, B, , Wire, , i, Source, , m, , Rail 2, , µ, , (a), i, , Ba 2, , (b), , b mg, , Ba 2, 2b mg, , Ba 2, Ba 2, (d), 4b mg, 8b mg, Consider the figure shown. A proton enters a uniform magnetic, (c), , B, , (a), , mg cos, , (b), , B, , mg sin, , field region with velocity v, , mg cos, , 6iˆ m/s. Thickness of magnetic field, , r, , where r is radius of rotation of the proton in the, 2, magnetic field. A large plane mirror in x-z plane is moving with, constant velocity 3iˆ 2 ˆj m/s. Velocity of the image of proton in, mirror when the proton comes out of the field is, , 2B, , region is, , 2 mg cos, mg sin, mg cos, (d), B, B, An electron and a positron are projected in a transverse magnetic, field (of strength B) with speed v each. If width of magnetic field, region is kr (where k is a constant, r is radius of revolution of each, of the particle). Time spent by each of the particles in magnetic, field is one-fourth of the time period of revolution of particles for, k equal to (Assume elastic collision, if particles collide), , (c), , 29., , mg sin, , 32., , 2mg sin, , y, r/2, x, , 1.5 r, , Mirror, , (a), , Electron, , Positron, , (c), , kr, , Answer Key, , 27, , (d), , 28, , (c), , Sol. from page 384, , 31, , (c), , 32, , (b), , z, , 3 3 iˆ 5 ˆj m/s, , 3 3 3 iˆ, , 29, , (b), , 5 ˆj m/s, , (b), , (d), , 3 3 iˆ 7 ˆj m/s, , 3 3 3 iˆ, , 30, , 7 ˆj m/s, , (c)

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374, 33., , ELECTRICITY & MAGNETISM, , A particle (mass m and charge q) is at rest at origin. An electric field, , x, , E 10kˆ units and magnetic field B –8iˆ 6 ˆj units is switched, on in the region. Speed of the particle as function of its z-coordinate, is, 10qz, m, , (a), , 20qz, m, , (b), , v1, , R, a, , 30qz, 40qz, (d), m, m, A charged particle (of charge q and mass m) is projected with, velocity v kˆ at a point R as shown. Electric and magnetic fields in, , S, z, , (c), , 34., , b, , 1, , v2, , the region are E0kˆ and B0 ˆj respectively. Velocity at another, , (a), , 2qE0b, , m v12, , v2 2, , (b), , 2E0a, , v2iˆ (as shown). Choose the correct, , (c), , 2qE0b, , m v2 2, , v12, , (d), , 2qE0 a, , point S in the region is, alternate, , 33, , Answer Key, , (b), , 34, , m v2 2, , v12, , m v12, , v2 2, , (c), , Sol. from page 384, , E xercise 5. 2, , MCQ Type 2, Multiple correct options, 1., , 2., , 3., , The radius of curvature of the path of a charged particle moving in, a static uniform magnetic field is :, (a) directly proportional to the magnitude of the charge or the, particle, (b) directly proportional to the magnitude of the linear momentum of the particle, (c) directly proportional to the kinetic energy of the particle, (d) inversely proportional to the magnitude of the magnetic field, If a charged particle kept at rest experiences an electromagnetic, force,, (a) the electric field must be there, (b) the magnetic field must be there, (c) the magnetic field may or may not be there, (d) the electric field may or may not be there., A charged particle moves in a gravity free space without change in, velocity. Which of the following is/are possible ?, (a), , 4., , 5., , E, , 0, B, , 0, , (b), , E, , 0, B, , E must be perpendicular to B, , (b), , v must be perpendicular to E, , Sol. from page 388, , (b), , q B, m, path length of the particle in region II is maximum when, , the particle enters the region III only if its velocity v, , q B, m, (d) time spent in region II is same for any velocity v as long as, the particle returns to region I., A particle of charge +q and mass m moving under the influence of, a uniform electric field Eiˆ and a uniform magnetic field Bkˆ follows, , velocity v, , 6., , 0, , trajectory from P to Q as shown in figure., The velocities at P and Q are viˆ, and 2vjˆ respectively. Which of, the following statement(s) is/are, correct?, , (c) v must be perpendicular to B, (d) E must be equal to vB., A particle of mass m and charge q, moving with velocity v enters, region II normal to the boundary as shown in the figure. Region II, has a uniform magnetic field B perpendicular to the plane of the, paper. The length of the region II is . Choose the correct choice(s), , Answer Key, , the particle enters region III only if its velocity v, , (c), , (c) E 0, B 0, (d) E 0, B 0., If a charged particle goes without any acceleration in a region, containing electric and magnetic fields,, (a), , q B, m, , (a), , (a), (b), (c), (d), , 1, , (b, d), , 2, , (a, c), , 5, , (a, c), , 6, , (a, b, d), , E, , 3 mv 2, 4 qa, , 3 mv3, 4 a, Rate of work done by electric field at P is zero, Rate of work done by both the fields at Q is zero, , Rate of work done by electric field at P is, , 3, , (b, c, d), , 4, , (a, b)

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Statement Questions, Read, (a), (b), (c), (d), , 1., , 7., , Statement -1, Statement - 2, Magnetic field can not change the speed of the charged particle., , 8., , Statement - 1, The free electrons in a conducting wire are in continuous thermal, motion. If such a wire, without current is placed in a magnetic, field, the force on the wire is zero., Statement - 2, , 9., , The each free electron in the wire experiences magnetic force., 3., , Statement - 1, The net magnetic force on a current carrying loop in magnetic field, is always zero., Statement - 2, , 10., , The resultant force on the loop is the vector sum of the forces on, its elements., 4., , Statement - 1, The kinetic energy of a charged particle in a perpendicular magnetic, field B is K. When the magnetic field becomes 3B, the kinetic, energy of the particle becomes 3K., Statement - 2, The magnetic force always acts perpendicular to the velocity vector, of the particle., , 5., , Exercise 5.3, , the two statements carefully to mark the correct option out of the options given below:, If both the statements are true and the statement - 2 is the correct explanation of statement - 1., If both the statements are true but statement - 2 is not the correct explanation of the statement - 1., If statement - 1 true but statement - 2 is false., If statement - 1 is false but statement - 2 is true., , Magnetic field can not accelerate the charged particle., , 2., , 375, , Statement - 1, , Statement - 1, A current carrying loop placed in a magnetic field must experience, a torque., Statement - 2, Torque on the loop is given by =MBsin ., Statement - 1, The net charge in a current carrying wire is zero and so magnetic, force on the wire in magnetic field is zero., Statement - 2, The force on a current carrying wire is given by F=Bi sin ., Statement - 1, A charged particle in a region passes undeviated. The region must, not have electric field, Statement - 2, The region may have electric field only or both electric and magnetic, field., Statement - 1, A charged particle moves in a uniform magnetic field. The velocity, of the particle at same instant makes an acute angle with the magnetic, field. The path of the particle is a helix with constant pitch., Statement - 2, The force on the particle is given by F, , 11., , q v, , B ., , Statement - 1, The figure shows the circular paths of two particles : electron and, proton that travel at the same speed in a uniform magnetic field, , B, which is directed into the page. The electron follows the path, of smaller radius., , Cyclotron does not accelerate electron., Statement - 2, Mass of the electron is very small., 6., , Statement - 1, Cyclotron is a device which is used to accelerate the positive, charged particles., Statement - 2, Cyclotron frequency depends upon the velocity of the charged, particle., , Statement - 2, The radius of path in the perpendicular magnetic field is given by, , r, , mv, ., qB, , Answer Key, , 1, , (d), , 2, , (a), , 3, , (d), , 4, , (d), , 5, , (a), , Sol. from page 389, , 7, , (d), , 8, , (d), , 9, , (d), , 10, , (b ), , 11, , (a), , 6, , (c)

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376, 12., , ELECTRICITY & MAGNETISM, , Statement - 1, The figure shows a current i through a wire in a uniform magnetic, , 13., , A charged particle moving through crossed fields with the forces, Fe and FB in opposition. If v, , field B, as well as the magnetic force FB acting on the wire. The, , E, , then particle moves, B, , undeviated., Statement - 2, , direction of field must be along negative y-axis., , If v, 14., , E, , the charged particle bends towards electric field., B, , Statement - 1, A proton of charge +e and of mass m enters a uniform magnetic, field B, , Bi with an initial velocity v, , voxiˆ voy ˆj. The x-, , component of proton velocity vox remains constant., Statement - 2, The x and y components of proton velocity : vox and voy change, but speed of the proton remains constant., , Statement - 2, The force FB, , i, , B., , 12, , Answer Key, , (a), , 13, , (b), , 14, , (d), , Sol. from page 389, , Passage & Matrix, , E xercise 5.4, , Passage for Q. 1 to Q.3., , Passage for Q. 4 to Q.6., , In a television tube, each of the electrons in the beam has a kinetic energy, of 12.0 keV. The tube is oriented so that the electrons move horizontally, from geomagnetic south to geomagnetic north. The vertical component, of earth's magnetic field points down and has a magnitude of 55.0 µT., , A certain commercial mass spectrometer is used to separate uranium ions, , 1., , 2., , The direction in which beam deflects :, (a), , east, , (b), , west, , (c), , north-east, , (d), , south-west, , of mass 3.92 10 25 kg and charge 3.20 10 19 C from related species., The ions are accelerated through a potential difference of 100 kV and, then pass into a uniform magnetic field, where they are bent in a path of, radius 1.0 m. After traveling through 180° and passing through a slit of, width 1.00 mm and height 1.00 cm, they are collected in a cup., 4., , The magnitude of the perpendicular magnetic field in the, separation:, , The acceleration of any electron due to the magnetic field is:, (a), , 3.14 1014 m/s 2, , (b), , 4.28 1014 m/s 2, , (c), , 5.56 1012 m/s 2, , (d), , 6.28 1014 m/s 2, , 5., 3., , 1.96 mm, , (b), , 2.98 mm, , (c), , 4.24 mm, , (d), , none of these, , Answer Key, , 1, , (a), , Sol. from page 389, , 6, , (c), , 2, , 6., , (d), , 495 mT, , (b), , 585 mT, , (c), , 625 mT, , (d), , 835 mT, , If the machine is used to separate out 100 mg of material per hour,, then the current of the desired ions in the machine is :, , The transverse deflection of the beam after travelling 20.0 cm, through the television tube :, (a), , (a), , (a), , 22.7 mA, , (b), , 33.51 mA, , (c), , 45.2 mA, , (d), , 66.7 mA, , The thermal energy produced in the cup in 1.00 h is :, (a) 4.07 MJ, (b) 5.30 MJ, (c) 8.17 MJ, (d) 10.00 MJ, , 3, , (b), , 4, , (a), , 5, , (a)

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, Passage for Q. 7 to Q.9., , 377, , qE ˆ, Bqt ˆ, t i v0 cos, j, m, m, , (c), , v, , (d), , none of these., , A particle of mass m and charge q is moving in a region where uniform,, constant electric and magnetic fields E and B are present. E and B are, parallel to each other. At time t = 0, the velocity v0 of the particle is, perpendicular to E. Assume that its speed is always < c, (the speed of, light in vacuum). Express all the required answers in terms of t, q, m, the, vectors E and B and their magnitudes v0, E and B., 7., , Suppose electric field is acting along positive direction of x and vx, is the x-component of the velocity of the particle at any time t,, , Paragraph for questions no. 10 and 11, A current loop ABCD is held fixed on the plane of the paper a as shown, in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are, joined by two straight wires AB and CD. A steady current I is flowing in, the loop. Angle made by AB and CD at the origin O is 30°. Another, straight thin wire with steady current I1 flowing out of the plane of the, paper is kept at the origin., , then vx is, , 8., , (a), , qE, t, m, , (b), , (c), , 2qE, t, m, , (d), , I1, O, , (c), , Eq, t, m, , vy, , 2, , vy, , Bt, v0, cos, 2, m, , vy, , (b), , vy, , (d), , v0 sin, , 10., , Bt, 2m, , v, , v0 cos, , 11., , Bt, m, (d), , (b) v, , qE ˆ, Bqt ˆ, t i v0 cos, j, m, m, , Answer Key, Sol. from page 389, , 7, , B qt ˆ, v0 sin, k, m, , (a), , 8, , zero, , C, , 0I b, , (b), , a, , 2 ab, , 0I b, , I1 I, /3 a b, 0 2 b a, 4, The magnitude of the net force on the loop is given by, 0 I I1 b, 2 ab, , (d), , D, , a, 0I 2 b a, /3 a b, (d), 4, 4 ab, Due to presence of the current I1 at the origin, (a) The force on AB and DC are zero, (b) The forces on AD and BC are zero, (c) The magnitude of the net force on the loop is given by, (c), , Bqt, v0 cos, m, , I, , The magnitude of the magnetic field (B) due to the loop ABCD at, the origin (O) is, (a), , The resultant velocity, v of the particle at any time t is given by, , (a), , 30°, b, , none of these, , If vy is the y-component of the velocity of the particle, then, (a), , 9., , E, B, , B, , A, , a, , 9, , (b), , a, , 10, , (b), , 11, , (b)

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378, , 12., , ELECTRICITY & MAGNETISM, , A square loop of side a and carrying current i as shown in the figure is placed in gravity free space having magnetic field B = B0 k . Now, match following :, , y, i, , x, , 13., , A., , Column-I, Torque on loop, , (p), , Column -II, is zero, , B., , Net force on loop, , (q), , is in direction (– k ), , C., D., , Potential energy of loop, Magnetic moment of loop, , (r), (s), , has minimum magnitudes, has maximum magnitudes, , A charged particle having charge q and mass m is to be subjected to a combination of constant uniform magnetic field ( B ) and a constant, uniform gravitational field (G). Apart from these field forces there exists no other force. Now match the column., Column-I, , Column -II, , (A), , The charged particle moves without change, in its direction., , (p), , It is possible that both B and G are zero., , (B), , The charged particle moves without change, in its velocity., , (q), , It is possible that both B and G are non zero., , (C), , The charged particle takes a circular path, , (r), , It is possible that B is zero and G is not zero., , (D), , The charged particle takes a parabolic path, , (s), , It is possible that B is non zero and G is zero., , Answer Key, Sol. from page 389, , 12, , A-p, s ; B-p ; C- s ; D-q, , 13, , A-p,q,r s ; B-p, q, s ; C- s ; D-r

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, 14., , 379, , Find correct match for the figure in column I as shown with the items given in column II :, Column I, Column II, B, I, , (A), , I, , (p), , F 0, , (q), , F, , 0,, , 0, , (r), , F, , 0,, , 0, , (s), , F, , 0,, , 0, , I, B, , (B), I, , R, , 2B, R, , (C), , I, , R, I, 2B, , B, , (D), R, I, , Answer Key, , 14, , A-q ; B-r ; C- s ; D-r, , Sol. from page 389, , Subjective Integer Type, , Exercise 5.5, Solution from page 391, , 1., , 104, , An alpha particle is accelerated by a potential difference of, V. Find the change in its direction of motion, if it enters normally, in a region of thickness 0.1 m having transverse magnetic induction, , - particle 6.4 10 27 kg ). Ans. 30°., Figure shows a convex lens of focal length 12 cm lying in a, uniform magnetic field B of magnitude 1.2 T parallel to its, principal axis. A particle having a charge 2.0×10–3C and mass, 2.0×10–5 kg is projected perpendicular to the plane of the diagram, with a speed of 4.8 m/s. The particle moves along a circle with its, centre on the principal axis at a distance of 18 cm from the lens., Show that the image of the particle goes along a circle and find the, radius of that circle., , 4., , 0.1 T. (Given mass of, , 2., , the velocity of the electron at the moment it reaches the opposite, plate ?, Ans. 16 km/s, An electron accelerated by a potential difference V = 1.0 kV, moves in a uniform magnetic field at an angle = 30° to the, vector B whose modulus is B = 29 mT. Find the pitch of the, helical trajectory of the electron., , 5., , 2, 2.0 cm., Ans. p 2 2 mV / eB cos, An electron moves through a uniform magnetic field given by, B Bx i% 3Bx ˆj. At a particular instant, the electron has the, , velocity v, it is 6.4 10, 6., , 3., , Ans. 8 cm., At the moment t = 0 an electron leaves one plate of a parallel, plate capacitor with a negligible velocity. An accelerating voltage,, varying as V = at, where a = 100 V/s, is applied between the, plates. The separation between the plates is = 5.0 cm. What is, , 2.0iˆ 4.0 ˆj m / s and the magnetic force acting on, 19, , kˆ N . Find Bx., , Ans. –2.0 T., , A wire of 62.0 cm length and 13.0 g mass is suspended by a pair, of flexible leads in a uniform magnetic field of magnitude 0.44 T, (figure). What are the magnitude and direction of the current, required to remove the tension in the supporting leads ?, , Ans. 467 m A towards right.

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380, , ELECTRICITY & MAGNETISM, , Subjective, , E xercise 5. 6, Solution from page 391, , 1., , A charge of 1.0 C moves with a speed of 2.0×106 m/s along the, positive x-axis. A magnetic field B of strength 0.20 ˆj 0.40kˆ T, exists in space. Find the magnetic force acting on the charge., Ans., , 2., , 0.8 ˆj 0.4kˆ N, , A proton, a deutron and an alpha particle moving with equal kinetic, energies enter perpendicularly into a magnetic field. If rp, rd and r, are the respective radii of the circular paths, find the ratio, , rp, r, , ., , Ans., , rp, rd, , and, , 6., , 1, ,1, 2, , 3., , A particle having a charge of 20 C and mass 20 g moves along a, circle of radius 5.0 cm under the action of a magnetic field B = 1.0, T. When the particle is at a point P, a uniform electric field is, switched on and it is found that the particle continues on the, tangent through P with a uniform velocity. Find the electric field., Ans. 0.5 V/m, the direction of field is radially outward at P., , 4., , A narrow beam of singly-charged carbon ions, moving at a constant, , Ans. (a) 14 cm (b) 4.7×10–8 s., An electron is released from the origin at a place where a uniform, electric field E and a uniform magnetic field B exist along the negative, y-axis and the negative z-axis respectively. Find the displacement, of the electron along the y-axis when its velocity becomes, perpendicular to the electric field for the first time., , Ans. y, , 4, , velocity of 6.0 10 m / s, is sent perpendicularly in a rectangular, region having uniform magnetic field B = 0.5 T (see figure). It is, found that two beams emerge from the field in the backward, direction, the separations from the incident beam being 3.0 cm and, 3.5 cm. Identify the isotopes present in the ion beam. Take the, mass of an ion = A 1.6 10, , 27, , 7., 8., , 2 Em, eB 2, , ., , A current loop of arbitrary shape lies in a uniform magnetic field B., Show that the net magnetic force acting on the loop is zero., A wire of length carries a current i along the x-axis. A magnetic, field exists which is given as B, , B0 iˆ, , ˆj kˆ T . Find the, , magnitude of the magnetic force acting on the wire., kg, where A is the mass number.., 9., , Ans. 2Boi, A rectangular wire-loop of width a is suspended from the insulated, pan of a spring balance as shown in figure. A current i exists in the, anticlockwise direction in the loop. A magnetic field B exists in the, lower region. Find the change in the tension of the spring if the, current in the loop is reversed., , Ans. 12C and 14C., 5., , A particle of mass m = 1.6×10–27 kg and charge q = 1.6×10–19C, moves at a speed of 1.0×107 m/s. It enters a region of uniform, magnetic field at a point E, as shown in figure. The field has a, strength of 1.0 T. (a) The magnetic field directed into the plane of, the paper. The particle leave the region of the field at the point F., Find the distance EF and the angle . (b) If the field is coming out, of the paper, find the time spent by the particle in the region of the, magnetic field after entering it at E., , Ans. 2Bia

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, 10., , (a), , A wire loop carrying a current I is placed in xy-plane as, shown in figure. If a particle of charge + Q and mass m is, , 16., , placed at the centre P and given a velocity v along NP. Find, its instantaneous acceleration., (b), , If an external uniform magnetic field B Biˆ is applied, find, the force and the torque acting on the loop due to this field., , 381, , A beam of electrons whose kinetic energy is K emerges from a, thin-foil “window” at the end of an accelerator tube. There is a, metal plate a distance d from this window and perpendicular to, the direction of the emerging beam (figure). Show that we can, prevent the beam from hitting the plate if we apply a uniform, magnetic field B such that B, , 2 mK, e 2d 2, , , in which m and e are the, , electron mass and charge. How should B be oriented ?, , Ans. (a) a, , 11., , 0.109 0 IQV, , at 30° with negative x-axis., ma, , (b) = 0.6136 a2IB along y-axis., A coil in the shape of an equilateral triangle of side 0.02 m is, suspended from a vertex such that it is hanging in a vertical plane, between the pole pieces of a permanent magnet producing a, horizontal magnetic field of 5×10–2 T. Find the couple acting on, the coil when a current of 0.1 A is passed through it and the, magnetic field is parallel to its plane., , 12., , Ans. 5 3 10, , Consider a solid sphere of mass m which has a charge q distributed, uniformly over its volume. The sphere is rotated about a diameter, with an angular speed . Show that the magnetic moment M and, , 18., , q, L., 2m, , Ans. 2 a i B sin (up)., Figure shows a rectangular 20-turn coil of wire, of dimensions 10, cm by 5.0 cm. It carries a current of 0.10 A and is hinged along one, long side. It is mounted in the xy-plane, at 30° to the direction of a, uniform magnetic field of magnitude 0.50 T. Find the magnitude, and direction of the torque acting on the coil about the hinge line., , A proton moves at a constant velocity of +50 m/s along an x-axis, through crossed electric and magnetic fields. The magnetic field is, B, , 15., , m., , Figure shows a wire ring of radius a that is perpendicular to the, general direction of a radially symmetric, diverging magnetic field., The magnetic field at the ring is everywhere of the same magnitude, B, and its direction at the ring everywhere makes an angle with a, normal to the plane of the ring. Find the magnitude and direction of, the force the field exerts on the ring if the ring carries a current i., , BiL2, BiL2, (b), 4, 16, , the angular momentum L of the sphere are related as M, , 14., , N, , A circular loop carrying a current i has wire of total length L. A, uniform magnetic field B exists parallel to the plane of the loop. (a), Find the torque on the loop., (b) If the same length of the wire is used to form a square loop,, what would be the torque ? Which is larger ?, Ans. (a), , 13., , 7, , 17., , 2.0 ˆj mT . What is the electric field?, , Ans., , V, 0.10 kˆ ., m, , Two concentric, circular wire loops, of radii 20.0 and 30.0 cm, are, located in the xy-plane; each carries a clockwise current of 7.00 A, (see figure)., , 19., , (a), (b), , Find the net magnetic dipole moment of this system., Repeat for reversed current in the inner loop., Ans. (a) 2.86 A–m2, (b) 1.10 A–m2., , Ans. 4.3×10–3 N-m; negative y., Two protons move parallel to each other with an equal velocity, v = 300 km/s. Find the ratio of forces of magnetic and electrical, interaction of the protons., Ans., , v2, c2, , 1 10, , 6

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382, 20., , 21., , ELECTRICITY & MAGNETISM, , In figure, the long straight wire carries a current of 30 A and the, rectangular loop carries a current of 20 A. Calculate the resultant, force acting on the loop. Assume that a = 1.0 cm, b = 8.0 cm, and, L = 30 cm., , Ans. 3.2 mN, towards the wire., Figure shows a wood cylinder of mass m = 0.250 kg and length L, = 0.100 m, with N = 10.0 turns of wire wrapped around it, longitudinally, so that the plane of the wire coil contains the axis of, the cylinder. What is the least current i through the coil that will, prevent the cylinder from rolling down a plane inclined at an angle, to the horizontal, in the presence of a vertical, uniform magnetic, field of magnitude 0.500 T, if the plane of the coil is parallel to the, inclined plane ?, , Ans. 2.45 A., , 22., , Figure shows a current loop ABCDEFA carrying a current 2 A. The, sides of the loop are parallel to the coordinate axes, with AB = 10.0, cm, BC = 20.0 cm, and FA = 5.0 cm. Calculate the magnitude and, direction of the magnetic dipole moment of this loop., , Ans. M, , 0.02 ˆj 0.04kˆ A m 2

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , 383, , S olutions Exercise 5.1 Level -1, 1., , 2., , 3., 4., , (d), , (b), , (d), (c), , The electric force on the positive charge particle is upward, and so magnetic force must be downwards. For this magnetic, field it is best represented in (4)., If velocity of the solid along y-axis, the electrons will, accumulate at the front face. So, this face will be at lower, potential., Use Fleming's left hand rule to get the answer., As, , mv, qB, , r, , Area,, , A =, , =, 5., 6., , P, qB, , We have, R, , B, , 18., , (d), , 19., , (c), , By Fleming's left hand rule, the force on the particle is along, OY., The forces on the loop are as shown. It shows that the loop, will move towards the wire (F1 > F2)., , RqB., P2, 2m, , 1 2, mv, 2, , 1, m1v12, 2, , qV and, , R1, R2, , m1v1 qB, m2v2 qB, , or, , m1, m2, , R12, , 1, m2v22, 2, , m1, m2, , F1, , i, , F2, , F, , qV, , 12, , 20., , (c), , 21., , (a), , 22., , (b), , The net force on current carrying loop in uniform magnetic, field will be zero., The force between two neighbouring turns will be of, `attractive nature, and so spring will get compressed, By Fleming's left hand rule, the force on the electron stream, is downwards., , i, , R22, , F = qvB sin, , F, qv, , Bmin =, , Pe eB, Pp eB, , (sin, , 10, 10, , 2me K, 2m p K, , 12, , 10, , 105, , Time period, T, , i1i2, , and so F, r, , 23., , (c), , F, , 24., , (c), , Force, F, , 25., , (c), , 26., , (b), , 27., , (c), , 28., , (d), , 1), 3, , 10 T , in z-direction., , me, mp, , As, mp > me, so, rp > re, (d), , 103 Wb m 2., , Electron streams, , =, , 12., , 10 4, 10, , F, , ( mv), , or, , E, v, , ( RqB)2, 2m, , m2v22, , Now,, , re, rp, , mAvA mBvB, , (b), , m1v12, , (b), , When charge particle is released at rest in electric and magnetic, field, the electric force moves it along the direction of electric, field., The work done by magnetic field for any displacement will, be zero, because it always acts perpendicular to the velocity, vector., , 17., , For equal charges,, , 11., , (a), , 2m, K, qB, , (c), , (d), , 15., , P2, qB, , 8., , 10., , For equal charge of the particles, radius, r ( mv) and so, , (c), , Electric field exerts force on electron in opposite direction of, projection., , r, , (b), , 16., , (d), , (b), , 14., , 1, , 2, , 7., , 9., , P eB, P eB, , (c), , P, qB, , mv, , or mv, qB, , Thus kinetic energy, K, , re, rp, , 13., , r2 =, , (a), (d), , velocity of the particle., , 2 m, , and so T does not depend on the, qB, , M, L, , ii, .12, 2 r, , 2F, ., 3, , 0, , 0, , 2, , ., , i2, ., d, , q, 2m, , By Fleming's left hand rule, each element of the loop, experiences radially outward force., Potential energy, U = – MB cos , and so, it greatest when, = 180° or U = – MB cos 180° = MB., The angle between magnetic field produced by the wires and, velocity of the charge particle will be zero, and so, F = qvB sin 0° = 0.

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384, 29., , 30., , (d), , (c), , ELECTRICITY & MAGNETISM, The electron will experiences a downward force and so, electron will move along negative y-axis, speed of the electron, remains constant in magnetic field, F = q (v, , B), 19, , = 1.6 10, , [4 105 (sin 30 iˆ cos30 kˆ) 4 kˆ], , E, ., B, Magnetic force on the proton is zero. The electric force will, accelerate the proton., Both types of ions experience force along –y, direction., or, , 37., , (a), , 38., , (c), , v =, , z, , = –12.8 × 10–14 ĵN, 31., 32., 33., , (a), , The angle between v and B is zero and so,, , (b), (d), , F = qvB sin 0° = 0, Use Fleming's left hand rule., If F4 is the force on segment QP, then, F4, , = 0, , and, , F4, , =, , or, , F4 =, , F1 F2, , F3, , B, , ( F1 F2, , F3 ), , F1 )2, , ( F3, , 39., , 35., , (c), , r=, , 36., , (c), , 2, , (1.67 10, , 27, , ) (2 106 )sin 30, , 1.6 10, 1.67 10, , 1.6 10, , 19, , 19, , 0.104, , 27, , 2, , 0.104, , 10, , 7, , s., , For undeflected charged particle,, Fe, , = 0, , B ) Eq, , = 0, , Fm, , or, , q (v, , (b), , F22 ., , Charge particle moves along a circular path under magnetic, field only., , 2 m, qB, , x, , E, , F, , (a), , T=, , +, –, , F, , 34., , mv sin 30, qB, , y, , The radius of path, r = (b – a). If v is the required velocity,, then, r = (b a), , or, , 0.1 m, 40., , (b), , 41., 42., , (a), (a), , v =, , mv, qB, , qB(b a), m, , There is only magnetic force of attraction between A and B., While in x and y there is more electric repulsion., Use Fleming's left hand rule., When current is set up in the rings, they produce magnetic, field along the axis. So one ring exerts torque on the other,, = mB sin 90°., , S olutions Exercise5.1 Level -2, 1., , 2., , (c), , (a), , F1, , q (v, , B), , q [(2iˆ 3 ˆj ) 4 kˆ ] q ( 8 ˆj 12iˆ), , F2, , q (v, , B), , q [(3iˆ 2 ˆj ) ( 4kˆ)], , F3, , q (v, , B), , q [(3 ˆj, , F4, , q (v, , B), , q [(20 iˆ) ( 4iˆ)] 0, , 2kˆ ) (4iˆ)], , q (8 ˆj 12iˆ), q ( 12kˆ 8 ˆj ), , It shows F1 = F2 = F3 and F4 = 0, For x = a to 2a ;, F, , q (vˆ B ), , q (v0iˆ B0 ˆj ), , and, , q [v0iˆ (– B0 ˆj )], , 3., , (a), , Radius,, , r =, , rP =, , 2mk, ,, qB, , =, , 2 (4m) k, (2q ) B, , rP, , 5., , (b), , For non-circular path, there must be presence of E -field, and B-field. For the given path the magnetic force is initially, along –y- axis, and so magnetic field must have a component, along z-axis and may have a component along x-axis., , 6., 2mk, qB, , r, , = mB sin = iAB sin . Area of circular loop is maximum for, given perimeter, and so will be maximum in case (d)., , (c), , It represents graph as in (a) is correct., P, qB, , 2 (2m ) k, ,, qB, , (d), , qv0 B0 kˆ, , qv0 B0kˆ, , =, , 4., , For x = 2a to 3a ;, F, , r, , F, , or, , 7., , (b), , 8., , (c), , =, , FAB, , FBC, , FCA, , F =, , Bi 1 Bi 2 Bi 1, = 4 Bi = 4 × 0.5 × 5 = 10 N, The path of the charged particle in this case will be cycloid., See theory of the chapter., F1, , =, , ejˆ, , Bz = –1, , e [iˆ ( Bxiˆ By ˆj Bz kˆ)]

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, v =, and, Now, , 2 gR sin, , sin, , F = qvB, , mv 2, N – (F + mgsin ) =, R, , N =, , F + mgsin +, , = qB, N =, , 27., , (d), , r, , 2 gR sin, , mv 2, R, , + mgsin + 2mg sin, , 30., , (c), , Fm, , 3mgsin + qB 2 gR sin, , Fe, , i1, , in, gs, (m, , 2, , q v, , B, , =, , q 2iˆ 3 ˆj 4kˆ, , =, , q, , 6kˆ 8 ˆj, , 0, Fm, , =, , –q, , 6kˆ 8 ˆj, , m T, =, , the direction of motion becomes opposite, qB, 2, to initial and so applied field may be, , 31., , ii, 2 012, 2, r, , =, , (c), , E =, 6iˆ 8 ˆj units, For wire not to slide on the rails,, f, , =, , Fmagnetic, , mg, , =, , Bil, , =, , Bil, mg, , ), , T, , T, , T, , (Bi ), To just move the block up,, T = mg sin + mg cos, and, 2T = Bi, i, , 2mg sin, , 2 mg cos, , or, , a – 2bt =, , or, , t =, , or, , =, , min, , B, 32., , (b), , bt 2, , B at, , =, , mg, d, =0, dt, , For minimum value of ,, , 29., , 2iˆ, , After, , or, os, , kr, 2, r, , =, , Fm Fe, , =, , gc, µmm, +, , =, , k =, , i, Magnetic field at the centre of the shell = 2 0 1, 2, r, Force per unit length of the wire, F = Bi2 × 1, , (c), , 4, , x, r, , –i1, , B, I2, , 28., , sin, , or, , =, , 387, , 0, a, 2b, , Ba 2, 4bmg, , v, , (b), , /4, r/2, , kr, 2, , kr, 2, , sin, , =, , r, 2, r, , 1, 2

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388, , ELECTRICITY & MAGNETISM, =, , rad, 6, So velocity of image of proton, , or, , 33. (b), , = 6 cos iˆ 6sin ˆj, 6, 6, , v, , az =, , Fz, m, , =, , 3 3 iˆ 7 ˆj m/s, , =, , Eq, m, , 10q, m, , v 2 = 0 + 2az, , Now, , = 3 3 iˆ 3 ˆj m/s, or, , Velocity of object w.r.t.minor(perpendicular to minor), , v0 mirror, , =, =, , 3 3 iˆ 3 ˆj, , 34., , 2 ˆj, , W =, or, , 3 3iˆ 5 ˆj m/s, , Velocity of image w.r.t. mirror, =, , (c), , E0 q × b =, , K, 1, m v22, 2, , 2, 2E0qb = m v2, , or, , 3 3 iˆ 5 ˆj m/s, , 20qz, m, , 2az =, , v =, , v12, v12, , Velocity of image w.r.t. ground observer, =, , 3 3 iˆ 5 ˆj, , 2 ˆj, , S olutions Exercise-5.2, 1., , (b, d)Radius of charged particle in magnetic field,, r =, , 1, B, (a, c) The charged particle at rest, experiences force due to electric, field., r, , 2., 3., , For v, , q B, , the particle will enter into region III., m, , For v, , q B, , the particle follows the path as shown., m, , (mv) and r, , (b, c, d) For without change in velocity, either E, B, , 4., , mv, , and so, qB, , 0, or E, , v, , 0 and, , Region I, , 0, so that Fe = Fm., , (a, b) For without acceleration, velocity of the particle must be, constant. For this Fe = Fm. The possible case may be shown, in figure., , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , v, , Region II, , 6., , (a, b, d) The work is only done by electric field, and so, Eq × 2a =, , y, E, or, , v, z, 5., , B, , x, , E =, , r =, , or, , v =, , q B, m, , 1, m (2v )2 v 2, 2, 3 mv 2, ., 4 qa, , Rate of work done at P,, = Fv cos 0°, = Eqv, , (a, c) If r is the radius of path, the, , mv, qB, , Region III, , =, , 3 mv 2, 4 qa, , qv, , 3 mv3, 4 a, , Rate of work done at Q,, = F × 2 v × cos 90° = 0

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389, , MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, , Solutions Exercise-5.3, 1., , (d), , 2., , (a), , 3., , (d), , 4., , (d), , 5., 6., 7., , (a), (c), (d), , 8., , (d), , Magnetic field can not change the speed, but can change the, velocity, and so can cause acceleration., Because of random motion of the free electrons, the force on, the electron is also randomly oriented and so net force on the, conductor becomes zero., The magnetic force on the loop is only zero when it placed in, uniform field., Magnetic field can not change the kinetic energy of the, particle, because magnetic force is always perpendicular to, its velocity vector., The mass of electron has relativity effect., The value of torque depends on . For = 0, = MBx, sin 0° = 0., The magnetic force acts only on moving electrons, and so net, force on the conductor is non-zero., , 9., , (d), , 10., , (b), , If charged particle projected in the line of electric field, it, goes undeviated. Or particle is projected in both the field, such that qE = qvB, then it also goes undeviated., Self explanatory., , 11., , (a), , The radius of path, r, , 12., , (a), , than mass of proton, so radius of electron is smaller., Use Fleming's left hand rule., , 13., , (b), , For v, , 14., , (d), , mv, . As mass of electron is smaller, qB, , E, E, , Fm = Fe or q vB = Eq. For v, , magnetic, B, B, force Fm becomes smaller than electric force., In magnetic field, the speed of charged particles remains, constant., , Solutions Exercise-5.4, Passage for Q. 1 to Q. 3, 1., (a) By Fleming's left hand rule, the force on electrons will be, along east., 2., , 1 2, mv, 2, , (d), , Passage for Q. 4 to Q. 6, 4., , (a), , 2K, m, , v =, , = K, , 2 3.2 10, , =, or, , 2 12 103 1.6 10, , 2K, m, , v =, , mv, qB, , Now, , 31, , 6.48 10 7, , 1.6 10, , 19, , 55 10, , v2, r, , (6.48 10 7 ) 2, 6.7, 14, , 3., , (b), , y, x, , mv, qB, , B =, , mv, qr, , 5., , (a), , =, , Current,, , 6., , (c), , 3.92 10, , 25, , 3.20 10, , 4.04 105, 19, , 1, , = 495 × 10–3 T., The number of ions in the machine, , m/s2, , = 6.23 10, If y is the required deflection, then, (r – y)2 + x2 = r 2, Here,, x = 20 cm, and r = 6.7 m., After substituting the values and simplifying, we get, y = 2.98 mm, , R, , r =, , 6, , = 6.7 m, Acceleration of the electron, a =, , 25, , = 4.04 × 105 m/s, , 9.1 1531, , 9.1 10, , 100 103, , 3.92 10, , 19, , = 6.48 × 10+7 m/s., Radius of path, r =, , 19, , i =, , 100 10, , 6, , 3.92 10, , 25, , q, t, , 25.51 1019, , 25.51 1019 3.20 10, 3600, , = 22.7 × 10–3 A., Charge flows in one hour,, q = 25.51 × 1019 × 3.20 × 10–19, = 81.6 C., Now work done, W = Vq, = (100 × 103) × 81.6, = 8.17 × 106 J., , 19

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390, , ELECTRICITY & MAGNETISM, , Passage for Q. 7 to Q. 9 :, , ax t, , vy, , v0 cos, , v0 cos, , and vz, , v0 sin, , v0 sin, , Also v, , t, , Eq, t., m, , vx, , 7., , Bqt, m, Bqt, ., m, , vx iˆ v y ˆj vz kˆ . ], , 8. (b), 9. (b), Passage for Q. 10 to Q. 12 :, See examples., 13. A-p, q, r, s; B-p, q, s; C-s; D-r, 14. A-q; B-r; C-p; D-s, , Bqt, m, , dF, , Id, , B;, , µ, , B, , S olutions Exercise-5.5, 1., , The situation is shown in figure., I =, , 1 2, mv, 2, , Given,, , =, , qV, , 3., , Electric field,, , E =, , v, , or, , 0.1 m, , =, , Ee, m, , dv, dt, , =, , 100t e, m, t, , dv, , or, , =, , 0, , =, , 2 (2 1.6 10, , 19, , 6.4 10, , 27, , =, , 106, , r =, , mv, qB, , =, , ) 104, , =, , 2., , m/s, , r =, , dx, dt, , Again, , 27, 19, , 2 (1.6 10, , x, r, , 0.1, 0.2, , By lens formula,, or, Now, , 1, v, , 1, 18, , 1, 2, , x, , For, , x =, , Ans., , 2 10, , v, I, O, , 10–2, , v, u, , =, , =, , t =, , 1.2, , e, , …(i), , 50e 2, t dt, m, , 50et 3, 3m, , …(ii), , , we have, 50et 3, 3m, , 3m 2, 50e, , 1/ 3, , Substituting this value in equation (i), we get, , m, , 1, 12, , = 36 cm, =, , 3, , =, , or, , (2 10 ) (4.8), , =, , 50e 2, t, m, , 0, , mv, qB, , = 4×, , =, , 0, , ) 0.1, , 5, , =, , 50e 2, t, m, , t, , dx, , 106, , 100t /, m, , 100te, dt, m, , =, , x, , 6.4 10, , = 30°., , Radius of the circle,, , 0, , v, , = 0.2 m, , or, , Ans., , 100t, , a, , v, , 2qV, m, , v =, , sin, , V, , 4 8 cm, , The acceleration of electron,, , v, , Radius of path,, , v, 36, O=, u, 18, , v, , 4., , =, , 50e 3m 2, m, 50e, , If v is the velocity of the electron, then, 1 2, mv, 2, , = eV, , 2/3, , 16 km/s Ans.

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391, , MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, v, , 5., , 2eV, m, , =, , or, T, , v cos, , Pitch =, =, , 2 m, eB, , =, , 2, , v cos, , 2 m, eB, , 6., , 2eV, cos, m, , 2mV, , cos, , eB 2, , Ans., , F, , =, , q(v B), , 6.4 10–19 kˆ, , =, , – e (2.0iˆ 4.0 ˆj ) ( Bx ˆj 3B y ˆj ), , Bx, , =, , –2.0 T, , Magnetic force,, , Ans., , The tension in the leads will be zero, if, weight of wire = magnetic force in upwards direction, or, mg = Bi, or, , mg, B, , =, , i, , 467mA towards right., , S olutions Exercise-5.6, 1., , 2., , F, , We know that,, , P, , We have,, , rd, , and, 3., , q ( v B), , =, , 1 10 6 [2 106 iˆ (0.20 ˆj 0.40kˆ)], , =, , –0.8 ˆj 0.4kˆ N, , =, r, , rp, , rp, r, , E, , mv, qB, , =, , 2mK, qB, , P, qB, , 2mK / qB, , =, , =, , 2(2m) K / qB, 2 mK / qB, , =, , (b), , 1, 2, 1, , 2(4m ) K /(2q ) B, , =, , qE, , =, , vB, , =, We have, , or, , r, m, , =, , m, , =, =, , A, , =, , The radius of the path, mv, qB, , EF, , =, , 2r sin 45, , =, , 0.14 m, , 6, , t, , Ans., 6., , =, , ) 1, , 0.1 m, , 1, , 2 0.1, , 2, , Ans., , 2, , 1.6 10, , 1.6 10, , 19, , 27, , 1, , 6.28 10, , 18, , s, , 3T, = 4.7 × 10–8 s, 4, , Ans., , Given, E – Ejˆ and B – Bkˆ ., The net force of the electron, vy, , v, , x, , 12, , 9, , 0.5 V/m., , =, , qE q (v B), , =, , ( e)(– Ejˆ) (– e)[(vxiˆ v y ˆj ) ( Bkˆ)], , =, , eEjˆ ev x Bjˆ ev y Biˆ, , =, , ev y Biˆ e( E v x B) ˆj, , a, , =, , F, m, , ax, , =, , dvx, dt, , ay, , =, , F, , Ans., , mv, qB, , rqB, v, , 0.015 1.6 10, , 6.0 10, 26, , 19, , 0.5, , 4, , The acceleration, , kg, , m, 1.6 10, , 19, , (1.6 10, , vx, , 20 10, , 2 10, , ) (1 107 ), , When magnetic field is reverse in direction, the particle covers angle 270° inside magnetic field., , y, , 0.05 20 10, , =, , =, , 2 m, qB, , 27, , (1.6 10, , r, , Time period, T =, , rqB, B, m, , =, , =, , (a), , 2mK, , For the given condition electric force on the particle is equal to, magnetic force. Thus, qvB, , 4., , 5., , =, , 27, , 12.5, , It will be 12. So the isotopes are of mass number 12 and 14. These, are 12C and 14C., , and, , dVy, dt, , ev y B, m, , iˆ, , e, ( E vx B) ˆj, m, , ev y B, m, , e, ( E vx B ), m, , Differentiating equation (ii) w.r.t , we get, , … (i), … (ii)

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MAGNETIC FORCE ON MOVING CHARGES & CONDUCTOR, =, , 1, 0.02 0.02, 2, , =, , 1.7 3 10, , Torque,, , 12., , (a), , Given,, , =, 0.1, , =, , 5 3 10, , 2 R, , =, , L, , R, , =, , 3 10, , Torque,, , iAB, , i ( A2 – A1 ), , =, , 1.10 A- m2, , 1 2, mv, 2, , =, , v, , =, , r, , =, , B, , =, , 5 10 sin 90, , Ans., , L, B, 4, , For square of side a,a, , =, , L/4, L, 4, , a2, , =, , iAB, , =, , 2, , L, 4, , 2, , BiL, N-m, 4, , mv, eB, , d, , L, 16, , L, 4, , L, 16, , 17., , i, , =, , BiL2, N-m, 16, , 2mK, e2d 2, , =, , q, (4 / 3) R3, m, (4 / 3) R3, , ., , b sin, , and By, , B cos, , F, , B, , Ans., The force due to Bx is, 2 a, , sphere, take a disc of volume dV, so we can write, , 2, , e2d 2, , The magnetic field can be resolved into two components; Bx and, By, These are, Bx, , =, , =, , 2mK, , B, 2, , 2K, m, ed, , m, , mv, ed, , Thus to prevent the beam from hitting, we have, , Ans., , M, q, for a uniformly charged disc is found equal to, . From the, L, 2m, , M, L, , 2K, m, , =, , i, , =, , L, 2, , 2, , 2, , 13., , K, , 2, , R2, , =, , Torque, , =, , L, 2, , A =, , A, , M, , If v is the speed of the electron, then, , We have, , The area of the loop,, , Area,, , 4, , N-m, , 7, , 2, , (b), , 16., , m2, , iAB sin, , =, , =, , 4, , (b), , 3, 2, , 393, , dV, dV, , F, , q, ., 2m, , Bx idl, , =, 0, , =, , Bx 2 ai, , =, , 2 aiB sin, , Clearly for the sphere of total charge q and mass m, M, , 14., , q, L., 2m, , For constant speed, net force on the particle is zero. So, F, , 15., , =, , =, , 0 = q( E v B ), , or, , 0, , =, , +e [ E 50iˆ 2 10 3 ˆj ], , or, , 0, , =, , [ E 0.10kˆ], , (a), , E, , =, , M, , =, =, , 0.10 kˆ V / m, , i ( A1, , A2 ), , 2.86 A, , m2, , The force due to B y is zero., Ans., , 18., , The area of loop, A 0.1 0.05 5 10 3 m2, The torque,, , =, , NiAB sin, , =, , 20 0.10 (5 10 3 ) 0.5sin 60, , =, , 4.3 × 10–3 N-m, , Ans.

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394, 19., , ELECTRICITY & MAGNETISM, , We know that, Fm, Fe, , =, , v, , i, , 2, , 20., , (3 108 ) 2, , 1 10, , 6, , Ans., 22., , =, , F1 F2, , =, , µ0 (30 20) L, a, 2, , =, 21., , (300 10 ), , The force between wire and loop, F, , 3.2 mN, , NiAB sin, , =, , weight, , =, , mg sin, , (30 20) L, ( a b), , Ans., , For the equilibrium of the cylinder, we have, magnetic, , mgR, NAB, , =, , 0.25 9.8 R, 10 ( L 2 R ) 0.5, , =, , 2.45 A, , e2, 3 2, , =, , =, , R, , Ans., , The given loop can be spitited into two loops of area,, , and, , A1, , =, , 0.20 × 0.1 = 0.02 m2, , A2, , =, , 0.20 × 0.05 = 0.01 m2, , The net magnetic moment, M, , =, , i[ A1 ( kˆ ), , =, , 2[0.02( k ) 0.01 ˆj ], , =, , [0.02 ˆj 0.04 kˆ] A-m 2, , A2 ˆj ], , Ans.

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396, , ELECTRICITY & MAGNETISM, , 6.1 MAGNETIC, , FIELD OF MOVING CHARGE, , We know that a point charge q, at rest in the observer's inertial frame, produces an, electric field along the radius vector and is given by, , E, , =, , 1, 4, , q, 0, , r3, , r., , If the charge is moving relative to the observer's inertial frame, it produces a magnetic, field in addition to electric field. The magnitude of which is proportional to the speed of, the charge relative to the observer provided (v < c). The magnetic field vector B at the, point P, a distant r from the charge q moving with a velocity v is found to be, , B, Fig. 6.1, , =, , µ0 q, (v r ). … (1), 4 r3, , The direction of B is thus perpendicular to the plane of v and r . It is in the direction, of advance of a right handed screw rotated from v to r . Its magnitude is given by, B, , =, , µ0 qv sin, 4, r2, , … (2), , The following points should be remembered regarding with the magnetic field :, 1., , The magnetic field B is zero at all points on a line on which charge moves. That is, , 2., , when = 0 or = 180°, B = 0., It is maximum in the plane perpendicular to the v and through the charge, as, sin, , 3., , 1 , at all points in this plane :, , B remains unaltered in magnitude at all points on the circumference of circle, passing through P and lying in a plane perpendicular to v with its centre on the, velocity direction., , 4., , The direction of B is given by either :, (a) Maxwell's right hand screw rule : If the direction of current through the, conductor or the direction of velocity of positive charge is represented by the, linear motion of the screw motion, then the direction of the magnetic field can, be represented by the direction of rotation of the screw., (b) Right hand claps rule : If a current carrying conductor is clapsed in the right, hand so that the thumb indicates the current direction, then the direction of, the magnetic field is represented by the finger tips round the wire., , Electromagnetic field, Fig. 6.2, , So far, we have considered electric and magnetic fields separately, without establishing, any clear relation between them. This could be done only because the two fields were, static. In other cases, however, it is impossible. It will be shown that electric and magnetic, fields must always be considered together as a single total electromagnetic field. In other, words, it turns out that electric and magnetic fields are in certain sense the components, of a single physical object which we call the electromagnetic field. The division of the, electromagnetic field into electric and magnetic fields is to relative nature since it depends, to a very large extent to the reference system in which the phenomenon are considered.

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MAGNETIC EFFECT, , OF, , CURRENT, , Force between moving charges, The force acting on a charge q2, moving with velocity v2 in a magnetic field produced by, charge q1 moving with a velocity v1 is, , F21, , =, , q2 (v2 B1 ), , =, , q2 v2, , µ0 q1, 4 r3, , (v1 r ), , µ0 q1q2, [v2 (v1 r )] ., 4 r3, The magnitude of force which they exert on each other, =, , Fm, For, , v1, , µ0 q1 q2 v1v2, 4, r2, = v2 = v, , =, , Fig. 6.3, , µ0 q1q2 2, v, .....(i), 4 r2, In addition to the magnitude force, there is an electric force between them, whose, magnitude is given by, Fm, , =, , q1q2, , 1, , ., ......(ii), r2, This force is of repulsive nature. On dividing equation (i) by (ii), we have, , Fe, , =, , Fm, Fe, , =, , 4, , 0, , v 2 µ0, , 0, , 1, , As, , c, Fm, Fe, , Since v < c, and so Fm, repulsive nature., , 6.2 THE, , Fe . As Fm, , =, , =, , µ0, v2, c2, , ., 0, , ., , .....(3), , Fe , so the net force between the charges is of, , BIOT SAVARTS LAW, , Biot-Savarts law is the fundamental law of the magnetics. It gives the magnetic field by, a small current element and is based on the experimental facts. Let us consider a small, element d, , of a wire of cross-sectional area A carrying current i. If n is the number of, , charge carries per unit volume, each of charge q and moving with a drift velocity vd ,, then, i = neAvd, The quantity of charge flowing in time dt,, dq = idt = neAvd(dt), Since vd (dt ) = d ,, , dq, , = neAd ., , The magnetic field due to the current element d, dB, , =, , at any point P at a distance r, , µ0 dq(vd r ), 4, r3, , Fig. 6.4, , 397

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398, , ELECTRICITY & MAGNETISM, , or, , dB, , =, , µ0 (neAd )(vd, 4, r3, , r), , =, , µ0, (d r ), (neAvd ), 4, r3, , =, , µ0 id r, 4, r3, , … (1), , This is called Biot-Savarts law., The field due to the whole current carrying conductor is given by, , or, , B, , =, , B, , =, , dB, , µ0, 4, , i(d, , µ0 i r, . … (2), 4 r3, , r), r, , 3, , Magnetic field due to a straight current carrying conductor, Consider a straight wire carrying current i. Its ends subtend angles, at which field is to be determined., Take a small element d, element is, , of the wire at a distance, , dB, , =, , 1 and, , 2 at, , point P, , from O. The magnetic field due to, , µ0 id sin(90, 4, x2, , ), , . … (i), , The direction of field at P is perpendicular to the plane of the diagram and going into it., The direction of the field is the same for all elements of the wire and hence net field due, to the wire is obtained by integrating equation (i)., From the diagram,, , =, , tan, , =, , r tan, , d, , =, , r sec 2 d ., , dB, , =, , µ0 i (r sec 2 d ), cos, 4, x2, , r, , or, and, Putting in equation (i),, Fig. 6.5, , Also, , x, r, , dB, , Total field, , B, , = sec, , or x = r sec, , =, , µ0 i, cos d, 4 r, , =, , µ0 i, 4 r, , 1, , cos d, 2, , or, Fig. 6.6, , B, , =, , µ0 i, (sin, 4 r, , 1, , sin, , 2), , We can draw magnetic field lines in the same way that of electric field lines. A tangent to, a magnetic field line gives the direction of the magnetic field existing at that point. For a, long straight wire, the field lines are circles with their centres on the wire.

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MAGNETIC EFFECT, Special cases :, (a) Field due to a long straight wire, 1, , 2, , and, , 2, , B =, , (b), , OF, , CURRENT, , Field along the end of the wire, 1 = 0 and 2 =, , 2, µ0 i, ., 2 r, , /2, B =, , µ0 i, ., 4 r, , Force between two parallel current carrying wires, Consider two long straight wires kept parallel to each other at a distance r and carrying, currents i1 and i2 respectively in the same direction., Magnetic force on the small element d of the wire 2 due to the magnetic field of 1 is, dF, , = i2 (d, , dF, , = i2 (d ) B1 sin 90, , B1, , =, , µ0 i1, 2 r, , dF, , =, , µ0 i1i2 ( d ), 2, r, , =, , µ0 i1i2, 2 r, , (a), , B1 ), , The magnitude of the force, , where, , dF, d, , or, , Thus force per unit length of wire 2 due to 1 is, , (b) Variation of B with r., , µ0 i1i2, . The same amount of force 1, 2 r, , exerts on 2., The direction of force on wires are towards each other that is attraction (By FLHR)., If direction of currents in wires are opposite, the force between them will be repulsive., , (c), Fig. 6.7, , Fig. 6.9, , Note:, , There is no electrical interaction between the wires, because they have no, , net charges., , Field due to a circular current carrying coil, Consider a circular loop of radius a carrying current i. We have to find the magnetic field, at a point P on the axis of the loop at a distance x from the centre of the loop. Consider, a current element i d, , of the wire. The magnetic field at P due to this element is, dB, , =, , µ0 d rˆ, .i, 4, r2, , As d is perpendicular to the plane of the paper, and so d, paper, as shown in figure., The magnitude of the field, , rˆ must lie in the plane of, , Fig. 6.8, , 399

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400, , ELECTRICITY & MAGNETISM, dB, , =, , µ0 id, 4 r2, , Fig. 6.10, Another element on the diametrically opposite point produces the same amount of field, on the point P. The resultant field due to these elements = 2dB sin ., The effective field of one element at P,, Field due to complete loop, B, , = dB sin, , =, , =, , or, Since, , µ0, 4, , 2 a, 0, , id, r2, , sin, , =, , µ0 i 2 a a, 4, r, r2, , B, , =, , µ0 2i a 2, 4, r3, , r, , =, , B, As, , dB sin, , i( a2 ), , =, , a2, , x2, , µ0 ia 2, 2(a 2, , x 2 )3 / 2, , = iA = M, where A is the area of the loop, = magnetic moment of the loop,, , µ0, 2M, ., 4 ( a 2 x 2 )3 / 2, Field at a point far away from the centrex > > a,, , B, , =, , µ0 2 M, 4 x3, From equation (1), we can get field at the centre of loop, x = 0, µ0i, ., B =, 2a, If there are N turns in the loop, µ0 Ni, B =, 2a, B, , … (1), , =, , … (2), , … (3), , … (4)

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MAGNETIC EFFECT, , OF, , CURRENT, , Note:, 1., 2., 3., , Equations (3) and (4) are derived for circular loop, but they can be used for other, shape of the loop also which has symmetry about the axis of the loop., In equation (4), N may be a whole number or fraction or decimal number., For the figure shown, N, , 2, , Fig. 6.11, , i, 2, ., 2a, The magnetic field at a point not on the axis of the loop is mathematically difficult, to calculate. The field lines for the circular loop are not circles, but they are closed, curves that link the conductor., , B, , 4., , =, , µ0, , =, , Field between two similar coaxial circular loops, Let us consider two loops, each having N turns and carrying current i are placed at a, distance 2d apart., , Fig. 6.12, , Fig. 6.14, Assuming the current is flowing in the same direction in each coil, the magnetic field at, a short distance x from midway point O, B, , =, , µ0 Nia 2, 2, (a 2, , 1, , 1, , x12 )3 / 2, , µ0 Nia 2, =, 2, [a 2, , (a, , 2, , x2 2 )3 / 2, , 1, (d, , 1, 2 3/ 2, , x) ], , [a, , 2, , The field will be uniform between the loops, if, (d, µ0 Nia 2, ( 3), 2, 2, [a (d, , or, Now, , x), 2 5/ 2, , x) ], , (d + x) [a2 + (d – x)2]5/2, [a 2, , (d, , 3, , x )2 ]5 / 2, , [a, , x) 2 ]3 / 2, , (d, dB, dx, , 2, , 0 i.e.,, , (d, , x), , (d, , x ) 2 ]5 / 2, , 0, , = (d – x) [a2 + (d + x)2]5/2, = [a 2, , d2, , x2, , 2 xd ]5 / 2, , d2, , 2 xd ]5/ 2, , Since x is small, so neglecting x2, we have, = [a 2, , ......(i), , Fig. 6.13, , 401

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402, , ELECTRICITY & MAGNETISM, =, , 2xd, a, [a 2, , Similarly, , 2, , d 2 )5 / 2 1, , (a 2, , d 2 )5 / 2 1, , 5/ 2, , 2 xd, ( a2, , d2), , 5 xd, a, , 2, , a, , 2, , d2, , << 1, , d2, , x)2 ]5 / 2, , (d, , (a 2, , (a 2, , =, , 5 xd, , d 2 )5 / 2 1, , d2, , Substituting these values in equation (i), we get, (d, , 5 xd, , x) 1, a, , 2, , d, , 5d 2, , or, , (a 2, , and, , B, , a, , =, , 2, , µ0 Nia 2, 2, 2, , 1, a, , or, , B, , =, , d2, , d = a/2, or a = 2d, , 1,, , d2), , 5 xd, , x) 1, , (d, , =, , 2, , 8µ0 Ni, 5 5a, , 2, , a, 2, , 2, , 3/ 2, , ., , … (2), , Fig. 6.15 Variation of magnetic field between the loops., , Ex. 1 A wire carrying current i has the configuration shown, in fig. 6.16. Two semi-infinite straight sections, both tangent to the, same circle, are connected by a circular arc, of central angle ,, along the circumference of the circle, with all sections lying in the, same plane. What must be in order for B to be zero at the centre, of the circle?, Sol., , B1 = 2, , µ0 i, ,, 4 R, , out of the plane, and field due to curved part of the wire, , The field due to the straight parts of the wire, , B2 =, , µ0, , 2, 2R, , i, , into the plane of the wire., , The resultant field at the centre to be zero, B1 = B2, , Fig. 6.16, , or, , 2, , µ0 i, 4 R, , µ0, =, , 2, 2R, , i, = 2 rad, , Ans.

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MAGNETIC EFFECT, , OF, , CURRENT, , 6.3 SOLENOID, A solenoid is a wire wound in a closely spaced spiral over a hollow cylindrical nonconducting core. The wire is coated with an insulating material so that the adjacent turns, physically touch each other, but they are electrically insulated., dN = ndx, If n is the number of turns per unit length, each carrying a current i, uniformly wound, round a cylinder of radius a, then the number of turns in length dx is ndx. Thus the, magnetic field at the axial point P due to the element, dB, , =, , (a), , µ0 ( ndx)i, 2( a, , 2, , 2 3/ 2, , x ), , The direction of magnetic field is along the axis of the solenoid and in the sense of, advance of a right handed screw. From geometry, we have, x, and, , Total field, , = a cot (180° – ) = – a cot, , dx, , =, , acosec2 d, , dB, , =, , µ0 ni sin d, 2, , =, , µ0 ni, 2, , B, , and hence, (b), , 2, , sin d, 1, , or, , B, , =, , µ0 ni, – cos, 2, , 2, 1, , =, , µ0 ni, [cos, 2, , cos, , 1, , (c), Fig. 6.17, 2 ]., , … (1), , Special cases :, Case 1 : Solenoid is of infinite length and the point chosen is at the middle, 1 = 0, 2 =, B, , = µ0 ni, , Fig. 6.18, , Case 2 : Solenoid is of infinite length and the point is at the end of the solenoid, 1, , /2 ,, , 2, , B, , =, , µ 0 ni, 2, , Fig. 6.19, , Note:, 1., , If the length of the solenoid is large compared with its radius, the internal field, near its centre is very nearly uniform and parallel to axis, and the external field, near the centre is very small. In practice we take the magnetic field inside very, tightly wound long solenoid is uniform everywhere and zero outside it., , 2., , If some material medium is present inside solenoid, the magnetic field inside it is, B = µr µ0 ni ; where µr is the relative permeability of the medium., , Fig. 6.20, , 403

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404, , ELECTRICITY & MAGNETISM, , 6.4 TOROID, , OR ANCHOR RING, , It is a solenoid of small radius bent round to form a toroid. In an ideal toroid, the field is, confined entirely within the core and is uniform. The value of magnetic field at any point, on the mean circumferential line is given by, B, , = µ0 ni, , If N is the total turns in the toroid, then, n, , N, ,, 2 R, , = µ0, , N, i, 2 R, , B, , = µ0, , N, i, 2 R, , B, , =, , Fig. 6.20, , or, , =, , µ0 Ni, ., 2 R, , where R is the mean radius of the ring., , Ex. 2, , B =, , A steady current is set up in a cubical network of resistive, wires, connected as in fig. 6.22. Find magnetic field at the centre of, the cube., , =, , R cos, , Thus, , B =, , n, , Fig. 6.22, , A current I flows along a thin wire shaped as a regular, polygon with n sides which can be inscribed into a circle of radius, R. Find the magnetic induction at the centre of the polygon. Analyse, the obtained expression at n, ., , Sol. In a regular polygon of n-sides, each of its side subtends an angle, , sin, , µ0 I, 4, , r =, , For n, , Ex. 3, , n, , where, , =, , Sol. The currents in the wires distribute symmetrically about the, dotted diagonal. And so currents of resistances of one face cancel the, magnetic field of its front face. Thus, Bnet = 0., Ans., , nB1, , sin, , n, , n, , r, , n, , µ0, I, 4, , ., , 2sin / n, R cos / n, , µ0nI, tan, 2 R, n, , Ans., , , we can write, B =, , µ0 I, tan / n, lim, 2R n, /n, , µ0 I, Ans., 2R, Polygon of infinite sides is a circle, and so its magnetic induction at the, centre is equal to that due to a circular loop., =, , Ex. 4 A current I = 5.0 A flows along a thin wire shaped as, shown in fig. 6.24. The radius of a curved part of the wire is equal to, R = 120 mm, the angle 2 = 90°. Find the magnetic induction of the, field at the point O., , 2, at the centre of the polygon. The angle, . If B1 is the field, n, n, produced by each side of the polygon, then total magnetic induction, , 3, turn. If B1 and B2, 4, are the magnetic inductions due to the curved and straight parts of the, wire, then, , Fig. 6.23, , Fig. 6.24, , Sol., , The curved part of the wire is equal to

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MAGNETIC EFFECT, , and, , B1 =, , µ0 3/ 4 I, 2R, , B2 =, , µ0 (sin 45 sin 45 ), I, ,, 4, r, , where, , r =, B2 =, =, , OF, , CURRENT, , 405, , Ex. 6, , Find the magnetic induction of the field at the point O if, the wire carrying a current I has the shape shown in fig. 6.26 a, b,, c. The radius of the curved part of wire is R, the linear parts of the, wire are very long., , 3µ0 I, ,, 8R, , R sin 45, , µ0 I 2sin 45, 4 R sin 45, µ0 I, ., 2 R, , Thus total magnetic induction,, B = B1 + B2, =, , µ0 I 3, R 8, , 1, 2, , On substituting the values, we get, B = 28 µT., , Ans., , Ex. 5, , Find the magnetic induction of the field at the point O, in a current carrying wire has the shape shown in fig. 6.25 a, b, c., The radius of the curved part of the wire is R, the linear parts are, assumed to be very long., , Fig. 6.26, , Sol., (a), , Fig. 6.25, , Sol.(a), , Straight part of the wire produces no magnetic induction at, O. So magnetic induction is only due to the curved part., , The magnetic induction due to straight part is, , Bstraight = 2, , 1, I, µ0 I, 2, B =, Ans., 2R, 4R, The magnetic induction due to upper straight wire at O will be, , µ0 I, ( kˆ ) , and due to curved part is, 4 R, , µ0, , (b), , zero and due to lower straight part is, , Bcurved, , µ0 I, .The magnetic induction, 4 R, , B, , =, (c), , The total induction, , Bstraight, , =, , 2, , =, , 2Bstraight + B, , =, , µ I, 2 0, 4 R, , =, , µ0 I, [2, 4 R, , µ0, , (b), , µ0 I, 4 R, , =, , Bstraight, , =, , curved, , 1, I, 2, 2R, , µ0 I, ( iˆ), 4R, , iˆ 2 kˆ, , µ0 I, ( kˆ), 4 R, , Ans., , Bcurved, µ0 I ˆ, ( i), 4 R, , µ0, , 1, I, ( iˆ), 2, 2R, , µ0 I, [( 1)iˆ kˆ] ., Ans., 4 R, In this case current in the curved conductor is divided into two, parts. The length of upper curved part is thrice that of hidden, =, , (c), , ], , B, , Bcurved, , µ0 I, ( kˆ), 4 R, , =, Ans., , In this case all parts of the wire produce non zero magnetic, induction at O., , 1, I, ( iˆ), 2, 2R, , =, , Bstraight + B curved, , µ0 I, 3, 1, 4 R, 2, , µ0, , Thus total magnetic induction, , 3, µ0, I, 4, due to curved part is, . Thus total induction, 2R, B =, , =, , Ans., , lower part and so currents in them are, , I, 3I, and, respectively.., 4, 4

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406, , ELECTRICITY & MAGNETISM, , Thus, = Bstraight, , B, , =, , µ0 I, ( kˆ), 4 R, , =, , µ0I ˆ, (j, 4 R, , =, , Bcurved, , 3 I, µ0, µ0 I, 4 4 ( iˆ), ( ˆj ) +, 4 R, 2R, , µ0, , kˆ ), , 1 3I, 4, 4 i, 2R, Ans., , Ex. 7, , A non-conducting sphere of radius R charged uniformly, with surface density rotates with an angular velocity about the, axis passing through its centre. Find the magnetic induction at the, centre of the sphere., , Sol. Take an element in the form of ring of angular width d at a, position . The charge on ring, , =, , Ex. 8, , µ0, 2, 1, 3, , R cos, , cos3, 3, , /2, 0, , R, , 0, , Ans., , A straight segment OC (of length L meter) of a circuit, , carrying a current i A is placed along the x-axis (see fig. 6.28). Two, infinity long straight wire A and B, each extending from, z = – to + , are fixed at y = –a metre and y = + a metre respectively, as shown in fig. 6.28. If the wire A and B each carry a current i A into, the plane of the paper, obtain the expression for the force acting on, the segment OC. What will be the force on OC if the current in the, wire B is reversed., , dq =, (dA), (2 R sin ) Rd, Due to its rotation, the equivalent current, i =, , (2 R sin ) Rd, (2 / ), , dq, =, dt, , R2 sin d, =, The magnetic induction at the centre of the sphere, , Fig. 6.28, , Fig. 6.29, , Sol., Fig. 6.27, dB =, , =, , The magnetic field due to the wire at a distance x from O can be obtained, as follows:, If B1 and B2 are the magnetic fields of wire A and B respectively then, resultant field at P, , µ0 2 M, . 3, 4, x, , B =, , µ0 2idA, . 3, 4, R, R 2 sin d ), , =, , µ0 2(, ., 4, , =, , µ0, 2, , R sin 3 d, , µ0, 2, , R, , B =, , =, , =, , µ0, 2, µ0, 2, , R3, , {( R sin )2}, , sin 3 d, , =, , 2, , µ 0i, cos, 2 r, , =, , 2, , 0 i, 2 r, , =, , dF, , µ0, , µ0 ix, . 2, r, , ix, (x, , 2, , a2 ), , i( d, , =, , idx B( ˆj), , /2, , (1 cos2 )sin d, 0, , =, , /2, , (1 cos 2 ) d cos, 0, , x, r, , =, , 0, , R, , As (B1 = B2), , along y-axis., , The force on the small length dx of the conductor OC, , /2, , R, , 2B1 cos, , 0i, , B), , 2, , xdx, x2, , a2, , ( kˆ )

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MAGNETIC EFFECT, The force on the whole conductor, , F, , =, , 2, 0i, , L, , ( kˆ ), 0, , =, , 6.5 AMPERE'S, , 0i, , 2, , =, , xdx, x2, , 2, , L, , a )0, , 2, , 2, , a2, =, , ( kˆ) ln ( x 2, , 0i, , 0i, , 2, , 2, , OF, , CURRENT, , ( kˆ ) ln( L2, , a 2 ) lna 2, , L2, ( kˆ )ln, , a2, a2, , LAW, , We know that the lines of magnetic field are continuous and do not arise from any source, in the way as lines of electric field originate from charges. These thus form loops without, any beginning or end. This property may be used to go through the nature of magnetic, field and for calculating the field in certain situation of high symmetry., µ0 i, . The field lines are concentric, 2 r, around the wire in the plane of the paper if the wire is perpendicular to this plane. The line, integral around the path of radius r, starting at any point and returning to the same point, is ;, , The field of a long round current carrying wire B, , B.d, , =, , B d cos 0, , =, , B d =, , µ0 i, ., 2 r, 2 r, , Fig. 6.30, , = µ0i., Although the above result is derived for the special case of the field of a long straight, conductor, the law is true for conductors and path of any shape. Thus the line integral of, the magnetic field B around any closed path is equal to µ0 times the net current across, the area bounded by the path. Hence, , B.d, , =, , µ0 iin ., , It is called Ampere's law. It plays the same role in magnetics as Gauss's law plays in, electrostatics., , Note:, 1., , 2., , The magnetic field B on the left hand side in Ampere's law is the resultant field, due to all the currents existing anywhere while on the right hand side the current, iin is due to the conductors inside the close loop., Let us consider a closed plane curve as shown in figure. If direction of integration, is taken along the path as shown then i1 and i3 will be positive and i2 will be, negative., Thus the total current crossing the loop is (i1 – i2 + i3). Any current outside the, area is not included in writing the right hand side of the equation. Thus, B.d, , 3., , =, , Fig. 6.31, , µ0 (i1 i2 i3 ), , Consider two current carrying conductors, carrying currents i1 and i2 in the, same direction and a close path as shown in fig. 6.32. For the close path, B.d, , = µ0 (i1 i2 ), Fig. 6.32, , ., , 407, , Ans.

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408, , ELECTRICITY & MAGNETISM, Application of Ampere's law, In some cases of practical importance, symmetry considerations make it possible to use, Ampere's law to compute the magnetic field caused by a certain current carrying, conductor. Few guiding principles, analogous to those stated for Gauss's law are following., 1., , If B is tangent to the integration path everywhere and has the same magnitude on, each point of the path, then, = µ0i ., B, Here is the length of the close path., , 2., , If B is everywhere perpendicular to the path or some portion of the path, then that, portion of the path makes no contribution to the line integral., , 3., , In the integral, , Fig. 6.33, , B.d , B is always the resultant magnetic field at each point of, , the path. In general this field is caused partly by currents linked by the path and, partly by currents outside. Even when no current is linked by the path the field at, points on the path need not be zero. In case, however,, 4., , B.d is always zero., , Some judgement is required in choosing an integration path. Two useful guiding, principles are that the point or points at which the field is to be determined must lie, on the path, and that the path must have enough symmetry so that the integral can, be evaluated., , Magnetic field due to long cylindrical wire, Fig. 6.34, , For r R :, The magnetic field at each point of circular path surrounded the conductor is tangential., Therefore for circular path 1, B.d, , Fig. 6.35, , =, , µ0 i, , or, , Bd cos 0, , = µ0 i, , or, , B d, , = µ0i, , or, , B 2 r, , = µ0i, , or, , B, , =, , µ0 i, 2 r, , For r < R :, If the current is distributed uniformly through the cross-section of the wire, then the, current inside the path is, i in, Now, , B.d, , =, , i, , r2, , R2, , R2, , = µ0iin, = µ0, , ir 2, , or, , B×2 r, , or, , B, , =, , µ0 ir, 2 R2, , Bmax, , =, , µ0 i, ., 2 R, , For r = R, , ir 2, , R2, , Long solenoid, For the field at point inside the solenoid, let us consider a rectangular path ABCDA as the, path of integration. This path is particularly simple as :, (a) Outside the solenoid, along CD, we can assume field to be zero as it is many times, weaker than that inside the solenoid

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MAGNETIC EFFECT, , OF, , CURRENT, , 409, , D, , B d, , = 0, , C, , (b), , Along BC and DA, B is either zero (for outside parts of BC and DA) or perpendicular, to B (for the parts inside the solenoid), A, , C, , D, , B, , (c), , Fig. 6.36, , B d =0, , B d, , Inside the solenoid B is constant and is along AB., B, , = B, , B d, A, , Now apply Ampere's Law for closed loop ABCDA, we get, B d, B, , C, , B d, , or, , D, , B d, , A, , =, A, , B d, , B, , C, , or, , B d = 0iin, D, , B, , As iin, , 0 iin, , n i, , =, , 0, , =, , 0ni., , n i, B, , Thin hollow current carrying tube, For x, , R,, , Bmax, , For x < R,, , iin, B, , µ0 i, ., 2 r, = 0,, = 0., , =, , Fig. 6.37, , Ex. 9, , Each of the eight conductors in fig. 6.38 carries 2.0 A of, current into or out of the page. Two paths are indicated for the line, integral, , B.d . What is the value of the integral for the path (a) at, , carries the current k i0.For those with odd k, the current is out of, the page; for those with even k, it is into the page., Evaluate, , B d, , along the closed path in the direction shown., , the left and (b) at the right ?, , Fig. 6.38, , Sol., (a), , By Ampere law, , B.d, , = µ0iin ;, , in close loop iin = (–2 + 2 – 2) A = –2A, B.d, , (b), , =, , µ0 ( –2) = –2µ0., , Ans., , In this loop iin = (2 – 2 + 2 – 2) = 0,, B.d, , Ex. 10, , =, , 0., , Ans., , Eight wires cut the page perpendicularly at the points, shown in fig. 6.39. A wire labeled with the integer k (k = 1, 2, ....8), , Fig. 6.39, , Sol., , In accordance with Ampere's law, B d, , =, , 0 iin, , =, , 5 0i0 ., , Ans.

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410, , ELECTRICITY & MAGNETISM, , Ex. 11 Inside a homogeneous long, , straight wire of circular, cross-section, there is a circular cylindrical cavity whose axis is, parallel to the conductor axis and displaced relative to it by a distance, . A direct current of density ĵ flows along the wire. Find magnetic, , 2 kR 3, 3, And current inside the close path of radius r, , =, , r, , induction B inside the cavity.., Sol. By the principle of superposition, the required quantity can be, given by, , iin =, , =, For r < R :, , or, , B.d, , =, , B 2 r =, , which gives, , B =, , r, , j 2 rdr, 0, , kr 2 rdr, 0, , 2 kr 3, 3, , ir 3, R3, , µ0iin, , µ0, , ir 3, R3, , ,, , µ 0i r 2, 2 R3, , Ans., , Fig. 6.40, B =, , B0, , It is more useful to write Ampere's law as follows :, , B',, , where B0 is the magnetic field of the conductor without cavity, while B ', is the magnetic field of the conductor which has removed., Here, , B0 =, , µ0 ir, 2 a2, , µ0 j ( a 2 ) r, 2, a2, , µ0 r, j, 2, , This expression can be represented in vector form, , B0 =, , µ0 ˆ, ( j r) ., 2, , Similarly, , B' =, , µ0 ˆ, ( j r '), 2, , Thus, , B =, , B0, , From the figure, , r, , =, , B =, , B', , µ0 ˆ, [ j (r, 2, , r',, , µ0 ˆ, (j, 2, , =, , µ0 j .dA, , Ex. 13, , Fig. 6.42 show a cross-section of a large metal sheet, carrying an electric current along its axis. The current in a strip of, width d is kd where k is constant. Find the magnetic field at a, point P at a distance x from the metal sheet., , r ')], , r –r', , )., , B.d, , ,, Ans., , Ex. 12, , If current density in the conducting wire is proportional, to the distance r from the axis of the conductor, then find magnetic, field at the position r < R, where R is the radius of cross section of, the conductor., , Sol. Let current density j = kr, where r is the distance from the axis, of the wire. If current in the wire is i, then, , Sol., , Fig. 6.42, , Consider two strips R and S of the sheet situated, symmetrically on the two sides of P. The magnetic field at P above sheet, and below sheet is parallel to sheet as shown in figure. There is no field, perpendicular to the sheet., , Fig. 6.41, R, , j, , i =, , 2 rdr, , 0, R, , kr 2 rdr, , =, 0, , Fig. 6.43, Now applying Ampere's law to the close path 1-2-3-4-1 as shown in fig., 6.43 ;

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MAGNETIC EFFECT, B.d, 2, , 3, , B.d, , 4, , B.d, , 1, , 2, , or, , =, , 0 B, , B.d, 0, , or, , µ0 ( k ), , =, , B =, , µ0 k, , Ans., , 1., 2., , Magnetic field in this case in independent of x., Two large metal current carrying sheets: The magnetic field at, different positions is shown in figure., , B1, , B2, , =, , µ0 i ( a b), . ln, ., a, 2 b, , µ0 k, 2, , b), , Ans., , Note:, , µ0k, ., 2, , Note:, 1., 2., , µ0 i, . lnx (aa, 2 b, , 4, , 411, , CURRENT, , =, , 1, , B.d, 3, , B, , µ0 iin, , OF, , B1, , B2, , 3., , Unlike the gravitational and electric field, there is no scalar potential, associated with the magnetic field., µ0 = 4 × 10–7 Tm/A is an exact number and not an empirical, constant., Ampere's law is not independent of the Biot-Savart law. It can be, derived from the Biot-Savart law. Its relationship to the BiotSavart law is similar to the relationship between Gauss's law and, Coulomb's law., , Ex. 15 Consider the current carrying loop shown in, , µ 0k, 2, , fig. 6.45, formed of radial line and segments of circles whose centres are at, point P. Find the magnitude and direction of B at point P., , B, , B1 B2, , µ0 k, , B, , B2, , B, , B1 B2, , 0, , B, , B1 B2, , µ0 k, , B, , B1 B2, , µ 0k, , B, , B1 B2, , 0, , Ex. 14, , B1, , 0, , Sol. Magnetic field by straight parts of the loop is zero because point, P lies on their axis. The field produced by curved parts is, , Fig. 6.44 shows a cross-section of a long thin ribbon of, , B =, , width b that is carrying a uniformly distributed total current i into, the page. Calculate the magnitude and direction of the magnetic, , =, , field B at a point P in the plane of the ribbon at a distance 'a' from, its edge., , As, , n1 =, B =, , Fig. 6.44, , Sol., , Take a small element of thickness dx at a distance x from P,, , the current in the element, di =, , i, dx ,, b, , and magnetic field at P, dB =, , µ0 di, ., 2 x, , =, , The total field, , B =, , Fig. 6.45, Ba – Bb, , µ0n1i, 2a, , µ0n2i, 2b, , 60, 360, , 1, 6, , 0i 1, 12 a, , (a b), a, , dx, x, , 1, b, , out of the page Ans., , Ex. 16 A loop, carrying a current i, lying in the plane of the, paper, is in the field of a long straight wire with constant current i0, (inward) as shown in fig. 6.46. Find the torque acting on the loop., , µ0 idx / b, ., x, 2, , µ0 i, ., 2 b, , n2, , Fig. 6.46

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412, , ELECTRICITY & MAGNETISM, , Sol. The field due to current carrying wire is tangential to every point, on the circular portion of the loop and hence the forces acting on these, segments are zero., Now consider two small elements of length dr at a distance r from the, axis symmetrically as shown in fig. 6.47., , Fig. 6.48, Force on charge q at origin, , F, , Fig. 6.47, The magnitude of the force experienced by each element is, dF = B i dr, =, , µ0 i0, idr, ., 2 r, , On element 1 it is into the page and on 2 it is out of the page, , d, , =, =, , dF, , 2r sin, , (b), , µ0i0i, dr, 2 r, , Now total torque,, =, , µ0i0i sin, , and, , b, , dr, a, , =, , µ0i0i, , sin (b, , a )., , Ans., , Ex. 17 A circular loop of radius R is bent along a diameter and, given a shape as shown in the figure. One of the semicircles (KNM), lies in the xz- plane and other one (KLM) in the yz- plane with their, centres at the origin current i is flowing through each of the, semicircles as shown in fig. 6.48., (a) A particle of charge q is released at the origin with a velocity, , v, , v0iˆ . Find the instantaneous force F on the particle., , q(v B), , =, , q (–v0iˆ), , =, , µ0qiv0, [– i ( iˆ, 4R, , =, , –, , µ0 i ˆ, ( i, 4R, , ˆj ), , ˆj )], , µ 0qiv0 ˆ, k, 4R, , Ans., , In a uniform field the force on the curved loop is F, where, , 2r sin, , =, , B) ,, , i(, , vector from one end of loop to the other end., , FKLM, , =, , i( KM, , FKNM, , =, , i[2 R ( kˆ ), , F, , =, , FKLM, , =, , 4BiRjˆ., , B), , i[2 R ( kˆ), , Bjˆ ], , 2 BiRiˆ, , Bjˆ ] = 2BiRiˆ, , FKNM, Ans., , Ex. 18, , A circular loop of radius r carrying a current i is held at, the centre of another circular loop of radius R ( >> r) carrying a, current I. The plane of the smaller loop makes an angle 30° with, that of the larger loop. If the smaller loop is held fixed in this, position by applying a single force at a point on its periphery. What, would be the minimum magnitude of this force ?, , Sol.The magnetic field at the centre of the smaller loop,, , B, , µ0 I, ., 2R, , Assume that space is gravity free., (b), , If an external uniform magnetic field Bjˆ is applied,, determine the forces F1 and F2 on the semicircles KLM and, KNM due to this field and the net force F on the loop., , Sol., , (a) The magnetic field due to the loop at the origin, B0, , =, =, , =, , BKLM + BKNM, , µ 0i, ( iˆ), 4R, µ0i ˆ, ( i, 4R, , ˆj ), , µ0i0 ˆ, ( j), 4R, , Fig. 6.49, According to right hand screw rule, it is along as shown in fig. 6.50. Since, smaller loop is placed at its centre and assuming that field is uniform all, over it, the torque exerted by magnitude field

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MAGNETIC EFFECT, =, , MB sin, , =, , (i r 2 ), , 2vx dvx, , µ0 I, sin 30, 2R, , which gives, , 2v y dv y, , F =, , =, , 413, , v y dv y, , Ex. 19, , A long straight wire carries a current i. A particle having, a positive charge q and mass m, kept at a distance x0 from the wire, is projected towards it with a speed v. Find the minimum separation, between the wire and the particle., , dx, x, , x, , dv y, , =, , k, , Initially at, x = x0, vy = 0, At minimum separation from wire, vx = 0, vy = –v, x, , Thus, x0, , dx, x, , v, , v, , nx x, , =, , x, x0, , =, , v, k, , x =, , x0 e, , =, , x0e, , 0, , n, , or, , k, , 0, , vy, , x, , or, , dv y, , =, , Sol., , Let the particle be initially at P. The magnetic field B at any, point to the right of the wire is along the negative z-axis. The magnetic, force on the particle is, therefore, in the xy- plane. As there is no initial, velocity along the z-axis, the motion will be in the xy- plane. Also its, speed remain unchanged. As the field is not uniform, so the path of the, particle not remain as circle., , ...(iii), , vy, , k, , =, , dx, , Ans., , v y dv y, , From equations (i), (ii) and (iii), we have, , µ0 iIr 2, 4R, , µ0 iIr, ., 4R, , 0, , v x dv x =, , or, or, , CURRENT, , On differentiating, , µ 0 iIr 2, =, 4R, If F is the force at the periphery then, to keep the loop at its position, F×r =, , OF, , or, , k, , 0, , v, k, 2 mv, 0 qi, , ., , Ans., , Ex. 20 A copper wire with cross-sectional area S, bent to make, three sides of a square can turn about a horizontal axis OO . The, wire is located in uniform vertical field. Find the magnetic induction, if on passing a current i through the wire the latter deflects by an, angle ., , Fig. 6.50, We have,, , v, , =, , B =, , and, , vx i v y j, i, x, , 0, , 2, , Sol., , k, , Let side of square is a. The magnetic field B makes 180o with OP and 0o, with QO therefore forces on these sides are zero. The force on side PQ, , The force on the particle, F, , =, =, , qv, , where, , qv y, , Also, As, , Fx, m, , =, , k, , ax =, , vx 2, , 0i, , 2 x, , ax =, , k =, , 0i, , q vx i v y j, , =, , Thus, , B, , 2 x, , k, , qv x 0i, j, 2 x, , i, , qv y, , = Bia, acts out of page and ^ to the plane of the figure. The torque, exerted by field, , 0i, , 2 mx, vy, , ....(i), , x, , =, , 0 qi, , Bia, , a cos, , Let mass of each side is m, then restoring torque, , 2 m, dvx, dt, , Fig. 6.51, , dvx dx, dx dt, , v y 2 = v2 (constant), , vx, , dvx, dx, , ...(ii), , res, , a, sin, 2, = 2mg a sin, , =, , mg, , =, , 2 Sa a sin, , mg, , a sin, , mg, , a, sin, 2

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414, , ELECTRICITY & MAGNETISM, part CD, but varies from C to D., , For equilibrium,, =, , 2 S ga 2 sin, , B =, , 2S g, tan ., i, , Bia 2 cos, or, , Ex. 21, , Ans., , A current of 10A flows around a closed path in a circuit, , which is in the horizontal plane as shown in fig. 6.52. The circuit, consists of eight alternating arcs of radii r 1 = 0.08 m and, r2 = 0.12 m each. Each arc subtends the same angle at the centre., (a) Find the magnetic field produced by this circuit at the centre., (b) An infinitely long straight wire carrying a current of 10A is, passed through the centre of the above circuit vertically with, the direction of the current being into the plane of the circuit., What is the force acting on the wire at the centre due to the, current in the circuit ?, What is the force acting on the arc ACD and the straight, segment CD due to the current at the centre ?, , Sol.(a), , Fig. 6.53, Let at any distance x from the centre., B =, , The force on length dx of CD, dF =, , Bi (dx), , i, idx, x, , 0, , =, , 2, , The magnetic field due to straight parts of the loop is zero,, , because centre O lies on their axis., , r2, , Total force, , 0, , F =, r1, , or, , 0, , =, , 1, i, 2, 2 r1, , 0, , 1, i, 2, 2 r2, , =, , 4, , 1, r1, , 1, r2 ,, , (d), , the direction of the field is out of the page, i.e. along z-axis, =, , =, , 4, , 10, 4, , 7, , 65.45 10, , 10, , 6, , T., , 1, 0.08, , 0 2, , i n, , =, , 2, , =, , 2 10, , =, , 8.1 10, , between the particles if v, (c), , 0i, , 2, , i2, dx, x, , r2, r1, , 7, , 10, 6, , N., , 2, , n, , 0.12, 0.08, Ans., , Ex. 22 Two particles each having a mass m are placed at a, separation d in a uniform magnetic field B as shown in fig. 6.54., They have opposite charges of equal magnitude q. At time t = 0, the, particles are projected towards each other, each with a speed of v., Suppose the Coulomb force between charges is switched off., (a) Find the maximum value of vm of the projection speed so that, the two particles do not collide., (b) What would be the minimum and maximum separation, , Fig. 6.52, The total field at the centre, B = B1 + B2, , (b), , i, x, , 0, , 2, , vm, ?, 2, , At what instant will a collision occur between the particles, if, v = 2vm ?, Suppose v = 2vm and collision between the particles is, completely inelastic. Describe the motion after the collision., , Sol.(a), , 1, 0.12, Ans., , As the field of the circuit at centre and conductor are anti parallel,, therefore force on it is zero. Further, field due to current carrying, conductor at centre will be tangentially to each point of AC., Therefore force of part AC is zero. While field is perpendicular to, , Fig. 6.54

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MAGNETIC EFFECT, They projected with such a speed vm so that each moves in a path, of radius r, , d, . If they will not collide, then, 2, r =, , or, (b), , vm =, , For, , v =, =, , r =, , d, 2, qBd, 2m, , mv sin + mv sin, , vm, 2, , v, , qBd, ,, 4m, , mv, qB, , and, (c), , xmax =, , 415, , When collision is perfectly inelastic, the particles stick together., , Momentum before collision =, , Ans., , qBd, m, 4m, qB, , CURRENT, , The resultant momentum of the combined mass in horizontal, direction becomes zero. Therefore the combined mass moves along, a straight line drawn upward in the plane of figure through the, point of collision. Because magnetic field does not exert force on, it, qnet = 0., , mvm, qB, , Both the particles move on the path of radius, From the figure xmin =, , (d), , OF, , d, 4, , =, , (2m)v, , =, , v sin, , =, , v, , =, , 2vm, 2, , momentum after collision, , 1, 2, , v, 2, vm ., , Ans., , Ex. 23 Given fig. 6.56 shows a coil bent with all edges of length, 1m and carrying a current of 1A. There exists in space a uniform, magnetic field of 2T in the positive y-direction. Find the torque on, the loop., , d, ., 4, , d, 2, , Sol. The forces on the edges FG and BC are zero. The forces on the, d, 2, , d, , other edges are :, , 3d, 2, , Ans., , When particle velocity, v =, , 2vm, , r =, , mv, qB, , qBd, ,, m, , m qBd, m, qB, , d, , Fig. 6.56, , Fig. 6.55, , d, They will collide at a horizontal distance, ., 2, , sin, or, , =, , d, d, 2, 2, =, d, r, , 2 m, qB, , FCD =, , Bi, , i ,, , F HG =, , Bi, , i ,, , F EF =, , Bi i ., , The forces on edges AH and DE are equal and opposite and so constitutes, a net torque. Thus, , radian, 6, Therefore each particle travel for time t before collision, , T, 2, , Bi i ,, , The torque of the forces FAB and FHG is zero and that of FCD and FEF is, zero., , 1, 2, , =, , t =, , FAB =, , 6, 2, , FAH =, , m, 6qB, , Torque,, Ans., , Bi, , k and FDE, , Bi k, , =, , FAH i, , Bi, , i, , =, , 2 1 12, , 2i N-m.

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416, , ELECTRICITY & MAGNETISM, , Ex. 24, (a), , magnetic, , or, , Current i enters into a loop as shown in figure. Find magnetic, field at the centre of the loop., , iAB sin 90, , =, , weight, , =, , mg × r, , i =, (c), , mgr, ., AB, , Ans., , For second conductor be at rest, Fmagnetic, , i2, h, , mg, Fig. 6.57, (b), , (c), , (d), , i1, , A circular current carrying circular loop of mass m and, radius r is placed on horizontal surface in a uniform magnetic, field B acts parallel to the plane of the ring. Find current in, the loop, so that it will tilt about the edge., A long current carrying conductor carrying current i, is fixed, on the surface. Another conductor of mass per unit length m, is at a vertical distance h and parallel to first conductor. Find, current i2 in the second conductor so that it remains at rest., A conductor of length and mass m carrying current i is in, on smooth inclined plane of inclination . A magnetic field B, is acting vertically. Find value of current i, so that it remains, in equilibrium on the plane., , Fig. 6.61, Weight per unit length = magnetic force on unit length, or, , mg =, , i2 =, , 0, , 2, , ., , i1i2, h, , mgh, 0, , 2, (d), , Ans., , ., i1, , If i is the current in the conductor, then it experience a horizontal, force of magnitude B i . so for its equilibrium., , B, i, , Bi, , i, , mg, sin, , Fig. 6.62, , Bi cos, , Fig. 6.58, , Sol., (a), , or, , The currents in two parts of the loop is shown in figure., (i) The magnetic field at the centre of the loop,, , B=, , 3 i, 4 4, –, 2r, , 0, , 1 3i, 4 4, 2r, , 90°, 3i/4, , (b), , mg tan, B, , i, , =0, Fig. 6.59, (ii) Solution is same as of part (i). B = 0, The loop will tilt about the edge if,, , ., , Ans., , A uniform constant magnetic field B is directed at an, angle of 45o to the x-axis in xy-plane. PQRS is a rigid square wire, frame carrying a steady current i0, with its centre at the origin O., At time t = 0, the frame is at rest in the position shown in the figure,, with its side parallel to x and y axis. Each side of the frame is of, mass M and length L., , r, , i, , i =, , Ex. 25, , i/4, 0, , = mg sin, , Ans., , n^, , magnetic, , r, B, , mg, Fig. 6.60, Fig. 6.63

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MAGNETIC EFFECT, (a), (b), , What is the torque about O acting on the frame due to, magnetic field ?, Find the angle by which the frame rotates under the action of, this torque in a short interval of time t, and the axis about, which this rotation occurs ( t is so short that any variation in, the torque during this interval may be neglected)., , Sol.(a), , 417, , CURRENT, , loop is held in the xy-plane and a current I is passed through it. The, loop is now released and is found to stay in the horizontal position, in equilibrium., , Given, magnetic field, B =, A =, , B, i, 2, , B cos 45 i sin 45 j, , j, , L2 k, , =, , M B, , =, , i A B, , =, , i0 L2 B, , i0 L2 k, , i, , 2, , B, 2, , i, , j ., , j, (a), (b), (c), , Ans., , i, (b), , OF, , It is clear from above result that axis of rotation is =, , j, , Sol.(a), , 4, , ML2, 12, , M, , L, 2, , 2, , the weight and, , 4, ML2, 3, , I SQ =, We know that, , =, , =, , 2, ML2, 3, , =, , 0, , mag, , =, , Here, , mg, , =, , mg, , and, , mag, , =, , M B, , =, , I A B, , =, , I ai, , ...(i), , mg, , a, j, 2, , bj, , =, , i0 BL2, 2, ML2, 3, , 3 i0 B, 2 M, , I ab, , =, , k, , =, , 1, 2, , =, , 1 3 i0 B, 2 2 M, , =, , 3 i0 B 2, t ., 4 M, , t, , 3abI j., , mg, , 3abI j =, , 2, , I =, , t2, , Ans., , (b), , A rectangular loop PQRS made from a uniform wire, has length a, width b and mass m. It is free to rotate about the arm, PQ, which remains hinged along a horizontal line taken as the yaxis (see fig. 6.64). Take vertically upward direction as the z-axis. A, , 4 k B O exists in the region. The, , (c), , a, j, 2, , mg, 6bB0, , Ans., , Positive sign with I indicates that direction of current in PQ is, correctly assumed., Force on the arm RS is given by, , I, , F =, , Ex. 26, , 3i, , 3i 4k B0, , Substituting these values in equation (i), we get, , Thus angular rotation, , uniform magnetic field B, , 3i 4k B0, , (Assuming direction of current in the loop clockwise.), , I, , I, , mg is the torque exerted by, , mag is the torque due to magnetic force, then, mag, , mg, , By perpendicular axis theorem, 2 ISQ = I., , I0, 2, , For the equilibrium of the loop in horizontal plane, the net, , torque on it must be zero. Thus if, , 2, , i.e., SQ., Moment of inertia of the frame about axis of rotation SQ :, Moment of inertia of the frame about an axis passing through O, and perpendicular to the plane of the frame., IO =, , Fig. 6.64, What is the direction of the current in PQ ?, Find the magnetic force on the arm RS., Find the expression of I in terms of BO, a, b and m., , B, , =, , I, , bj, , =, , I B0, , It has been obtained in part (a)., , 3i 4k B0, , 4i 3k N, , Ans.

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418, , ELECTRICITY & MAGNETISM, , Ex. 27, , Ring of radius R having uniformly distributed charge, Q is mounted on a rod suspended by two identical strings as shown, in fig. 6.65. The tension in strings in equilibrium is T 0. Now a, vertical magnetic field is switched on and the ring is rotated at, constant angular velocity . Find the maximum with which the, ring can be rotated if the strings can withstand a maximum tension, , The torque experiences now, =, , MB sin90°, , =, , iAB., , If T1 and T2 are the tensions in the strings now , then, , T1 T2, , 3T0, 2, or, Also, , D, 2, , =, , T1 T2 =, T1 + T2 =, , iAB, , 2iAB, D, , ...(ii), , mg, , ...(iii), , Solving equations (ii) and (iii), we get, T1 =, , mg, 2, , iAB, D, , T2 =, , mg, 2, , iAB, D, , T1 =, , 3T0, 2, , (Given)., , Fig. 6.65, , Sol., , Initially, , 2T 0 =, , mg, , If be the frequency corresponding to the breaking of the string, then, current in the ring, i =, , Q, T, , and, , ...(i), , Q, 2, , Q, 2, , As T1 > T2,, , 3T0, 2, , Hence, , =, , Magnetic moment of the loop, M =, , iA, , Fig. 6.66, , or, , T0 =, , or, , =, , 2T0, 2, , Q, 2 D, , R2, , B, , QBR 2, D, , DT0, QBR2, , ., , Ans.

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MAGNETIC EFFECT, , OF, , 419, , CURRENT, , Review of formulae & Important Points, 1., , B =, , 2., , qv sin, , 0, , 0, , 4, , 0 Ni, 2a, , B =, , The magnetic field on the axis of the circular current carrying, coil, , q1q2 v1v2, , ., , r2, , Biot-Savart law :, The magnetic field due to the current element d, point P at a distance r is given by, dB, , 4., , Magnetic field at the centre of the circular current, carrying coil, , ,, , r2, , 4, , and its direction can be obtained by right hand screw rule., The magnetic force between two moving charges q1 and, q2 is given by, , Fm =, 3., , 6., , Magnetic field of moving charge :, , =, , 0, , id, , r, r3, , 4, , at any, , 2 a2, , 7., , i, sin, 4 r, 0, , 2, , x2, , 3/ 2, , Magnetic field due to a solenoid, , ., 0 ni, , B =, , Magnetic field due to a current carrying wire :, B =, , 0 ia, , B =, , 2, , cos, , 1, , cos, , 2, , Magnetic field due to a long solenoid at its centre, 1, , sin, , 2, , ., , B =, , 0ni., , Magnetic field of a toroid of radius R, B =, 8., , Ampere s law : It states that, Bd, , 0i, ,, 2 r, , at the middle region of the wire., and, 5., , B =, , 0i, , 4 r, , at the end of the wire., , The force between two parallel current carrying wires :, dF, d, , =, , 0 i1i2, , 2, , r, , ., , =, , 0 iin., , The line integral in this equation is evaluated around a, closed loop called Amperian loop. The current iin is the, net current encircled by the loop., , Magnetic field due to a long straight wire, B =, , 0 Ni, 2 R, , 9., , The magnetic field inside a current carrying tube is zero., , 10., , The magnetic field due to a magnetic dipole on its axis, , B, , =, , 0, , 4, , 2M, x3

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420, , ELECTRICITY & MAGNETISM, , Exercise 6.1, , MCQ Type 1, , Only one option correct, 1., The figure shows three long, straight, parallel, equally spaced wires, with identical currents either into or out of the page. If FA, FB and, FC are the magnitudes of forces on them respectively, then, , 2., , (a), , FA, , FB, , FC, , (b), , FA, , FB, , FC, , (c), , FB, , FC, , FA, , (d), , FC, , FA, , FB, , The figure shows three equal currents i (two parallel and one, antiparallel) and four closed loops. Which loop has greatest value, of, , (a) inside the inner conductor, (b) inside the outer conductor, (c) in between the two conductor, (d) outside the cable., A current-carrying straight wire is kept along the axis of a circular, loop carrying a current. The straight wire, (a) will exert an inward force on the circular loop, (b) will exert an outward force on the circular loop, (c) will not exert any force on the circular loop, (d) will exert a force on the circular loop parallel to itself., , 5., , The magnetic field at the origin due to a current element id, , 6., , placed at a position r is, , Bd ?, , (a), , (c), 7., , 3., , (a) 1, (b) 2, (c) 3, (d) 4, The figure shows four arrangements of circular loops of radius r or, 2r, centered on vertical axes (perpendicular to the loops) and, carrying identical currents in the direction indicated, in which, arrangement, the magnitude of the net magnetic field at the dot is, the greatest :, , id, , r, r, , 0, , (b), , 3, , 0, , 4, , id, , r, r, , 3, , r id, , 4, , (d), , r3, , none of these, , Two infinitely long conducting wires carry currents, as shown in, figure. If the horizontal wire carrying current produces a magnetic, field B at the point P(–2d,–d,0), then the resultant magnetic field, at this point is, (a), (b), (c), (d), , 8., , 2B kˆ, 2 B kˆ, , 2 B kˆ, zero, , A length L of wire carries a steady current I. It is bent first to form, a circular plane coil of one turn. The same length is now bent more, sharply to give a double loop of smaller radius. The magnetic field, at the centre caused by the same current is, (a) a quarter of its first value, (b) unaltered, (c) four times of its first value, (d) a half of its first value, A vertical straight conductor carries a current vertically upwards., A point P lies to the east of it at a small distance and another point, Q lies to the west at the same distance. The magnetic field at P is, (a) greater than at Q, (b) same as at Q, (c) less than at Q, (d) greater or less than at Q depending upon the strength of the, current, , 9., , 4., , 0, , 4, , (a) I, (b) II, (c) III, (d) IV, In a coaxial, straight cable, the central conductor and the outer, conductor carry equal currents in opposite directions. The magnetic, field is zero, , Answer Key, , 1, , (c), , 2, , (c), , 3, , (d), , 4, , (d), , Sol. from page 437, , 6, , (a), , 7, , (a), , 8, , (c), , 9, , (b), , 5, , (c)

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MAGNETIC EFFECT, 10., , An infinitely long straight conductor is bent into the shape as, shown in the figure. It carries a current of i ampere and the radius, of the circular loop is r metre. Then the magnetic induction at its, centre will be, , (a), 11., , 0, , 2i, r, , 1, , 2i, 4 r, 0, , (b), , 15., , 1, , (c), 12., , 13., , 4, 0i, , 4, , 1, R2, , 0i, , (b), , R1 R2, , 1, R1, , 4, , 17., , 0i, , (d), , 1, R2, , R1 R2, , 4, , A vertical wire kept in zx-plane carries a current from Q to P (see, figure). The magnetic field due to current will have the direction at, the origin O along, (a), , OX, , (b), , OX, , (c), , OY, , (d), , OY, , (a), , 0i, 4r, , (c), , 0i, 2 r, , (c), , 1, , (b), , 0i, 2r, , (d), , 0i, 2 r, , 18., , / m2 at the centre of the coils will be, , An electron moving in a circular orbit of radius r makes n rotation, per second. The magnetic field produced at the centre has a, magnitude of, (a), , 14., , 1, R1, , 0 ne, , 0n, , (b), , 2r, 0 ne, 2 r, , (d), , 2, , 421, , 1, , The earth s magnetic field at a given point is 0.5 10 5 Wb-m–2., This field is to be annulled by magnetic induction at the center of, a circular conducting loop of radius 5.0 cm. The current required to, be flown in the loop is nearly, (a) 0.2 A, (b) 0.4 A, (c) 4 A, (d) 40 A, The magnetic field due to a current carrying circular loop of radius, 3 cm at a point on the axis at a distance of 4 cm from the centre is, 54 T. What will be its value at the centre of the loop, (a) 250 T, (b) 150 T, (c) 125 T, (d) 75 T, Two concentric coils each of radius equal to 2 cm are placed at, right angles to each other. 3 ampere and 4 ampere are the currents, flowing in each coil respectively. The magnetic induction in weber, , 16., , 0i, , CURRENT, , A part of a long wire carrying a current i is bent into a circle of, radius r as shown in figure. The net magnetic field at the centre O, of the circular loop is, , (c) zero, (d) infinite, The magnetic induction at the centre O in the figure shown is, , (a), , OF, , 0, , 4, , 10 7 Wb / A.m, , (a) 5×10–5, (b) 7×10–5, –5, (c) 12×10, (d) 10–5, Wires 1 and 2 carrying currents i1 and i2 respectively are inclined, at an angle to each other. What is the force on a small element d, of wire 2 at a distance of r from wire 1 (as shown in figure) due to, the magnetic field of wire 1, , 19., , e, , 2r, zero, , Two long straight wires are set parallel to each other. Each carries, a current i in the same direction and the separation between them, is 2r. The intensity of the magnetic field midway between them is, (a), , 0i / r, , (b), , 4 0i / r, , (c), , zero, , (d), , (a), , (c), , 0i / 4 r, , 0, , 2 r, 0, , 2 r, , i1i2 d tan, , (b), , i1i2 d cos, , (d), , 0, , 2 r, 0, , 4 r, , i1i2 d sin, i1i2 d sin, , Answer Key, , 10, , (b), , 11, , (a), , 12, , (d), , 13, , (a), , 14, , (c), , Sol. from page 437, , 15, , (c), , 16, , (b), , 17, , (a), , 18, , (a), , 19, , (c)

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422, 20., , ELECTRICITY & MAGNETISM, , Which of the following graphs shows the variation of magnetic, induction B with distance r from a long wire carrying current, , (a), , (a), , Ie R, Ic, , (b), , Ic R, Ie, , (c), , Ic, Ie R, , Wire, R, , Ic, H, , (b), , (d), (c), , 23., , (d), , Ie, Straight, , Ie, Ic R, , A charged particle moving in x-y plane enters a region of uniform, magnetic field of intensity B, , 21., , A wire of length L metre carries a current I ampere, is bent in the, form of a circle. The magnitude of its magnetic moment in MKS, units is :, , B0 k at P(0, 0) with velocity, , v1, , vi and leaves the magnetic field at Q(a, b) with velocity, , v2, , v 'i, , 3 v ' j . Choose the correct statement(s):, , 2, , (a), , (c), 22., , 4, , IL2, , (b), , IL2, , (d), , 2IL, , P, , IL2, 4, , Circular loop of a wire and a long straight wire carry currents, Ic and Ie, respectively as shown in the figure. Assuming that these, are placed in the same plane. The magnetic fields will be zero at the, centre of the loop when the separation H is, , Sol. from page 437, , (c), , 21, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , × Q×, , ×, , v, , (b) Deviation suffered by the particle as it moves from P to Q, equals 60°, , (d), , 20, , ×v1 ×, , 2, x, (a) The particle can't be positron, , (c), , Answer Key, , y, , (d), , 3a, v1, , b, , v2, , 22, , (a), , 23, , (b)

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MAGNETIC EFFECT, , OF, , CURRENT, , 423, , Level -2, Only one option correct, 1., , (d), , B d, , 2., , (a), , I, , (b), , II, , (c), , III, , (d), , IV, , 4., , (a), , PQRS is a square loop made of uniform conducting wire the current, enters the loop at P and leaves at S. Then the magnetic field will be, , (b), , (d), 5., , 1 :, , 2, , 2, , 1, 3, :, 4, 4, , 1, 2, , An infinitely long conductor PQR is bent to form a right angle as, shown. A current I flows through PQR. The magnetic field due to, this current at the point M is B1. Now another infinitely long, straight conductor QS is connected at Q so that the current is I/2, in QR as well as in QS, The current in PQ remaining unchanged., The magnetic field at M is now B2. The ratio B1 / B2 is given by, , (c), , 3., , : :3, 2 2 4, , (c), , Figure shows a uniform magnetic field B and four straight-line, paths of equal lengths. Which path has the greatest value of, , 1, 2, 1, 2, 3, 2, , A non-planar loop of conducting wire carrying a current I is placed, as shown in the figure. Each of the straight sections of the loop is, of length 2a. The magnetic field due to this loop at the point P (a,, 0, a) points in the direction, , (a) maximum at the centre of the loop, (b) zero at the centre of loop, (c) zero at all points inside the loop, (d) zero at all points outside of the loop, The magnetic field at the centre of a circular coil of radius r is, times that due to a long straight wire at a distance r from it, for, equal currents. Figure here shows three cases : in all cases the, circular part has radius r and straight ones are infinitely long. For, same current the B field at the centre P in cases 1, 2, 3 have the, ratio, , 1, (a), , 1, , ˆj kˆ, , 2, , (b), , ˆj kˆ iˆ, , 3, , 1 ˆ ˆ ˆ, 1 ˆ ˆ, i j k, i k, (d), 3, 2, A long straight wire along the z-axis carries a current I in the, (c), , 6., , negative z direction. The magnetic field vector B at a point having, coordinates (x, y) in the z = 0 plane is, 0I, , (a), , (b), , 2, , 2, , :, , 2, , 1 :, , :, , 2, , 3, 4, , (a), , 1, 2, 3, 4, , 1 :, , 2, 0I, , 1, 2, , (c), , Answer Key, , 1, , (a), , 2, , (b), , Sol. from page 438, , 5, , (d), , 6, , (a), , 2, , yiˆ, , xjˆ, , x2, , y2, , xjˆ, , yiˆ, , x2, , 3, , 0I, , (b), , x2, , 2, 0I, , (d), , y2, , (a), , xiˆ, , xiˆ, x2, , 2, , 4, , yjˆ, y2, yjˆ, y2, , (c)

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424, 7., , ELECTRICITY & MAGNETISM, , A steady current i flows in a small square loop of wire of side L in, a horizontal plane. The loop is now folded about its middle such, , 10., , that half of it lies in a vertical plane. Let M1 and M 2 respectively, denote the magnetic moments due to the current loop before and, after folding. Then, (a), , M2, , (b), , M 1 and M 2 are in the same direction, , 0, , 11., , M1, (c), 8., , M1, , 2, , (d), , M2, , Two long parallel wires are at a distance 2d apart. They carry, steady equal currents flowing out of the plane of the paper, as, shown. The variation of the magnetic field B along the line xx is, given by, 12., , (a), , (a), , proportional to r, , (b), , inversely proportional to r, , (c), , directly proportional to, , (d), , independent of r, , A thin uniform ring of radius R carrying uniform charge Q and, mass M rotates about its axis with angular velocity . The ratio of, its magnetic moment and angular momentum is :, , 1, 2, , M2, , An infinitely long wire has a charge per unit length. The wire, moves along its axis with a velocity v (v << c, the velocity of, light). The ratio of the magnetic field to the electric field at a point, r from the wire is :, , (b), , (a), , Q, M, , (b), , M, Q, , (c), , Q, 2M, , (d), , M, 2Q, , A slab of resistance R is inserted between two parallel plates of a, capacitor charged to Q0. The capacitor is discharged through the, solid. A magnetic field is present throughout and is out of the plane, of the paper. The total momentum given to the slab after complete, discharge is (given that the capacitance of the capacitor is C and, the plates are separated by a distance d)., Q0, (a) CQ B, C, 0, , (c), , 9., , (d), , Two parallel beams of protons and electrons, carrying equal, currents are fixed at a separation d. The protons and electrons, move in opposite directions. P is a point on a line joining the, beams, at a distance x from any one beam. The magnetic field at P, is B. If B is plotted against x, which of the following best represents, the resulting curve, , 13., , (b), , dBQ0, , (c), , zero, , (d), , infinity, , B, , d, , Q0, , A charge q is moving with a velocity v1, , 1iˆ m/s at a point in a, , magnetic field and experiences a force F1, , q(–1 ˆj 1kˆ) N . If the, , charge is moving with a velocity v2, experiences a force F2, , 1 ˆj m/s at the same point, it, , q(1iˆ – 1kˆ) N . The magnetic induction B, , at that point is :, (a), , (b), , 14., , (c), , (d), , (a), , (1iˆ 1 ˆj 1kˆ)Wb/m 2, , (b), , (1iˆ – 1 ˆj 1kˆ )Wb/m 2, , (c), , (–1iˆ 1 ˆj – 1kˆ )Wb/m 2, , (d), , (1iˆ 1 ˆj – 1kˆ )Wb/m 2, , A long wire carries a steady current. It is bent into a circle of one, turn and the magnetic field at the centre of the coil is B. It is then, bent into a circular loop of n turns. The magnetic field at the centre, of the coil for same current will be, (a), , nB, , (b), , n2B, , (c), , 2nB, , (d), , 2n2B, , Answer Key, , 7, , (c), , 8, , (b), , 9, , (c), , Sol. from page 438, , 12, , (b), , 13, , (a), , 14, , (b), , 10, , (d), , 11, , (c)

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MAGNETIC EFFECT, 15., , Infinite number of straight wires each carrying current I are equally, placed as shown in the figure. Adjacent wires have current in, opposite direction. Net magnetic field at point P is, a, 30°, 30° 1 2 3 4 5, , 16., , 4, , (c), , 4, , n2 ˆ, k, 3a, , I, , 0, , (a), , I, , 0, , n4, 3a, , kˆ, , (a), , (b), , x, , Disc, , z, , B, , n4 ˆ, k, 3a, , I, , 0, , (b), , 4, , (d), , zero, , Fixed axis, , 1, 2, , 2 0i, , d, , 2 d, , 1, 2, , 0i, 1, 2 d, , (d), , 20., , P, , , clockwise, , (b), , q r 2 B sin, 4, , , anticlockwise, , (c), , q r 2 B sin, 2, , , anticlockwise, , M 2 , then which of the following is incorrect?, (a), , M1, , (b), , M1. M1 M 2, , (c), , M1, , (d), , M 1 . M 2 can never be zero, , 2, 0, , 2, , 0, , (d) none, AB and CD are long straight conductors, distance d apart, carrying, a current I. The magnetic field at the midpoint of BC is, , I, , B, , 21., , C, , ^j, i^, , k^, , A, , (c), , 0I ˆ, k, 2 d, 0I, , 4 d, , (b), , D, (b), , kˆ, , (d), , 0I, , d, 0I, , 8 d, , M 1 . M 2 M 2 may be zero, , 0 NI, , (a), , I, , 0, , A coil having N turns is wound tightly in the form of a spiral with, inner and outer radii a and b respectively. When a current I passes, through the coil, the magnetic field at the centre is, , d, I, , (a), , M2, , 0, , (b), (c), , q r 2 B sin, , clockwise, 4, A current carrying ring is bent along its diameter such that its one, half has magnetic moment M 1 and other half has magnetic moment, , A hollow cylinder having infinite length and carrying uniform current, per unit length along the circumference as shown. Magnetic field, inside the cylinder is, (a), , 18., , q r 2 B sin, 2, , (d), , 0i, , 45°, , 17., , (a), , 90°, , i, , 2 d, , (c), , 425, , A disc (of radius r) carrying positive charge q is rotating with, angular speed in a uniform magnetic field B about a fixed axis (as, shown in figure), such that angle made by axis of disc with magnetic, field is . Torque applied by axis on the disc is, , Find the magnetic field at P due to the arrangement shown, 0i, 1, 2 d, , CURRENT, , y, , a, P, , 19., , OF, , kˆ, , (c), , kˆ, , (d), , b, 2, , NI, , 0, , a, , a, , NI, 2 b a, 0, , 0I, , 2 b, , a, , Answer Key, , 15, , (b), , 16, , (a), , 17, , Sol. from page 438, , 19, , (d), , 20, , (d), , 21, , b, n, a, n, , b, , b, a, , (b), (c), , 18, , (b)

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426, 22., , ELECTRICITY & MAGNETISM, , An infinite cylinder of radius r with surface charge density is, rotated about its central axis with angular speed . What is the, magnetic field at any point inside the cylinder?, (a), , 0, , r2, , (b), , (c), , 0, , r, , (d), , 0, , 24., , A conducting wire bent in the form of a parabola y2 = x carrying, a current i = 1A as shown in the figure. This wire is placed in, a magnetic field B, of force is, , r, , 2kˆ tesla. The unit vector in the direction, , y, , 23., , 2, 0, , a, , r, , A star shaped loop (with = length of each section) carries current, i. Magnetic field at the centroid of the loop is, , x=4, x=1, , O, , x, , b, , (a), , (c), (a), (c), , 3 0i, , 3, , 3, , 0i, , (b), , 3 0i, 2, , (d), , 3, , 25., , 3, , 3iˆ 4 ˆj, 5, iˆ, , iˆ, , (b), , 2 ˆj, 5, , ˆj, 2, , (d) none of these, , Circular regions (1) and (2) have current densities J and – J, respectively, such that their region of intersection carries no, current. Magnetic field in their region of intersection is, , 0i, , 2, , (2), , (1), r2, , r1, d, (a), (b), (c), (d), , Answer Key, Sol. from page 438, , 22, , (c), , 23, , (c), , uniform, proportional to (r1 + r2) – d, uniform, proportional to d, non-uniform, zero, , 24, , (b), , 25, , (b)

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MAGNETIC EFFECT, , OF, , MCQ Type 2, Multiple correct options, 1., Two coaxial solenoids 1 and 2 of the same length are set so that one, is inside the other. The number of turns per unit length are n1 and, n2. The current i1 and i2 are flowing in opposite directions. The, magnetic field inside the inner solenoid is zero. This is possible, when, , 2., , (a), , i1, , i2 and n1, , n2, , (b), , i1, , i2 and n1, , n2, , (c), , i1, , i2 and n1, , n2, , (d), , i1n1, , 5., , i2 n2, , Consider the magnetic field produced by a finitely long current, carrying wire. Then, (a) the lines of field will be concentric circles with centres on the, wire., (b) there can be two points in the same plane where magnetic, fields are same, (c) there can be large number of points where the magnetic field, is same in magnitude., (d) the magnetic field at a point is inversely proportional to the, distance of the point from the wire., , 6., , 427, , CURRENT, , Exercise 6.2, , (a), , B = 0 for all points on the x-axis, , (b), , At all points on the y-axis, excluding the origin, B has only a, z-component., , (c), , At all points on the z-axis, excluding the origin, B has only a, y-component., , (d), , B cannot have an x-component, , An electron is moving along the positive x-axis. You want to apply, a magnetic field for a short time so that the electron may reverse its, direction and move parallel to the negative x-axis. This can be done, by applying the magnetic field along,, (a), , y-axis, , (b), , z- axis, , (c), , y-axis only, , (d), , z-axis only, , Inside a region of non uniform magnetic field given by the formula, , B, , B0 x kˆ, a particle of mass M and charge q is projected, , horizontally as shown. It is given that qB0 / M, , 1. At any instant, ajˆ bkˆ. If, , the velocity vector of the particle is found to (v / 2) iˆ, the initial value of the velocity is v m/s, then, y, , 3., , 4., , Consider three quantities x = E/B, y =, , 1/, , 0, , 0, , and z, , The direction of magnetic, , ., , CR, Here l is the length of a wire, C is a capacitance and R is a resistance., All other symbols have standard meanings. Then :, (a) x, y have the same dimensions, (b) y, z have the same dimensions, (c) z, x have the same dimensions, (d) none of the three pairs have the same dimensions., Two long thin, parallel conductors carrying equal currents in the, same direction are fixed parallel to the x-axis, one passing through, y = a and the other through y = –a. The resultant magnetic field due, to the two conductors at any point is B. Which of the following are, correct?, , v m/s, , x, , (a), , a, , (b), , a, , (c), , x coordinate of the particle at the instant when its velocity in, , v / 2 and b = 0, 3 v / 2 and b = 0, , the x-direction becomes equal to zero, is 2v, z, , ., , (d) x coordinate of the particle at the instant when its velocity in, the x-direction becomes equal to zero, is 3v, , –a, O, , i, , 1/ 2, , a, , 1/ 2, , y, , i, , x, , Answer Key, , 1, , (c, d), , 2, , (a, c, d), , Sol. from page 440, , 5, , (a, b), , 6, , (b, c), , 3, , (a, b, c), , 4, , (a, d)

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428, , ELECTRICITY & MAGNETISM, , Statement Questions, Read, (a), (b), (c), (d), , 1., , 2., , 3., , the two statements carefully to mark the correct option out of the options given below:, If both the statements are true and the statement - 2 is the correct explanation of statement - 1., If both the statements are true but statement - 2 is not the correct explanation of the statement - 1., If statement - 1 true but statement - 2 is false., If statement - 1 is false but statement - 2 is true., , Statement - 1, Alternating current shows no magnetic effect., Statement - 2, The magnetic field produced by alternating current in complete, cycle is zero., Statement - 1, A direct current flows through a thin conductor produces magnetic, field only outside the conductor., Statement - 2, There is no flow of charge carriers inside the conductor., Statement - 1, A current i flows along the long straight and thin walled pipe. Then, magnetic field at any point inside the pipe is zero., Statement - 2 It is according to, , 4., , Exercise 6.3, , B.d, , 6., , Statement - 1, Figure shows cross-sections of two long straight wires; the lefthand wire carries current in directly out of the page. If the magnetic, field due to the two currents is to be zero at P, the current i2 must, be into the page., , Statement - 2, i2 > i1., 7., , Statement - 1, Figure shows a current carrying circular loop. The magnetic field, at the centre of loop is zero., , 0i ., , Statement - 1, Ampere s law used for the closed loop shown in figure is written, as, , Bd, , 0 i1, , i2 . Right, , side of it does not include i3,, Statement - 2, , because it produces no magnetic field at the loop., , Magnetic field at the centre of loop is given by, , B, Statement - 2, The line integral of magnetic field produced by i3 over the close, loop is zero., 5., , 8., , 0 ni, , 2R, , ., , Statement - 1, Figure shows two identical currents i and a close path encircling, them. For the close path, , B d, , 2i., , Statement - 1, The magnetic field due to a very large current carrying loop is zero, at its centre., Statement - 2, Magnetic field at the centre of loop is, B, , 0i, , ., , 2R, Statement - 2, For the close path, , Bd, , 0., , Answer Key, , 1, , (d), , 2, , (c), , 3, , (a), , 4, , (d), , Sol. from page 441, , 5, , (a), , 6, , (a), , 7, , (a), , 8., , (d)

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MAGNETIC EFFECT, 9., , 10., , Statement - 1, The force between two parallel current carrying conductors carrying, currents in same direction is attractive because there is no electrical, interaction between them., Statement - 2, The force between two electrons streams moving in the same, direction repulsive because there is no magnetic interaction between, them., Statement - 1, Two square conducting loops carry currents 3A and 1A as shown, in figure., For the path 1,, , Bd, , 2, , 12., , 0., , 13., , OF, , 429, , CURRENT, , Statement - 2, For the close path, the magnetic field at each point on the path is, zero., Statement - 1, The magnetic field of a long solenoid is 0ni at its centre., Statement - 2, The magnetic field at any point outside the solenoid on its bisector, is zero., Statement - 1, Two long current carrying wires are placed perpendicular to each, other. The net force exerted by one wire on the other is zero., , Statement - 2, For the path 2,, 11., , B d, , 0., , Statement - 1, Figure shows a current carrying conductor and a close path. For, the close path, , Bd, , Statement - 2, The net torque exerted by one wire on the other is zero., , 0., , Answer Key, , 9, , (c), , Sol. from page 441, , 13, , (c), , 10, , (b), , 11, , (c), , 12, , (b)

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430, , ELECTRICITY & MAGNETISM, , Passage & Matrix, , 4., , The magnetic field due to this solenoid at the centre of its axis is :, , Passage for Q. 1 to Q. 3, A coaxial line carries the same current i up the inside conductor of radius, a as down the outer conductor of inner radius b and outer radius c., , (a), , 0i, 2a, , (c), , 0, , (b), , ni, , 1, 5., , 1., , (c), 2., , 0, , (b), , ia, , (d), , r2, , 4, , 0, , (a), , ir, a2, , 2, 0, , 2, , (c), 6., , i, r, , (c), , zero, , 0, , 2, , (b), , i, r, , (d), , 0, , 4, 0, , i, r, , The magnetic field at a distance r from the axis, for b < r < c is :, , (a), , 2, , i, r, , 2, , i, r2, 1 2, r, c, , 0, , (c), , 0, , (b), , b2, , 0, , 2, , a2, , 1, , 4a2, , 2, , 2, , 0n i, , (b), , 2, 0 ni, 2a, , (d), , 0ni, 0 ni, , a, , (a), , zero, , (c), , 0ni, , (b), , 0n i, , 2, , (d) infinite, , Passage for Q. 7 to Q. 9, A charge particle of mass m and charge q is projected on a rough, horizontal xy-plane. Both electric and magnetic fields are given by, , ir, a2, , 2, , 0n i, , (d), , The magnetic field due to the solenoid at the centre of its axis in the, situation a << is :, , The magnetic field at a distance r from the axis, for a < r < b is :, (a), , 3., , zero, , ni, 2a, , 0, , The magnetic field due to the solenoid at the centre of its axis in the, situation a >> is :, , The magnetic field at a distance r from the axis, for r < a is :, (a), , Exercise 6.4, , 10kˆ N / C and magnetic field B, , E, , 5kˆ tesla are present in the, , r2, , b2, , region. The particle enters into the magnetic field at (4, 0, 0) m with, , c2, , b2, , a velocity 50 ˆj m/s. The particle starts into a curved path on the, plane. If coefficient of friction, , (d) none of these, , b2, , 1, 3, , between particle and plane,, , then (qE = 2mg, g = 10 m/s2) :, 7., , Radius of curvature of the path followed by particle, initially,, is, (a) 5 m, (b) 2.5 m, (c) 1.25 m, (d) 10 m, The time after which particle comes to rest, is, (a) 5s, (b) 4s, (c) 3s, (d) 1s, Total work done by electric force on the particle is, (a) 250 J, (b) zero, (c) 125 J, (d) none, , Passage for Q. 4 to Q. 6, Take a tightly-wound solenoid of radius a and length . The number of, turns per unit length in it is n. It carries current i. Consider a small, element of length dx of the solenoid at a distance x from one end. This, contains ndx turns and can be assumed as a current carrying loop. The, magnetic field due to whole solenoid will be the sum of magnetic field due, to such elements., , 8., , 9., , Answer Key, , 1, , (b), , 2, , (c), , 3, , (c), , 4, , (d), , Sol. from page 441, , 6, , (c), , 7, , (a), , 8, , (a), , 9, , (b), , 5, , (c)

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MAGNETIC EFFECT, 11., , Passage for Q. 10 to Q. 12, An electron is moving with velocity v1iˆ v2 ˆj, , (v1, , (a), , origin in the presence of electric and magnetic fields Ejˆ & Bjˆ, respectively., 10., , (a), , (c), , 13., , m v12, , v22, , (b), , 2eE, mv22, eE, , 12., , mv22, 2eE, , (a), , 2 mE, (d), , (c), , eB 2, , 431, , nE, B, , v2, , (b) v2, , 2 nE, B, , E, nE, (d) v2, nB, B, The number of revolutions the charge does before its y coordinate, becomes negative is:, (c), , The maximum value of the y-coordinate during the motion of the, electron is, , CURRENT, , The condition such that the electron passes the origin again is, (where n is a natural number), , 0) at the, , 0, v2, , OF, , v2, , B v12, , v22, , (b), , E, , E, Bv2, , (d), , B 2 v12, , v22, , E2, Bv2, E, , Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column-II. In each case, a point M and line, PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given, charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be, the magnetic field at M and be the magnetic moment of the system in this condition. Assume each rotating charge to the equivalent to a steady, current., , Column-I, A., , Column-II, +, , E=0, , B., , V, , C., , B=0, , (p), , – Q, +, , Charges are at the corners of a regular hexagon. M is at the centre of the, hexagon. PQ is perpendicular to the plane of the hexagon., , + – +, , Charges are on a line perpendicular to PQ at equal intervals. M is the, mid-point between the two innermost charges., , –, M, , 0, , –, , P+, , D., , – + –, , (q), , M, Q, +, , +, , –, , P, , +, , –, , +, , Charges are placed on two coplanar insulating rings at equal intervals. M, is the common centre of the rings. PQ is perpendicular to the plane of, the rings., , –, M, , P, –, , +, , +, , Charges are placed at the corners of a rectangle of sides a and 2a and at, the mid points of the longer sides. M is at the centre of the rectangle. PQ, is parallel to the longer sides., , Q, –, , P, , +, , (t, , Q, , – M –, , (r), , (s), , P, , –, +, , M–, , –, , Charges are placed on two coplanar, identical insulating rings at equal, intervals. M is the mid-point between the centres of the rings. PQ is, perpendicular to the line joining the centres and coplanar to the rings., , Q, , Answer Key, , 10, , Sol. from page 441, , 13, , (b), , 11, , (a), , A-p, r, s ; B - r, s ; C- p, q, t ; D- r, s, , 12, , (d)

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432, 14., , ELECTRICITY & MAGNETISM, Two wires each carrying a steady current I are shown in four configurations in Column I. Some of the resulting effects are described in Column, II. Match the statements in Column I with the statements in column II and indicate your answer by darkening appropriate bubbles in the, 4 × 4 matrix given in the ORS., , Column I, A., , Column II, , Point P is situated midway between, , (p) The magnetic fields (B) at P due to the currents, , the wires., , in the wires are in the same direction., , P, B., , Point P is situated at the mid-point of the, , (q) The magnetic fields (B) at P due to the currents, , line joining the centers of the circular, , in the wires are in opposite directions., , wires, which have same radii., , P, , C., , Point P is situated at the mid-point of the, , (r) There is no magnetic field at P., , line joining the centers of the circular, wires, which have same radii., , P, , D., , Point P is situated at the common center, , (s) The wires repel each other., , of the wires., , P, , 15., , Match entries in Column I to all possible entries in Column II., Column I, A., , Column II, , Motion of a free positively charged particle in uniform, , (p) Uniform motion, , magnetic field only, B., , Motion of a free negatively charged particle in electric, , C., , Motion of a particle for which velocity is given to be, , (q) Trajectory is circular, , field only, , V, D., , k sin tiˆ cos ktjˆ, , (r) Trajectory is helical, , kkˆ . Here k is an integer, , Superposition of simple harmonic motions given by :, , (s) Trajectory is parabolic, , x = Asin t, y = Bsin2 t. Here A and B are constants, (t) Trajectory is elliptical, , Answer Key, , 14, , A - q, r ; B - p ; C - p, r ; D- q, s, , Sol. from page 441, , 15, , A-p, q, r ; B - q, r, s, t ; C- p, r ; D- s, , 16, , A- r, t ; B - r, t ; C- r, s ; D- r, t

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MAGNETIC EFFECT, 16., , Column – I, I2, A., , OF, , CURRENT, , 433, , Column – II, , 1 2, , P, , I1, , (p), , 1 and 2 attract each other, , (q), , 1 and 2 repel each other, , (r), , 1 and 2 do not exert any force on each other., , (s), , Magnetic field at point P due to wires 1 and 2 are, perpendicular to each other., , (t), , Magnetic field at point P due to wires 1 and 2 are in, same direction., , I2, P, I1, , B., , 2, 1, , I2, 2, , C., , P, I1 1, 2, I2, , D., , P, , 1, I1, , Answer Key, , Wires 1 and 2, are in same, plane without, touching each, other, , 16, , A- r, t ; B - r, t ; C- r, s ; D- r, t, , Sol. from page 441, , Subjective Integer Type, , Exercise 6.5, Solution from page 444, , 1., , 2., , A long, straight wire carries a current i. Let B1 be the magnetic field, at a point P at a distance d from the wire. Consider a section of, length of this wire such that the point P lies on a perpendicular, bisector of the section. Let B2 be the magnetic field at this point, due to this section only. Find the value of d/ so that B2 differs, from B1 by 1%., Ans : 0.07., Four long copper wires are parallel to each other, their crosssections forming the corners of a square with sides a = 20 cm. A 20, A current exists in each wire in the direction shown in figure. What, , (i), , An electron moving with a velocity of 4×10+5 m/s along the, positive x-direction experiences a force of magnitude, 3.2×10–20 N at the point R. Find the value of I., , (ii), , Find all the positions at which a third long parallel wire, carrying a current of magnitude 2.5 A may be placed so that, the magnetic induction at R is zero. Ans : (i) 4A (ii) ± 1m, , are the magnitude and direction of B at the centre of the square ?, , 4., , A 200-turn solenoid having a length of 25 cm and a diameter of 10, cm carries a current of 0.30 A. Calculate the magnitude of the, magnetic field B inside the solenoid., , 3., , Ans : 80 T up the page., Two long parallel wires carrying current 2.5 and I ampere in the, same direction (directed into plane of the paper) are held at P and, Q respectively such that they are perpendicular to the plane of, the paper. The points P and Q are located at a distance of 5 metre, and 2 metre respectively from a collinear point R (fig.)., , 5., , Ans : 0.30 mT., , A toroid having a square cross section, 5.00 cm on a side, and an, inner radius of 15.0 cm has 500 turns and carries a current of 0.80, A. What is the magnetic field inside the toroid at, (a) the inner radius and (b) the outer radius of the toroid?, Ans : (a) 533 T (b) 400 T.

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434, , ELECTRICITY & MAGNETISM, , Subjective, , Exercise 6.6, Solution from page 445, , 1., , Consider a long, straight wire of cross-sectional area A carrying, current i. Let there be n free electrons per unit volume. An observer, places himself on a trolley moving in the direction to the current, , i, neA, , with a speed v, , and separated from the wire by a, , 8., , distance r. Find the magnetic field as seen by the observer., Ans : B, 2., , In Bohr model of hydrogen atom, the electron circulates around, nucleus on a path of radius 0.51Å at a frequency of 6.8×1015 rev/, s. Calculate the magnetic field induction set up at the centre of, the orbit. What is the effective dipole moment ?, Ans : 14, , 3., , 4., , Wb, m, , 2, , , 9.0 10, , A m2., , 2 2 0i, ., a, , 9., , 4, 4 x2, , 0, , ia 2, , a 2 4 x2, , 2a 2, , 1/ 2, , Ans : 10–4 Wb/m2., In the circuit of figure, the curved segments are arcs of circles of, radii a and b with common centre P. The straight segments are, along radii. Find the magnetic field B at the point P, assuming a, current i in the circuit., , A square loop of wire of edge length a carries current i. Show that, the magnitude of the magnetic field produced at a point on the, axis of the loop and a distance x from its centre is, , B, 6., , 24, , At a certain position, earth s magnetic field of 39 T is horizontal, and directed due north. Suppose the net field is zero exactly 8.0, cm above a long, straight, horizontal wire that carries a constant, current. What are (a) the magnitude and (b) the direction of the, current ?, Ans : (a) 16 A (b) west to east., A square loop of wire of edge length a carries current i. Show, that, the magnetic field at the centre of loop is B, , 5., , 0i, ., 2 r, , horizontal axis passing through the mid point of the rod with, the same angular frequency. Calculate the magnetic field at a, point on the axis at a distance of 0.4 m from the centre of rod., Ans : (a) 11.3 gauss (b) 22.6 gauss., A pair of stationary and infinitely long bent wires are placed in the, xy-plane as shown in figure. The wires carry currents of i = 10A, each as shown., The segments L and M are along the x-axis. The segments P and Q, are parallel to the y-axis such that OS = OR = 0.02 m. Find the, magnitude and direction of the magnetic induction at the origin O., , ., , In figure, a straight wire of length L carries current i. Show that, the magnitude of magnetic field at P, a perpendicular distance R, from one end of the wire is, , B, , 0i, , 4 R L2, , L, R, , 2 1/ 2, , ., , Ans :, 10., , 0i, , 4, , 1, a, , 1, , out of page., b, , Figure shows the cross-section of a long conducting cylinder with, inner radius a = 2.0 cm and outer radius b = 4.0 cm. The cylinder, carries a current out of the page, and the current density in the, cross-section is given by J = cr2, with c = 3.0×106 A/m4 and r in, metre. What is the magnetic field B at a point that is 3.0 cm from, the central axis of the cylinder ?, , 7., , (i), , (ii), , A charge of 1 coulomb is placed at one end of a nonconducting rod of length 0.6 m. The rod is rotated in a, vertical plane about a horizontal axis passing through the, other end of the rod with angular frequency 104 rad/s., Find the magnetic field at a point on the axis of rotation at, a distance of 0.8 m from the centre of the path., Now half the charge is removed from one end and placed on, the other end. The rod is rotated in a vertical plane about, , Ans : B = 2.0×10–5 T.

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MAGNETIC EFFECT, 11., , A long wire carrying a current i is bent to form a plane angle . Find, the magnetic field B at a point on the bisector of this angle situated, at a distance x from the vertex., Ans :, , 12., , 13., , 0i, , 2 x, , cot, , 16., , 4, , Find the magnitude and direction of magnetic field induction at, point O due to current carrying wire shaped in figure. The current, I = 5.0 A, R = 12.0 cm and angle = 90°., , r, J 0 . Find the magnetic, a, , field inside the wire., Ans :, , 0 J0r, , 3a, , 2, , ., , Figure shows a cross-section of a long cylindrical conductor, of radius a containing a long cylindrical hole of radius b., The axes of the cylinder and hole are parallel and are a distance d, apart; a current i is uniformly distributed over the shaded area., , (a), , 1, 5, , 1, 3 13, , , out of page., , Four long, straight wires, each carrying a current of 5.0 A, are, placed in a plane as shown in figure. The points of intersection, form a square of side 5.0 cm., (a) Find the magnetic field at the centre P of the square., (b) Q1, Q2, Q3 and Q4 are points situated on the diagonals of the, square and at a distance from P that is equal to the length of, the diagonal of the square. Find the magnetic fields at these, points., , Ans : (a) Zero (b) Q1 : 1.1×10–4 T, , Q2 : zero; Q3 : 1.1×10–4 T, ; and, Q4 : zero., 17. A long straight wire carries a current of 10 A directed along the, negative y-axis, as shown in figure. A uniform magnetic field B0 of, magnitude 10–6 T is directed parallel to the x-axis. What is the, resultant magnetic field at the following points ?, (a) x = 0, z = 2m;, (b) x = 2m, z = 0;, (c) x = 0, z = –0.5 m., , Use superposition to show that the magnetic field at the, centre of the hole is, , B, , 0 id, 2, 2, , 2 a, , ., , b, , (b), (c), 15., , 0i, a, , 435, , Ans : 2.8×10–5 T., The current density inside a long, solid cylindrical wire of radius a, is in the direction of the central axis and varies linearly with radial, distance r from the axis according to J, , 14., , 2, Ans :, , CURRENT, , OF, , Discuss the two special cases b = 0 and d = 0., Use Ampere s law to show that the magnetic field in the hole, is uniform., Figure shows a square loop of edge a made of a uniform wire. A, current i enters the loop at the point A and leaves it at the point C., Find the magnetic field at the point P which is on the perpendicular, bisector of AB at a distance, , 18., , a, from it., 4, 19., , Ans : (a) zero (b) 1.41×10–6 T, 45° in xy-plane (c) 5.00×10–6 T,, +x-direction., A long solenoid with 10.0 turns/cm and a radius of 7.00 cm carries, a current of 20.0 mA. A current of 6.00 A exists in a straight, conductor located along the central axis of the solenoid. (a) At, what radial distance from the axis-will the direction of the resulting, magnetic field be at 45.0° to the axial direction ? (b) What is the, magnitude of the magnetic field there ?, Ans : (a) 4.77 cm; (b) 35.5 T., A capacitor of capacitance 100 F is connected to a battery of 20, volt for a long time and then disconnected from it. It is now connected, across a long solenoid having 4000 turns per metre. It is found that, the potential difference across the capacitor drops to 90% of its, maximum value in 2.0 second. Estimate the average magnetic field, produced at the centre of the solenoid during the period, Ans : 16, , 10, , 8, , T.

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436, 20., , ELECTRICITY & MAGNETISM, , A current I flows along a lengthy thin-walled tube of radius R with, longitudinal slit of width h. Find the induction of the magnetic, field inside the tube under the condition h << R., Ans : B ;, , 21., , 0 hI, , 4, , 2, , Rr, , 23., , Find the magnetic moment of a thin round loop with current if the, radius of the loop is equal to R and the magnetic induction at its, , , where r is the distance from the cut., 24., , 0 nI, , Ans : (a) B, 0I, 2, , b, 0, , 1, , r, , 1, , 2, , ., , R, , Find the current density as a function of distance r from the axis of, a radially symmetrical parallel stream of electrons if the magnetic, induction inside the stream varies at B = br , where b and are, positive constant., Ans : j r, , ., , A very long straight solenoid has a cross-section radius R and n, turns per unit length. A direct current I flows through the solenoid., Suppose that x is the distance from the end of the solenoid,, measured along its axis. Find :, (a) the magnetic induction B on the axis as a function of x; draw, an approximate plot of B vs the ratio x/R;, (b) the distance x0 to the point on the axis at which the value of, , 1% from that in the middle section of the, , B differs by, solenoid., , 22., , 2 R3 B, 0, , A current I flows in a long straight wire with cross-section having, the form of a thin half-ring of radius R (figure). Find the induction, of the magnetic field at the point O., , Ans : B, , Ans : M, , centre is equal to B., , ., , x, , 1, , x2, , R2, , and x < 0 inside the solenoid., , (b), , x0, , R1 2, 2, , 1, , ; 5 R., , , where x > 0 outside the solenoid

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MAGNETIC EFFECT, , OF, , CURRENT, , 437, , Solutions Exercise 6.1 Level-1, 1., , (c), , If r is the separation between A and B or B and C, then force, per unit length, , i2, r, , 0, , FA, , 2, , i2, , FB, 2r, , i2, r, , 0, , 2, , i2, , FC, r, , 0, , 2, , i2, r, , i2, 2r, , 12., , (d), , 13., , (a), , (c), , In loop 1 :, loop 2 :, , 4., , (d), , 0 (i, , B.d, , loop 4 :, (d), , B.d, , 0(, , 6., , (c), , i i), , 0i, , i), , (c), , B, , 15., , (c), , B, , 16., , (b), , i), , 2 0i, , 17., , (c), , B1, 9., 10., , (b), (b), , B, , 0i, , 2r, , 4, , (a), , 0i, and B2, 2R, , B, , i, ., 2 r, , 1, i, 0, 2, 2 R1, 0i, , 4, , (a), , (4), 2 2d, 0, , 1, R1, , 0i, , 2r, , 10 7 i, 2 0.05, i = 0.4 A., , 5, , 4, , 2(a, , 2, , 0ia, , 19., , (c), , 20., , (c), , 21., , (d), , R 2., 0 (2)i, , 2r, , 0 (2)i, 2R 2, , 2, , .......... (i), , 2 32, , r ), , Bcentre, , B, , B12, , 250 10 6 T ., , B22, 2, , 2, , 4, 2(2 ), 0, , = 5 × 10–5 T.., , 4 B1., , ii, . 12, 2 r, 0 i1i2, ., (d cos )., 2 r, 1, As B, , so it represents rectangular hyperbola., r, F, , L, , 0, , 22., , (a), , 1]., , I ( R2 ), , iA, , I, , L, 2, , 2, , IL2, ., 4, , Bcircular = B straight, or, , 1, i, 0, 2 ., 2 R2, , 0ic, 2R, , H, , 23., , L, 2, , 2 R or R, , Now M, , 1, ., R2, , , out of plane of paper.., , 2r, , 0.5 10, , 0, , B., , 0, , 2i, [, r, , 0i, , 3, 2(2 ), , As the distance of the points P and Q is equal from the wire, and so magnetic field is also equal., , 0, , 11., , 2 2 r , or r, , i, r, , 0, , So, resultant magnetic field = B + B = 2 B along –ve z-axis., If R and r are the radius of bigger and smaller loop respectively,, then 2 R, , i, r, , Bcentre, , d, , Magnetic field due to 4 A wire, B, , 8., , 2, , ., , 0i, .......... (ii), 2a, On dividing equation (ii) by (i), and putting the known values,, we get, , 0 . (2) ., , 2, , ., , 0, , 2, , B, and, , i) 0, , or, B = 0., The magnetic field of the straight wire is parallel to circular, current and so force on it will be zero., , Given B, , 0, , (a), , (a), (a), , i, r, , B, or, , 18., 7., , ., , 0, , 2, , 0 en, ., 2r, , 0i, , If r is the distance of dot from centre of any coil, and B, B' are, the fields due to larger and smaller coil respectively, then, B1 = B + B = 2B, B2 = B – B = 0, B3 = (B – B) + (B' – B') = 0,, B4 = B + B' + B' = B + 2B'., Clearly the field is greatest in case of IV., Outside the cable,, 0 (i, , 14., , qn en, 0i, 2r, , i) 0, , 0 (i, , B.d, , 5., , 0 (i, , B.d, , loop 3 :, , 3., , B.d, , q, T, , i, Now, B, , Clearly, FB > FC > FA., 2., , Using right hand screw rule, the direction of magnetic field is, along OY'., , (b), , tan, or, , 0, , 2, , ., , ie, H, , v1, , ie R, ., ic, , vy, vx, = 60°, , 3, 1, , vy, , 3, , v2, vx

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438, , ELECTRICITY & MAGNETISM, , Solutions Exercise6.1 Level -2, 1., 2., , B1d, , (a), , For, , (b), , magnetic field is greatest in case I., Magnetic field at the centre of the loop,, , B, , 0, , 3, , 4, , to be greatest, the projected length along, , ., , B sin iˆ B cos ˆj, , Now B, , 0, , ., , 0, , ., , 2, , i 4, (sin 45, r, , sin 45 ), 2, , 0 3i 4, ., (sin 45, 4, r, , sin 45 ), , 0, , 7., , I y iˆ, r r, , xˆ, j, r, , I, , xjˆ ), , r2, , ( yiˆ, , ˆ ˆ, 0 ( yi xj ), I, ., 2 (x2 y2 ), M1 = iA = iL2, , (c), , L2, i, 2, , M2, , 2, , L2, i, 2, , 2, , 2, , (a), , If B is the magnetic field of straight conductor at P, then, , B, , B2, 2, , B1, , 5., , 2, , 0 I, . ,, 4 r, , 3 B, 4, , B, , B3, 2, , So B1 : B2 : B3, 4., , :, , 2, , :, , 0 I 2, ., 2, r, , 3, 4, , B, 2, , L, , ., , L/2, , 1, ., 2, , I 2, ., 4, r, , 3, 2, , 0, , (c), , B1, , (d), , B1 2, ., B2 3, The equivalent loop is shown in figure. If B is the magnetic, field of each loop, then, B, Biˆ Bkˆ B (iˆ kˆ), , B2, , M1, M2, , I, ., 4 r, 0, , 8., , (b), , iˆ kˆ, ., 2, , So unit vector of resultant field, , z, , i, . , it represents a rectangular hyperbola. The, 2 r, combined graph is shown in figure., B, B, B, B, +, , (c), , Similar to Q. 8., , 10., , (d), , B, , 11., , (a), , B, , (c), , x, , Magnetic field due to long straight wire,, , d, , 9., , So,, , i, , 2., 0, , As B, , resultant, , 6., , ., , Li, , L i, 3., , iL2, , 12., , (b), , 0, , 2, B, E, , ., , i, and E, r, , 2, , 0, , r, , k .r ., , Q, Q, A, T, 2, L = I = M R2, , R2, , M Q R2 2, L, MR 2, Momentum,, , Q, ., 2M, , M, , I, . ., 2 r, 0, , iA, , P, , y, , d, , Q R2, ., 2, , Fdt, ( Bid ) dt, Bd idt, , = BdQ0., , r, y, , x, , 13., , B, , (a), , F, or, , x, , and, , q (v, , B), , q ( ˆj kˆ), , q iˆ ( Bxiˆ B y ˆj Bz kˆ ), , ..... (i), , q (iˆ kˆ), , q ˆj ( Bxiˆ B y ˆj Bz kˆ ), , ..... (ii)

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MAGNETIC EFFECT, On simplifying above equations, we get, , (iˆ, , B, 14., , (b), , 2 R, , 0i, , Now, B, 15., , (b), , 0, , B, , 2R, , 0ni, , r, , (d), , Rn, , n B., , Also their magnetic moment are perpendicular to each other,, then M 1 . M 2, , 1, 2a cos30, , 2sin 30, 1, 0, .I, 1, 4, 2, a cos30, , 1, ....., 3, , 1, 3a cos30, , ....., , kˆ, , 21., , and, , a, , i, 2, ., sin( 45 ) sin 90, 4 d cos 45, , B, , 1, , 0i, , 22., , (c), , 90° i, , i, , 23., , (c), , 45°, , 0, , 2, , 2, , 0, , 0 i, sin 60, 4 r, , (b), , B, , 2, , 0, , 4, , ., , I, ( kˆ), d 2, , 0I, , ., , 0, , ., , 4, , ., , d, , 0, , 4, , = 6, , (b), , i, r, , (d), , 6, , 0, , 4, , ., , i, sin 30, r, , 3 1, , 3 1, , 0i, , 3, , ., , ˆj, , y, , ., , a, F, , 45°, 1m, , r, , 45°, , dM B sin, , =, , x, , 3m, , 0, , r, , =B, , r 3dr, , 2m, , r4, 4, , b, , sin, 0, , =B, =(, , r2), , sin, , r 2 B sin, 4, , sin 30, , 3, 2, , iˆ, , rdr, , B, , At x = 1, y = 1 and at x = 4.y = 2., The given wire can be replaced by a wire joining the two, ends; a and b. So unit vector in the direction of force, i cos 45, j sin 45, , kˆ, , 2 rdr, dq, =, dt, 2, dm = (di) × r2 =, r3dr, , The current, di =, , sin 60, , i, , 2, , 19., , r, , 3l, 2, , r, , = 6, , 24., , dr, , dq, 2 rl, = r, 2 /, t, Using Amphere’s law, we can write, B = 0i, or B = 0( r ), B= 0 r, , = 3, , 18., , r, , i, , B= 6, , P, The cylinder is cut into two equal halfs, as shown. Each half, produces magnetic field inside is 0 2. So,, 0, , a, b, , b, 0 NI, n, ., 2(b a ), a, , 1 ., , 2, , B, , N, drI, b a, 2r, , 0, , B, , 0, , 2 d, , (b), , 2r, b, , n4 ˆ, .I, k, 4, 3a, , 17., , 0 ( dN ) I, , Now dB, , kˆ, , 0, , N, dr ., b a, , dN, , (c), , 0, , (a), , M2, , 2, , sin 30 ), , 1, a cos30, , 16., , q r 2 B sin, 4, Both the half of equal area and so, , M1, 0ni, , and B, , I (sin 30, , 4, , 20., , R, ., n, , r, , 439, , CURRENT, , =, , ˆj kˆ) Wb m 2 ., , n(2 r ),, , OF

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440, , ELECTRICITY & MAGNETISM, , Alternative sol., (b), , ab, , Ob, , 4iˆ 2 ˆj iˆ, , Oa, , y, , ˆj, , 3iˆ 3 ˆj, 25., , a (1, 1), , d2, , J, , x=4, , J, , x, , x=1, , O, , d1, , (b), , ˆ, B.di, 0iin, B × 2 r = 0(j × r2), , or, or, , Fˆ, , i (lˆ, , B ) 1(3iˆ 3 ˆj ) ( 2 kˆ ), , 6 ˆj 6iˆ, 6 2, , iˆ, , 0 Jr, , B1, , B2, , 2, If d 1 and d 2 are the distances from the centres of the regions,, then, , b (4, 2), F, , B, , (6 ˆj 6iˆ), , B, , ˆj, 2, , =, , 0J d, 1, , =, , 0J d, , Solutions Exercise-6.2, 1., , 2., , (c, d) The magnetic field inside the inner solenoid is zero if,, Bouter = Binner, or, µ0n1i1 = µ0n2i2, or, n1i1 = n2i2, Also if i1 = i2, then n1 = n2., (a, b, c) See figure,, , 1, . points 1, 2, 3, ..... have equal amount of, r, magnetic field., , Also, B, , 6., , 2, 2, , (b, c) As the magnetic field is acting along z-axis, and particle, initial velocity is along x-axis, so particle will remain in xyplane. Also its speed must be constant. Thus, b = 0, and v =, , v x2, , v, , v y2, , 3., , v, 2, , a, , =, , 3, v., 2, , qvB, , =, , M v, , or, , qv (– B0 x), , =, , Mv, , or, , – dv, , Now,, , dv, dx, , dv, dx, , (a, b, c) E/B has dimensions of velocity. CR has dimensions of time, and so, , 5., , a2, , v, , 0, , 4., , 2, , =, , v2, , or, , i, 3, 12, , d2, , has dimensions of velocity. So x, y and z all the, CR, three have dimensions of velocity., (a, d) On x-axis two conductors produce equal and opposite, fields, so net field becomes zero., (a, b) Electron stream can reverse the direction either through xyplane of through xz-plane., According to Fleming's left hand rule the magnetic field, either along z-axis or along y-axis., , z, , Electron stream, , y, x, , x, , =, , qB0, xdx, M 0, , qB0 x 2, M, 2, , v, , or, , v, , =, , or, , v, , =, , x, , =, , 1, , x2, 2, , 2v ., , x

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MAGNETIC EFFECT, , OF, , CURRENT, , 441, , Solutions Exercise-6.3, 1., , (d), , Current shows magnetic effect, but in one cmplete cycle the, net magnetic field beocmes zero., , 2., , (c), , The magnetic field inside conductor is given by, 0, , B=, , 3., , (a), , 2, , ., , The magnetic field of two equal halfs of the loop is equal and, opposite and so B, , 8., , (d), , ir, , For the given close path,, B.d, , a2, , 0., , 0 (i, , =, , i), , 0., , 9., , (c), , In case of two electron streams the electric repulsion is greater, than magnetic attraction., , In thin walled tube, iin = 0 and so, , 10., , (b), , For path 1,, , B.d, , 0(, , For path 2,, , B.d, , 0 (3 – 3 – 1), , or, , = µ0 × 0, , (d), , The magnetic field at any point on the closed loop is due to, all the three currents, but line integral of i3 over the closed, loop will be zero., , 5., , (a), , Magnetic field at the centre of circular loop is given by, 0i, , as R, 2R, , B, , , and so B, , 0., , 11., , (c), , 0, , ., , B.d, , We know that ,, B.d, , –2, –, , 0., , 0., , 0iin, , 12., , (b), , See theory of the chapter, , 13., , (c), , Each wire will experiences, equal and opposite forces and so, net force on them becomes zero, but these forces constitues, a couple whose torque is non zero., , F, , i1, i2, =, r1, r2 ., , As r2, , B, F, , r1 , and so i2, , i1 ., , Solutions Exercise-6.4, Passage for Q. 1 to Q. 3, 1., , B.d, , (b), , or B, , B=, 2., , (c), , 0, , 2, , i, r2, or B = 2 . r 1 2, c, , 0iin, , i, , 2 r =, , ., , 0, , a, , r2, , 2, , a, , b2, , Passage for Q. 4 to Q. 6, , 4., , ., , 2, , b2, , ndx, , ir, , (d), , dB =, , 0 ( ndx)i, 2 3/ 2, , 2(a 2, , x ), , ,, , a, x, , For a < r < b :, 0i, , B 2 r =, , where x = – a cot, , 0 i, . ., or B =, 2 r, , 3., , (c), , B×2 r=, , 0, , i, , a, , 2, , For, b < r < c :, , i (r 2, (c, , 2, , b2 ), 2, , b ), , 0 and so, , 0 . The magnetic field on any point on the close, , i1, 0 i2, ., r1 = 2 r2, , or, , . Since iin, , loop is zero., , For magnetic field to be zero at P, , 2, , 3 1), , B = 0., , 4., , (a), , (a), , For thin conductor, the point is on the axis of the conductor, and so r = 0, B = 0., , B.d, , 6., , 7., , 1

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442, , ELECTRICITY & MAGNETISM, B=, , 2, , Putting cos, , Time period of revolution,, , t, , 1, , [cos, , cos, , 1, , 2], , nT, , 2, , or t, , 2v2, ay, , 2v2, Ee, m, , so n, , 2 m, eB, , 2v2, Ee / m, , 2, , /2, , 0 ni, 2, , we get B, , 2, , ,, , nE, ., B, , v2, 12., , 0 ni, 2, , 1 4a /, , 2, , 4a, , (d), , We have, v2 =, , 2, , and so n, , 6., , (c), , (c), , In case a >>, , 2, , ,, , 2, , For a << ,, , 4a, , 0 ni, , 4a 2 ; 2a and so B, , and so B, , 2, , 2a, , 0ni, , nE / B, , v2 B, E, , ., 13., , (p), , ., , +, , –, , Q, , –, , +, , M, , y, Passage for Q. 7 to Q. 9, , P+, , E = –10 k^ N/C, , ^, , (a), , r=, , mv, =, qB, , m 50, 2mg, 5, 10, , v = 50 m/s, 4m, , x, E–, , =5m, 8., , (a), , The retardation due to friction, , f, N, =, m, m, , a=, , ( mg, , 2 mg ), , 3g, , 1, 3, , Here | E | | E | . The symmetry of the situation shows that E =, 0 at M., , 3 10 10 m/s 2, , Therefore (A) is the correct option., The electric potential due to all the charges at M is zero., , Now, 0 = u – at,, , u, t=, a, , or, (b), , 50, 10, , Therefore (B) is incorrect option., , 5s ., , When the system of charges is rotated about line PM, the net, current will be zero., , There is no displacement in the direction of electric force and, so work done by it is equal to zero., , Passage for Q. 10 to Q. 12, 10., , (b), , u 2y, , y, , 2, , (C) is the correct option., =0, , (D) is incorrect option., , y, , 2a y y, , 2, or 0 = v2, , Therefore the magnetic field at M is zero., When magnetic field is zero, then, , Using third equation of motion, we have, , v 2y, , E–, E+, 120°, 120°, E+ 120° E+, E–, , ( mg Eq ), m, , m, , 9., , –, , The electric field at M due to the charges at the corners of regular, hexagon is as shown, , B = – 5k, 7., , 1, a yt 2, 2, , v2t, , 2, , /2, , /2, 2, , a, , 5., , 2 m, and 0, eB, , n, , /2, 1, , a, , and cos, , (a), , sin d, , 2, , 0 ni, , =, , 11., , 2, , 0 ni, , (q), , Ee, y, m, , mv22, 2eE ., , – + – + – +, M, Q, , v2, v1, Ee, , P, , x, The electric field due to the inner most positive and negative charges

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MAGNETIC EFFECT, at M is E1, , 2 k, , CURRENT, , OF, , 443, , (A) is the correct option., , q, , towards left. The electic field due to the, , r2, , next positive and negative charges at M is E2, , 2 k, , q, (2 r )2, , The potential at M due to the charges is, , V, , towards, , k, , q, a/2, , q, 4, a/2, , q, 5a, 4, , 0, , right. The electric field due to the outermost positive and negative, (B) is the correct option., charges at M is E3, , 2 k, , q, , When the whole system is set into rotation with a constant angular, velocity about the line PQ we get three loops in which current is, flowing., , towards left. Clearly the vector, , (3r )2, , sum of these three electric field is not zero., , The magnetic field due to these currents produce a resultant, magnetic field at M which is not equal to zero. Therefore a net, magnetic dipole moment will be produced., , (A) is incorrect option., The electric potential due to the charges at M, , k, , q q, r r, , q, 2r, , q, 2r, , q, 3r, , q, 3r, , (C) is an incorrect option., , 0, , (D) is correct option., , (B) is incorrect option., , P, +, , The net current due to the innermost positive and negative charges, is zero. Similarly the net current due to other charges in pairs is, zero. Therefore the magnetic field at M is zero. Also the magnetic, moment is zero., , (t), , +, , M–, , –, , Q, , (C) is the correct option, , There will be a net electric field due to the arrangement of charges, at M towards the right side., , (D) is incorrect option., (r), , +, , –, , (A) is an incorrect option., , +, , Q, , –, –, , M, , –, , The electric potential at M will can out in pairs by positive and, negative charges, due to symmetrical arrangement of charges., , +, , (B) is an incorrect option., , P, , When the system of charges rotates about PQ, the net current is, zero due to symmetrical arrangement of charges. Therefore B = 0, and = 0, , +, The net electric field due to negative charges in the inner circle is, zero. Similarly the net electric field due to positive charges in the, outercircle is zero., The electric potential due to negative charges at M is different, from the electric potential due to positive charges at M. Therefore, the electric potential at M is not equal to zero., (B) is the correct option., When the system of charges rotate, we get a current I1 due to, negative charges and another current I due to positive charges. The, magnitude of the magnetic at M due to the currents is different., Therefore B 0 and, 0., (C) is incorrect option, , 2a, a, , M, , P, –, , +, , Statement -2 : The magnetic field at P due to current flowing in AB, is perpendicular to the plane of paper acting vertically downward., And the magnetic field at P due to current flowing in CD is, perpendicular to the plane of paper acting vertically upwards., Therefore, q is correct., As P is the mid point, the two magnetic fields, cancel out each, other. Therefore, r is correct., B:p, , C : q, r, , –, , +, , A : q, r, , Statement -2 : The magnetic field at P due to current in loop A is, along the axial line towards right. Similarly, the magnetic field at P, due to current in loop B is also along the axial line towards right., , (D) is the correct option., , –, , (D) is the incorrect option., 14., , (A) is the correct option., , (s), , (C) is the correct option., , Q, –, , The electric field at M due to all the charges is zero because the, electric field due to different charges cancel out in pairs., , Statement -2 : The magnetic field due to current in loop A at P is, equal and opposite to the magnetic field due to current in loop B at, P., D : q, s, Statement -2 : The direction of magnetic field at P due to current

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444, , ELECTRICITY & MAGNETISM, , in loop A is perpendicular to the plane of paper directed vertically, upwards., , Given; y = B sin 2 t and x, , The direction of magnetic field at P due to current in loop B is, perpendicular to the plane of paper directed vertically downward., Since the current are in opposite direction the wires repel each, other., 15., , x, A, , y= B, , A-p, q, r : In uniform magnetic field, the motion of particle is, , B, , 2, , x2, , or, , y=, , or, , A2, x2 = B y ., , A2, , A sin t, , uniform and circle of helical depend on the angle between v and, B., , B- q, r, s, t :, Electric field exerts the force on the charged particle so its motion, no remains uniform. Depending on the nature of electric field, the, path of the particle may be parabolic, etc., , It represents a parabola., 16., , Using right hand screw rule to get the direction of resultant field., , C-p, r :, ( k sin t ) 2, , v, , ( k cos t ) 2, , k2 =, , A - r,t ; B - r,t ; C - r,s ; D - r,t, For A, B and C– Magnetic field at location of 1 due to 2 is parallel, or antiparallel to current in 1, so force experienced by 1 due to, magnetic field of 2 is zero. For D–Force experienced by upper, half of 2 due to 1 is along left, while on lower half it is towards, right. So, net force of interaction between the two is zero., , 2 k (constant), , D- s :, , Direction of magnetic field at P can be found by using RHPR No. 1., , Solutions Exercise-6.5, 1., and, , B1, , =, , B2, , =, , Given, , 2., , sin ], , =, , d2, , B, 1, , =, , 0.93 , cos, , µ0 ni, (cos, 2, 2, , 4, 2B, , 2B, , a, , B, , r, , 5., , 20, = 4 (2 10 ), 0.20, , v, , =, , cos, , 1, , 10, cm, , 1, , (a), , 80µT up the page. Ans., , µ0 I1, ., 2, 5, , I2, ( ˆj ), 2, , 4 10 5 iˆ m/s, , Force, F, =, ( e)[v B], After substituting the values and simplifying, we get, , (b), , 10, , 7, , =, , 200, 0.25, 2, , 0.3 (2 0.93), , = 0.3 mA, The inner radius is, , Ans., , ri, , =, , 0.15m, so, , B, , =, , µ0 Ni (4, 2 ri =, , =, , 5.33 10 4 T, , =, , µ0 Ni, 2 r0, , The magnetic field at point R,, =, , 1), , 2), , 25 cm, , 7, , B, , cos, , 1, , = cos(180, , 2, , µ0 i, ., 2 r, , µ0 i, =4, 2 a, , (i), , 0, , 0, , a, , µ0, i, =2 2 2 ., a/ 2, , 3., , =, , B3, , 0.01, , = 2 2B, =2 2, , B1 B2, , 4 A and I3 2.5 A, x = 1 m., , After substituting and simplifying, we get, = 0.07, Ans., d/, The magnetic field due to all the wire is shown in figure. So, Btotal, , I3, x, , I2, 2, , =, , Ans., The magnetic field at the centre of the solenoid is given by, , Here cos, , B1 B2, B1, , µ0 I1, 2, 5, , Here, I1 2.5 A , I 2, 4., , /2, 2, , 4A ., , B, , or, , 2sin, , /2, , =, , (ii), , P, , µ0 i, ., 4 d, , =, , sin, , i, , µ0 i, . [sin, 4 d, , =, Here, , I2, , µ0 i, ., 2 d, , B, , =, , (4, , 10 7 )(500)(0.80), 2 0.25, , 10 7 )(500)(0.80), 2 (0.15 0.05), , 4.00 × 10–4 T, , Ans.

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MAGNETIC EFFECT, , 445, , CURRENT, , OF, , Solutions Exercise-6.6, 1., , 2., , Previously the magnetic field is due to the current causes due to, movement of the free electrons. To an observer in a trolley moving, in the direction of current the current is due to electrons becomes, zero but it is now due to movement of positive ions. So magnetic, field, B, , =, , µ0 i, . ., 2 r, , The current,, , i, , =, , q, t, , The magnetic field,, , B, , =, , µ0i, 2r, , Magnetic moment,, , M, , =, , 14, , =, , iA, , =, 3., , a/2, , µ0i, , =, , i, , =, , µ0i, , a, , = 2 R. 2, a 4R 2, , 4x, , ef ( r 2 ), , R, , 24 A-m 2, , x, , a/2, i, , BH, , Btotal, (39 10 6 ) 0.08, 2 10, , =, , 4B cos, , =, , 4 Ba, 2R, , 7, , a2, , (4 x, , 2, , 2, , a )(4 x 2, , 2a 2 )1/ 2, , Ans., , The magnetic field due to a straight wire is given by, =, , B, , where, r = R, sin, µ0 i, . (sin 45, 4 r, , 4 x2, , 4 µ0ia 2, , =, , 45°, 45°, , 4, , 4 Ba, , Ans., , i, , =, , ., , Ans., , 6., , B, , 2a 2, , B, , 16 A from west to eas, The magnetic field at the centre of the loop,, , r, , a 2 4 x2, , P, , Wb/m2, , BH r, ( µ0 / 2 ), , 2, , Now magnetic field due to the all the four sides of the square can, be calculated as :, , ef ., µ0 (ef ), 2r, , 9 10, , µ0 i, . sin dx, 4 r2, , =, , B, , The magnetic field of earth at that point becomes zero, when wire, produces equal and opposite field. So, µ0 i, ., 2 r, , 4., , e, T, , a/ 2, , L, 1, , L2, , R2, , B, , =, , sin 45 ), , µ0 I, . (sin 1 sin 2 ), 4 r, , sin 2, , 0, , µ0, L, ., i, 2, 4 R, L R2, , Ans., , Here, r a / 2 ,, After simplifying, we get B =, 5., , P, , 2 2µ0i, a, , The magnetic field due to one side of the square can be calculated, as:, =, , dB, , µ0 i, . sin dx, 4 r2, , 7., , (i), , 0.6 m, , The equivalent current i, , =, , q, T, , q, 2 /, , q, 2, , The rotating rod with charge at its end behaves like a circular, loop. So, R, x, dx, , r, , B, B, , =, , µ0, iR 2, 2, 2 (R, x 2 )3 / 2, , Here, R =0.6 m and x = 0.8 m.

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446, , ELECTRICITY & MAGNETISM, After substituting the values and simplifying we get,, B, , (ii), , 8., , =, , 11.3 ×, , Magnetic field is given by, , T, , 22.6 × 10–4 T, , B, , =, , B, , =, , BL, , =, , 0, , Ans., , r, , The magnetic field, BP, , BQ, , µ0 i, ., 4 r, , µ0 i, ., 2 r, , µ0 i, ., 4 r, , 2 10, , 7, , i, , 0, , B, 10, 0.02, , Magnetic field due to segment of a circular loop is given by, B, , =, , Bnet, , =, , µ0 i, 4 a, , 12., , µ0 i 1, 4 a a, , 1, b, , B, , µ0 i, . (sin, 4 r, , =, , 2, , =, , 2, , =, , µ0i [1 cos / 2], µ0i, cot / 4, =, 2 x sin / 2, 2 x, , =, , Bcurved, , out of page, , µ0, 4, , µ0, , =, , Using Ampere's law, we have, B.d, , =, , Here, , µ0iin, , r, , =, , sin, , 1, , 2), , i, [sin(90, x sin / 2, , / 2) sin 90 ], , Ans., , Bstraight, , 3, I, 4, 2R, , µ0 I, . (sin 45, 4 r, , sin 45 ), , R cos 45, , After substituting the values and solving, we get, B, , dr, , 13., r, , 2, , /2 x, /2, , BM, , 10–4 Wb/m2, , =, , 10., , 11., , In this case i q / 2 , R 0.3m , x 0.4 m, , =, , 9., , 10–4, , =, , 5, , 2.8 10, , T, , Ans., , Using Ampere's law, we have, , a, , B.d, , =, , µ0iin, , =, , µ0 JdA, , b, r, , or, , B 2 r, , 0, r, , Here, , iin, , r, , idA, , =, , µ0, , =, , 0, , r, a, , J0, 0, , 2 rdr, , r2, , cr 2 2 rdr, , =, , B, , r1, , 14., , B 2 r1, , =, , 2 c, , =, , µ0, , ( r24, , r14 ), 4, , c ( r24 r14 ), 2, , (a), , µ0 J 0 r 2, ., 3a, , =, , The area of shaded portion, , A, , (a 2 b 2 ), , =, , The current in whole cross-section of the conductor, , c (r24 r14 ), 2, , I, , or, , B, , =, , =, =, , µ0c 4, ( r2, 4 r1, (4, , r14 ), , =, , i a2, (a, , 2, , ia 2, 2, , b ), , (a, , 2, , b2 ), , The magnetic field due to cylindrical conductor at a distance, r, inside, , 10 7 )(3 106 )[0.034, 4 0.03, , 2.0 × 10–5 T, , Ans., , 0.024 ], , Ans., , B, , =, , µ0 Ir, ., 2 a2

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MAGNETIC EFFECT, , µ0id, , =, , (b), 15., , 2 ( a 2 b2 ), , Q4 ., , 17., , ., , The magnetic field at a distance of 2m from the wire, B, , µ0id, , For the case b = 0; B, , Similarly at Q3 1.1 10 4 T into the page. It will be zero at Q2 and, , 2, 2, 2, µ0 ia /( a b ) d, ., 2, 2, a, , =, , , and for d = 0, B = 0, , 2 a2, , (a), , µ0 i, ., 2 r, , 10, 2, , 7, , 2 10, , 10 6 T, , The net magnetic field at x = 0, z = 2m, B, , = BAB BDC, (b), , B, , C i, , D, , =, , ., , The magnetic field of side AD and BC are equal and opposite so, they cancel each other., Bnet, , 447, , CURRENT, , OF, , (c), , 3a/4, , 18., , i2, , a/4, , =, , B1iˆ B2iˆ, , =, , 10 6 iˆ 10 6 iˆ 0, , =, , 10 6 iˆ 10 6 kˆ, , =, , 2 10 6 T, , Ans., , Do this part similarly., , The field of the solenoid at the point is parallel to the solenoid axis, and the field of the wire is perpendicular to the solenoid axis. The, total field makes and angle of 45° with the axis if these two fields, have equal magnitudes. Therefore, , i, A, , B, , i/2, , µ0 i / 2, ., (sin, 4, a/4, , =, , a/2, , Here, sin, a/2, , 2, , (a / 4) 2, , sin ), , i/2, (sin, 3a / 4, , a/2, , 2, , 3a / 4, , =, , µ0 iwire, ., 2, r, , r, , =, , iwrie, 2 nisol, , sin ), , a/2, , , sin, , µ0 nisolenoid, , The resultant field, Bresultant =, , 2, , B, , 16., , (a), (b), , 1, , 1, , 5, , 3 13, , , out of page Ans., , The magnetic fields of parallel wires cancel each other, so net, magnetic field at P will be zero., , 19., , We know that,, or, , 2µnisolenoid, , =, , 3.55 × 10–5 T, , V, , =, , V0e, , t/, , 0.90V0, , =, , V0e, , 2/, , =, , 19.23 s, , =, , CR, , The magnetic field at, µ0, , Q1 = 2, Q1 a/2, i, a/2, , a, , i, i, 2, 2, a/2, 3a / 2, , As, , Q2, , i, , 2B, , =, , After substituting the values, and simplifying, we get, 2 µ0i, =, a, , 4.77 10 2 m, , R, , =, , i0, , =, , B, , =, , 19.23, C, , 100 10, , 19.23 10 4, , 6, , i, P, , 20, , 1.04 10, , 19.23 104, , R, , 4, , i, , Magnetic field, Q4, , Q3, , µ0 ni, , µ0 n(ie, , t/i, , ), , 2, , Bdt, , 8µ0i, =, 3 a, , = 1.1 × 10–4 T, out of page, , Bav, , =, , 0, , 2, , 16, , 10 8 T, , Ans., , A

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450, , ELECTRICITY & MAGNETISM, , 7.1 PERMANENT, , MAGNET, , : AN, , INTRODUCTION, , In the previous chapter, we have seen that a current carrying loop, produces magnetic, field B =, 4, , 0, , 2M, r3, , at an axial point. Here M iA is the magnetic moment of the loop. Vector A represents, the area vector of the current loop. Current loop in magnetic field experiences a, torque, , M, , B . We also know that the electric field of the dipole is E, , 1, 4, , 2P, 0, , r3, , on, , Fig. 7.1, , the axis of the dipole and it experiences a torque, P E., The above similarity of current loop with the electric dipole makes the following, hypothetical model :, (1) There are two types of charges; positive magnetic charge (+m) and negative, , Fig. 7.2, , magnetic charge (–m). A magnetic charge m placed in a magnetic field B experiences, a force, , F, , (2), , (3), (4), , ...(1), , =, , 0 m, 4 r2, , ...(2), , at a distance r from it., A magnetic dipole is formed when two equal and opposite charges are placed at, some separation., Magnetic dipole moment,, M = m., A current loop of area A carrying current i may be replaced by a magnetic dipole of, moment M = m, , Fig. 7.4, , mB, , The force on a positive magnetic charge is along the field B , and the force on, negative charge is opposite to the field., A magnetic charge m produces a magnetic field, B, , Fig. 7.3, , =, , iA., , Bar magnet : Bar magnet is the permanent magnetic dipole., (i), , Magnetic length, , 0.84 geometric length, (ii) Magnetic charge or pole strength = m, SI unit of m is A-m or N/T., (iii) Magnetic moment M = Pole strength magnetic length, or, SI unit of M = A-m2 or J/T., , M, , =, , m, , Field of bar magnet is strong at its ends and negligibly small, at its middle., If we break a bar magnet, it seems, to isolate in a single pole or monopole. However, we, cannot get an isolated pole even if we break the magnet down to its individual atoms and, then to its electrons and nuclei. Each fragment has a north pole and south pole. Thus the, simplest magnetic structure that can exist is a magnetic dipole. Magnetic monopoles do, not exist., , 7.2 COULOMB'S, , LAW IN MAGNETISM, , According to Coulomb, the magnetic force between two magnetic charges m1 and m2, placed at a separation r is given by, F, , =, , 0, , 4, , m1m2, r2, , .

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, Magnetic field due to a magnetic pole: If m0 is the magnetic charge, , and F is the force exerted by another magnetic charge m, then magnetic field at a distance, r from m can be defined as, , B, , =, , F, m0, , Fig. 7.6, , mm0, , 0, , r2, , 4, , =, , 0, , m0, , 4, , m, r2, , ., , Magnetic potential, For a pole strength m, the field at a distance r is, B, , =, , m, , 0, , ,, , r2, , 4, , and radially away from the pole. The potential at a distance r is given by,, r, , V, , B dr, , =, r, , =, , =, Also, , B, , m, , 0, , 0, , 4, , dr, , r2, , 4, m, r, , dV, dr, , =, , Magnetic potential due to a dipole, Consider a magnetic dipole of moment M. If m is the pole strength and is the distance, between the poles, then M = m . If r1 and r2 are the distances of point P from the poles,, then, , and, , r1, , =, , r, , r2, , =, , r, , V, , = VN + VS, , 2, 2, , cos, cos, , Magnetic potential at P,, , =, , =, , =, , 0, , 4, , m, r1, , 0m, , 4, , m, r2, 1, , r, , 0m, , 4, , Fig. 7.7, , 2, , 1, , cos, cos, , r2, , 2, , 4, , cos 2, , r, , 2, , cos, , 451

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452, , ELECTRICITY & MAGNETISM, Putting m = M, and neglecting 2 in comparison to r, we get, V, , =, , M cos, , 0, , ., , r2, , 4, , Magnetic field of a dipole, The magnetic field at any point P at an angular position at a distance r from the centre, of the dipole can be obtained by resolving it in to two components. One component Br, which is along the radius vector and the other B which is perpendicular to radius vector., The radial component of magnetic field, Br, , =, , dV, dr, , =, , d, 0 M cos, dr 4, r2, , (a), , =, and, , B, , (b), , 2 M cos, , 0, , r3, , 4, , = –, , dV, dr, , =, , 1 dV, r d, , =, , 1 d, r d, , Fig. 7.8, , =, , ...(1), , dV, rd, , 0, , M cos, r2, , 4, , M sin, , 0, , ...(2), , r3, , 4, , The resultant magnetic field, B, , Br 2, , =, , B, , 2, , On substituting the values of Br and B , we get, B, If it makes angle, , =, , M 3cos 2, , 0, , 4, , r, , tan, , =, , B, Br, , tan, , =, , tan, ., 2, , Fig. 7.10, , ...(3), , ...(4), , Special cases :, Case 1:, , 1, , with r , then, , or, , Fig. 7.9, , 3, , On the axis of dipole; = 0,, , and, Case 2 : On the equator of dipole,, , V, , =, , B, , =, , M, , 0, , r2, , 4, 0, , 4, , = 0, 90 ,, V = 0, , ., , ,, , 2M, r3, , ,

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, B, and, , =, , 0, , M, , r3, = 180o., 4, , Field of a bar magnet, Case 1 : On the axis of the magnet or end on position : For a magnet of length ,, magnetic field on the axis of the magnet at a distance r from its centre., B, , =, , 2Mr, , 0, , 4, r, , 0, , ,, , 4, , 2M, , ., 4 r3, Case 2 : On the equator of the magnet or broad side on position :, for, , r, B, , 2 2, , 2, , B, , =, , M, , 0, , 3, 2 2, , 4, r2, , for, , r, B, , 0, , 4, , ,, , 4, , M, r3, , Torque on a magnet in magnetic field, When a magnet is placed in an uniform magnetic field, its poles experience equal and, opposite forces and so net force Fnet on the magnet is zero. But the magnet experiences, a torque given by,, = mB sin, = MB sin, In vector form, we can write, = M B, Work done in rotating the magnet : The torque exerted by the agent for any angle is, MB sin . Thus work done in rotating the magnet from 1 to 2,, 2, , W, , 2, , d, , =, , M B sin d, , 1, , or, , W, , =, , Fig. 7.11, , 1, , MB cos, , 1, , cos, , 2, , ., , Potential energy in magnetic field, If we take potential energy at 1, , or, , 90 to be zero, then potential energy at 2 =, , MB cos90, , U, , =, , U, , = – MBcos, , cos, , Magnetic force between two short magnetic dipoles, (i), , (ii), , Consider two short dipoles M 1 and M 2 are placed on the same axis at a separation, r. The force of interaction between them., F, , =, , F, , =, , 0, , 6 M1 M 2, , Fig. 7.12, , r4, When dipoles are placed on mutually perpendicular axes, the force of interaction, between them, 4, , 0, , 4, , 3M1M 2, r4, , Fig. 7.13, , 453

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454, , ELECTRICITY & MAGNETISM, , Ex. 1, , (a) Indicate the positions and the nature of equilibrium, of a number of magnetic needles arranged in a straight line at, equal distances from one another., (b) A strong horseshoe magnet is closed by an iron plate A fig., (a). The weight of the plate corresponds to the lifting force of, the magnet, and the magnet can easily hold the plate. If the, poles of the magnet are now touched on the sides with a plate, B made of soft iron, the plate A will drop at once. If the plate, B is removed the magnet will again be capable of holding the, plate A. Explain this phenomenon., , (e), , Two identical horseshoe magnets are linked by their opposite, poles as shown in fig.7.14(e) One of the magnets has round it, a coil A whose ends are connected to a galvanometer G. If the, magnets are detached, the pointer of the galvanometer will, be deflected at this instant through a certain angle. If the, magnets are connected again the pointer of the galvanometer, will also be deflected, but this time in the opposite direction., Indicate the causes for the deflection of the pointer of the, galvanometer., , (f), , Permalloy can be magnetized appreciably in the terrestrial, magnetic field and does not possess residual magnetism, i.e.,, it is the softest material as far as magnetism is concerned., How will a magnetic needle on a vertical axis near a long bar, made of this alloy behave if :, (i), , the bar is vertical fig.7.15 (f);, , (ii) the bar is horizontally placed along the magnetic, meridian;, , Fig. 7.14, (c), , (d), , A long rod made of soft iron is secured in a vertical position., If a strong magnet A is brought to the top of the rod as shown, in fig.7.14c, (i) The rod will be magnetized so intensely as to retain at, its other end several small pieces of iron. If the same, magnet A is applied to the side of the rod near the bottom, end fig.7.14c, (ii) the magnetization will be weak and the pieces of iron, will fall. Explain why the magnet A acts differently in, these two cases., A strong magnet of magneto alloy can hold a chain consisting, of several cylinders made of soft iron fig.7.14(d). What will, happen to the cylinders if a similar magnet is brought up, from below to this chain ? The magnets are arranged with, their like poles facing. What will happen to the cylinders if, the magnets have their opposite poles facing ?, , Fig. 7.15, (iii) the bar is in a horizontal plane perpendicular to the, magnetic meridian., Will the behaviour of the needle change in these three, cases when the bar is turned ?, (g), , A small thin iron nail is suspended from a light fire-proof, thread. A strong electromagnet is placed near the nail, fig.7.15(g). The flame from a powerful gas burner is directed, precisely between the nail and the magnet and licks the nail, when it is deflected by the magnet. If the windings of the, electromagnet are energized the nail will be at once deflected, into the flame and will then be ejected from it to assume its, original position. After a lapse of time the nail will again be, drawn to the magnet., Explain what causes these periodic oscillations of the nail., , Sol., (a), , The arrangement shown in Fig.(i) corresponds to the position of, unstable equilibrium and that in Fig.(ii) corresponds to stable, equilibrium.

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , 455, , In the second case, when the pole of the opposite sign is drawn, closer, the "strength" of the chain will increase as the lower magnet, is brought near. As soon as the second magnet comes in contact, with the lower cylinder, the magnet will be attracted to the chain, and will remain hanging on it., (e), , The magnets are detached owing to a sharp decrease in the number, of the lines of force passing inside each magnet. At the moment of, detachment an e.m.f. is induced in the coil due to the reduction in, the number of the lines of force., , (f), , (i), , The bar will be magnetized due to the action of the vertical, component of the terrestrial magnetic field and the magnetic, needle will be attracted as it is brought near the ends of the, bar., , (ii), , The bar will be magnetized by the action of the horizontal, component of the terrestrial magnetic field. The magnetic, needle will always turn towards the nearest end of the bar., , Fig. 7.16, (b), , (c), , (d), , Upon contact with plate B, part of the magnetic lines of force are, short-circuited through this plate (Fig. (b)). The number of the, lines of force penetrating through plate A sharply decreases. As a, result the interaction between the magnet and plate A also decreases, and the latter drops., In the second case most of the magnetic lines of force are shortcircuited inside the part of the rod adjoining the magnet (Fig. 7.17., (c)) and the rod cannot, therefore, be magnetized as intensely as in, the first case., In the first case, as the lower magnet is brought nearer, the cylinders, will be detached one after another from the chain and attracted to, the lower magnet., , (iii) The bar will not be magnetized by the magnetic field of the, Earth and the needle will not change its position when the, bar is brought close., It is assumed in all the three cases that when the needle is, brought close to the bar the distance between them is still, large enough for the additional magnetization of the bar due, to the external magnetic field of the needle to be neglected., If the bar is turned, the behaviour of the needle will not, change in any of the three cases., (g), , (b), , (c), , Iron loses its magnetic properties at quite a high temperature and, behaves as any other non-magnetic substance (copper, glass, etc.)., When the nail is heated in the flame of a burner to this temperature, the interaction of the magnet and the nail abruptly decreases, the, nail returns to the initial position, leaves the flame and gets cool., The magnetic properties of the nail thus cooled are regained, the, forces of interaction between the nail and the magnet increase and, the nail is again drawn to the magnet., , Fig. 7.17, , 7.3 EARTH'S, , MAGNETISM, , It is a well known fact that freely suspended magnet or current, carrying solenoid rests in specific direction, called magnetic, meridian. It shows that earth has its own magnetic field. The, modern theory about earth magnetic field is that, the earth, rotates about an axis and has the surrounding ionised region, due to interaction of cosmic rays. Due to rotation of earth the, surrounding ionised region gives rise to strong electric current, which causes magnetic field. Its value on earth surface is 1, gauss., , Fig. 7.18, , Elements of earth's magnetic field, To know about earth's magnetic field, we need three, informations. They are :, (1), Magnetic declination ( ): Angle between geographical and, magnetic meridian is known as angle of declination. It has, an average value 17.5o., Fig. 7.19

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456, , ELECTRICITY & MAGNETISM, (2), , Angle of inclination or dip ( ) : It is the angle between the magnetic field of earth, and the horizontal at that place. It is zero at magnetic equator and 90o at poles., In the magnetic northern hemisphere, the vertical component of earth's field points, downward., , (3), , Horizontal component of earth's magnetic field (BH) : At any place other than, magnetic poles, there is horizontal component of field., and vertical component, , BH, , = Bcos, , ...(i), , BV, , = Bsin, , ...(ii), , By squaring and adding equations (i) and (ii), we get, B, , =, , 2, BH, , BV2, , ....(1), , Dividing equation (ii) by (i), we get, tan, , =, , BV, BH, , ....(2), , True dip and apparent dip, The angle of dip in magnetic meridian plane is called true dip ( ), and angle of dip in, different plane from magnetic meridian plane is called apparent dip., We know that true dip, tan, , =, , BV, BH, , ...(i), , On any other plane at an angle from the meridian plane, the horizontal component of, earth magnetic field will be BH cos while vertical component remain as such. Thus, apparent dip,, Fig. 7.20, , tan, , =, , BV, BH cos, , ...(ii), , =, , tan, cos, , ....(iii), , From equations (i) and (ii), we get, tan, , Compass needle and dip needle, Compass needle gives the direction of BH and the dip needle gives direction of B ., Magnetic map :It is found that many places have the same value of magnetic elements., The lines drawn by joining all places on the earth having same value of magnetic element, form magnetic map., Isogonic line :This is the line joining the places of equal angles of declination., Agonic line :This is the line which passes through places having zero declination., Magnetic meridian itself is a agonic line., Isoclinic line :This is the line joining the points of equal dip., Aclinic line :This is the line joining the places of zero dip. Magnetic equator is an aclinic, line., Fig. 7.21, , Isodynamic line:This is the line joining the places of equal value of horizontal components, of earth magnetic field.

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457, , PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , Ex. 2, , If 1 and 2 are the angles of dip observed in two vertical, planes right angles to each other and be the true angle of dip,, show that cot2 = cot2 1 + cot2 2., , m, , 0, , BN =, , 2, , 4, , along NP, , r2, , Sol., , We have,, , tan, , =, , BV, BH, , ...(i), , The value of vertical component remain same but horizontal component, in two perpendicular planes are BH cos, , and BH sin, , as shown in, , fig. 7.22., , Fig. 7.23, Its component along north is BN cos On substituting the value of , it, is, , m, , 0, , =, , 2, , 4, , r, , 2, , r2, , mr, , 0, , =, , r, 2, , 4, , 2, , 3, 2, , r2, , For neutral point at P, we have, BH =, =, Fig. 7.22, tan 1 =, , or, , cos, , and, , =, , tan 2 =, , or, , sin, , =, , tan, cos, , BV, BH cos, tan, tan, , =, , BN cos, , BS, , m, , 0, , 4, , 10, , r, , 7, , 2, , =, tan, sin, , BV, BH sin, tan, tan, , 3, 2, , 0.06, , 0.062, 6, , 0.082, , T., , 0.062, , 3, 2, , Ans., , Ex. 4, , Fig. 7.24 shows some of the equipotential surfaces of, the magnetic scalar potential. Find the magnetic field B at a point, in the region., , Sol., , ...(iii), , 2, , 22 10, , r2, , 1, , 1, , ...(ii), 1, , mr, , 2, , Squaring equations (ii) and (iii), and adding, we have, cos2, , or, , sin 2, , =, , 1, tan, , 2, , =, , tan 2, , 1, , 1, , tan 2, , tan 2, , 1, tan, , 1, , 2, , 1, , 2, , 1, , tan 2, , = cot2 1 + cot2 2., Proved, Ex. 3 A bar magnet of length 8 cm and having a pole strength, of 1.0 A-m is placed vertically on a horizontal table with the south, pole on the table. A neutral point is found on the table at a distance, 6.0 cm north of the magnet. Calculate the earth's horizontal, magnetic field., or, , cot2, , Fig. 7.24, , 2, , We know that, , =, =, , Sol., , The magnetic field at P due to south pole, BS =, and due to north pole, , 0, , 4, , m, r2, , B =, , =, , Ex. 5, , V2 V1, , V, r, , r, , V1 V2, r, , 0.2 0.1, , 10, , 4, , 0.1sin 30, 2.0 10, , 4, , T.., , Ans., , A magnetic dipole of magnetic moment 0.72 2 A-m2, is placed horizontally with the north pole pointing towards east., Find the position of the neutral point if the horizontal component, of the earth's magnetic field is 18 T.

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458, , Sol., , ELECTRICITY & MAGNETISM, Given M, , From the figure, , 0.72 2 A-m, , Fig. 7.28, and the moment produced by the load suspended from the needle and is, , Fig. 7.25, = 90o, = (90° – ), , +, , equal to P, , 2, , . Thus for the equilibrium of the needle, , mBv, , Hence,, Fig. 7.26, We have,, , or, , tan, , =, , tan(90° – ) =, , and total field, , tan, 2, , =, , or, , tan2, , =, , 2, , =, , tan, , 2, , P, 2m, , Ans., , B =, , BH 2, , BV 2 ., , Ans., , Ex. 7, , Two long equally magnetized needles are freely, suspended by their like poles from a hook as shown in fig. 7.29. The, length of each needle is cm and the weight is W. In equilibrium, the needles make an angle, with each other. Determine the, magnetic pole strength of the needles. The magnetic pole strength, is concentrated at the ends of the needles., , tan, 2, , cot, , Bv =, , P, , The horizontal component of the terrestrial magnetic field, BH = Bv tan ,, , tan, 2, , or, , =, , 1, , 2, , Ans., , Let neutral point is at a distance r, then, BH = B, =, , 0, , or, , 18 10, , 3cos 2, , 1, , r3, , 4, , 10, 6, , M, , 7, , =, , 0.72 2 3, , 1, 3, , 2, , 1, , r3, Fig. 7.29, , On solving, , r =, , 0.20 m., , Ans., , Ex. 6, , In order to keep the needle (see fig. 7.28) in a horizontal, position a load P is suspended from its top end. Find the magnitudes, of the horizontal and vertical components of the intensity of the, terrestrial magnetic field. Calculate the total intensity of the, terrestrial magnetic field. The dip angle is . The pole strength is, m., , Sol., The magnetic needle will be acted by two forces tending to turn it, their, moments are : the moment of the force due to the terrestrial magnetic, field acting on the poles of the needle, = (mBv), , Sol., Let us consider one of the needles. It will be acted by the weight W, which, acts through the centre of gravity B and the repulsive force between the, poles F, , m2, , acted through the point A. For the needle to be in, a2, equilibrium the sum of the moments of the forces acting on the needle, should be equal to zero, i.e.,, , P sin, 2, 2, Here, , =, , F =, , F cos, , m2, a2, , 2, , , where a, , 2 sin, , 2, , .

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , P, sin, 2, 2, , Thus, , m 2 cos, , =, , m =, , Ex. 8, , 4, , 2, , sin, , sin, , 2, , 459, , 2, , 2, , 2, , 2 P tan, , 2, , ., , Ans., , A magnetic needle with length and poles of pole strength, , + m is attached to a wooden bar of length L (see fig. 7.30) and placed, in a uniform magnetic field of intensity B. The bar and needle can, revolve around point O. Find the magnitude of the moment of force, that will cause the rotation of the bar around point O if the bar, makes an angle with the direction of the lines of force of the, magnetic field., , Fig. 7.30, Hence, the net moment of the force acting on the bar is, =, , mB, , =, , mB sin, , L sin, , mBL sin, , m B sin, , = MBsin ., , Ans., , Sol., , Note: The earth's magnetic field is not due to a large bar, , The moment of forces about O, acting upon each pole of the needle will, be equal to mB ( + L) sin and –mBL sin ., , magnet inside it because the earth's core is hot and molten. It may, be due to the convection currents around the equator. As to what, dynamo effect sustains this current, and why the earth's field, reverses polarity every million years or so, we do not know., , 7.4 MOVING, , COIL GALVANOMETER, , It is used to detect the direction of direct current. It is based on the interaction between, current carrying loop with the magnetic field. In a moving coil galvanometer, the coil is, suspended between the pole pieces of a strong magnet. The poles are made cylindrical, and a soft iron cylindrical core is placed within the coil. This makes the field radial. In, such a field the plane of the loop always remains parallel to the field, and = 90o. The, deflecting torque,, o, def = NiBA sin90, = NiBA, ...(i), This torque deflects the coil. As soon as coil starts rotating a restoring torque is set up, in the suspension wire. If is the angle of twist in the suspension wire and C is the, torsional rigidity, then restoring torque, ...(ii), rest = C, From equations (i) and (ii), we have, NiBA = C, , or, where k, , i, , =, , i, , =, , C, NBA, k ,, , ....(1), , C, is known as galvanometer constant., NBA, , Fig. 7.31, , Current sensitivity (s) : Current sensitivity is defined as, , Therefore, , S, , =, , d, or simply, di, i, , S, , =, , NBA, C, , .....(2)

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460, , ELECTRICITY & MAGNETISM, , 7.5 TANGENT, , LAW, , When a compass needle is placed in two perpendicular fields, B and BH , the needle, rests at an angle, , with BH . In equilibrium, , B, BH, , Fig. 7.32, , = tan, , B, , = B tan, , This is known as the tangent law of perpendicular fields., , 7.6 TANGENT, , GALVANOMETER, , Tangent galvanometer is an instrument to measure direct current. It is based on tangent, law., The current to be measured is passed through coil set in M.M. produces a magnetic field, 0 Ni, . In addition to, 2R, this field, there is earth magnetic field with horizontal component BH, so the compass, needle placed at the centre of the galvanometer, defects in the direction of resultant field., Thus, B = BH tan, , at the centre that is perpendicular to the plane of the coil, B =, , 0 Ni, , or, , Fig. 7.33, , where k =, , 2R, , =, , BH tan, 2 RBH, tan, 0N, , i, , =, , i, , = k tan, , 2 RBH, is called reduction factor.., 0N, , Sensitivity :Good sensitivity means that change in deflection is large for a given, fractional change in current. We have, or, or, , or, , i, di, di, i, , d, di, , = k tan, = k sec2 d, =, , k sec 2 d, k tan, d, , =, , sin cos, , =, , 2d, sin 2, , =, , sin 2, ., 2i, , 45, For maximum sensitivity sin 2 1, or, So the tangent galvanometer is most sensitive when the deflection is around 45 o., , 7.7 VIBRATION, , MAGNETOMETER, , Vibration magnetometer is used for comparison of magnetic fields at different places and, also to compare magnetic moments of the magnets. First of all set the magnetometer with, its magnet along the meridian.

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, Then the experimental bar magnet is deflected from its mean, position by external magnet and left free to oscillate in BH ., The restoring torque acts on the magnet for small angle, = –MBH sin, or, Compare with, , =, , MBH, [for small , sin, I, – 2 , we get, MBH, I, , =, , and, , T, , =, , ], , Fig. 7.34, , I, MBH, , 2, , ....(1), , where I is the moment of inertia of the magnet, which is given by, , W 2, a b2, 12, mass of the magnet., I, , W, 1., , =, , Comparison of horizontal components of magnetic field at two places, If T1 and T2 are the time periods of motion of experimental magnet at two different, places, then, T1, BH1, BH 2, , 2., , Fig. 7.35, , =, , =, , 2, , M, and T2, BH1, , T22, , ., , T12, , 2, , M, BH 2, , ...(2), , Comparison of magnetic moments of the magnets, Suppose M1 and M2 are the magnetic moments of the magnets. First place the, magnets on the pan with like poles together, let time period is T1 and then unlike, poles together, let time period is T2, then, , and, , T1, , =, , 2, , T2, , =, , 2, , I1 I 2, M1 M 2 B H, I1, , I2, , M1 M 2 B H, , From above equations, we get, , 7.8 DEFLECTION, , M1, M2, , =, , T22 T12, T22 T12, , ., , ....(3), , MAGNETOMETER, , M, for a permanent bar magnet. Its working is based on tangent, BH, law. It consists of a small compass needle, which is pivoted at the centre of the, , It is used to determine, , M, can be measured in two standard positions of the, BH, magnetometer. One is called Tan-A position and the other is called Tan-B position., magnetometer. The quantity, , Fig. 7.36, , 461

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462, , ELECTRICITY & MAGNETISM, 1., , Tan-A position :, In this position the magnetometer is set perpendicular to magnetic meridian, so, that, magnetic field due to magnet is in axial position and perpendicular to earth's, field. Hence by tangent law, BHtan, = B, or, , BH tan, , M, BH, , =, , =, , 2Mr, , 0, , 4, r, , 2, , r, , 2, , 2, , 2, , 4, 2, , 2, , 4, , 4, , tan, , 2r, , 0, , ......(1), , Fig. 7.37, , 2., , Tan B position :, In this position the magnetometer is set in south north direction, so that the, magnetic field due to magnet is at its equator. Hence by tangent law, BHtan, = B, or, , BH tan, , =, , M, , 0, , 4, r2, , M, BH, , Fig. 7.38, , 7.9 GAUSS'S, , =, , 4, , r2, 0, , 3, 2 2, , 4, , 2, , 4, , 3, 2, , tan, , .....(2), , LAW IN MAGNETISM, , In electrostatics, we have studied the Gauss's law for electric field, i.e.,, E dA, , =, , qinside, 0, , where, , E dA is the electric flux and qinside is the net charge enclosed by the closed, , surface. On the similar way Gauss's law for magnetic field can be written as ;, B dA, where, , =, , µ0 minside, , B dA is the magnetic flux and minside is the net 'magnetic charge' inside the, , closed surface. In any closed surface minside = m – m = 0, as isolated pole does not exist, and so Gauss' law for magnetism, therefore states that, Fig. 7.39, , B dA, , = 0.

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , Ex. 9, , Two identical thin bar magnets, each of length and, pole strength m are placed at right angles to each other with the, north pole of one touching the south pole of the other. What is the, resultant magnetic moment of the system ?, , Sol. Previously,, f =, , 1, 2, , MBH, I, , =, , 1, 2, , M BH, , Sol., If M is the magnetic moment of each bar magnet, then M = m ., , and then,, , f, , Fig. 7.40, The resultant magnetic moment, M =, , 463, , B, , I, , Fig. 7.42, , 2M, , 2m ., , f '2, , Ex. 10, , f, , A magnetic needle is free to rotate in a vertical plane, which makes an angle 60o with the magnetic meridian. If the needle, stays in a direction making an angle of, , tan -1, , 2, 3, , 60, 40, , or, , with the, , horizontal, what would be the dip at that place ?, , Sol., , or, , Given, apparent dip, , or, , 2, , =, , BH B, BH, , =, , BH B, BH, , 2, , B, BH, , = 1.25, , B = 1.25 BH, 30 10 6 T, , = 1.25 24 T, , The oscillating magnet is in end- on position of the short magnet., Thus,, B, , or, , 30 10, , 6, , M, , =, =, , 0, , r3, , 30 10, , tan, , tan, , =, , or, , tan, , = tan, =, , Ex. 11 The, , 2, 3, , = 30o., , 2, , 30 10, , 6, , 0.20, , 10, , 7, , 2, , = 1.2, , tan, cos, , We have, , or, , =, , = 60o, , and, , r3, , 4, , 2, 3, , 1, , 6, , 0, , Fig. 7.41, =, , 2M ', , 4, , 3, , A-m2., , Ans., , Ex. 12 A long vertical wire carries a steady current of 10 A, flowing upwards through it at a place where the horizontal, component of earth magnetic field is 0.3 gauss. What is the, resultant horizontal magnetic field at a point 5 cm from the wire, due magnetic north of the wire ?, , cos, , 1, 2, , 1, 3, Ans., , frequency of oscillation of the magnet in an, oscillation magnetometer in the earth magnetic field is 40, oscillations per minute. A short bar magnet is placed to the north of, the magnetometer, at a separation of 20 cm from the oscillating, magnet, with its north pole pointing towards north. The frequency, of oscillation is found to increase 60 oscillations per minute., Calculate the magnetic moment of the short bar magnet., Horizontal component of the earth's magnetic field is 24 T., , Sol., The magnetic field due to current carrying conductor, B =, =, and, , 0, , 2, , i, r, , 2 10, , 5 10, , 0.4 10 4 T, , BH = 0.3 gauss, , 7, , 10, 2, , 0.4 gauss

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464, , ELECTRICITY & MAGNETISM, For the second magnet, C 270, , 30, , =, , ....(ii), , M 2 BH sin 30, , Dividing equation (i) by (ii), we get, M1, M2, , =, , 150, 240, , 5, 8, , Ans., , Ex. 14, , A thin magnet is cut into the equal parts by cutting it, perpendicular to its length. What is the new magnetic moment of, each part ? What is the time period of each part as compared to that, of original magnet if vibrated in the same magnetic field., , Fig. 7.43, Net field at P, , =, , B2, , =, , 0.42, , Sol., , BH2, , If is the initial length of the magnet, then magnetic moment M = m and, , = 0.5 gauss, , 2, , mass, 12, equal parts, the magnetic moment, , its moment of inertia I, , 0.32, , Ans., , Ex. 13, , A magnet is suspended in the magnetic meridian with, an untwisted wire. The upper end of the wire is rotated through, 180o to deflect the magnet by 30o from the magnetic meridian. Now, this magnet is replaced by another magnet and the upper end of the, wire has to be rotated through 270o to deflect the magnet by 30o, from magnetic meridian. Compare magnetic moments of the two, magnets., , Sol., , M =, , m, , . When the magnet is cut into two, , M, and, 2, , 2, , mass, 2, 2, 12, , I =, , 2, , I, ., 8, , If T and T are the initial and final time periods, then, , We know that,, =, where is the angle of twist and, the magnetic meridian., For first magnet, C 180, , 30, , =, , C, , MBH sin, is the deflection of the magnet from, , M1BH sin 30, , .....(i), , T =, , 2, , I, and T ', MB, , 2, , I, 8, M, H, 2, , On comparing above equations, we get, T' =, , 7.10 MAGNETISM, , T, ., 2, , AND ELECTRON, , Any material is magnetic because of the motions of electrons within it. There are two, motions of electrons: spin and orbital, each involving a magnetic dipole moment that, produces a magnetic field in the surrounding space. Explanation of these requires quantum, physics which is beyond the scope of the book, so here we shall discuss only that part, which can be understood by the previous knowledge of the subject., , Magnetic moment due to orbital motion of the electron, Suppose an electron moving at constant speed v in a circular path of radius r, counter, clockwise as shown in fig. 7.44. The conventional current i will be clockwise. The, magnitude of the orbital magnetic dipole moment of such a current loop, Morbital = i A,, where A is the area enclosed by the loop, which is r2, and, Fig. 7.44, , i, , =, , Charge, time, , =, , ev, ., 2 r, , e, 2 r, v

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, ev, evr, r2, 2 r, 2, The direction of this dipole moment is downward, by right-hand rule., The angular momentum, , Thus, , Morbital, , =, , L, , =, , As r and v are perpendicular, so, Lorbital, , 465, , ...(i), , m r v, , = mvr sin90o = mvr, , ...(ii), , Lorbital is directed upward., From equations (i) and (ii), we get, e, Lorbital, ...(1), 2m, This result obtained by classical analysis is in agreement with that given by quantum, physics., , M orbital, , =, , Magnetic moment due to spin of the electron, An electron has an angular momentum due to its spin motion, is called spin angular, momentum S . The magnetic dipole moment due to the spin motion of the electron is, given by, M spin, , =, , e, S., m, , ...(2), , 19, Here e is the elementary charge 1.60 10 C and m is the mass of an electron, , 9.11 10, , 31, , kg . The minus sign means that M spin and S are oppositely directed., , The resultant magnetic dipole moment is the vector sum of these dipole moments. Thus, , M, , =, =, , M orbital, , M spin, , e, Lorbital, 2m, , e, S, m, , e, Lorbital 2 S, ....(3), 2m, If the resultant magnetic dipole moment produces a magnetic field, then the material is, magnetic. The types of magnetic material will be discuss little later in this chapter., , or, , 7.11 MAGNETIC, , M, , =, , PROPERTIES OF THE MATERIALS, , Intensity of magnetisation, Each substance contains a large number of atoms. In general, the magnetic moments of, these atoms are randomly oriented and there is no net magnetic moment in any volume, of the material that contains thousands of atoms. However, when the material is kept in, an external magnetic field, atomic dipoles try to align parallel to the field. The degree of, alignment increases if the intensity of the applied field increases and also if the, temperature is decreased. With sufficient strong field, the alignment is nearly perfect., We then say that the material is magnetically saturated., When the atomic dipoles are aligned, partially or wholly, there is a net magnetic moment, in the direction of the field in any small volume of the material. Thus magnetisation is, defined as,, , I, , =, , Magnetic dipole moment, volume, , Fig. 7.45. Magnetic dipole moment due, to the spin motion of the electron

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466, , ELECTRICITY & MAGNETISM, or, , =, , I, , M, V, , ......(1), , Unit of magnetisation is A/m., Consider a bar magnet of pole strength m, length, intensity of magnetisation, Fig. 7.46, , and area of cross section A. The, , M m, m, ., V, A, A, Thus for a bar magnet, the intensity of magnetisation is defined as the pole strength per, unit face area., , I, , =, , Intensity of magnetising field, When any material is placed in magnetic field, it gets magnetised. The actual magnetic, field inside the material is the sum of the applied magnetic field and the magnetic field, due to magnetisation of the material. Thus magnetic field inside the material, , B = B0 Bm ,, where B0 is the magnetic field in the vacuum produced by magnetising field, and equal, to, , 0H, , ., , Here H is called intensity of magnetising field. Thus H can be defined as :, Fig. 7.47, H, , =, , B0, , 0 ni, , 0, , ni., , 0, , Bm is the magnetic field due to magnetisation of material, which is, Bm, , 0I ., , Magnetic susceptibility, Magnetic susceptibility is defined as the intensity of magnetisation per unit magnetising, field. Thus, , I, ,, ......(2), H, is a dimensionless quantity and may be positive or negative., =, , where, , Magnetic permeability, Magnetic permeability of the material is the measure of degree to which the material can, be permeated by magnetic field and is defined as the ratio of magnetic field in the material, to the magnetising field. Thus, =, , B, H, , =, , B0, , and, H, , or, , B, B, , =, =, , B0 Bm, 0H+ 0I, , or, , B, , =, , Also, , 0, , 0, , Here r is called relative permeability., , Relationship between, , r, , and, , We have,, , or, As, , 0H, , B, 0H, , = 1, , B/H, , =, , ,, , I, H, , 1, , I, H, , r.

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, = 1, , 0, , or, , r, , = 1+, , I, H, ., , ......(3), , Note:, The following points in regarding with B and H should be noted :, 1., , 2., 3., , H is the cause and B is the effect. Both measure the intensity of the magnetic field, and are vectors. B includes the presence of medium, B = H and B0 = 0H., B and H refer to external fields and not to magnet's own field., The above magnetic phenomenon is analogous to electrostatic phenomenon, just, as P, D and E in electrostatics are related to bound charges, free charges and all, charges respectively. Similarly Im, H and B are related to bound currents, free, currents and all currents. Im is analogous to dielectric polarisation P., , Classification of magnetic materials, 1., , 2., , 3., , Diamagnetic material : Suppose a material in which individual atom does not have, a net magnetic moment. When such a material is placed in a magnetic field, dipole, moments are induced in the atoms by the applied field. The magnetic field due to, induced magnetic moment opposes the original field. Thus the resultant field inside, the material is smaller than the applied field. This type of material is called, diamagnetic material., Magnetic moments are induced in all materials whenever a magnetic field is applied., Thus all materials have the property of diamagnetism. However, if there is a, permanent atomic magnetic moment, then paramagnetism or ferromagnetism is, much stronger than diamagnetism and the material does not show diamagnetic, property., Paramagnetic material : Now consider a material whose individual atoms have a, net magnetic moment. When such a substance is placed in magnetic field, an extra, magnetic field produces in the material in the direction of the field. The resultant, magnetic field in the material is then greater than the applied field. The tendency to, increase the magnetic field due to magnetisation of material is called paramagnetism,, and material is called paramagnetic material., Ferromagnetic material : In some materials, the permanent atomic magnetic, moments have strong tendency to align themselves, even without any external, field. These materials are called ferromagnetic materials. In every unmagnetised, ferromagnetic material, the atoms form domains inside the material. Different, domains, however, have different directions of magnetic moment and hence the, materials remain unmagnetised. On applying an external field, these domains rotate, and align in the direction of magnetic field., , (a), , (b), (c), Fig. 7.50, Because of the domain character of ferromagnetic materials, even if a small magnetic, field is applied, gives rise to large magnetisation. The resultant field is much larger, than the applied field in such a material., , Fig. 7.48, , Fig. 7.49, , 467

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468, , ELECTRICITY & MAGNETISM, Curie's law : With the increase of temperature, the randomisation of atomic magnetic, moments increases thereby decreasing the magnetisation I. The resultant magnetic field, B decreases, which means, decreases with temperature. The susceptibility of, paramagnetic materials,, =, , C, ,, T, , (Curie's law), , where C is Curie constant., , Curie temperature, The temperature at which ferromagnetic substance becomes paramagnetic is called Curie, temperature., At temperature above curie temperature the magnetic susceptibility of ferromagnetic, materials is given by, =, Here Tc, , C', ., T Tc, , Curie temperature, for iron it is 1043K., , Hysteresis, Hysteresis is shown only by ferromagnetic substances. In ferromagnetic materials, by, removing external field, the magnetic moment of some domains remain aligned in the, direction of previous applied magnetising field which results a residual magnetism. The, laging of intensity of magnetisation (I) to magnetising field is (H) is known as hysteresis., OA, Retentivity or residual magnetism, OB, Coercivity, , Fig. 7.51, Steel, , Soft iron, I, , I, , H, , H, , The area of hysteresis loop is large and thus, high energy loss., High retentivity, large coercivity., Less magnetic permeability, I and, Magnetisation, and, demagnetisation are, difficult., Used for making permanent magnets, and, thereby used in loudspeaker, microphone etc., , Note: 1., , The area of hysteresis loop is less and thus low, energy loss., Low retentivity, less coercivity, High magnetic permeability, I and, Magnetisation and demagnetisation are easy., Used for making electromagnet, and thereby used in, dynamo, transformer etc., , Diamagnetism is the universal property, it is present in all materials., , But it is weak and hard to detect if the substance is para or ferromagnetic., 2., , The phenomenon of magnetic hysteresis is similar to the elastic hysteresis : Strain, may not be proportional to stress; here H and I are not linearly related., , 3., , Electrostatic shielding can be possible by any metal. But magnetic shielding can, only possible by iron

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469, , PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, Comparative study of magnetic properties of materials :, Diamagnetic substances, Found in solids, liquids and, gases., Examples : Ag, Au, Br, Cu,, H2 O, NaCl, Sb etc., Universal property, Can be explained on the basis, of orbital motion of electrons, , Paramagnetic substances, Found in solids, liquids and, gases., Examples : Al, CuCl2 , Na, O2, Mn, Pt etc., Not universal, Can be explained on the basis, of spin and orbital motion of, electrons, placed, in strong, When, placed in strong When, field, magnetised, magnetic, field, magnetised magnetic, weekly in the direction weekly in the direction of field, B = B0 + Bm, opposite of field B = B0 –B m, , N, , S, , N, , Susceptibility, negative | | ~ 1, , Ferromagnetic substances, Found normally in solids., Examples: Co, Fe, Ni, Cd, Fe3 O 4, etc., Not universal, On the basis of domain theory, , When placed in weak magnetic, field,magnetised strongly in the, direction of field. B = B0+ B m, , S, , N, , S, , low, , and Susceptibility low but positive Susceptibility, ~1, positive ~ 102, 3, Relative permeability r < 1, r> 1, r >> 1 ( r ~ 10 ), , Diamagnetic, , Ex. 15, , x, , x, , T, , T, , and, , Ferromagnetic, , Paramagnetic, , x, , high, , T, , Tc, , Each atom of an iron bar 5 cm × 1 cm × 1 cm has a, =, , magnetic moment 1.8 × 10-23 A - m 2 . What will be the magnetic, moment of bar in the state of magnetic saturation ?, Density of iron = 7.78 103 kg / m3, Atomic weight of iron A = 58, Avogadro number, N, , Sol., The number of atoms per unit volume in a specimen, n =, , N, V, , N, A, , N, A, , 6.02 10 23, 56, , 28, 3, = 8.38 10 / m, Total number of atoms in the bar, , =, , = 6.02 1023 / g -mol., , 7.8 103, , nV, , 8.38 10 28, , 23, = 4.19 10, The saturated magnetic moment, =, Magnetic moment of each atom, , =, =, , 5 1 1, , 10, , 6, , total number of, atoms in the bar, , 1.8 10 23 4.19 10 23, 7.54 A-m2., , Ans.

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470, , ELECTRICITY & MAGNETISM, , Ex. 16, , A toroid having 500 turns of wire and a mean, circumferential length of 50 cm carries a current of 0.3A. The, relative permeability of the core is 600., (a) What is the magnetic field in the core ?, (b) What is the magnetic intensity ?, (c) What part of the magnetic field is due to surface currents ?, , Sol., (a), , Ni, 2 R, , r, , 7.94 105 A-m, , =, , I, H, , Ans (b), , Magnetic susceptibility, , 7.94 105, 2000, , 397, , Ans (c), , Ex. 18, , A bar magnet has a coercivity of 4 103 A / m . It is, desired to demagnetize by inserting it inside a solenoid 12 cm long, and having 60 turns. What current should be carried by the solenoid ?, , The magnetic field in the core of toroid, B =, , =, , Ni, 2 R, 0, , Sol., We have, , 600, , 4, , =, =, , (c), , B0 =, =, =, , Ans., , Ni, 2 R, 300 Am, , 1, , Ans., , 4 10 7 300, 0.0003768 T, B0 + Bm, B – B0, 0.2262 – 0.0003768, 0.2258 T, , 1.0 =, , which gives, , =, =, , Magnetisation,, , =, , (b), , =, We know that, , I =, , 4, , = 4 10 A / m, Magnetic intensity, due to north pole, , B, Hn =, , (c), , Ans., , 1 m, 4 r2, , 0, , 1, 4, , 3.6, 6 10, , 2, , 2, , 79.6 A/m, , The magnetic field B at the centre is, Fig. 7.54, , 1, 0.0005, , B =, , 0, , =, , 4, , I, , H, , Ans (a), , 10, , 7, , 4 10 4 159.2, , H, , 5.0 10 2 T, The field is towards north pole, , 0, , 4, , 0.90 10, , Ans (e), , =, , 1, =, , 7, , 2000 Am–1, , B, , 3.6, , Similarly, H s = 76.6 A/m, Resultant magnet intensity, H = Hn + Hs, = 159.2 A/m towards the south pole, Ans., , 398, , H =, , m, A, , Fig. 7.53, , 0.0005, , 10, , Ans., , 4, , 400 2.0, 0.40, , B, Magnetic intensity,, , I =, , =, , 4, , 8A, , Fig. 7.52, Ans., , Ni, 2 R, , 0, , 3, , (a), , 0.0005, r, , 4 10, 60, 0.12, , H and (c) magnetic field B at the centre of a bar magnet having, pole strength 3.6 A-m, magnetic length 12 cm and cross-sectional, area 0.90 cm2., , Sol., , or, , H, n, , Sol., , The current in the windings on a toroid is 2.0 A. There, are 400 turns and the mean circumferential length is 40 cm. With, the aid of a search coil and charge-measuring instrument, the, magnetic field is found to be 1.0 T. Calculate, (a) the magnetic intensity, (b) the magnetisation, (c) the magnetic susceptibility, (d) the equivalent surface current, and, (e) the relative permeability, , B =, , i =, , Ex. 19 Find (a) the magnetisation I, (b) the magnetic intensity, , Ex. 17, , We have,, , ni, , =, , 500 0.3, 0.50, H, 0, , We know that,, B =, Bm =, =, =, , H =, , 0.50, , H =, =, , 500 0.3, , 0.2262 T, , B, (b), , 7, , 10, , 10, , 7, , 2000, , Ans.

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471, , PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , Ex. 20 A moving coil galvanometer has a coil of area A and, , (c), , number of turns N. A magnetic field B is applied on it. The torque, acting on it is given by = k i where i is current through the coil. If, moment of inertia of the coil is I about the axis of rotation, (a) Find the value of k in terms of galvanometer parameters (N,, B, A)., (b) Find the value of torsional constant if current i0 produce, , or, , radian., 2, If a charge Q is passed almost instantaneously through coil,, find the maximum angular deflection in it., , (b), , The torque exerted by magnetic field on the coil of galvanometer, = N i AB, Given, = ki, k i = N i AB, or, k = NBA, Ans., If C is the torsional constant of the head, then, C, , or, , 2, , =, , 0, , 2NAB i0, , ...(i), , NABQ, I, Substituting this value in (i), we get, , or, , = N i0 AB, C =, , I, C, , =, , dL, dQ, AB, = N, dt, dt, or, dL = NAB dQ, On integrating, we get, L = NABQ, or, I = NABQ, , Sol., (a), , 0, , We know that, = N i AB, ...(ii), If L is the angular momentum of the coil, then L = I , and equation, (ii) can be written as, , angular deflection of, , (c), , If 0 is the maximum deflection, then by conservation of energy, 1 2, 1, C 0 =, I 2, 2, 2, , =, , Q, , NAB, 2i0 I, , Ans., , Review of formulae & Important Points, 1., , Bar magnet : If m is the magnetic charge and is the length of the, magnet, then magnetic moment of the magnet, M, , 2., , B =, , m, , =, , (ii), , 0, , F =, , =, , Magnetic field due to a magnetic charge m at a distance r, 0, , m, r2, , 4, , Magnetic field due to a short magnetic dipole M at a distance r, , and, (i), , tan, On the axis of the dipole,, , =, , 0, , 4, , tan, 2, =0, , M, r3, , 3cos 2, , 4, , M, r3, , M, , B, , 6., , Work done in increasing angle from 1 to 2, W = MB(cos 1–cos 2)., , 7., , The potential energy of magnet in magnetic field, U = –MBcos ., , 8., , Fig. 10.23, B =, , 0, , Torque on a magnet in magnetic field, , r2, , 4, , B =, 4., , = 90°, B =, , 5., , m1m2, , 2M, r3, , Coulomb s law in magnetism :, The magnetic force between two magnetic charges m1 and m2, placed at a separation r is given by, , 3., , On the equator,, , 0, , 4, , Elements of earth magnetic field :, (i), , Angle of declination ( ) : Angle between geographical meridian, and magnetic meridian is known as angle of declination., , (ii), , Angle of dip ( ) : The angle made by dip needle with the, horizontal is known as angle of dip. It is zero at the equator, and 90° at the poles., , (iii) Horizontal component of earth magnetic field : If B is the, resultant magnetic field at any place, then horizontal, component, , 1, , BH = B cos ., 9., , True dip and apparent dip : If is the true dip and is the, apparent dip in a plane making angle with the meridian plane,, then

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472, , ELECTRICITY & MAGNETISM, tan, , 10., , =, , (i), , Tangent galvanometer : If, then, , where, , 2 RBH, 0N, , (ii), , M, V, , =, , Intensity of magnetising field, , is the reading of the galvanometer,, , B0, , =, , H, , 0, , . Also H = ni., , (iii) Magnetic permeability, , 2 RBH, tan ,, 0N, , B, and, H, , =, , 0, , B0, H, , Also relative permeability, is called reduction factor.., r, , Vibration magnetometer : The time period of magnet of, magnetometer, T =, , 2, , I, MBH, , For magnets of magnetic moments M1 and M2, M1, M2, 13., , Intensity of magnetisation, I, , C, NBA, , i =, , 12., , Magnetic properties of materials :, , Moving coil galvanometer : If i is the current in galvanometer,, then, i =, , 11., , 14., , tan, cos, , =, , T2 2 T12, , = 0, , 0, , ., , is the susceptibility, then, r, , =, , 1, , ., , 15., , Diamagnetic : It is found in solids, liquids and gases. Ex. Ag, Au,, Bi, Cu, H2O, NaCl, Sb etc., , 16., , Paramagnetic : It is found in solids, liquids and gases. Ex. Al,, CuCl2, Na, O2, Mn, Pt etc., , 17., , Ferromagnetic : It is normally found in solids. Ex. Co, Fe, Ni,, Cd, Fe3O4 etc., , T2 2 T12, , Gauss s law in magnetism :, B dA, , (iv) If, , =

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , Exercise 7.1, , MCQ Type 1, , 7., , Only one option correct, 1., , An electron in an external magnetic field Bext has its angular, , spin flip so that L is then parallel with Bext , then, , 2., , (c), , (a) energy is supplied to the electron, (b) energy is lost by the electron, (c) energy neither supplied nor lost, (d) none of them., The magnetic dipoles in a diamagnetic material are represented, for, three situations. The three situations differ in the magnitude of a, magnetic field applied to the material. In which situation the, magnetisation of the material is the greatest :, , 8., , 4., , 5., , 6., , (a) A, (b) B, (c) C, (d) equal in A, B and C., The magnetic susceptibility is negative for, (a) diamagnetic substance, (b) paramagnetic material, (c) ferromagnetic material, (d) all the above., Electromagnets are made of soft iron because soft iron has, (a) low retentively and low coercivity, (b) low retentively and high coercivity, (c) high retentively and low coerrity, (d) high retentively and high coerrivity., A compass which is allowed to move in a horizontal plane is taken, to a geomagnetic pole. It, (a), , will stay in north-south direction only, , (b), , will stay in east-west direction only, , (c), , will stay in any position, , (d), , none of these, , A tangent galvanometer is connected directly to an ideal battery. If, the number of turns in the coil is doubled, the deflection will, (a), , increase, , (b) decrease, , (c), , remain same, , (d) any of these, , M, , (b), , M, , 2M, , (d) M, , A magnetic needle lying parallel to a magnetic field requires W, units of work to turn it through 60°. The torque required to, maintain the needle in this position will be, (a), , (b) W, , 3W, , 3, W, (d) 2W, 2, A magnet of magnetic moment M is rotated through 360° in a, magnetic field H, the work done will be, (a) MH, (b) 2MH, (c) 2 MH, (d) Zero, Force between two identical bar magnets whose centres are r metre, apart is 4.8 N, when their axes are in the same line. If separation is, increased to 2r, the force between them is, (a) 2.4 N, (b) 1.2 N, (c) 0.6 N, (d) 0.3 N, ˆ, A magnet of magnetic moment 50i A-m2 is placed along the x-axis, (c), , 9., , 10., , 3., , An iron rod of length L and magnetic moment M is bent in the form, of a semicircle. Now its magnetic moment will be, (a), , momentum L antiparallel to Bext . If the electron undergoes a, , 473, , 11., , in a magnetic field B, , 0.5iˆ 3.0 ˆj T . The torque acting on the, , magnet is, (a), , 12., , 175 kˆ N - m, , (b) 150 kˆ N - m, , (c) 75 kˆ N - m, (d) 25 37kˆ N- m, The magnetic field lines due to a bar magnet are correctly shown in, , (a), , (b), , (c), , (d), , Answer Key, , 1, , (a), , 2, , (b), , 3, , (a), , 4, , (a), , 5, , (c), , 6, , (c), , Sol. from page 481, , 7, , (b), , 8, , (a), , 9, , (d), , 10, , (d), , 11, , (b), , 12, , (d)

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474, 13., , 14., , 15., , 16., , 17., , 18., , 19., , ELECTRICITY & MAGNETISM, , A very small magnet is placed in the magnetic meridian with its, south pole pointing north. The null point is obtained 20 cm away, from the centre of the magnet. If the earth s magnetic field (horizontal, component) at this point be 0.3 gauss, the magnetic moment of the, magnet is, (a), , 8.0 102 e.m.u., , (b) 1.2 103 e.m.u., , (c), , 2.4 103 e.m.u., , (d) 3.6 103 e.m.u., , 20., , field intensity of 0.1 10 5 T . At another place, it takes 2.5 sec to, complete one vibration. The value of earth s horizontal field at, that place is, , 21., , At a place, if the earth s horizontal and vertical components of, magnetic fields are equal, then the angle of dip will be, (a), , 30°, , (b) 90°, , (c), , 45°, , (d) 0°, , length of the magnet, , (b), , pole strength of the magnet, , (c), , horizontal component of earth s magnetic field, , (d), , length of the suspension thread, , 22., , 4:9, , (b) 9 : 4, , (c), , 2:3, , (d) 3 : 2, , 23., , The period of oscillation of a magnet in vibration magnetometer is, 2 sec. The period of oscillation of a magnet whose magnetic moment, is four times that of the first magnet is, (a), , 1 sec, , (b) 4 sec, , (c), , 8 sec, , (d) 0.5 sec, , 24., , A magnetic needle suspended by a silk thread is vibrating in the, earth s magnetic field. If the temperature of the needle is increased, by 500°C, then, , 0.25 10 6 T, , (b) 0.36 10 6 T, , (c), , 0.66 10 8 T, , 6, (d) 1.2 10 T, , The magnetic needle of a tangent galvanometer is deflected at an, angle 30° due to a magnet. The horizontal component of earth s, , (a), , 1.96 10 4 T, , (b) 1.96 10 5 T, , (c), , 1.96 104 T, , (d) 1.96 105 T, , Relative permeability of iron is 5500, then its mangetic, susceptibility will be, (a), , Two magnets of same size and mass make respectively 10 and 15, oscillations per minute at certain place. The ratio of their magnetic, moments is, (a), , (a), , 4, magnetic field 0.34 10 T is along the plane of the coil. The, magnetic intensity is, , Time period of a freely suspended magnet does not depend upon, (a), , A magnet makes 40 oscillations per minute at a place having magnetic, , (b) 5499, 107, , (c), , 5500 ×, , (d) none of these, , (d), , paramagnetic and ferromagnetic materials, , Curie temperature is the temperature above which, (a), , a paramagnetic material becomes ferromagnetic, , (b), , a ferromagnetic material becomes paramagnetic, , (c), , a paramagnetic material becomes diamagnetic, , (d), , a ferromagnetic material becomes diamagnetic, , Liquid oxygen remains suspended between two poles faces of a, magnet because it is, (a), , diamagnetic, , (b) paramagnetic, , (c), , ferromagnetic, , (d) antiferromagnetic, , (a), , the time period decreases, , (b), , the time period remains unchanged, , A bar magnet is placed north south with its north pole due north., The points of zero magnetic field will be in which direction from, the centre of the magnet, , (c), , the time period increases, , (a), , north and south, , (d), , the needle stops vibrating, , (b), , east and west, , (c), , north-east and south-west, , (d), , north-west and south-east., , The time period of oscillation of a bar magnet suspended, horizontally along the magnetic meridian is T0. If this magnet is, replaced by another magnet of the same size and pole strength but, with double the mass, the new time period will be, (a), (c), , T0, 2, , (b), , 25., , 5501, , 26., , T0, 2, , (d) 2T 0, , 2T0, , Two identical magnetic dipoles of magnetic moments 1.0 A-m2, each, placed at a separation of 2m with their axis perpendicular to, each other. The resultant magnetic field at a point midway between, the dipoles is, 5 10 7 T, , (a), , 5 10 7 T, , (b), , (c), , 10–7 T, , (d) none of these, , Answer Key, , 13, , (b), , 14, , (c), , 15, , (d), , 16, , (a), , 17, , (a), , 18, , (c), , 19, , (c), , Sol. from page 481, , 20, , (b), , 21, , (b), , 22, , (b), , 23, , (b), , 24, , (b), , 25, , (b), , 26, , (b)

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475, , PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, 27., , 28., , A bar magnet of magnetic moment 3.0 A-m2 is placed in a uniform, magnetic field of 2 × 10–5 T. If each pole of the magnet experiences, a force of 6 × 10–4 N, the length of the magnet is, (a) 0.5 m, (b) 0.3 m, (c) 0.2 m, (d) 0.1 m, A current carrying coil is placed with its axis perpendicular to, N-S direction. Let horizontal component of earth’s magnetic field, be H0 and magnetic field inside the loop be H. If a magnet is, suspended inside the loop, it makes angle with H. Then =, (a), , (c), 29., , 30., , tan, , 1, , cos ec, , H0, H, 1, , 1, , (b) tan, , H, H0, , (d) cot, , 1, , 31., , 32., , (a), , is permanently magnetized, , (b), , remains ferromagnetic, , (c), , behaves like a diamagnetic material, , (d), , behaves like a paramagnetic material, , Needles N1, N2 and N3 are made of a ferromagnetic, a paramagnetic, and a diamagnetic substance respectively. A magnet when brought, close to them will, , H, H0, H0, H, , Demagnetisation of magnets can be done by, (a) rough handling, (b) heating, (c) magnetising in the opposite direction, (d) all the above, , When a ferromagnetic material is heated to temperature above its, Curie temperature, the material, , 33., , (a), , attract N1 strongly, N2 weakly and repel N3 weakly, , (b), , attract N1 strongly, but repel N2 and N3 weakly, , (c), , attract all three of them, , (d), , attract N1 and N2 strongly but repel N3., , Two identical short bar magnetics, each having magnetic moment, M, are placed a distance of 2d apart with axes perpendicular to, each other in a horizontal plane. The magnetic induction at a point, midway between them is, , If the angular momentum of an electron is J then the magnitude, of the magnetic moment will be, (a), (c), , eJ, m, , (a), (b), , eJ 2m, , (d), , eJ, 2m, , (c), , 2m, eJ, , 0, , 2, , 4, 2, , 0, , M, , (b), , d3, M, , (d), , d3, , Answer Key, , 27, , (d), , 28, , (a), , 29, , (d), , Sol. from page 481, , 31, , (d), , 32, , (a), , 33, , (d), , 0, , 3, , 4, 0, , 5, , 4, , 30, , M, d3, M, d3, , (b)

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476, , ELECTRICITY & MAGNETISM, , Level -2, 4., , Only one option correct, 1., , Figure (a) shows a pair of opposite spin orientations for an electron, in an external magnetic field Bext . Figure (b) gives three choices, for the graph of the potential energies associated with those, orientation as a function of magnitude of Bext . Choices B and C, consist of intersecting lines, choice A of parallel lines. Which is the, correct choice, , 5., , 6., (a), , 2., , (a), , A, A, , (b), , B, B, , (c), , C, C, , (d), , A, A; B, B., , Two short magnets of equal dipole moment M are fastened, perpendicularly at their centres. The magnitude of the magnetic, field at a distance d from the centre of the bisector of the right angle, is, (a), , (b), , (c), , (d), 3., , (b), , 0, , 2s, , (b), , 2/3s, , (c), , 2 3s, , (d), , 2 / 3s, , The true value of angle of dip at a place is 60°, the apparent dip in, a plane inclined at an angle of 30° with magnetic meridian is, (a), , tan, , 1, , (c), , tan, , 1, , 1, 2, 2, 3, , 1, , (b), , tan, , 2, , (d), , None of these, , A small coil C with N = 200 turns is mounted on one end of a, balance beam and introduced between the poles of an electromagnet, as shown in figure. The cross sectional area of coil is, A = 1.0 cm2, length of arm OA of the balance beam is = 30 cm., When there is no current in the coil the balance is in equilibrium., On passing a current I = 22 mA through the coil the equilibrium is, restored by putting the additional counter weight of mass m = 60, mg on the balance pan. Find the magnetic induction at the spot, where coil is located., , d3, , 4, , 2M, , 0, , d3, , 4, 0, , 2M, d3, , 4, , 0, , 4, , 2 2M, , 7., , d3, , (a), , nearly 2 aiB perpendicular to the plane of the loop, , (b), , 2 aiB in the plane of the loop, , (d), , (a), , M, , A very long bar magnet is placed with its north-pole coinciding, with the centre of a circular loop carrying an electric current i. The, magnetic field due to the magnet at a point on the periphery of the, loop is B. The radius of the loop is a. The force on the loop is, , (c), , The length of a magnet is large compared to its width and breadth., The time period of its oscillation in a vibration magnetometer is 2, s. The magnet is cut along its length into three equal parts and, three parts are then placed on each other with their like poles, together. The time period of this combination will be, , (a), , 0.4 T, , (b), , 0.3 T, , (c), , 0.2 T, , (d), , 0.1 T, , At a place on earth, horizontal component of earth’s magnetic, field is B1 and vertical component of earth’s magnetic field is B2. If, a magnetic needle is kept vertical, in a plane making angle with, the horizontal component of magnetic field, then square of time, period of oscillation of needle when slightly distributed is, proportional to, (a), , 1, B1 cos, , aiB along the magnet, , (b), , 1, B2, , (d), , infinite, , 1, , zero., , (c), , B1 cos, , 2, , B22, , Answer Key, , 1, , (b), , 2, , (d), , 3, , (a), , Sol. from page 482, , 5, , (b), , 6, , (a), , 7, , (c), , 4, , (b)

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , MCQ Type 2, Multiple correct options, 1., The figure shows two diamagnetic spheres located near the south, pole of a bar magnet. Then, , (a), (b), (c), , 2., , the force on sphere 1 is directed towards the magnet, the force on sphere 2 is directed away from the magnet, the magnetic dipole moment of sphere is directed towards, the magnet, (d) the magnetic dipole moment of sphere 2 is directed away, from the magnet., Figure shows a loop model (loop L) for a diamagnetic material., , 4., , 5., , Mark out the correct options., (a), , Diamagnetism occurs in all materials, , (b), , Diamagnetism results from the partial alignment of permanent, magnetic moment., , (c), , The magnetizing field intensity H is always zero in free, space., , (d), , The magnetic field of induced magnetic moment is opposite, to the applied field., , Which of the following statements are true about the magnetic, , (a), , 3., , The net dipole moment of the loop directed towards the, magnet, (b) The net dipole moment of the loop directed away from the, loop, (c) The loop gets attracted towards the magnet, (d) The loop gets repelled by the magnet., A ferromagnetic material is placed in an external magnetic field., The magnetic domain, (a) may increase in size, (b) may decrease in size, (c) retain their size, (d) nothing can be said., , E xercise 7. 2, , susceptibility, , (a), , 6., , 477, , m, , of paramagnetic substance, , value of m is inversely proportional to the absolute, temperature of the sample, , (b), , m, , is positive at all temperature, , (c), , m, , is negative at all temperature, , (d), , m, , does not depend on the temperature of the sample, , The current sensitivity of a moving coil galvanometer can be, increased by, (a), , increasing the magnetic field of the permanent magnet, , (b), , increasing the area of the deflecting coil, , (c), , increasing the number of turns in the coil, , (d), , increasing the restoring couple of the coil, , Answer Key, , 1, , (b, d), , 2, , (b, d), , Sol. from page 482, , 5, , (a, b), , 6, , (a, b, c), , 3, , (a, b), , 4, , (a, d)

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478, , ELECTRICITY & MAGNETISM, , Statement Questions, Read, (a), (b), (c), (d), 1., , 2., , the two statements carefully to mark the correct option out of the options given below:, If both the statements are true and the statement - 2 is the correct explanation of statement - 1., If both the statements are true but statement - 2 is not the correct explanation of the statement - 1., If statement - 1 true but statement - 2 is false., If statement - 1 is false but statement - 2 is true., , Statement - 1, , Statement - 2, , The poles of magnet can not be separated by breaking into two, pieces., , Reduction factor increases with increase of current., 8., , Paramagnetic substances possess the property of diamagnetism., , The magnetic moment will be reduced to half when a magnet is, broken into two equal pieces., , Statement - 2, Diamagnetism is the universal property, it is present in all, substances., , Statement - 1, 9., , Statement - 2, , Magnetic moment is given by M = iA., , At Curie point a ferromagnetic substance start behaving as a, paramagnetic substance., , Statement - 1, Gauss s law is not applicable in magnetism., , 10., , For making permanent magnets, steel is preferred over soft iron., , Isolated pole does not exist., , Statement - 2, , Statement - 1, , Steel possesses high retentivity., 11., , Soft iron is used as transformer core., , A bar magnet does not exert a torque on itself due to its own field., , Statement - 2, , Statement - 1, , Soft iron has low hysteresis loss., 12., , Statement - 2, , Statement - 1, , At a high temperature magnet losses its magnetism., , To protect any instrument from external magnetic field, it is put, inside an iron body., , Statement - 1, , Statement - 2, , The tangent galvanometer can be made more sensitive by increasing, the number of turns of its coil., , Iron has high permeability., , Statement - 2, Current through galvanometer is proportional to the number of, turns of the coil., 7., , Statement - 1, , Statement - 2, , The earth s magnetic field is due to iron present in its core., , 6., , Statement - 1, , Statement - 2, , We cannot think of magnetic field configuration with three poles., , 5., , Statement - 1, The ferromagnetic substance do not obey Curie s law., , Statement - 2, , 4., , Statement - 1, , Statement - 2, , When radius of a circular current carrying loop is doubled, its, magnetic moment becomes four times., , 3., , E xercise 7. 3, , Statement - 1, Reduction factor (k) of a tangent galvanometer helps in reducing, deflection to current., , 13., , Statement - 1, The sensitivity of a moving coil galvanometer is increased by, placing a suitable magnetic material as a core inside the coil., Statement - 2, Soft iron has high magnetic permeability and cannot be easily, magnetized or demagnetized., , Answer Key, , 1, , (b), , 2, , (a), , 3, , (a), , 4, , (b), , 5, , (d), , 6, , (b), , Sol. from page 483, , 8, , (a), , 9, , (b), , 10, , (a), , 11, , (a), , 12, , (a), , 13, , (c), , 7, , (c)

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , Subjective Integer Type, , 479, , E xercise 7. 4, Solution from page 483, , 1., , At a certain place the horizontal component of earth s magnetic, field is, , 4., , An iron rod of volume 10–4 m3 and relative permeability 1000 is, placed inside a long solenoid wound with 5 turns/cm. If a current, of 0.5 A is passed through the solenoid, find the magnetic moment, of the rod., Ans : 25 A-m2., , 5., , An ideal solenoid having 40 turns/cm has an aluminium core and, carries a current of 2.0 A. Calculate the magnetisation I developed, in the core and the magnetic field B at the centre. The, susceptibility of aluminium = 2.3×10–5. Ans : I = 0.18 A/m., , 6., , A tangent galvanometer shows a deflection of 45° when 10 mA of, current is passed through it. If the horizontal component of the, earth s magnetic field is BH = 3.6×10–6 T and radius of coil is 10, cm, find the number of turns in the coil., Ans : 570., , 3 times the vertical component. What is the angle of, , dip at that place ?, , Ans : 30°., A-m2, , 2., , A bar magnet of magnetic moment 2.0, is free to rotate, about a vertical axis passing through its centre. The magnet is, released from rest from east-west position. Find the kinetic energy, of the magnet as it takes north-south position. [Horizontal, component of earth s field is 25 T]., Ans : 50 J., , 3., , A magnetic needle is free to rotate in a vertical plane which makes, an angle of 60° with the magnetic meridian. If the needle stays in, a direction making an angle of tan, , 1, , 2, 3, , with the horizontal,, , what would be the dip at that place ?, , Ans : 30°., , Subjective, , E xercise 7.5, Solution from page 483, , 1., , A magnet is suspended in such a way that it oscillates in the, horizontal plane. If it makes 20 oscillations per minute at a place, where dip angle is 30° and 15 oscillations per minute at a place, where dip angle is 60°. Find the ratio of total earth magnetic field, at the two places., , 2., , Ans :, , 16, 9 3, , A moving-coil galvanometer has 100 turns and each turn has an, area 2.0 cm2. The magnetic field produced by the magnet is 0.01 T., The deflection in the coil is 0.05 radian when a current of 10 mA is, passed through it. Find the torsional constant of the suspension, wire. Ans : 4.0×10–5 N-m/rad., , 7., , Two long bar magnets are placed with their axes coinciding in such, a way that the north pole of the first magnet is 2.0 cm from the, south pole of the second. If both the magnets have a pole strength, of 10 A-m, find the force exerted by one magnet on the other., Ans : 2.5×10–2 N., , 8., , A bar magnet has a length 8 cm. The magnetic field at a point at a, distance 3 cm from the centre in the broad side-on position is, found to be 4×10–6 T. Find the pole strength of the magnet., , Find the magnetic field due to a dipole of magnetic moment 1.2 Am2 at a point 1 m away from it in a direction making an angle of 60°, with the dipole axis., Ans : 1.6 10, , 7, , T at an angle, line where, , with the radial, , tan, , 1, , 3, ., 2, , 3., , A magnet is 10 cm long and its pole strength is 12 A-m. Find the, magnitude of the magnetic field B at a point on its axis at a distance, 20 cm from it., Ans : 3.4×10–5 T., , 4., , A dip circle shows an apparent dip of 60° at a place where the true, dip is 45°. If the dip circle is rotated through 90°, what apparent, dip will it show ?, Ans : cot–1(0.816)., , 5., , 6., , The needle of a deflection galvanometer shows a deflection of 60°, due to a short bar magnet at a certain distance in tan A position. If, the distance is doubled, what will be the deflection ?, Ans : tan, , 1, , 3, 8, , Ans : 6×10–5 A-m., 9., , Assume that each iron atom has a permanent magnetic moment, equal to 2 Bohr magnetons (1 Bohr magneton = 9.27×10–24 Am2). The density of atoms in iron is 8.52 1028 atoms / m3., , 10., , (a), , Find the maximum magnetisation I in a long cylinder of iron., , (b), , Find the maximum magnetic field B on the axis inside the, cylinder., Ans : (a) 1.58×106 A/m (b) 2.0 T., , The susceptibility of annealed iron at saturation is 5500. Find the, permeability of annealed iron at saturation., Ans : 6.9×10–3.

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480, 11., , ELECTRICITY & MAGNETISM, , Imagine rolling a sheet of paper into a cylinder and placing a bar, magnet near its end as shown in figure. (a) Sketch the magnetic, field lines that pass through the surface of the cylinder., (b) What can you say about the sign of B.dA for every area dA, on the surface ?, (c), , Does this contradict Gauss s law for magnetism ?, , 13., , A short magnet oscillates in an oscillation magnetometer with a, time period of 0.10s where the earth s horizontal magnetic field is, 24 T. A downward current of 18 A is established in a vertical wire, placed 20 cm east of the magnet. Find the new time period., Ans : 0.076 s., , 14., , A deflection magnetometer is placed with its arm in north-south, direction. How and where should a short magnet having, , Ans : (b) sign is minus; (c) no, there is compensating positive, flux through open end near magnet., 12., , A compass needle made of pure iron (with density 7900 kg/m3), has a length L of 3.0 cm, a width of 1.00 mm, and a thickness of, 0.50 mm. The magnitude of the magnetic dipole moment of an iron, atom is M Fe, , 2.1 10, , 23, , J / T . If the magnetisation of the needle, , is equivalent to the alignment of 10% of the atoms in the needle,, what is the magnitude of the needle s magnetic dipole moment M ?, Ans : 2.7×10–3 J/T., , M, BH, , 40, , A-m2 be placed so that the needle can stay in any position ?, Ans : 2.0 cm from the needle, north pole pointing towards south.

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , 481, , S olutions Exercise 7.1Level -1, 1., , (a), , U i = – MB cos 0° = – MB, U f = – MB cos 180° = MB, U = Uf – Ui = MB – (–MB), = 2MB., The largest number of dipole moment is in case B and so, greatest magnetisation is of B., and, , 2., , (b), , 3., 4., 5., , (a), (a), (c), , 6., , i =, , V, R, , V, (2 R ), From (i) and (ii),, Given :, Now,, , ', , 2 RBH, tan, 0N, , =, , (a), and, , 2 RBH, tan ', 0 (2 N ), , =, , (d), , 2M, , 12., , (d), , … (ii), , 14., , (c), , 15., , (d), , 16., , (a), , 17., , 3W ., , (c), , 19., , (c), , 0, , 6 M 1M 2, 4, , r, , r24, r14, , 2r, r, , F2 =, , F1, 16, , 4.8, 16, , =, , M, , B, , =, , 150 kˆ N -m, , =, , 0.3 N., , (b), , 60, 60, 6 s and T2, 10, 15, For identical magnets,, T1, , M1, M2, , =, , T1, T2, , =, , T2, , =, , 4s, , T22, , 42, , T12, , 62, , 16, 36, , 4, ., 9, , M2, M1, M1, M2, , T1, , 2, , 1, 4, , 1 sec., , Due to change in temperature, magnetic moment of magnet, decreases, and so time period of oscillations will increase., , I1, , m 2, and I 2, 12, , 2, , (2 m), 12, , 2 I1., , Now ,, , T1, T2, , =, , I1, I2, , or, , T0, T2, , =, , 1, 2, , T2, , =, , 2 T0 ., , T1, , 60, 1.5 s and T2, 40, , T1, T2, , We know that,, , 50 iˆ (0.5 iˆ 3 ˆj )., , The field lines emerge from north pole and enters into magnet through south pole., , = 0.3, , 1, , 4, , 20., , r3, , BH, , = 0.3, (20)3, M = 1.2 × 103 emu., , (a), , 18., , 2M, , 2M, , 7, , BV, BH, = 45°, , tan, , ., , ., , 4, , F1, F2, , (b), , 10, , … (i), , U i = – MB cos 0° = –MB, and, U f = – MB cos 360° = –MB, Now,, W = Uf – Ui = – MB – (– MB) = 0., Force between two short bar magnets is given by, , or, , 0, , 4, , or, , W = MB cos 60°, = MB sin 60°, , F =, , 11., , S, , or, , = m × 2R = m 2, , =, , 10., , N, , R or R, , =, , (d), , E, , B =, , ., M = m, , M', , 9., , r, , 2RBH, tan, 0N, , When number of turns are doubled, resistance of the coil is, also doubled, so, , 8., , W, , BH, , (c), , (b), , N, , S, , (b), , At geomagnetic poles, there is no horizontal component of, earth field and so compass needle may stay at any position., , or, , 7., , 13., , or, , B2 =, =, , T12, T22, , B1, , 2.5 s, , B2, B1, , =, 1.5, 2.5, , 0.36 × 10–6 T, , 2, , (0.1 10 5 )

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482, 21., , (b), , ELECTRICITY & MAGNETISM, 27., , We know that, , or, , 22., , (b), , (b), (b, (b), , 26., , (b), , =, , F = mB, 6 × 10–4 = m × 2 × 10–5, m = 30 A-m, , tan, , B =, =, =, , BH tan, 0.34 × 10–4 tan 30°, 1.96 × 10–5 T, , µr =, x =, =, , 1+x, µr – 1, 5500 – 1 = 5499, , and, , B2, , Now, B, , 0, , ., , 4, 0, , 4, , 2M, , ., , r, , 3, , M, r, , B12, , 13, 1, , 7, , 10, , 3, , 2 1, , 7, , 10, , (a), , tan, , 29., 30., , (d), (b), , We know that,, , M, J, , 3, , 1, , B22, , 2 10 7 T, , 10 7 T, , 31., 32., , (d), (a), , 33., , (d), , 5 10 7 T ., , H0, , 0.1 m, , H0, H, , 28., , M, B1, , 3, 30, , M, m, , Now,, , We know that,, or, , 23., 24., 25., , (d), or, , B, BH, , q, 2m, e, ., 2m, , J, , B1, , H, , 0, , 4, , ., , 2M, d, , B12, , Now, B, , B22, , 5, , 4 d3, , 0, , 4, , M, , 0, , and B2, , 3, , ., , M, d3, , ., , ., , Solutions Exercise 7.1Level -2, 1., , 2., , (b), , (d), , U = – MB cos, For, = 0,, U = – MB, and for = 180°, U = + MB., These represent straight inclined lines., The resultant magnetic moment M, 0, , Magnetic field, B, , 3., , (a), , 4, , ., , I, , 2M, , d, , 2, , T, T, , Now,, , 2M, d3, , T, , 5., , I, ., 9, , 6., , M, M, I, I, , T, , tan, cos, , 2, , tan 60, cos30, , 3, 3 2, , 2, s., 3, , 1, 9, , 2, , tan, , (a), , tan 1 (2)., For equilibrium of the beam balance, m g × = N iA B, , Bi 2 a, , 0, , I, I, , M, , (b), , 2 a, , Bi, , 3, 12, , Also magnetic moment, M, , 0 2 2M, ., ., 4, d3, The loop experiences the force due to north pole only. So,, , F, , 3m3., , S, , mg, ., NiA, After substituting the given values and simplifying, we get, B = 0.4 T., Resultant magnetic field in the plane, B1, B, , N, , a, , 7., , (c), , B, 4., , (b), , The moment of inertia of original magnet I, , B=, , m 2, 12, , 2, , B1 cos, , Time period,, , After dividing into three equal parts,, , T, , B22, , 2, , I, MB, , B1cos, , B2, , Solutions Exercise 7.2, 1., , 2., , (b, d) Diamagnetic material gets repelled by magnet, and so both, the spheres will be repelled by the magnet. The dipole, moment of spheres also directed away from the magnet., (b, d) The near face of the loop behaves like south pole and far, face as north pole. So loop will be repelled by the magnet., N, , S, , 3., 4., , (a, b) The size of the domain may increase or decrease depending, on the orientation of the domain w.r.t. magnetic field., (a, d), , 5., , (a, b), , 6., , (a, b, c), , m, , C, ., (T TC )

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PERMANENT MAGNET AND MAGNETIC PROPERTIES OF SUBSTANCES, , 483, , S olutions Exercise 7.3, 1., , (b), , When a magnet is cut into pieces, each piece becomes new, m, 2, , magnet. M, 2., , M, ., 2, , (a), , M =, , r2, , iA i, , = i, (2r )2 4 M ., M, The smallest element of the magnetism is dipole and so, m in = m – m = 0, and Gauss's law, and, , 3., , (a), , B.dA, , 4., 5., , (b), (d), , 6., , (b), , 0., , 0 min, , =, , 7., , (c), , 8., 9., , (a), (b), , 10., 11., 12., , (a), (a), (a), , 13., , (c), , Reduction factor of tangent galvanometer does not depend, on current., Statement-2 is the explanation of statement-1., At Curic temperature, ferromagnetic substance changes into, paramagnetic substance., Because of high permeability of the iron, the entire magnetic, field will pass through iron, and so rest space becomes free, from magnetic field., Sensitivity of galvanometer,, s, , Magnetic field of earth is due to moving charged particles in, the atmosphere. With increase in temperature, the magnetic, moment of magnet decreases., In tangent galvanometer,, , 2 RBH, tan , clearly i, 0N, , i, , Sensitivity, s, , 0N, , tan, i, , i, , 2 RBH, , tan, , i, , tan, i, , 0N, ., 2 RBH, , If a magnetic material is placed inside coil of galvanometer,, then, , ., , s, , ., , r 0N, ., 2 RBH, , S olutions Exercise-7.4, 1., , Angle of dip,, , tan, , BV, BH, , =, , 1, , =, 2., , W, , =, , 3, , 4., , = 30°, = K, , N, BH, , 0, , d, , or, , Ans., , We have, , S, , MB sin d, , = Kf, , 0, , 90, , or, 3., , Kf, , = – MB(– cos 0 cos 90 ), = MB = 2 × 25 = 50 µ J, , Given apparent dip,, ', We know that, apparent dip, , = tan, , 1, , tan, cos, , =, , tan, , = tan ' cos, , T1, , =, , 60, 20, , 2, , I, MBH1, , … (i), , At place II,, , T2, , =, , 60, 15, , 2, , I, MBH 2, , … (ii), , 4, 3, , 2, , =, , BH1, BH 2, , 6., , We know that, , I =, , Ans., , M =, =, =, H =, =, I =, =, i, , =, , M, V, , IV, (2.5 × 105) × (10–4), 25 A-m2, Ans, ni = (4000) × (2), 8 × 103 A/m, µrH = (2.3 × 10–5) × (8 × 103), 0.18, Ans., 2 RBH, tan, µ0 N, , After substituting, the values and simplifying, we get, N = 570., , At place I,, , From equations (i) and (ii), we have, , We know that,, , Ans., , S olutions Exercise-7.5, 1., , 5., , 2/ 3, , tan ', , 3, , = 2.5 105 A/m, , Ki, , 0, , 1, , cos 60, , = 30°, H = ni = (500) 0.5, = 250 A/m, , 90°, , = Kf, , 3, , I = µr H 1000 250, , 90, , or, , or, We know that,, , 2, , or, As, , BH1, BH 2, B, , B1, B2, , =, , Ans., , 16, 9, , = BH / cos ,, =, =, , BH1, BH 2, , 16, 9, , cos 60, cos 30, , 1/ 2, 3 /2, , 16, 9 3, , Ans.

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484, , ELECTRICITY & MAGNETISM, , 2., , B, , µ0 M 3cos 2, ., 4, r3, , =, , B, , = 10, r = 1m, 60°, , 3., , 1.2, , 7, , 1, , 3cos 2 60, , 3, , tan 60, 2, , 3, 2, , µ0, ., 4, , cot 2, , = cot, , 1, , = cot 2 60, , cot 2 45, , 11., , µ = µr µ0 (5501) (4 10 7 ), = 6.9 × 10–3, Ans, (b) The field is entering into the surface so flux is negative., (c) Explanation is given in the answer., The volume of the needle,, , 4, , (0.1) 2, 4, , 2, , cot, , 12., , V, , 2, , cot 2, , The mass of the needle, , , we have, , 1, , tan, , r13, , = (2)3, , 2, , M =, , 3, 8, , =, , N, , C, , =, =, , S, , 13., , N, , 7., , C, NBA, , =, 4.0 10, For long bar magnet, we can write,, , 1, (1.27 1021 ) (2.1 10 –23 ), 10, , T, MBH1, , = 2, , B =, , Ans., … (i), , 5, , µ0, 4, , = 1.8 10, , N-m/rad, , Now, , µ0 m m, = 10, 4, r2, , 7, , (10)2, , r2, , 2, , BH 2, , T2, , 0.02 2, , 3/ 2, , ; M, , (18), 0.20, , = BH1, , 6, , T, , B = 42 × 10–6 T, I, MBH 2, , = 2, , … (ii), , Using equations (i) and (ii) , and substituting the values, we get, 14., , T2 = 0.076 s, For the needle to stay in any position, B, , M, , µ0 i, ., 2 r, , –7, = (2 10 ), , 4, , = 2.5 10 2 N, Ans., Magnetic field due to a bar magnet in the broad-side on position is, given by, m ., , 4, , After substituting the values and simplifying we get, (a), , T1, , (10 10 ) (100) (0.01) (2 10 ), 0.05, , B =, , 9., , We know that, , iNBA, , F =, , 8., , 6.02 1023, , BH1 = 24 10 6 T, Where, The magnetic field produced by, wire, , 3, , =, , 56 10, , = 2.7 × 10–3 J/T, , Ans., , 2 cm, i, , 4, , 3, , = 1.27 × 1021, , For moving coil galvanometer, we have, S, , 1.185 10, , The needle's dipole moment, , tan 60, tan 2, , or, , = (3 10 2 ) (1 10 3 ) (0.5 10 3 ), = 1.5 × 10–8 m3, = V, , =, , r23, , =, , 2, , = 1+x, = 1 + 5500 = 5501, , = 7900 1.5 10 8, = 1.183 × 10–4 kg, The number of atoms in the needle, , 2, , cot 2 = 0.816, Ans., For deflection galvanometer, with short magnet, we can write, tan, tan, , µr, , Ans., , 2, , Ans., , We know that, , 2, , 2, , 10–5T, , 2, , ), , 10., , 2 (12 0.1) 0.20, , 7, , = 3.4 ×, Using the relation, , = (4 10 7 ) (1.58 106 ), = 2.0 T, , 2 Mr, r2, , 24, , = 1.58 106 A/m, The magnetic field due to magnetisation, , Ans., , The magnetic field at the axis of the magnet is given by,, =, , (8.52 1028 ) (2 9.27 10, 1, , Bm = µ0 I, , (0.2) 2, , 6., , (b), , 1, , tan, 2, , =, , tan, , = 10, , 5., , 1, , = 1.6 10 7 T, , B, , 4., , I =, , B = 6 10 5 A -m, The total magnetic moment per unit volume. i.e.,, magnetisation, , or, , µ0 2 M, . 3, 4, r, , Ans., , = BH, = BH, , r, , =, , µ0 2 M, ., 4n BH, , 1/ 3, , After substituting the values and simplifying we get,, r = 2.0 cm

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486, , ELECTRICITY & MAGNETISM, , 8.1 ELECTROMAGNETIC, , INDUCTION, , : AN, , INTRODUCTION, , In 1831, Michael Faraday suggested that if electricity moving in a wire produces, magnetism, then opposite might be true; a magnet moving near a wire could produce, electricity. He moved a magnet in and out of a coil of wire, and electricity flowed in the, coil. This is called electromagnetic induction. The electric current flows only while the, magnetic field moves or varies. If the magnet and coil are still, no current flows in the coil., Electromagnetic induction is used in hundreds of machines and devices, like electric, motors, generators etc. The energy needed to turn the magnet is provided by energy, sources such as steam, moving water or the wind etc., , 8.2 MAGNETIC, , FLUX, , In fluid mechanics, we defined a quantity; rate of flow, Q = vAcos , in electrostatics; an, electric flux, e = EAcos . On the same way here we can define magnetic flux. The, magnetic flux through small element of surface dA is defined as;, d B, , Fig. 8.1, , = B dA, = (Bcos )dA, , = B dA, The total magnetic flux through the surface, B, , =, , B dA ., , When B is uniform over a plane surface with total area A,, , More about, (a), , B, , =, , B A, , BA cos ., , B, , 1., 2., 3., , We may take dA positive, pointing out of the surface., Magnetic flux is a scalar quantity, but it may be negative., Its SI unit is tesla-m2, which is called the weber. Thus, 1 weber = 1 Wb = 1T-m2., , 4., , If the elemental area of dA is right angles to the field lines, then, , d B, dA, Thus, the magnetic field is equal to the flux per unit surface area at right angle to, the magnetic field, and so it is also called magnetic flux density. Its unit is weber/m2, (Wb/m2)., B, , (b), Fig. 8.2, , =, , 5., , Fig. 8.3, , 8.3 FARADAY'S, , LAW OF, , EMI, , According to Faraday, whenever there is change in magnetic flux linked with the circuit,, there induces an emf in the circuit. The rate of change of magnetic flux is equal to the, induced emf. Thus, , d B, ., dt, Here negative sign indicates that induced emf opposes the change in flux., e, , =

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ELECTROMAGNETIC INDUCTION, Total magnetic flux (Magnetic-flux linkage), If a closed loop in which an emf is induced contains not one but N turns, then induced, emf e will be equal to the sum of the emfs induced in each turn. And if the magnetic flux, associated by each turn is the same and equal to B, the total flux Total through the, surface extended over such a complex loop can be represented by, Tota l = N B ., This quantity is called the total magnetic flux, or the magnetic flux-linkage. In this case,, the emf induced in the loop is defined by the formula, e, , =, , d, , Total, , N, , dt, , d B, ., dt, , Note:, 1., , When the field B varies with time as well as the configuration or arrangement of, the loop in the field, the induced emf should be calculated by taking in to, consideration of two factors. So induced emf can be written as :, e, , E d, , =, , t, , v, , B, , d, , Here the first term is due to the time variation of the magnetic field, while the, second is due to the motion of the loop., 2., , When magnetic field changes with both time and space:, e=, , 3., , d B, dt, , B, , B, , t, , x, , x, t, , A, , B, t, , B, v, x, , An emf is always induced whenever there is change in magnetic flux in the circuit., But current is induced only in closed circuit., , 8.4 LENZ'S, , LAW, , Lenz's law is used to get the direction of induced current. Which means that we can, apply it directly only to a closed conducting loop. According to Lenz s law an induced, current in a closed conducting loop will appear in such a direction that opposes the, change which produces it., , More about Lenz's law, Whether we push the magnet towards the loop or pull it out from the loop, we shall, always experience a resisting force and thus will have to do work. From the principle of, conservation of energy, this work must exactly equal to the thermal energy that appears, in the coil because there are only two energy transfer that take place in this isolated, system. The faster we move the magnet, the more rapidly we do work, and thus greater, the rate of production of thermal energy in the coil. If we cut the loop and then do, experiment, there will be no induced current, no thermal energy, no resisting force on the, magnet and no work requires to move it., To understand Lenz s law, let us apply it to a specific case; namely, the first of Faraday s, experiments., , 487

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488, , ELECTRICITY & MAGNETISM, Lenz's law 1 :, Let us consider a magnet with its north pole facing a closed conducting loop moves, towards it or with the south face moves away from it., When magnet moves towards the closed loop, the near face of loop becomes north, so, that it repels the incoming magnet (according to Lenz s law), and therefore the induced, current in the loop will induce in counter clockwise direction., If the magnet moves away from the closed loop with its N-pole facing it, the facing side, of the loop behaves like south pole and opposes the receding magnet. The direction of, induced current in the loop will be clockwise., , Lenz's law 2 :, , Fig. 8.4, , When the magnet moves with north pole towards the closed conducting loop, the flux, through the loop increases, the induced current, therefore will be in such a direction; its, magnetic field produces flux in opposite to the flux of the magnet., When magnet with its north pole moves away from the loop, the flux through the loop, decreases. The induced current in the loop, try to compensate this decrease in flux and, therefore will be in the direction as shown in the figure., , Note:, 1., 2., , In the experiment, we have moved the magnet, but the same effect will be observed, if we move the loop and keep the magnet at rest., The induced magnetic field does not oppose the magnetic field of the magnet but, it opposes the change in this field., , Flux change and induced emf, The magnetic flux, = BAcos has three variables, and so flux will change if either of, them changes. Thus :, Fig. 8.5, , (i), , When B alone is changing,, , e, , =, , A cos, , dB, ., dt, , (ii), , When A alone is changing,, , e, , =, , B cos, , dA, ., dt, , (iii) When alone is changing,, , e, , =, , BA, , d cos, ., dt, , In case of closed loop, a current will induce. If R is the resistance of the loop, then, induced current, , i in, , =, , e, R, , d, dt, R, , The induced charge in the loop, , dq, , or, , q, , = iin dt, , =, , R, , d, dt, R, , dt, , d, R

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ELECTROMAGNETIC INDUCTION, , Ex. 1, , A closed conducting loop is placed in a uniform magnetic, field as shown in fig. 8.6. Find the direction of induced current in, the loop when, , Fig. 8.6, (i), , magnetic field is increasing with time, , (ii), , magnetic field is decreasing with time., , Ex. 4, , (i), (ii), (iii), (iv), , 489, , In which case the flux changes and an emf induces?, , Fig. 8.9, If magnet is rotated about its axis in the loop., Loop is placed at rest in uniform field., A closed loop moves in uniform field., Loop is placed with its axis parallel to a current carrying, conductor., , Sol., , Sol., , The flux change in each case is zero, and so induced emf is zero., , (i), , Ex. 5 Figure 8.10 shows a conducting loop consisting of a half, circle of radius r = 0.20 m and three straight sections. The half-, , (ii), , In case when field increases, the flux of which increases into the, page and so the field of the induced current will be in upward, direction. For this the current in the loop will be in counterclockwise, direction., The direction of induced current will be in clockwise., , Ex. 2 A closed conducting loop is placed in uniform magnetic, field which points into the plane of the loop. What will be the, direction of induced current when loop is expanding., , circles lies in a uniform magnetic field B that is directed out of the, page; the field magnitude is given by B = t2 + 2t + 5, with B in tesla, and t in second. An ideal battery with emf = 2V is connected to the, loop. The resistance of the loop is 2, , Fig. 8.7, , Sol., When area of the loop increases, the flux will increase into the page, and, so the field of induced current must be out of the page. For this the, direction of induced current will be counterclockwise., , Ex. 3, , Two closed conducting loops are placed parallel to each, other. The current in one of the loops starts increasing. Will the, loops attract or repel ?, , ., , Fig. 8.10, (a), , What are the magnitude and direction of the emf induced, around the loop by field at t = 10s ?, , (b), , What is the current in the loop at t = 10s ?, , Sol., (a), , The flux of the field through the half loop (upto the field), =, , r2 2, t, 2, , BA, , 2t, , 5, , According to Faraday's law, |e| =, , Fig. 8.8, , =, , Sol., When current in first loop increases, the flux associated in second loop is, also increases. To compensate this increase in flux the loops must move, away from each other; means they will repel each other., , =, , r2 d 2, (t, 2 dt, , d, dt, , r2, 2t, 2, , 0.2, , 2, , = 1.38 V., , 2t, , 5), , 2, , 2 10 2, 2, Ans.

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490, (b), , ELECTRICITY & MAGNETISM, , The induced emf causes the current in clockwise direction around, the loop; the battery emf tends to drive a current in counter, clockwise direction. Because bat is greater than ind, and so the, net emf net is counterclockwise. Thus, net, , i =, , =, , R, , bat, , As, , d, dt, , =, , |e| =, , ind, , (ii), , R, , 2 1.38, 2, , Br 2, 2, , ., , Ans., , When loop enters into the magnetic field, the flux into it increases., By Lenz s law the induced current in the loop will be anticlockwise, T, the loop, 2, starts coming out from the magnetic field, and so the direction of, current reverses, i.e., it becomes clockwise., , so as to compensate the increasing flux. After each, , = 0.31 A, , Ans., , Ex. 6, , Space is divided by the line AD into two regions. Region, I is field free and region II has a uniform magnetic field B directed, into the plane of the paper. ACD is a semicircular conducting loop, of radius r with centre at O, the plane of the loop being in the plane, of the paper. The loop is now made to rotate with a constant angular, velocity about an axis passing through O and perpendicular to, plane of paper. The effective resistance of the loop is R., , (iii), , The magnitude of the induced emf e, , Br 2, 2, , is constant. It, , 2, changes in sign each after T, 2, , 2, , . The graph between, , induced emf and time is shown in fig. 8.12., , Fig. 8.12, , Ex. 7, , A plane loop shown in fig. 8.13 is shaped as two squares, with sides a = 0.20 m and b = 0.10 m and is introduced into a, uniform magnetic field at right angles to the loop s plane., Fig. 8.11, Obtain an expression for the magnitude of induced current, in the loop., (ii) Show the direction of current when the loop is entering into, region II., (iii) Plot a graph between the induced emf and the time of rotation, for two periods of rotation., (i), , Sol., (i), , Suppose the loop is rotated by small angle d in time dt. As it is, rotated with constant angular velocity, so d = dt. The area of the, loop into the field, dA, , 1, r rd, 2, , r2, d . The magnetic flux, 2, , = BdA, =, , B, , Sol., The loops are connected in such a way that if the current is clockwise in, I loop, then it will be anticlockwise in the II loop (see figure). The emf in, loop I, , through this area, d, , Fig. 8.13, The magnetic induction varies with time as B = B0 sin t where B0, = 10mT and = 100 rad/s. Find the amplitude of current induced in, the loop if the resistance per unit length is equal to = 50 milli, ohm/m. The inductance of the loop is negligible., , r2, d, 2, , e1 =, , According to Faraday s law, induced emf, e =, , d, dt, , Br 2 d, 2 dt, , d, dt, , d Ba 2, dt, d, B0 sin t, dt, , =, , a2, , =, , B0a 2 cos t

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ELECTROMAGNETIC INDUCTION, , Ex. 9, , Similarly, the emf in the II loop,, , e2 =, =, , 491, , A magnet falls into (a) closed conducting loop; (b) open, conducting loop. Discuss the motion of the magnet., , d Bb 2, , b2, , dt, , d, B0 sin t, dt, , B0b 2 cos t, , The net emf in the circuit, enet = e1 – e2, =, , B0 a 2, , b2, , Fig. 8.15, , cos t., , Given resistance per unit length of the loop wire is , thus total resistance, R = 4(a + b) ., , Sol., (a), , When magnet falls into the loop, the induced current in the, loop will be in such a direction that the facing side of the, loop opposes the incoming and outgoing magnet. So when, north pole approaches towards the loop, the facing side of, the loop becomes north and the acceleration of the magnet, becomes less than g., , (b), , If there is a cut in the loop and magnet falls into it, there will, be no induced current in the loop and hence the acceleration, of the magnet remains equal to g., , The induced current, i =, , =, , =, , enet, R, , B0 a 2, , b2, , cos t, , 4 a b, , B0 a b, , cos t, , 4, , Ex. 10, , The amplitude of the current, i0 =, , Magnet falls into a long conducting tube. Discuss, , its motion., , Sol., , B0 a b, 4, , On substituting the values given, i0 = 0.5 A., , Ans., , Ex. 8, , Figure 8.14 illustrated plane figures made of thin, conductors are located in a uniform magnetic field directed away, from a reader beyond the plane of the drawing. The magnetic, induction starts diminishing. Find how the currents induced in, these loops are directed., , When magnet falls into the tube, the flux into it changes and so current, will induce along its periphery. Because of the induced current the, acceleration of falling magnet becomes less than g. The acceleration, continuously decreases and becomes zero. There after magnet falls with, constant velocity. This can be understand as; the tube can be assumed to, be madeup of number of rings. Each ring decreases the acceleration of, falling magnet and ultimately acceleration of magnet becomes zero. It, means initially with the increasing of velocity of magnet, flux changes, with increasing rate, and induced current increases in magnitude, which, opposes the motion with greater force. After falling some distance the, rate of change of flux becomes constant, which is enough to provide, retarding force equal to weight of the magnet., , Fig. 8.14, , Sol., As magnetic field is decreasing into the plane of the figure, so induced, current will compensate this decrease by flowing in clockwise direction., Thus, (a), , in round conductor the current flows clockwise, there is no current, in the connector., , (b), , in the outside conductor, clockwise;, , (c), , in both round conductors, clockwise : no current in the connector., , (d), , in the left-hand of the figure eight, clockwise., Fig. 8.16

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492, , ELECTRICITY & MAGNETISM, , 8.5 MECHANISM, , OF, , EMI, , Consider a conducting rod of length, , moving with constant velocity v which is, , perpendicular to a uniform magnetic field B directed into the plane of paper. Let the rod, is moving toward right as shown in fig. 8.17. The free electrons also move to the right as, they are trapped within the rod., The magnetic field exerts force on the free electrons, Fm, , e v, , B so they move, , toward the end Q within the rod. The end P of the rod becomes positively charged while, end Q becomes negatively charged, hence an electric field E is set up within the rod, which exerts force on the free electrons in opposite to magnetic force. At equilibrium, , Fe, or, , eE, , e v, , or, , Fm, , = 0, , B, , = 0, , E, , =, , v, , e, , =, , E d, , e, , =, , v, , e, , =, , vi B, , e, , = vB ., , B, , The induced emf across the rod, Fig. 8.17, or, , B d, , ...(1), , In the case under consideration, k, , d j, , ...(2), , Induced emf/current in moving conductor slides along a, stationary U-shaped conductor, Method 1 :When conductor moves perpendicular to magnetic field, the charge carriers, inside it experience magnetic force (qvB). When a charge q moves from Q to P through, a distance , the work done by the force F is W = F = qvB, The emf is the work done per unit charge, e, , Fig. 8.18, , =, , W, q, , vB ., , Method II : By Faraday's law, Let conductor PQ slides a distance x in time t, the change in magnetic flux in this time, = B A = B( x), and, , |e|, , =, , t, , B, , x, t, , = B v, i in, , Bv, ., R, , Induced current, , 2., , Magnetic force on the conductor : Conductor PQ experiences a force opposite to, the direction of motion, Fm, , =, , e, R, , 1., , = Bi in, , B, , Bv, R, , B 2v, R, , 2

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493, , ELECTROMAGNETIC INDUCTION, 3., , Power dissipated in moving the conductor :, P agent, , 4., , dW, dt, , =, , Fagent v, , B2 v, R, , 2, , v, , B2 v2, R, , 2, , Electrical power : Electrical power dissipated through the resistance, 2, = i R, , P thermal, , Bv, R, , 2, , R, , B 2v 2, R, , 2, , More about direction of induced current, Rule 1 : The direction of induced current in a closed circuit with moving conductor, can be obtained as follows : Place a positive charge on moving conductor, and then find, direction of force on it by Fleming left hand rule. The charged particle will move in the, direction of force, which will be the direction of induced current., Rule 2 : Direction of induced current can also be obtained by Fleming right hand rule., According to this, if middle finger, fore finger and thumb of right hand are held mutually, perpendicular, then fore finger represents the direction of magnetic field, thumb represents, the direction of velocity of the conductor; and middle finger will represent the direction, of induced current., , Ex. 11 Find the induced emf across the ends of the conducting Ex. 12, rod in the following situations. The length of the conductor is ., , A closed loop of wire consists of a pair of equal, semicircles, of radius 3.7 m, lying in mutually perpendicular planes., The loop was formed by folding a plane circular loop along a, diameter until the two halves became perpendicular. A uniform, , magnetic field B of magnitude 76 mT is directed perpendicular to, the fold diameter and makes equal angles (= 45o) with the planes of, the semicircles as shown in fig. 8.20. The magnetic field is reduced, to zero at a uniform rate during a time interval of 4.5 ms. Determine, the magnitude of the induced emf and the direction of the induced, current in the loop during this interval., , Fig. 8.19, , Sol., (a), , In this case we can make v perpendicular to length of the rod or, perpendicular to v. Thus, e =, , (b), , B vsin, , or Bv sin, , = Bv sin, Ans., If we take the plane of motion of the rod as xy, then, , vi, , e =, , Bj, , Fig. 8.20, , Sol., , The change in flux in the given time interval t, , r 2 B cos 45, , =, , i, , By Faraday s law, the induced emf, =, , vB, , k i, , (c), , e =, , vi Bi, , (d), , e =, , v j Bi, , =, (e), , If PQ line makes, , 0, , j, , e =, , t, , 0, 3.7, , vB, , k i, , i, , =, , 0, , =, , with the velocity vector, then, e =, , r 2 B cos 45, t, , B v sin, , PQ ., , 2, , 76 10, , 4.5 10, 51 10 3 V, , 3, , 1, 2, , 3, , 51 mV, , The direction of the current is clockwise (see figure)., , Ans.

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494, , ELECTRICITY & MAGNETISM, , Ex. 13 The dotted circle in fig. 8.21 shows the region in which, , How will the magnetic field affect the motion of the pendulum, ? What is the direction of the currents in the circuit of the, pendulum ?, , a permanent magnetic field B is localised (it is directed, perpendicularly to the plane of the figure). This region in encircled, by a fixed metallic ring R. By moving the sliding contacts to the, other side of the ring, we introduce the magnetic flux into the, closed contour containing a galvanometer G (1-initial position, 2final position). Will the galvanometer show a current pulse ?, , (c), , Fig. 8.21, , (d), , Sol., There will be no induced current. There is no current since in this case, both, , dB, and Lorentz force are equal to zero : the magnetic field B is, dt, , (e), , Fig. 8.22 (c), Fig. 8.22(d), A copper wire connected to a closed circuit is surrounded by a, thick iron shell fig. (c) and introduced together with the shell, into the space between the poles of an electromagnet. The, iron shell acts as a magnetic screen for the wire., Will an e.m.f. be induced in the wire ?, An aircraft flies along the meridian. Will the potentials at the, ends of its wings be the same ?, Will the potential difference change if the aircraft flies in, any other direction with the same velocity ?, A rectangular wire frame rotates with a constant velocity, around one of its sides parallel to a current carrying rectilinear, conductor nearby fig. (e)., , constant and the closed loop moves in the region where there is no, magnetic field., , Ex. 14 (a)A small rectangular wire frame falls freely in, , the, space between the wide poles of a sufficiently strong electromagnet, fig. (a)., , (f), , (b), , Fig. 8.22(a), Show the direction of the currents induced in the frame when, the middle of the frame passes through the positions A, B, and C. How will the frame move in these sections ?, A small pendulum consisting of a metal thread, a ball and a, sharp point immersed in a cup of mercury fig. (b) makes part, of an electric circuit. The pendulum is placed in the space, between the broad poles of an electromagnet and swings in, the plane perpendicular to the lines of force of the magnetic, field. During the oscillations the sharp point of the pendulum, point of the pendulum remains immersed in mercury., , Sol., (a), , (b), , (c), Fig. 8.22(b), , Fig. 8.22(e), Indicate the positions in which the maximum and the, minimum e.m.f.s will be induced in the frame., Two circular conductors are perpendicular to each other as, shown in fig. (f)., Will a current be induced in the conductor A if the current is, changed in the circuit B ?, As the frame passes through the position A the current will flow, counterclockwise. When passing through the position B there is, no change in flux through the coil and hence no induced current. In, the position C the current will flow clockwise., As the pendulum oscillates, the periodic changes in the area enclosed, by the circuit will induce currents in the latter. The induced currents, will be directed so that their magnetic field compensates for the, change in the flux of the magnetic lines of force penetrating the area, of the circuit., When the pendulum swings so as to increase the area enclosed by, the circuit the current will flow counterclockwise, and if the motion, is such as to decrease the area enclosed by the circuit, the current, will be clockwise. The interaction of the magnetic field of induced, currents with the field of a permanent magnet will further damp, the oscillations of the pendulum., An emf will be induced because the introduction of the wire into, the space between the poles of the magnet will change the flux, passing through the area enclosed by the circuit.

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ELECTROMAGNETIC INDUCTION, (d), , There will be p.d. across the ends of the wing. If Bv is the vertical, component of earth field and v the horizontal velocity of the, aircraft, then, , (e), , The emf will be minimum when the frame arranges itself in the, plane passing through the rectilinear conductor., , (f), , There will be no induced current., , 495, , (b), , Closed conducting loop coming out from the magnetic field.: When, one of its sides QR comes out of the field, the emf across PQ, QR, and RS are zero. But there is induced emf across PS, which is e =, Bv . So the net induced emf across the loop is also e. The direction, of induced current in the loop will be clockwise., , (c), , Closed conducting loop coming out of the uniform magnetic field,, but velocity vector is perpendicular to one of its diagonals., , e = Bv v ., , Ex. 15, , Let us consider a closed conducting loop is moving in a, uniform magnetic field, which is perpendicular to plane of the, loop. Discuss about induced emf in the loop., , Sol., Method I : (a)The magnetic flux through the loop is B = BA. Till the, loop remain entirely inside the field, the flux will not change and so the, net induced emf in the loop will be zero., Fig. 8.25, The induced emf across the loop e = Bvy. As the loop is moving to, the right, y increases and therefore induced emf increases and, becomes maximum emax, , Bv, , PR . There after induced emf, , starts decreasing and becomes zero when entire loop comes out of, the field., Fig. 8.23, Method II : We can split the loop into its four sides as shown in fig., 8.24. The induced emf across side PQ and SR is zero, while induced emf, across PS and QR is e = Bv . For the closed loop PQRS, the net emf, becomes, e – e = 0., , (d), , In the case shown in fig. 8.26, the induced emf across the ends of, the conductor, e =, , Bv, , 1 sin 1, , 2 sin 2, , Fig. 8.26, Fig. 8.24, , 8.6 ROTATING, 1., , CONDUCTOR, , Rod rotating in uniform magnetic field, Consider a rod of length is rotating about an axis passing through one of its ends, with constant angular velocity in an uniform magnetic field B as shown in fig., 8.27., Induced emf across the element, (de) = Bvx (dx), = B( x)dx, Emf across the entire rod, e, , B, , =, , xdx, Fig. 8.27, , 0, , e, , = VP – VQ =, , 2, , B, 2

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496, , ELECTRICITY & MAGNETISM, 2., , Cycle wheel, Flux cutting by each metal spoke is same. Each spoke becomes cell of emf, 2, , B, , e, , . All such cells are in parallel fasnion, therefore enet = e. Each point on, 2, the periphery of wheel has same potential., Fig. 8.28, , 3., , Faraday disc dynamo : A metal disc can be assumed to be madeup of number of, , B R2, . All such, 2, , radial conductors. The emf induced across each conductor is e, conductors behave like a number of cells in parallel. Therefore, e net, Fig. 8.29, , =, , B R2, ., 2, , e, , Note: The induced emf in a rotating conductor does not depend on the shape of, the conductor., , Ex. 16, , A conducting rod PQ is rotated in a magnetic field, about an axis passing through O. The one end of the rod is at a, distance a and other end is at a distance b from O. Find induced emf, across the ends of the rod., , Sol., , The induced emf across the element of length dx is, de = Bvx dx = B( x) dx., , B, , OB, 2, , VO VA, 2, , B, 2, , VO – VA =, , B, 2, , 2, , 4R2, , 2, , ...(i), , 4R2, , 2, , ., , 2, , B, 4R2 2B R2 ., Ans., 2, The result obtained will remain same if conductor AB is straight., =, , B, , x dx, a, , B, , b2, , a2, , 2, , ., , Ans., , In fig. 8.31 shows a bent rod rotating about its end O in, , a plane perpendicular to the magnetic field B . The part OA of the, rod is non-conducting while the part AB is conducting. Find the, induced emf between the ends A and B., , Sol. Suppose straight length of the rod is, , . The distance of the end B, , from O, OB2 =, , VO – VB =, , VA – VB =, , b, , Ex. 17, , Here, , VO VB, , B, B, 2, OA, 2, 2, Substituting these values in equation (i), we get, , Fig. 8.30, , =, , VA – VB =, , and, , The emf across the whole rod, e =, , The potential difference between A and B can be calculated as :, , OA2, , 2R, , 2, , 2, , 2R, , Ex. 18 A uniform wire of resistance per unit length is bent, into a semicircle of radius a. The wire rotates with angular velocity, in a horizontal plane about a vertical axis passing through C. A, uniform magnetic field B exists in space in a direction perpendicular, to paper inwards., (a), (b), , Calculate potential difference between points A and D. Which, point is at higher potential ?, If points A and D are connected by a conducting wire of zero, resistance, then find the potential difference between A and, C., , 2, , Fig. 8.32, Fig. 8.31

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ELECTROMAGNETIC INDUCTION, , Sol., (a), , (b), , The distance of end A from C,, , 2a sin, , AC =, , 497, , When A and D are connected together, the current starts flowing, from A towards B. The resistance of the loop, , 2, , The distance of end D from C,, , 2a cos, , CD =, , 2, , The potential differences :, , and, , VC – VA =, , B, 2, , VC – VD =, , B, CD, 2, , ThusVA – VD, , 2, , =, , VC VD, , VC VA, , =, , B, 2, , AC 2, , =, , B, 2, , 2a cos, , B, , 2a, , =, =, , 8.7 INDUCED, , Fig. 8.33, R = (length of wire AD), =, a ., Thus current in the closed loop,, , 2, , AC, , CD 2, , i =, , 2 B a 2 cos, 2 B a cos, . Ans., a, The equivalent circuit is shown in fig. 8.33 (c)., The resistance R1 = (a ) ., The p.d. between A and C now becomes, VA – VC = e1+ iR1, =, , 2, , 2, , 2a sin, , 2, , 2, , 2, 2, = 2 B a sin, , 2, , cos, , 2, , V A VD, R, , 2, , 2, , sin, , 2, , 2B a2 cos, , 2 B a cos, 2, , Ans., , =, , 2 B a 2 sin 2, , 2, , cos, , ELECTRIC FIELD, , Consider a metal ring is placed in a uniform external magnetic field as shown in fig. 8.34, (a). The field extends upto cylindrical region of radius R. If we increase the intensity of, the field, the magnetic flux through the ring will then change and by Faraday s law an, induced emf and thus induced current will flow in the ring. If there is a current in the metal, ring, an electric field must be present at various points within the ring, and it must have, been produced by the changing magnetic flux. This induced electric field En is just as, real as an electric field by static charges. Thus we can say that changing magnetic field, produces an electric field., If we replace the metal ring by a hypothetical circular path of radius r and magnetic field, is increasing at the constant rate dB/dt, the electric field induces at various points, around the circular path. Due to symmetry it will be tangent to each point of the path (see, fig. 8.34 b)., , Another form of Faraday's law, If En is the induced electric field, then work done by the force of this field in moving a, test charge q0 around the path is (q0E) (2 r)., Suppose e is the induced emf, then work done on the test charge in one revolution is eq0., Thus we can write, , or, , eq0, , =, , q0 En, , e, , =, , En 2 r, , =, , E d, , 2 r, , More generally En(2 r) can be written as, e, , a, , 2, , Fig. 8.34, , Ans.

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498, , ELECTRICITY & MAGNETISM, =, , d B, ,, dt, , En d, , =, , d B, dt, , En d, , =, , d B, dt, , According to the Faraday s law, , e, , Magnitude of En, We have;, (1), , For r < R :, En, or, , Fig. 8.35, , (2), , 2 r, , =, , r2, , d B, dt, , En, , =, , r dB, 2 dt, , For r R :, En, , or, , 2 r, , =, , R2, , d B, dt, , En, , =, , R2, 2r, , dB, ., dt, , Difference between electric potential and induced emf, Electric field can be produced by static charge or by changing magnetic field. And, electric field produced by any means exert forces on charged particles. The electric field, produced by static charges never form closed loop, but induced electric fields form, closed loop. The field produced by a static charge is of conservative nature while induced, electric field is of non-conservative nature, and so its line integral over a closed path is, not zero. Thus we can say that electric potential has meaning only for fields produced by, static charges; it has no meaning for induced electric field., , Note:, 1., 2., , Lenz s law is consistent with the principle of conservation of energy., The induced emf in Faraday s law does not have a chemical or electrostatic, origin. The work done in carrying a charge around the closed loop is not zero. If, , En is the induced electric field associated with an induced emf, then, E d, , 0. Thus here induced electric field is not a conservative field., , Ex. 19, , A conducting rod is placed in a changing magnetic field, as shown in fig. 8.36. Find induced emf across its length., , Sol., Let us consider a rod of length is placed in magnetic field, which is, perpendicular to the plane of paper and pointing into it. The field is, changing at the constant rate of dB/dt., Now consider a point A in the rod, it is at a distance r, , d, cos, , from the, , centre of field. The induced electric field at A,, En, , =, , r dB, =, 2 dt, , d, 2cos, , dB, dt, , Fig. 8.36

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ELECTROMAGNETIC INDUCTION, , Sol., , The component of field along the rod, E =, , At any instant,, , 499, , = t and the flux through the loop, , d dB, 2 dt, , En cos, , Potential difference between ends of the rod, e =, , d, 2, , E, , dB, dt, , Case (1) : If rod is placed along the diameter, d = 0,, , e = 0., , Case (2) : If the rod is outside the changing field, E =, , Ex. 20, , Fig. 8.38, B = NBA cos, = NBa2 cos t, , R 2 dB, cos ., 2d dt, , Figure 8.37 shows a conducting loop a b c d e f a made of, , Induced emf,, , six segments ab, bc, cd, dc, ef and fa, each of length . Each segment, makes a right angle with the next so that a b c is in the xy- plane, c, d e in xy- plane and e f a in the yz- plane. A uniform magnetic field, B exists along the x-axis. If the magnetic field changes at a rate, dB, , find the emf induced in the loop., dt, , Since magnetic field is along x-axis, there is flux only across the face, 2, , d NBa 2 cos t, , =, =, , dt, NBa2, , sin t, , Ans., , Note:, , Sol., oafeo, B = B, , d B, dt, , e =, , If the same loop is rotated about an axis parallel to one of its sides and, passing through centre of square, the induced emf will be same. i.e.,, , . The flux across any other face is zero., , Fig. 8.39, , Ex. 22, , Figure 8.40 shows a straight, long wire carrying a, current i and a rod of length coplanar with the wire and, perpendicular to it. The rod moves with a constant velocity v in a, direction parallel to the wire. The distance of the wire from the, centre of the rod is x. Find the motional emf induced in the rod., , Fig. 8.37, The induced emf,, , Sol., e =, , =, , d B, dt, 2, , dB, ., dt, , d B, , The magnetic field due to wire at a distance x from the wire,, , 2, , dt, , Ans., , Ex. 21 A square loop of edge 'a' having N turns is rotated with, a uniform angular velocity about one of its diagonals which is, kept fixed in a horizontal position. A uniform magnetic field B, exists in the vertical direction. Find the emf induced in the coil., , Fig. 8.40, Bx =, , 0, , 2, , i, x

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500, , ELECTRICITY & MAGNETISM, , The induced emf across the ends of the element dx,, de = Bxvdx, =, , As, , i, v dx, x, , 0, , 2, , i = CBa ., The resisting force on wire exerted by magnetic field, , Emf induced across whole length of the rod, x, , 2, , Now by Newton s second law, F–Frest = m a, , 2, , or, x, , =, , i0v, dx, x, , 0, , 2, , 0, , Fres = Bi = B(CBa ), = B2Ca 2, , 2, , e =, x, , dv, is the acceleration a,, dt, , 2, , x, , a =, , dx, x, , iv, , F – B2Ca 2 = m a, , Ex. 24, , 2, , F, m, , B 2C, , 2, , ., , Ans., , Figure 8.42 shows a rectangular conducting loop of, , resistance R, width L, and length b being pulled at constant speed is, =, , 0, , 2, , iv n x, , x, x, , 2, , x, 0, , =, , =, , 2, , iv n, x, , 0, , 2, , iv n, , through a region of width d in which a uniform magnetic field B, is set up by an electromagnet., , 2, , 2x, 2x, , (a), , Plot the flux B through the loop as a function of the position, x of the right side of the loop. Assume that L = 40 mm, b = 10, cm, d = 15 cm, R = 1.6, , 2, 2, , ., , , B = 2.0 T, and v =1.0 m/s., , (b), , Plot the induced emf as a function of the position of the loop., , (c), , Plot the rate of production of thermal energy in the loop as a, function of the position of loop., , Ans., , Ex. 23 A wire of mass m and length, , can freely slide on a pair, of parallel, smooth, horizontal rails placed in a vertical magnetic, field B (fig. 8.41). The rails are connected by a capacitor of, capacitance C. The electric resistance of the rails and the wire is, zero. If a constant force F acts on the wire as shown in the figure,, find the acceleration of the wire., , Fig. 8.42, , Sol., , (a), Fig. 8.43, , Fig. 8.41, , Sol., Let any instant the velocity of wire is v, the induced emf,, e = Bv, and charge on capacitor, q = Ce, = CBv ,, and current in the wire, i =, , dq, dt, , =, , CB, , d, CBv, dt, , dv, dt, , (b), , Fig. 8.44, (c), , ., Fig. 8.45

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ELECTROMAGNETIC INDUCTION, (a), , The magnet flux is zero when the loop is not in the field; it is B =, BL x, when loop is entering into the field. It increases linearly and, becomes maximum when x = b, i.e.,, , B, , = BL b = 2, , 40 10, , 3, , 2, , 8 mWb., , BL b, , x d, , 10 10, , When the loop is leaving the field, it is, , B, , 501, , Sol., Take an element of thickness dx, its area dA = Hdx. The magnetic flux, through this area, d, , =, , BdA cos0, , B dA, , BdA, , = B(Hdx), = 4t2x2 Hdx, The total flux through the entire loop, =, , 4t 2 H, , d, , 3.0, , x 2dx, , 0, , Fig. 8.46, (b), , =, , d B, , which can be written as, dt, , The induced emf is equal to, , d B, dt, , e =, , d B dx, dx dt, , d B, is the slope of curve (a)., dx, From 0 to 10 cm, , d B, dx, , =, , 8, 0.10, , 80 1, , e =, From 10 cm to 15 cm, , 80 mV ., , (c), , We have, , P =, , Pmax =, and, , Ex. 25, , 72t 2 (H = 2.0 m), 0, , =, t =, e =, =, , d 72t, d, dt, dt, 144 t, 0.10 s,, 144 0.10, 14.4 V, , 2, , Ans., , Ex. 26, , A square loop of side 12 cm with its side parallel to x, and y axis is moved with a velocity of 8 cm/s in the positive xdirection in a magnetic field pointing towards positive z-direction., The field has a gradient of 10–3 T/cm. Find the magnitude of the, induced emf if field changes at the rate of 0.1 T/s., , = –80, 80 1, , e =, , 3.0, , Direction of induced emf : The flux of B through the loop is into the, page and is increasing in magnitude with time. According to Lenz s law,, the field of the induced current must oppose this increase and so is, directed out of the page. Therefore the direction of induced current or, emf is counterclockwise in the loop (see figure)., , d B, = 0, dx, e = 0., From 15 cm to 25 cm, d B, dx, , e =, At, , 80, , x3, 3, , Now by Faraday s law, the magnitude of induced emf, , d B, v, dx, , where, , 4t 2 H, , Sol., , 80mV, , We know that, the induced emf in this case is given by, , e2, R, , |e| =, , 80 10, 1.6, , Pmin = 0., , 3, , =, , 2, , 4 mW ,, , B, t, , 144 10, , v, , B, x, , 4, , 0.1 8 10, , = 155 mV., Ans., , Figure 8.47 shows a rectangular loop of wire immersed, , in a non-uniform and varying magnetic field B that is perpendicular, to and directed into the page. The field s magnitude is given by B =, 4t2x2, with B in tesla, t in second, and x in metre. The loop has width, W = 3.0 m and height H = 2.0 m. What are the magnitude and, direction of the induced emf around the loop at t = 0.10 s ?, , Fig. 8.47, , A, , 3, , Ans., , Ex. 27 A wire PQ of length , mass m and resistance R slides on, a smooth, thick pair of metallic rails joined at the bottom as shown, in fig. 8.48. The plane of the rails makes an angle with the, horizontal. A vertical magnetic field B exists in the region. If the, wire slides on the rails at a constant speed v. Show that, B =, , mgRsin, v 2cos2, , Fig. 8.48, , .

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502, , ELECTRICITY & MAGNETISM, , Sol., The induced emf across the wire PQ, e =, , Bv sin(90, = Bv cos, , ), , (b), , e = B(a sin30° × 2)u, = Bau, enet = 2e = 2B a u, The induced current in the frame, , The current in the wire,, , i =, i =, , e, R, , Bv cos, R, , (c), , The magnetic force on the wire, F = Bi ,, its direction is as shown in fig. 8.48, The wire moves with constant velocity,, mg sin = F cos, or, mg sin = Bi cos, or, , mg sin, , =, , B =, , mgR sin, v, , 2, , cos 2, , 1, , q =, , ., , Proved., , Ex. 28, , Figure 8.49 shows a metallic square frame of edge a in, a vertical plane. A uniform magnetic field B exists in the space in a, direction perpendicular to the plane of the figure. Two boys pull, the opposite corners of the square to deform it into a rhombus., They start pulling the corners at t = 0 and displace the corners at a, uniform speed u., (a) Find the induced emf in the frame at the instant when the, angles at these corners reduce to 60°., (b) Find the induced current in the frame at this instant if the, total resistance of the frame is R., (c) Find the total charge which flows through a side of the frame, by the time the square is deformed into a straight line., , 2, , R, R, 2), =, BA, =, B(a, 1, 2 = 0, , where,, , cos, , Ans., , Total charge flows,, q =, , Bv cos, R, , B, , 2 Bau, R, , enet, R, , Ans., , Ba 2, ., R, , Ans., , Ex. 29 The rectangular wire frame, shown in fig. 8.51 has a, width , mass m, resistance R and a large length. A uniform magnetic, field B exists to the left of the frame. A constant force F starts, pushing the frame into the magnetic field at t = 0., (a) Find the acceleration of the frame when its speed has, increased to v., (b) Show that after some time the frame will move with a constant, velocity till the whole frame enters into the magnetic field., Find the velocity, (c) Show that the velocity at time t is given by, v =, , v0 1 e, , Ft / mv0, , Fig. 8.51, , Sol., (a), Fig. 8.49, , e = Bv , and the induced current i =, , Sol., (a), , The induced emf in the loop, when its left side enters into the field, , e, , the restoring force exerted, R, , by magnetic field, , The situation is shown in fig. 8.50., , F =, , Bi, , =, , B, , B, , e, R, , Bv, R, , Bv, R, , Now by Newton s II law, F – F = ma, or, , F, , B 2v, R, , 2, , = ma, , a =, Fig. 8.50, The loop can be split into two identical parts. The emf across each, part will be, , (b), , FR, , v 2B2, mR, , Ans., , With the increase in speed, the acceleration of frame decreases. At, a particular value of v0 the acceleration of the frame becomes zero., There after frame will move with constant velocity

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ELECTROMAGNETIC INDUCTION, 0 =, , which gives, (c), , Let us consider an element of dy at a distance y from the apex. The, magnetic flux associated, d B = B(2x × dy), ...(i), We have, y = c x2, , FR, 2 2, , B, , We have,, , v, , =, , FR v B, mR, , or, , 0, , B, mR, , or, , or, or, , or, , t =, , 2y, a, , or, where, , FR v 2 B 2, mR, , =, , FR, mR, , B, t, mR, , or, , v, v0, , 1, , v, v0, , =, , B, t, mR, , =, , Ft, mv0, , =, , v =, , e, , dy, dt, , = B × 2x × at, B, , 2, , |e| =, , By, , 8a, c, , y, c, , a, , 2y, a, , Ans., , Ex. 31 The magnetic field in the cylindrical region is shown in, , B, t, mR, , fig. 8.53 increases at a constant rate of 20.0 mT/s. Each side of the, square loop abcd and defa has a length of 1.00 cm and a resistance, of 4.00 . Find the current (magnitude and sense) in the wire ad if, (a) the switch S1 is closed but S2 is open, (b) S1 is open but S2 is closed,, (c) both S1 and S2 are open, (d) both S1 and S2 are closed., , 2 2, , FR, , B 2x, , 2 2, , =, , 2 2, , B, FR, , n 1, , =, , =, n, , ...(ii), , = B × 2x × v, v = at, d B, dt, , 2 2, , or n, , 1 2, at, 2, , 0, , FR v B, mR, , FR 1 v, , y =, , d B, dt, , 2 2, , n, , In time t1, , From equation (i), we get, , = –t, , 2 2, , n, , =, , v, , v 2B2, mR, , FR, , dt, , =, , 2 2, , n, , FR v 2 B 2, mR, , y, c, , which givesx, , t, , dv, , 0, , or, , Sol., , v 2B2, mR, , v = v0 =, , dv, dt, , or, , FR, , 503, , Ft / mv0, , v0 1 e, , Ft / mv0, , Proved, , Ex. 30 A wire bent as a parabola y = c x2 is located in a uniform, magnetic field of induction B, the vector B being perpendicular to, the plane -xy. At the moment t = 0 a connector starts sliding, translationwise from the parabola apex with a constant acceleration, a. Find the emf of electromagnetic induction in the loop thus formed, as a function of y., , Fig. 8.53, , Sol., (a), , Switch S1 is closed and S2 is open : The flux is confined in loop, adef., The induced emf in the loop, |e| =, , Fig. 8.52, , d B, dt, , d BA, dt, , dB, dt, , =, , A, , =, , 1 1 10, , = 2 × 10–6 V, , 4, , 20 10, , 3

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504, , ELECTRICITY & MAGNETISM, , The current in wire ad,, i =, , B0 y, v0 dy, L, , =, , e, R, , The induced emf across whole length of the rod, , where R is the resistance of the loop = 4 4 16, , L, , i =, , e, R, , 2 10, 16, , 6, , = 1.25 × 10–7 A, , B0, v0 y dy, L, 0, , e =, Ans., , B0v0 L, 2, , =, , (b), , (c), , (d), , The direction of current in ad is to oppose the increasing flux., Thus it is from a to d., S1 is open but S2 is closed., Now flux is associated with closed loop abcd therefore induced, emf will occur in it., i = 1.25 × 10–7 A from d to a Ans., When both the switches are open : No flux is associated with any, loop., e = 0 and i = 0., When both the switches are closed : The flux is associated with, both the loops, and there is induced emf in both the loops. The, equivalent circuit is as shown. Since e1 = e2 = e., , Ans., , Ex. 33 A rectangular metallic loop of length and width b is, placed coplanerly with a long wire carrying a current i. The loop is, moved perpendicular to the wire with a speed v in the plane, containing the wire and the loop. Calculate the emf induced in the, loop when the rear end of the loop is at a distance a from the wire., , Fig. 8.56, , Sol., Method I : Consider an element of width dx in the loop. The magnetic, flux across the element, d B = Bx (bdx), , Fig. 8.54, The p.d. across a and d = 0, Current in wire ad, i, , = 0., , B0, y, L, where L is a fixed length. A conducting rod of length L lies along yaxis between the origin and the point (0, L, 0). If the rod moves with, , Ex. 32 The magnetic field in a region is given by, , a velocity v, , B, , k, , 2, , The total flux across the loop, x, , B, , v0i , find the emf induced between the ends of the, , rod., , 0, , =, , 2, , =, , Sol., Consider a small element of width dy at a distance y from the origin. The, induced emf across it, , 2, , x, , 0, , ib n x x, , 0, , ib, , 0ib, , 2, 0ib, , =, , 2, , nx, , d, dt, , n x, , nx, , d, dx, , n x, , nx, , 0ib, , 1, x, , 2, 2, , =, , n x, , 0ib, , e =, =, , dx, x, , d B, dt, , e =, =, , when x = a,e, , ib, x, , 2, , =, Now,, , Fig. 8.55, de = B v (dy), , i, bdx, x, , 0, , =, , 1, a, , 0ib v, 2 a a+, , 1, a, , 1, x, , v, , v, Ans., , dx, dt

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505, , ELECTROMAGNETIC INDUCTION, Method II : When x = a , the position of the loop is shown in fig. 8.57., The magnetic field at left arm of the loop, , The same current will flow across OA from A to O., Therefore the force exerted by magnetic field on the rod, Fm = B i a, =, , 0 i, 2 a, Similarly, magnetic field at right arm of the loop, , B1 =, , =, , i, , 0, , 2, , B 2 a3, 2R, , The rod is to be rotated with constant angular velocity so the, torque of net force about centre of loop is equal to zero. i.e.,, (F – Fm) × OD = 0, or, F = Fm, , Fig. 8.57, , B2 =, , B a2, a, 2R, , B, , B 2 a3, ., 2R, , Ans., , a, , Now induced emf across left and right arms are, , Fig. 8.58, e1 = B1v b and e2 = B2v b, enet = e1– e2, = vb [B1– B2], i, 0 i, 0, = vb 2 a 2 a, 0i, , =, , b v, , ., , 2 a a, , (b), , Ans., , Ex. 34 (a)Figure 8.59 shows a conducting circular loop of radius, , (b), , a placed in a uniform, perpendicular magnetic field B. A thick, metal rod OA is pivoted at the centre O. The other end of the, rod touches the loop at A. The centre O and a fixed point C on, the loop are connected by a wire OC of resistance R. A force is, applied at the middle point of the rod OA perpendicularly, so, that the rod rotates clockwise at a uniform angular velocity, . Find force., Suppose the wire connecting O and C has zero resistance but, the circular loop has a resistance R uniformly distributed, along its length. The rod OA is made to rotate with uniform, angular speed as shown in the figure. Find the current in, the rod when, , AOC, , Fig. 8.60, The effective resistance between A & C, , 90 ., , R =, , i =, , =, , i=, , e, R, , B a, 2R, , 2, , e, R', , B a2, 2, 3R, 16, , 8 B a2, ., 3, R, , Ans., , At any instant, the total magnetic flux through the coil is, = NBA cos t, = N(B0 sin t) A cos t, =, , When rod rotates with constant angular velocity, the induced emf, across its ends, B a2, e = VO – VA =, 2, The current in the wire OC,, , 3R, 16, , Ex. 35 A coil of N turns with the cross-sectional area A is placed, inside a long solenoid. The coil is rotated at a constant angular, velocity, around the axis coinciding with its diameter and, perpendicular to the axis of the solenoid. The magnetic field in the, solenoid varies according to the law B = B0 sin t. Find the emf, induced in the coil, if at the instant t = 0 the coil axis coincided with, the axis of the solenoid., , Sol., (a), , R, 4, R, 4, , Now current in the rod OA, , Sol., Fig. 8.59, , 3R, 4, 3R, 4, , By Faraday s law, , e =, , 1, NB0 A sin 2 t, 2, , d, dt, , d, , 1, NB0 A sin 2 t, 2, dt, , 1, NB0 A 2 cos 2 t, 2, = –NB0A cos2 t. Ans., , =

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506, , ELECTRICITY & MAGNETISM, , Ex. 36, 20, , A wire frame of area 3.92×10–4 m2 and resistance, , is suspended freely from a 0.392 m long thread. There is a, , Here, , uniform magnetic field of 0.784 tesla and the plane of wire-frame, is perpendicular to the magnetic field. The frame is made to, oscillate under gravity by displacing it through 2×10–2 m from its, initial position along the direction of magnetic field. The plane of, the frame is always along the direction of thread and does not, rotate about it. What is the induced emf in wire frame as a function, of time ? Also find the maximum current in the frame., , Sol., , 9.8, 0.392, , =, , e, , =, , 1, 2, , 0.784, , 5 s 1,, 2 10 2, 0.392, , x0, , 0, , and, , At the intant, when thread makes an angle with the vertical, the magnetic, flux through the frame, , g, , =, , 3.92 10 4, , 2 10 2, sin10t, 0.392, , 5, , = 2 × 10–6 sin10t., Maximum induced emf, emax = 2 × 10–6 V, and, , imax =, , emax, R, , 2 10, 20, , 6, , 10, , 7, , A Ans., , Ex. 37, , An infinitesimally small bar magnet of dipole moment, M is pointing and moving with the speed v in the x-direction. A, small closed circular conducting loop of radius a and negligible, self-inductance lies in the yz-plane with its centre at x = 0, and its, axis coinciding with the x-axis. Find the force oppositing the motion, of the magnet, if the resistance of the loop is R. Assume that the, distance x of the magnet from the centre of the loop is much greater, than a., , Fig. 8.61, = B A cos, The induced emf, e =, , d, dt, , d, dt, , BA sin, , Sol., Suppose magnet is at a distance x from the centre of the loop. The, magnetic field due to the magnet at the centre of the loop, , ...(i), , For small , sin, ., The restoring torque on the frame, = –mg( sin ), , d2, , or, , dt 2, , =, , =, , =, , mg, I, , mg, m, , 2, , Fig. 8.62, , g, , 0 2M, ., 4 x3, The magnetic flux due to the magnet, , B =, , g, , Putting, , d2, , =, , 2, we get, , = BA =, 2, , = 0, dt 2, This represents SHM, and so, =, 0 sin t, Substituting value of in equation (i), we have, e =, , BA, , 0 sin, , =, , BA, , =, , 1, BA, 2, , 0, , t, , sin t, 0, , 2, , ...(ii), , 0, , 4, , 2M, x3, , a2, , Induced emf in the loop, d, dt, , e =, d, dt, , 0 sin, , 0, , cos t, , t, , As, sin 2 t, , 0, , =, dx, dt, , Ma 2, 2, , = v,, e =, , 3 0 Ma 2v, 2x4, , d, 0 2M, dt 4 x3, , 3 dx, x 4 dt, , a2

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ELECTROMAGNETIC INDUCTION, Induced current,, , i =, , y = vt., , 3 0 Ma 2v, e, =, R, 2x4R, , The corresponding x value is, x = y tan45° = y., , If Mloop is the magnetic moment of the loop, then, Mloop = i A, =, , 3 0 Ma 2v, , =, , 3, , 0 Ma, 4, , 4, , The distance, , (a), , The opposing force, , F =, , or, , 6 M magnet M loop, , =, , = BA = B (xy), , B, , = B (vt) × vt, = Bv2t2., , At t = 3.0 s,, =, , B, , x4, 0, , 6M, , 4, , x4, , 9, 4, , 2, , 0, , 3, , xy, , Thus the flux through the triangle, , v, , 4, , 1, 2x y, 2, , The area of the triangle OPQ =, , 2x R, 0, , PQ = 2x = 2y, = 2vt., , a2, , 2x4R, , 507, , 4, , 0 Ma v, , (b), , 2 x4 R, , M 2 a 4v ., R x8, , B, , =, , 0.35, , 5.20, , 2, , 32.0, , 85.17Wb, , The induced emf around the triangle, e =, , Ans., , Ex. 38, , =, , Two straight conducting rails from a right angle where, their ends are joined. A conducting bar in contact with the rails, starts at the vertex at time t = 0 and moves with a constant velocity, of 5.20 m/s along them, as shown in fig. 8.63. A magnetic field with, B = 0.35 T is directed out of the page. Calculate, , d B, dt, , d Bv 2t 2, dt, , = Bv2 × 2t, = 2B v2t, =, , 2 0.35, , 2, , 5.20 t, , = 18.93 t, At, (c), , t = 3s,, , Induced emf, , e = 18.93 × 3 = 56.79 V, e = 18.93 t., , On comparing with e = atn, we get n = 1, , Ans., , Ex. 39, , (a), (b), (c), , Fig. 8.63, The flux through the triangle formed by the rails and bar at, t = 3.0 s and, the emf around the triangle at that time., If we write the emf e = atn, where a and n are constants, what, is the value of n ?, , A shaped as a semi-circle of radius a rotates about an, axis OO with an angular velocity in a uniform magnetic field of, induction B (see fig. 8.65). The rotation axis is perpendicular to the, field direction. The total resistance of the circuit is equal to R., Neglecting the magnetic field of induced current, find the mean, amount of thermal power being generated in the loop during a, rotation period., , Sol., Suppose joint of the rails is the origin of the coordinate axis. The distance, moved by the bar in time t,, , Fig. 8.65, , Sol., , a2, ., 2, If is the angle between magnetic field and normal to the loop at any time, t, then, =, t,, and flux, = BA cos t, The area of the loop,, , A =, , =, Fig. 8.64, , B, , a2, cos t, 2

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508, , ELECTRICITY & MAGNETISM, , According to Faraday s law, , T, , e =, =, , Pdt, , d, dt, , B, , Thus mean power, , a2, 2, , e, =, R, The intantaneous power generated, Induced current, , P = ei =, , sin t, , B, , i =, , Pmean =, , B, , a, 2, , 2, , T, a2, , B, , a2, sin t, 2R, 2, , 0, , 2, , =, , 2, , =, , sin t, R, , 8.8 SELF, , 2, , T, , 1, sin 2 t dt, R 0, T, , a2, , B, 2, , 2, , 1, . Ans., 2R, , INDUCTANCE, , We can define an inductor as an arrangement that can be used to produce a known, magnetic field. We can express the connection between capacitors and inductors as, follows :, Capacitor is to electric field, , C, , Inductor is to magnetic field, , q, V, , SI unit of C is C/V, , L, The product N, , N, i, , is called flux, , linkage., 1 C/V = 1 farad, , SI unit of inductance is T. m2/A., We call this as the henry (H), , Sumbol :, , 8.9 SELF, , Symbol :, , INDUCTION, , Whenever there is change in magnetic flux,electromagnetic induction will take place., If there is a time varying current in a coil the magnetic field of this current will also, change. And this leads to the variation of the magnetic flux through the coil, and hence, to the appearance of an induced emf., Fig. 8.66, , Thus, the variation of current in a circuit will cause an induced emf in this circuit. This, phenomenon is called self induction., In the absence of any ferromagnetics in the space surrounding the coil, the total flux, through the coil are proportional to the current i, and we can write, N B = Li, ...(1), where L is called the coefficient of self induction of the circuit. The inductance L depends, on the shape and size of the loop as well as on the magnetic properties of the surrounding, medium. The unit of inductance is henry.

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ELECTROMAGNETIC INDUCTION, Emf of self-induction, According to Faraday s law, a variation in the current of the circuit causes an induced, emf. Thus on differentiating equation (1), we get, , d N, , =, , dt, or, , e, , di, dt, , L, , L di, dt, , =, , ...(2), , Thus when current in the coil increases, the induced emf in the coil opposes the applied, emf, and so net emf of the circuit, Ldi, dt, , net, , =, , e, , net, , =, , Ldi, dt, , Fig. 8.67, , If current in the coil decreases, then the induced emf in the coil favours the applied emf,, and so net emf of the circuit, net, , L di, dt, , e, , =, , Steps for finding L, 1., 2., 3., 4., , Fig. 8.68, , Assume current i in the coil., Determine the magnetic field due to this current in the coil., Obtain flux linkage N B., Compare with N = Li to get L., , Calculation of self inductance, 1., , Circular coil :, Let us consider a circular coil of radius r and containing N-turns. Suppose it carries, a current i. The magnetic field due to this current, B, And total flux, , N, , B, , =, , 2r, = NBA, , =, , or, , =, , Now compare with N, , N, , 0 Ni, 2r, , 0, , N 2r i, 2, , 0, , N 2r, 2, , r2, , B = L i, we get, , L, , Note:, , 0 Ni, , =, , Inductance may be viewed as electrical inertia. It is analogous to inertia in, mechanics. It does not oppose the current, but it opposes the change in, current., , 509

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510, 2., , ELECTRICITY & MAGNETISM, , Solenoid : Consider a long solenoid of cross-sectional area A. Take a length near, the centre of this solenoid. The number of flux linkages for this section of the, solenoid is, N B = (n ) (BA), or, = n ( 0ni) A, Now compare with, N B = Li, we get, L = 0 n2 A, Inductance per unit length for a long solenoid near its centre, , L, 3., , =, , 0n, , Fig. 8.69, , 2A, , Toroid of circular cross-section :, Consider an air-core toroid of cross-sectional area A and mean radius r is closely, wound with N turns of wire. We neglect the variation of B across the crosssection, assuming its average value to the very nearly equal to be value at the, centre of the cross-section. Then the flux linkage, N B = N (BA), =, , 0 Ni, , N, , A, , 2 r, , Now compare with N B = Li, we get, L, , 4., , 0N, , =, , 2, , A, , 2 r, Inductance of toroid of rectangular cross-section : The magnetic field (assuming, uniform) at a distance r from the centre of toroid,, B, , =, , B, , =, , Fig. 8.70, , 0 Ni, , 2 r, where i is the current in toroid. The flux B over the entire cross-section must be, found out by integration., If h(dr) is the area of the elementary strip shown, then we have,, , B dA, b, , B h dr, , =, a, b, , =, a, , =, =, , Total flux linkage, , Therefore, , or, , N B, , N, , B, , i, , L, , =, , =, , =, , 0 Ni, , h dr, , 2 r, 0 Nih, , 2, , a, , 0 Ni, , h, , 2, 0N, , b, , 2, , dr, r, , n, , Fig. 8.71, , b, a, , ih n, , 2, 0N, , 2, , h n, , b, a, , 2, 0N, , 2, , h n, , 2, , b, a, , b, a

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ELECTROMAGNETIC INDUCTION, Energy in an inductor, A changing current in an inductor causes an emf, the source supplying the current must, maintain a potential difference between its terminals and hence must supply energy to, the inductor. When an inductor carries an instantaneous current i which is changing at, di, the rate dt , the induced emf is equal to L di , and the power P supplied to the inductor is, dt, di, Li, P = ei =, dt, The energy dU supplied in time dt is Pdt, or, dU = Pdt = Li di, and the total energy supplied while the current increases from zero to a final value I is, I, , U, , =, , L idi, 0, , or, , U, , =, , 1 2, LI ., 2, , Note: After the current has reached its final steady value, di 0, the power input, , dt, is zero. The energy that has been supplied to the inductor is used to establish the, magnetic field in and around the inductor, where it is stored as a form of potential energy, as long as the current is maintained. When current is reduced to zero, this energy is, returned to the circuit which supplied it. If the current is suddenly interrupted by opening a switch, the energy may be dissipated in an arc across the switch contacts., The energy can be considered as associated with the magnetic field itself, and a, relationship can be developed which is analogous to that obtained for electric field, energy., The self inductance of solenoid,, L = 0n2 A, , and, , U, , =, , 1 2, LI, 2, , =, , 1, 2, , =, , =, , 0n, , 0 nI, , 2, , 2, , A I2, , 2, , A, , 0, , B2, V, 2 0, , Thus the magnetic energy per unit volume, u, , 2., , =, , B2, ., 2 0, , In case of AC of frequency f and amplitude i0, the rate of change in current from, i, i0 to – i0 will be ;, i0, 2i0, 4i0, di, 4 fi0, t, dt T / 2, T, –i0, T/2, , 511

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512, , ELECTRICITY & MAGNETISM, Electrical and magnetic quantities, Electrical, 1., , Definition, C, , 2., , Dimensions, Constants, , 4., , Energy stored, , Ex. 40, , A 12 H inductor carries a steady current of 2.0 A. How, can a 60 V self-induced emf be made to appear in the inductor ?, , Sol. We know that, , L, , di, =, dt, , |e|, L, , 1, 2, , ue, , (a), , B, , i, a length, , L =µ0, µ0, , 1.26 µH / m, , UB, , 1 2, Li, 2, , uB, , B2, 2µ0, , = RC, , =, , L, R, , The induced emf in the inductor is given by, , L, , di, dt, , di, = +5 A/s,, dt, , For, , e = –1 × 5 = –5V., The net emf of the circuit, net = 10 – 5 = 5 V, , Ex. 41, , (b), , N, , L, , Sol., , 5 A/s., , A resistor of 1 and an inductor of 1 H are connected, in series across a source of 10V. Suppose the current in the circuit, is changing by the added device. Find the net emf of the circuit, when, , E2, , 0, , e =, , 60, 12, , q2, 2C, , 1, CV 2, 2, , Ue, , Time constant, , Thus current in the inductor should change at the rate of 5 A/s., , (a), , 8.85 pF / m, , di, dt, , |e| =, , di, = + 5 A/s,, dt, , 0, 0, , 5. Energy density, 6., , q, V, a length, , C, , 3., , Magnetic, , (b), , di, = – 5 A/s., dt, , Ans., , di, = –5 A/s,, dt, , For, , e = –1 × (–5) = 5V, Thus the net emf of the circuit, net = 10 + 5 = 15 V, , Ans., , Fig. 8.72, , 8.10 RL - DC, , CIRCUIT, , Consider a circuit having inductor and resistor in series with a direct current source of, emf ., As soon as the switch S is closed, the current in the resistor starts increasing. If the, inductor were not present, the current would rise rapidly to a steady value, , R, , . Because, , di, appears in the circuit; from Lenz s, dt, law, this emf opposes the rise of the current, which means it opposes the battery emf., Therefore, , of the inductor, however, self induced emf e, , e net, , =, , L, , L, , di, dt, , = iR, , or, , L, , di, dt, , =, , –iR, , or, , di, iR, , =, , dt, L, , Now by loop rule, , Fig. 8.73, , L, , di, dt, , ...(i)

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ELECTROMAGNETIC INDUCTION, Integrating both sides of above equation, we get, i, , di, iR, , 0, , n, , or, , or, , i, , iR, , =, , R, , n, , iR, , or, , n, , or, , n 1, , 0, , n, , =, , iR, , =, , i, , =, , R, /R, L/R, , Substituting, , n 1, , dt, L, , =, , i, i0, i, , t, L, R, t, L, R, , t, L, R, , .....(ii), , = i0, maximum current and, = , time constant in equation (ii), we have, =, , or, , t, L, , t, , t/, ), = i0 (1 e, , … (1), , Inductive time constant (t) :, For, , t, , =, , , we have, , i, , 1, = i0 1 e, , Fig. 8.74, , = 0.638 i0 0.64 i0, Thus the inductive time constant can be defined as the time in which rising current in the, circuit becomes 0.64 times the maximum current., , Energy stored, For any current i, the energy stored in the inductor is, U, , =, , 1 2, Li, 2, , 1, L i0 1 e, 2, , =, , 1 2, L i0 1 e, 2, , t, , 2, , 2, , t, , t 2, , or, , U, , = U0 1 e, , ., , … (2), , Decay of current, After establishing current in the circuit (i0 = / R), the source is disconnected from the, circuit. And the circuit is short circuited. The current in the circuit starts decaying. The, energy stored in inductor will change into thermal energy. The equation for decaying, current can be obtained by putting = 0 in previous equation (i), we have, , L, , di, dt, , = iR, , 513

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514, , ELECTRICITY & MAGNETISM, di, i, , or, , i, , Integrating above equation, we get, i0, , di, =, i, , i, , ni i, , or, , dt, L, R, , =, , dt, L, R, , t, , =, , 0, , Fig. 8.75, or, , ni – n i0, , n, , or, , i, i0, , t, , =, , t, R, , =, , t, , or, , i, , = i0 e, , … (1), , 1, , For t = , i i0 e, 0.37 i0 . Thus the inductive time constant can also be defined as, the time in which the decaying current becomes 0.37 times the maximum current., , Fig. 8.76, , Energy at any time :, , U, , 1 2, Li, 2, , =, , 1 2, Li0 e, 2, , =, U, , = U e, 0, , t 2, , 1, L i0 e, 2, , t, ', , 2t, , ; ', , … (2), , 2, , More about inductor, At t = 0, i = i0 (1–e0) = 0 and at t, , 2., , Ex. 42 A coil has an inductance of 53 mH and a resistance of, 0.35, ., (a) If a 12 V emf is applied across the coil, how much energy is, stored in the magnetic field after the current has built up to, its maximum value ?, (b) After how many time constants will half the maximum energy, be stored in the magnetic field ?, , Sol.(a), , (b), , i0 =, , R, The maximum energy stored, , 1 2, Li0, U0 =, 2, = 31 J, , U =, , 3, , 34.3, , Ans., , 2, , U0 1 e, , For U = U0 / 2, we have, U0, 2, , or, 2, , = i0, , t, , or, , 34.3 A, , 1, 53 10, 2, , i0 1 e, , We know that, energy at any time in the coil is given by, , The maximum current in the coil, 12, 0.35, , , i, , . It means inductor, R, offers infinite resistance initially and zero resistance after very long time : It becomes, ineffective when current becomes steady in the circuit., Inductor does not oppose the current, but it opposes the change in current. It, causes delay in growing or decaying of current. Therefore it also called electrical, inertia., , 1., , t, , e, t, , t, , =, , U0 1 e, , =, , 1, , 1, 2, , 2, , 0.293, , = 1.23, t = 1.23, , Ans.

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ELECTROMAGNETIC INDUCTION, , 8.11 THE LC-OSCILLATIONS, Consider a LC- circuit shown in fig. 8.77, a resistanceless inductor is connected between, the terminals of a charged capacitor. At the instant when connections are made, the, capacitor starts to discharge through the inductor. At a later instant, the capacitor has, completely discharged and the potential difference between its terminals has decreased, to zero. The current in the inductor has meanwhile establishes a magnetic field in the, space around it. This magnetic field now decreases, inducing an emf in the inductor in, the same direction as the current. The current therefore persists, although with decreasing, magnitude, until the magnetic field has disappeared and the capacitor has been charged, in the opposite sense to its initial polarity. The process now repeats itself in the reverse, direction, and in the absence of any energy losses, the charges on the capacitor surge, back and forth indefinitely. This process is called electrical oscillations. From the energy, state point, the oscillations of an electrical circuit consist of a transfer of energy back, and forth from electric field of capacitor to the magnetic field of the inductor, the total, energy associated with the circuit remaining constant. This is analogous to the transfer, of energy in an oscillating mechanical system from kinetic to potential, and vice versa., Let a capacitor C is given an initial charge Q and, at t = 0 is connected to the inductor of, , 1 2, L i at any time corresponds, 2, to the kinetic energy of the oscillating mass-spring system. The potential energy of the, , self inductance L. The magnetic energy of the inductor, , capacitor, , q2, 2C, , corresponds to the potential energy of the spring, , these energies of the system, is equal to initial energy,, , 1 2, kx . The sum of, 2, , Q2, . Therefore by conservation, 2C, , of energy, , 1, q2, Q2, L i2, =, ., 2, 2C, 2C, Differentiating above equation w.r.t. time, we get, 1, L 2i, 2, , di, dt, , 1, 2C, L, , or, , dq, = 0, dt, , 2q, di, dt, , q, C, , As, , Fig. 8.77, , = 0, i, , =, , dq, ,, dt, , d 2q, , q, = 0, ...(1), LC, dt, Compare above equation with differential equation of oscillations of mass-spring system,, i.e.,, 2, , d2x, dt, , 2, , 2, , x, , = 0, we get, , 1, ,, LC, This is called natural frequency of the LC circuit., =, , Time period, Also,, , T, q, , = 2 LC, = Q cos( t + ), , ...(2), , 515

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516, , ELECTRICITY & MAGNETISM, Comparison between electrical oscillations with the oscillations of mass spring system, , and, Fig. 8.78, , Fig. 8.79, , 8.12 DAMPED, , UE, , =, , q2, 2C, , UB, , =, , 1 2, Li, 2, , OSCILLATIONS IN AN, , Q2, cos 2, 2C, Q2, sin 2, 2C, , RLC, , t, , t, , CIRCUIT, , In the previous case we have assumed that the LC- circuit contains no resistance. This is an, idealisation, of the circuit for every real inductor there is resistance associated with the, windings, and there may be resistance in the connecting wires as well. The effect is to, dissipate the electromagnetic energy and convert it to heat; thus, resistance in an electric, circuit plays a role analogous to that of friction in the mechanical system., Suppose an inductor of self-inductance L and a resistor of resistance R are connected in, series with the capacitor. If the capacitor is initially charged, it starts discharging at the, instant the connections are made, but because of i2R losses in the resistor, the energy of, the inductor when capacitor is completely discharged is less than the original energy of the, capacitor. In the same way, the energy of the capacitor where the magnetic field has collapsed, is still smaller, and so on. If the resistance R is relatively small, the circuit oscillates, but with, damped harmonic motion. As R is increased, the oscillations die out more rapidly. At a, sufficiently large value of R, the circuit no longer oscillates and is said to be critically, damped. For still larger resistance it is overdamped., Using loop rule to the circuit yields the equation :, Rate of dissipation of electromagnetic energy = power generated as heat, or, or, , d q2, dt 2C, 2q dq, 2C dt, , or, , or, , d 2q, dt, , 2, , L, 2, , 1, L i2, 2, , = –i2 R, , di, dt, , = –i2 R, , q, d 2q, i Li 2, C, dt, , = –i2 R, , 2i, , R dq, L dt, , q, LC, , = 0, , ...(1)

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ELECTROMAGNETIC INDUCTION, , 517, , The above equation is analogous to damped harmonic oscillator of mechanical system., i.e.,, , d2x, 2, , b, , dx, dt, , dt, The solution of equation (1) can be, , kx, , Here,, , = 0, , q, , = Qe, , d, , =, , Rt, 2L, , cos, , 1, LC, , R, 2L, , dt, , ...(2), , 2, , , is called damed frequency, , Critical damping, 4L the quantity under the, C, underroot becomes zero, and the case is called critical damping., As R increases, , becomes smaller and smaller; when R 2, , 8.13 MUTUAL, , Fig. 8.80 Variation of charge of capacitor, with time., , INDUCTION, , Consider two coils placed close together, a steady current i in one coil will set up a, magnetic flux B linking with the other coil. If we change i with time, an emf will appear, in the second coil. This is called mutual induction., , Note:, For ease of representation, the two coils shown in fig. 8.81 are not actually drawn as, close-packet : In close-packet coils the turns have the same flux through them., Consider two circular closed packed coils are placed near each other and sharing a, common central axis. There is steady current i1 in coil 1, set up by the battery in the, external circuit. This current produces a magnetic field B1, which produces magnetic, flux N2 21 in coil 2, linked by its N2 turns., We can define the mutual inductance M21 of coil 2 with respect to 1 as, M 21, , =, , N 2 21, i1, , or it can be written as, If i1 changes with time, then, , M21 i1, , Fig. 8.81, , = N 2 21., , dii, d 21, = N2, dt, dt, According to Faraday s law, the right side of the equation is equal to –e2. Thus, M 21, , e2, , M 21, , dii, dt, , =, , M12, , di2, dt, , =, , M, , =, , By doing similar treatment for first coil, we get, e1, By experiments, M12 = M21 = M. Thus, e1, , di2, dt, , ...(1)

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518, , ELECTRICITY & MAGNETISM, and, , e2, , =, , M, , di1, dt, , ...(2), , The SI unit of M is henry., The coefficient of mutual induction depends on the shape, size, and mutual arrangement, of the coils, as well as the magnetic permeability of the medium surrounding the coils., , Reciprocity theorem, Calculations show (and experiments confirm) that in the absence of material medium, between the coils, the coefficients M12 and M21 are equal :, M12 = M21, This property of mutual inductance is called the reciprocity theorem. Because of this, reason, we do not have to distinguish between M12 and M21 and can simply speak of, the mutual inductance of two circuits., , Steps for finding M, 1., 2., 3., 4., , Imagine current i1 in the first coil., Determine magnetic field due to i1 in the second coil., Obtain flux linkage, N2 2 = N2 (B1A2)., Compare with, N2 2 = M i1 and find M., , Mutual induction between circular coils, Consider two circular close-packed coils, the smaller (radius R2, with turns N2) being, coaxial with the larger (radius R1, with N1 turns) and in the same plane. Assuming R1 >>, R2. Imagine a current i1 in the larger coil, the value of B 1 at the centre of coil,, B1 =, , 0 N1i1, , 2 R1, , Because we have assumed that R1 > > R2, we may take B1 to the magnetic field at all, points within the boundary of the smaller coil. Therefore the number of flux linkage for, the smaller coil is, N 2 2 1 = N2(B1) ( R22), =, Now compare with, Fig. 8.82, , N2, , 21, , M, , 0 N1i1, , N2, , 2R1, , R22, , = M i1, we get, =, , 0, , N1 N 2 R2 2, 2 R1, , Mutual induction between solenoid and coil, A long solenoid of length and cross-sectional area A is closely wound with N1 turns of, wire. A small coil of N2 turns surrounds at its centre. A current i1 in the solenoid set up, a magnetic field B at its centre of magnitude, B, Fig. 8.83, , =, , 0 ni1, , 0, , N1 i1, , The flux through the central section is equal to BA, and since all of this flux links with the, small coil, then, N 2 2 1 = N2(B1A2), =, , N2, , 0 N1 i1, , A

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ELECTROMAGNETIC INDUCTION, e1, , =, , M, , di, ', dt, , e2, , =, , L2, , di, dt, , e2, , =, , M, , di, dt, , The total induced emf, , =, , and, , Thus total inductance, , LTotal, , di, L1 L2, dt, = L1 + L2 – 2M, , 2M, , Ex. 43, , Two circular loops 1 and 2 whose centres coincide lie, in a plane (see fig. 8.89). The radii of the loops are a 1 and a 2., Current i flows in loop 1. Find the magnetic flux 2 associated by, loop 2, if a1< < a2., , 521, , Fig. 8.88, ....(2), the rectangular contour but along the boundary of the half-plane,, enveloping it at infinity. The magnetic field created by this current in the, region of the rectangular loop has a simple configuration as this is the, field of a straight current. Hence we can easily find the magnetic flux, through the rectangular contour, , =, a b, , Fig. 8.89, , Sol., , 2, , a, , The direct calculation of the flux 2 is clearly a rather complicated problem, since the configuration of the field itself is complicated. However, the, application of the reciprocity theorem greatly simplifies the solution of, the problem. Let us pass the same current i through loop 2. Then the, magnetic flux 1 created by this current through loop 1 can be easily, found, provided that a1 < < a2 : it is sufficient to multiply the magnetic, 0i, 2a2, , field B at the centre of the loop B, Thus, , 0, , =, , 2, , =, =, =, , by the area a12 of the circle., , =, , 0i, , i, dx, x, , n, , 2, , a, , b, a, , ., , Ans., , Ex. 45, , In the circuit shown in fig. 8.91, the emf of the source,, its internal resistance r and the inductances L 1 and L 2 of, superconducting coils are known. Find the currents established in, the coils after key K has been closed., , 1, , B2 A1, 0i, , 2a2, , a12 ., , Ans., , Ex. 44 A loop with current i has the shape of a rectangle. Find, the magnetic flux through the hatched half-plane (see fig. 8.90), whose boundary is at a given distance from the contour. Assume, that this half-plane and the loop are in the same plane., , Fig. 8.91, , Sol. Suppose i1 and i2 are the currents in the respective coils and i is, the current drawn from the battery, then, i =, , r, = i1 + i2, ...(i), Inductors L1 and L2 are in parallel, and so e1 = e2, Fig. 8.90, , Sol., In this case, the magnetic field of current i has a complex configuration,, and hence it is very difficult to calculate directly the flux in which we, are interested. However, the solution can be considerably simplified by, using the reciprocity theorem. Suppose that current i flows not around, , or, , L1, , di1, dt, , =, , L2, , di2, dt, , On integrating both sides, we have, L1 i 1 = L2 i 2, Solving equations (i) and (ii), we get, i1 =, , ...(ii), , L2, r L1, , L2

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522, , ELECTRICITY & MAGNETISM, , and, , i2 =, , L1, r L1, , Ans., , L2, , Ex. 46, , A small cylindrical magnet M is placed at the centre of, a thin coil of radius a, containing N turns (see figure). The coil is, connected to a galvanometer. The resistance of the circuit is R., After the magnet had been rapidly removed from the coil, a charge, q passed through the galvanometer. Find the magnetic moment of, the magnet., , Sol., , For coil B,, , LB =, , NB, i, , B, , 16000, , 0.8 10, , 3, , 5, , = 0.26 H, For the mutual induction, we can write, M =, , We know that, the charge flow, , NB, iA, , B, , N B 0.6, , A, , iA, , 16000 0.6 0.6 10, =, , 3, , 5, , =, , 1.15 H, , k =, , M, L1 L2, , Thus coefficient of coupling, Fig. 8.92, q =, where, , R, , =, =, q =, , =, , ,, , f, , ,, , ...(i), , R, , is the magnetic flux through the coil at the beginning of the process., The quantity can not be determined directly. This difficulty, however,, can be overcome by using the reciprocity theorem. We mentally replace, the magnet by a small current loop creating in the surrounding space the, same magnetic field as that of the magnet. If the area of the loop is A and, current in it is i, then its magnetic moment M = iA. According to reciprocity, theorem M12 = M21 and the problem is reduced to determine the magnetic, flux through the area A of the loop, which creates the same current i, but, flowing in the coil. Assuming the field to be uniform within the loop, we, have, =, Substituting the value of, , q =, , iA =, , L1 =, , 2500 mH. Thus, , 2500 =, , L1 + L2 + 2M, 1200 + 800 + 2M, , 2500 =, , LTotal =, , iA, , 2 aR, , 1200 mH, L2 = 800 mH., , LTotal =, , 250 mH., , In fig. 8.93 (b), , 2aqR, 0N, , Ans., , Ex. 47, , Two coils, A of 12500 turns and B of 16000 turns lie in, parallel planes so that 60% of flux produced in A links with B. It is, found that a current of 5A in A produces a flux of 0.6 mWb while the, same current in B produces 0.8 mWb. Determine (i) mutual, induction, (ii) coefficient of coupling., , Sol., For coil A,, , Sol.Given,, In fig. 8.93 (a),, , M =, , in equation (i), we have, 0N, , Ans., , Fig. 8.93, , or, , 0 Ni, A ...(ii), 2a, , BA =, , 0.586, , Ex. 48 Two coils with terminals T1, T2 and T3, T4 respectively, are placed side by side. When measured separately, the inductance, of the first coil is 1200 mH and that of second is 800 mH. With T2, joined to T3, the inductance between T1 and T4 is 2500 mH. What is, the mutual inductance between the two coils ? Also, determine the, inductance between T1 and T3 when T2 is joined to T4., , 0, , i, , 1.15, 0.15 0.26, , L1 + L2 – 2M, , =, , 1200 + 800 – 2 × 250, , =, , 1500 mH, , Ans., , Note:, In accordance with the total inductance, the total energy of the two coils, is the sum of energies due to their self induction and mutual induction, between them : Thus, U =, , 1, L1i12, 2, , 1, L2 i22, 2, , Mi1i2 ,, , when both the coils carrying currents in same direction, and, LA =, =, , NA, i, , A, , 0.15 H, , 12500, , 0.6 10, 5, , 3, , U =, , 1, L1i12, 2, , when current are in opposite directions., , 1, L2 i22, 2, , Mi1i2 ,

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ELECTROMAGNETIC INDUCTION, , 523, , Ex. 49 A rectangular conducting loop in the vertical xz-plane Ex. 50, has length L, width W, mass M and resistance R. It is dropped, lengthwise from rest. At t = 0 the bottom of the loop is at a height h, above the horizontal x-axis. There is a uniform magnetic field B, perpendicular to xz-plane, below the x-axis. The bottom and top of, the loop cross this axis at t = t1 and t = t2 respectively. Obtain the, expression for the velocity of the loop for the time t1, , t, , t2 ., , A thermocole vessel contains 0.5 kg of distilled water, at 30°C. A metal coil of area 5 × 10–3 m2, number of turns 100, mass, , 0.06 kg and resistance 1.6, is lying horizontally at the bottom of, the vessel. A uniform, time varying magnetic field is setup to pass, vertically through the coil at time t = 0. The field is first increased, from zero to 0.8 T at a constant rate between 0 and 0.2 s and then, decreased to zero at the same rate between 0.2 and 0.4 s. This cycle, is repeated 12000 times. Make sketches of the current through the, coil and the power dissipated in the coil as functions of time for the, first two cycles. Clearly indicate the magnitudes of the quantities, on the axes. Assume that no heat is lost to the vessel or the, surroundings. Determine the final temperature of the water under, thermal equilibrium. Specific heat of the metal = 500 J/kg-K and, specific heat of water = 4200 J/kg-K. Neglect the inductance of the, coil., , Sol., , The induced emf in the coil, , Fig. 8.96, , e =, , Sol., , For 0 t t1; there is no induced current and hence loop, falls freely. Thus its velocity is given by, v1 =, , For t1, , t, , gt1, , i =, , t2 ; the emf induces across the bottom side of the loop, which, , BvW, in counter clockwise direction., R, Due to this, it experiences an upward force,, 2, 2, BvW, W = B vW ., R, R, Now using Newton s second law for the motion of the loop, we have, , F = BiW = B, , mg, , B 2vW 2, R, dv, , or, , g, , B 2vW 2, mR, , =, , m, , =, , dt, , v1, , g, , B 2vW 2, mR, mR, , B 2W 2, , or, , n g, , n g, , B 2vW 2, mR, , g, , B 2vW 2, mR, , g, , B 2v1W 2, mR, , or, , or, , t, , dv, , v =, , mgR, B 2W 2, , NA, , dB, ., dt, , e, R, , NA dB, ., R dt, , dB, 0.8 0, =, dt, 0.2 0, Thus for the time interval 0 to 0.2 s,, Given, , 4 T / s., , 100 5 10, i =, , 1.6, , 3, , 4, , = –1.25 A., In the interval 0.2 to 0.4s, dB/dt is negative and so induced current will be, +1.25 A., Power dissipated, P = i2R = 1.252 × 1.6, = 2.5 W, The total energy dissipated in 12000 cycles, E = Pt, = 2.5 × (12000 × 0.4), = 12 × 103 J, , dv, dt, , On integrating above expression, we have, v, , d, dt, , Induced current, , 2 gh, , is BvW, and induced current i =, , N, , dt, , =, t1, , v, , B 2vW 2, mR, , n g, , = t – t1, v1, , B 2v1W 2, mR, , t t1, , = e, , v1, , mgR, B 2W 2, , =, , t t1, , B 2W 2, mR, , B 2W 2, mR, , t t1, , e, , Fig. 8.97, If T is the change in temperature of water, then by conservation of, energy, 12 × 103 = (m1c1+ m2c2) T, B 2W 2, mR, , Ans., , T =, , 12 103, m1c1 m2c2

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524, , ELECTRICITY & MAGNETISM, =, , Final temperature, , =, =, =, , (b), , 12 103, 0.5 4200 0.06 500, 5.6°C, Ti + T, 30 + 5.6 = 35.6°C, , Equation (i) can be written as, d = R(idt) + Ldi, or, d = R(dq) + Ldi, On integrating, we get, , d, , Ans., , Ex. 51 A metal bar AB can slide on two parallel thick metallic, , ii, , q, , f, , =, , R dq, , L di, , 0, , i, , 0, , ( f – i) = Rq + Li1, Charge flown during the time t = 0 to t = T will be, , rails separated by a distance .A resistance R and inductance L are, connected to the rails as shown in fig. 8.98. A long straight wire, carrying a constant current i0 is passed in the plane of the rails and, perpendicular to them as shown. The bar AB is held at rest at a, distance x0 from the long wire. At t = 0, it is made to slide on the, rails away from the wire. Answer the following questions :, , f, , q =, , Li1 ...(ii), R, , i, , R, , The change in flux can be obtained as, 2 x0, , f– i =, , 2 x0, , BdA, x0, , x0, , 0i0, , 2 x, , dx, , 0i0, n 2., 2, On substituting this value in equation (ii), we get, , =, , L i1, 0i0, n2, 2 R, R, When the bar is stopped, the induced emf becomes zero, and so, from equation (i), q =, , (c), Fig. 8.98, (a), , (b), , (c), , d, di, Find a relation among, i,, and dt , where i is the current, dt, in the circuit and is the flux of the magnetic field due to the, long wire through the circuit., It is observed that at time t = T, the metal bar AB is at a, distance of 2x0 from the long wire and the resistance R carries, a current i1. Obtain an expression for the net charge that has, flown through resistance R from t = 0 to t = T., The bar is suddenly stopped at time T. The current through, resistance R is found to be, , i1, at time 2T. Find the value of, 4, , L, in terms of the other given quantities., R, , di, If e is the induced emf and, is the rate of increase of current in, dt, the circuit, then by Kirchhoff s II law, we have, e iR, , L, , di, dt, , or, , or, , d, dt, , =, , 0, , di, dt, , e =, , iR, , L, , =, , iR, , L di, dt, , i1, 4, i1, , di, i, , =, , i1, , or, , or, , n, , R, dt, L, , R, L, , 2T, , dt, T, , n i i4, , =, , R, 2T, L, , 1, 4, , =, , RT, L, , 1, , or, , Sol., (a), , di, =, i, On integrating, we have, , n4 =, L, R, , =, , T, , RT, L, T, n4, , Ex. 52 Two infinitely, , Ans., , long parallel wires carrying current, i = i0 sin t in opposite directions are placed at a distance 3a apart., A square loop of side a of negligible resistance with a capacitor of, capacitance C is placed in the plane of wires as shown. Find the, maximum current in the square loop. Also sketch the graph showing, the variation of charge on the upper plate of the capacitor as a, function of time for one complete cycle taking anticlockwise, direction for the current in the loop as positive., , ...(i), , Fig. 100

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525, , ELECTROMAGNETIC INDUCTION, , Sol., Take an element of thickness dx at a distance of x from left wire. The, magnetic field at the position of element, B =, , =, , 0i, 2 x, , i, , 0, , 2, , 0i, , 3a, , 1, x, , 2, , x, , upward, , 1, 3a, , x, , The magnetic flux through the area of the element, d = B(dA) = B (adx), =, , 0i, , 1, x, , 2, , 1, 3a, , Ex. 53 A thin wire ring of radius a and resistance r is located, inside a long solenoid so that their axes coincide. The length of the, solenoid is equal to , its cross-sectional radius to b. At a certain, moment the solenoid was connected to a source of a constant voltage, V. The total resistance of the circuit is equal to R. Assuming the, inductance of the ring to be negligible, find the maximum value of, the radial force acting per unit length of the ring., , adx, , x, , The total magnetic flux through the loop, =, , =, =, , 0ia, , 2, 0ia, , 0ia, , 1, x, , a, , 1, 3a, , nx, , 2, n 2a, , 2, =, , 2a, , 0ia, , n 3a, , na, , Sol., , dx, , x, , x, , na, , 2a, a, , The self inductance of the solenoid, 2 2, L =, 0n b ,, where n is the number of turns per unit length of the solenoid. When, solenoid is connected to a source of a constant voltage V, a current is setup. The current is given by, , n 2a, , i =, , i =, =, , i0i0sin t,, 0, , tR, L, , The flux through the coil, , n 2a, , coil, , Given, , V, 1 e, R, , =, =, , |e| =, , i0 sin t a n 2a, , B Acoil, ( 0ni) a2, , d, , coil, , dt, , 0n, , di, dt, , a2, , The current in the coil, Induced emf, , |e| =, , d, dt, 0i0, , =, , a, , 2 di, 0n a, e, dt, i =, r, r, The force on unit length on the ring, F = B i × 1 = ( 0n i)i, , n 2a cos t, , The maximum value of induced emf, 0i0 a, , e0 =, , n 2a, =, , V, 1 e, 0n, R, , tR, , tR, L, , 0n, , a2, , r, , Ve L, R, , R, L, , Impedance of the capacitor, 2, , 1, Z =, C, Maximum value of the current, I0 =, The charge on the capacitor, q =, =, , e0, Z, , 0, , =, , 0i0 a, , 2, , C n 2a, , For maximum value of F,, Ans., or, , Ce, 0i0 a, , C n 2a, , cos t, , d, e, dt, , tR, L, , 1 e, , dF, dt, , tR, L, , q0 =, , According to the equation i = i0 sin t, current and hence flux in the loop, increases initially in upward direction, and so direction of induced current, makes upper plate of the capacitor positive., , 1 e, , tR, L, , = 0., , By doing this the maximum value of the term is found to be, 2, , C n 2a, , tR, L, , 0;, , Thus maximum value of charge, 0i0 a, , a 2V 2 n 2, e, rRL, , Fmax =, , =, , 0, , a2V 2n2, 4rLR, , 0a, , 2, 0, , 4r, , 1, . Thus, 4, , a 2V 2 n 2, , 0n, , 2, , b2 R, , 2 2, , V, , 4rRb2, , Ans.

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526, , ELECTRICITY & MAGNETISM, , 8.15 EDDYS CURRENT, So far we have considered only instances in which currents resulting from induced, emf s were confined to well-defined paths provided by the wires like coils or rings., However, one finds masses of metal located in changing magnetic field or moving in a, magnetic field, with the result induced currents circulate throughout the volume of the, metal. Because of their general circulatory nature, these are referred to as eddy currents., Consider a metal sheet entering into a magnetic field perpendicular to the plane of the, sheet. The magnetic flux through the sheet increases. The current induced in the body, of sheet in the form of eddys as shown in fig. 8.102., The direction of eddy currents are such that, they compensate the increasing flux (by, Lenz s law). Accordingly when metal sheet enters into the field, the direction of eddys, are counterclockwise. When it is leaving the field the direction of eddys will be clockwise., , 8.16 AC, Fig. 8.102, , GENERATOR OR DYNAMO, , It is used to convert mechanical energy into electrical energy., Construction : The main components of ac generator are :, (i) Armature coil : It consist of large number of turns of insulated copper wire wound, over iron core., (ii) Magnet : Strong permanent magnet (for small generator) or an electromagnet (for, large generator) with cylindrical poles in shape., (iii) Slip rings : The two ends of the armature coil are connected to two brass rings R1, and R2. These rings rotate along with the armature coil., (iv) Brushes : Two carbon brushes (B1 and B2), are pressed against the slip rings., These brushes are connected to the load through which the output is obtained., Principle : It works on the principle of electromagnetic induction. According to it when, a coil is rotated in magnetic field, an emf is induced in the coil. The coil may be rotated, by water energy, steam energy or oil energy. Let at any instant magnetic flux through, armature coil,, N B = NBA cos = NBA cos t, The induced emf, or, and induced current, , e, , d B, =NBA sin t, dt, = e0 sin t, where e0 = NBA ., , i, , =, , e, , =, , e, R, , Fig. 8.103 AC Generator, , e0, sin t, R, , i0 sin t

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ELECTROMAGNETIC INDUCTION, , 8.17 DC, , GENERATOR, , It produces direct current. It is possible by providing split rings or commutator in place, of slip rings. DC generator consists of :, (i) armature coil,, (ii) magnet,, (iii) split rings,, (iv) brushes., For DC generator, output current, i = | i0sin t |, , 8.18 DC, , MOTOR, , It is an electrical machine which converts electrical energy into mechanical energy., Principle : It is based on the fact that a current carrying coil placed in magnetic field, experiences a torque. Because of this torque the coil starts rotating., Construction : It consists of, (i) strong magnet,, (ii) armature,, (iii) split rings,, (iv) brushes., Back emf :, As soon as motor starts rotating, the flux changes in the coil, because of this change in, flux, an emf is induced in the coil which opposes the emf of the battery. This induced emf, is called back emf. The net emf of the circuit, e net = – e, where e = NBA sin t = k, Armature current, , i, , =, , Fig. 8.104 DC Generator, , e, , enet, =, R, , ., , R, , k, R, Armature current i will be maximum, when motor is just started i.e., = 0, imax = /R. And, minimum when motor picks full speed, , =, , i min, , k max, R, , =, , Efficiency of motor :, If is the potential applied and i is the current in armature, then, i = i2 R + mechanical power, Mechanical power, =, i – i2R = i ( – iR), = ie, (e is the back emf), Efficiency, , =, , Available mechanical power, input electrical power, , =, , ei, i, , e, , ., , is maximum, when (e i) is maximum, Let, , x, , = ei, , =, , R, e2, , e, , or, , e, , e, , R, , Fig. 8.105 DC Motor, , 527

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528, , ELECTRICITY & MAGNETISM, x to be maximum,, or, , dx, de, – 2e, , which gives, , = 0, = 0, , e, , =, , 50% ., , 2, , 8.19 TRANSFORMER, (1), (2), (3), , (4), , Transformer is used to change the voltage in ac circuits., It is based on principle of mutual induction., It works with ac only., It can not change the voltage without change in current. There will inverse change, in current. For ideal transformer, P out = Pin, or, VS iS = VP iP, VS, VP, , =, , d S, dt, , =, , d P, dt, eP, , iP, ...(i), iS, Same flux links in both primary and secondary, so the induced emf per turn is the, same in each., , i.e.,, , NP, , For primary coil, , VP, , =, , and for secondary coil, , VS, , = eS = N S, , VS, VP, , =, , eS, eP, , d P, ,, dt, , d S, dt, , NS, NP, , ...(ii), , From equations (i) and (ii), we get, VS, VP, , eS iP N S, K, ...(1), eP iS, NP, where K is called transformer ratio. K > 1 for step-up transformer and K < 1 for step, down transformer., , =, , Fig. 8.106, (5), , Symbol of transformer is, , ., , Losses in transformer, The output power of a transformer is necessarily less than the input power because of, unavoidable losses. These losses consist of i2R losses in the primary and secondary, coils, and hysteresis and eddy losses in the core. Hysteresis losses are minimised by the, use of iron core, and eddy losses are minimised by laminating the core. The electrical

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ELECTROMAGNETIC INDUCTION, , 529, , resistance between the surfaces of the laminations (by insulating varnish) effectively, confines the eddy currents to individual laminae. The resulting path is greatly increased,, with consequent increase in resistance. Hence, although the induced emf is not altered,, but the currents and their heating effects are minimised., Efficiency of a transformer ( ), (%), , =, , Pout, Pin, , 100, , Vs is, Vp ip, , ...(2), , 100, , Fig. 8.107, , In practice can be achieved from 90% to 99%., , Ex. 54, , A transformer has 500 primary turns and 10 secondary, , turns., (a) If VP is 120 V (rms), what is Vs with an open circuit ?, (b), , If the secondary now has a resistive load of 15, the currents in the primary and secondary ?, , , what are, , Given, NP = 500, Ns = 10, VP = 120 V., (a) For a transformer, , Vs, , (b), , =, , Ns, NP, , =, , VP, , =, , 10, 120, 500, , N =, , Sol., , Ns, NP, , =, , is, , =, , iP =, =, , 2.4 V, , We know that i =, , Ans., where, , i0 =, , di, dt, , =, , Ns, NP, , =, , , = 12 V,,, , i0 1 e, , i0, , ....(i), , 12, 6, , R, , di, =, dt, The time of i = 1A:, , L, , =, , L, R, , At t = 0,, , (b), , Ns, NP, , 2A, , t, , t, , R e, L, R, , e, , e, , 3, 6, , 3.2 m A, , Ans., , A toroidal solenoid has a mean radius of 0.12 m and a, cross-sectional area 20 × 10–4 m2. It is found that when the current, is 20 A, the energy stored is 0.1 J. How many turns the winding have, ?, , or, Now, , Sol., , 1 =, , 21 e, , t, , 1, 2, , =, , e, di, dt, , t, , =, =, , e, , L, 2 A/s, , For toroidal solenoid energy stored, =, , 1 2, Li, 2, , =, , 1, 2, , 0, , 4 A/s, , Ans., , 1, 2, t, , Ex. 55, , U, , L, , 12, 3, , L, , t, , e, , From equation (i), , 10, 500, , 0.16, , 20, , t, , (a), , 0.16 A, , is, , 4, , 12, 3, , 1, 2, Ans., , t, , (c), , N2 A 2, i, 2 r, , 2, , Ans., , We know that, , ip, , N 2 20 10, 0.12, , 387, , Given; H = 3 H, R = 6, , 2.4, 15, , Vs, Rs, , 7, , 1 2 10, 2, , Ex. 56, , The current in secondary coil, is =, , 0.1 =, , An inductor of inductance 3H and resistance 6, is, connected to the terminals of a battery of emf 12 V and of negligible, internal resistance. Find, (a) the initial rate of increase of current in the circuit,, (b) the rate of increase of current at the instant when the current, is 1A,, (c) the current 0.2 s after the circuit is closed,, (d) the final steady state current., , Sol., Vs, VP, , or, , i =, , i0 1 e, , =, , 2 1 e, , =, , 2 1 e, , =, , 0.659 A, , 0.2, (1/ 2), 0.4, , Ans.

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530, (d), , ELECTRICITY & MAGNETISM, , Steady currenti0, , =, , R, , 12, 6, , 2 A., , Ans., , Ex. 57 Consider the circuit shown in fig. 8.108. With switch S1, closed and the other two switches open, the circuit has a time, constant C. With switch S2 closed and the other two switches open,, the circuit has a time constant L. With switch S3 closed and other, two switches open, the circuit oscillates with a period T. Find T., , Fig. 8.110, , Fig. 8.108, , B =, , Sol., When only S1 is closed, the circuit is active through C and R,, C = CR, Similarly, , L, C L, , =, =, , Ex. 58, , i, , 1, x, , 1, d, , B =, , L/R, , 2, , LC, , =, , 2, , C, , =, , Ans., , L, , =, , 0, , 2, , 2, , i, , 0i, , n d, , d, , n, , 1, x, , a, , nx, , 2, , d a, , 0i, , B dx, , LC, , =, , x, , Now consider an element of width dx and length . The flux enclosed, between wires, , Now when S3 is closed, the circuit elements are L and C, T, , 0, , 2, , a, , x, , n, , a, , 1, d, , x, , dx, , d a, a, , a, d, , a, , In the given circuit containing an inductor and, , resistors. Find the time constant of the circuit., , =, , Sol., For the given circuit, the effective resistance across the inductor, , =, , The inductance, , L, , =, , 0i, , 2, 0i, , B, , d, , n, , 2, , a, a, , d, , n, , a, a, , 0, , i, , n, , d, , a, a, , Ans., , Ex. 60, (a), (b), (c), (d), , Fig. 8.109, R =, , =, , Ex. 59, , 6 12, 6 12, , 4, , L, R, , 1, s, 4, , 2, 8, , 8, , For the given circuit here, find the, energy stored in C, energy stored in L, current in each circuit element and, voltage across each circuit element., , Ans., , Two parallel wires whose centres are at a distance d, , apart carry equal currents in opposite directions. Neglecting the, flux within the wires themselves, find the inductance of a length of, such pair of wires, given that the radius of the wires is a., , Sol., The magnetic field at a distance x from left wire due to the current in the, two wires is, , Fig. 8.111, , Sol., After reaching steady state, capacitor stops the current. While inductor, offers zero resistance. The current in the right side of the battery will, pass through inductor., The above circuit can be reduced as follows :

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531, , ELECTROMAGNETIC INDUCTION, , Sol., , Current through resistor, i =, , To an inductor, , Ldi, dt, , =, , i, , Fig. 8.112, In close loop ABCFA,, 20 i1 =, i1 =, , 10, , 0.5 A, , and in close loop FCDEF,, 2i2 = 10, i2 = 5A, (a) The p.d. across the capacitor, =, , p.d. across 12, , = 12 i1 = 12, , 1, 2, , 1, 200 10, 2, = 3.6 mJ, The energy stored in inductor, , (b), , =, , i =, , For capacitor,, , q =, =, , Current, , i =, =, , Ex. 62, , 6V, , 30.0, (a), (b), (c), (d), , 1, CV 2, 2, , =, , or, , resistor, , Energy stored in capacitor, =, , =, , 0, , 1, A, 2, , 6, , 62, , Ans., , Kt3, t, , di, , or, , Kt 3, R, , V, R, , K 3, t dt, L0, Kt 4, 4L, CV, C K t3, , Ans., , d CKt 3, , dq, dt, , dt, , 3CKt2, , Ans., , In fig. 8.115, = 100 V, R1 = 10.0, , , R2 = 20.0, , , R3 =, , , and L = 2.00 H. Find the values of i1 and i2, immediately after switch S is closed;, a long time later;, immediately after switch S is opened again, a long time later., , Ans., , 1 2, Li, 2, , 1, 2, 3 10 3, 5, 2, = 37.5 m J, Ans., The current in each circuit elements are shown as follows., =, , (c), , Fig. 8.115, , Sol., (a), , Immediately after switch S is closed, inductor offers infinite, resistance, therefore current in its arm will be zero. For closed, loop ABEFA, i1 =, , (b), , (d), , R1, , R2, , 100, 10 20, , After a long time, inductor becomes ineffective. The effective circuit, becomes, , Fig. 8.113, The p.d. across each circuit element are shown in fig. 8.112, , Fig. 8.116, Fig. 8.114, , Ex. 61 A voltage waveshape of the form V(t) = Kt3 is applied to a, circuit element at time t = 0. Find current through a resistor R, an, inductor L and a capacitor C., , 3.33 A, , i1 =, , R1, , R2 R3, R2 R3

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532, , ELECTRICITY & MAGNETISM, =, , 100, 20 30, 10, 20 30, , =, , 4.55 A, , We have i1 = i2 + i3 and 20 i2 = 30 i3, , (d), , 2, i2, 3, , 2i2, 3, or, i2 = 2.73 A, i3 = 1.82 A, Immediately after switch S is opened, i1 = 0, and i2 = i3 = 1.82 A, 4.55 =, , (c), , i3, , i2, , Fig. 8.120, In close loop A B C D E F A,, – e –(i1 + i2) R + = 0, In close loop B C D E B,, – e + i2R = 0, From equation (ii), we have, i2 = e / R, substituting this value in equation (i), we get, – e –(i1+e / R)R +, = 0, or, –2e – i1R +, = 0, , Fig. 8.117, After a long time, the magnetic energy stored in inductor will, change into heal and therefore, i1 = i2 = 0., , Ex. 63, , In the circuit diagram shown in fig. 8.118, R = 10 , L, = 5H, = 20 V, i = 2A. This current is decreasing at a rate of 1.0 A/, s. Find Vab at this instant., , 2L, , or, i1, , di1, dt, , =, , 0, , n, , i1R, R, , ...(ii), , ( – i1R), t, , di1, i1R, , or, , ...(i), , dt, 2L, 0, , =, , i1, , =, 0, , t, 2L, , Fig. 8.118, , Sol., , The induced emf across inductor, , Fig. 8.119, e =, =, , Ldi, dt, , 5, , 1.0, , Fig. 8.121, , = 5 V, As the current is decreasing the inductor can be replaced by a source of, emf 5V in such a manner so that it compensate the decreasing current. It, to be like as shown in fig. 8.119., Now by loop rule, –Ri + e – – Vab = 0, or, Vab = Ri + – e, = 10 × 2 + 20 – 5, = 35 V, Ans., , Ex. 64, , Find the time dependence of the current flowing, through the inductance L of the circuit shown in fig. 8.120 after the, switch S is shorted at the moment t = 0., , Sol., , Let at any instant the current in the inductor is i1 and increasing, , at the rate of, , dii, . The induced emf across L e, dt, , L, , increasing current. So it to be as shown in the circuit., , di1, dt, , opposes the, , or, , or, , n, , i1R, n 1, , Rt, 2L, , n =, i1, , =, , R, or, , i1 =, , Rt, 2L, , R, , 1 e, , tR, 2L, , .

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ELECTROMAGNETIC INDUCTION, , 533, , Review of formulae & Important Points, 1., , Magnetic flux : Magnetic flux of the magnetic field B through, the normal area A is, B, , 2., , =, , B A, , U =, , BA cos, , If B is the magnetic field in the coil, then, , Faraday s law : Whenever there is change in magnetic flux linked, with circuit, there induces an emf in the circuit. The rate of change, of magnetic flux is equal to the induced emf. Thus, , d B, ,, dt, , =, , 1 2, Li ., 2, , U =, 10., , B2, 2 0, , Vol., , RL-DC circuit : For a circuit with time constant , the growing, current i in circuit at any time t, , here negative sign indicates that induced emf opposes the change, in flux., Induced charge in time t,, Q, 3., , =, , R, , Motional emf : When a metallic conductor moves in a magnetic, field B with velocity v , an emf induces across its ends. The, induced emf, , i =, , e = Bv sin ., where i0 =, , In general, it can be written as :, , v, , e =, 4., , 2, , 6., , 11., , ,, , Ed, , =, , d 2q, dt, , d B, dt, , Also, , e = NBA sin t, 7., , 8., , 12., , Self induction of circular coil of N turns and radius r, L =, , 9., , 0, , 2, , N r, ., 2, , Energy stored in an inductor : Energy in the inductor stores due, to magnetic field in it. For any current i in the inductor, the energy, stored, , t/, , 2, , q, , = 0,, , 1, ., LC, , =, T, , =, , 2, , LC ., , Electrical energy of capacitor will store in inductor after time T/4, and vice-versa., , Self induction : The induced emf in the coil itself due to change, in current in it is called self induction. For the coil of N turns, N B = Li,, where L is called self inductance., , 2, , where, , Induced emf in a coil rotating in uniform magnetic field, , i0 e, , LC-oscillations : When a capacitor C with a charge q is connected, to an inductor L, the energy of the circuit oscillates between C and, L. If is the angular frequency of oscillations, then, , Induced electric field : If En is the induced electric field, then by, Faraday s law, e =, , L, R, , i =, , 2, , B, , and, , t/, , The decay current in the circuit, with initial current i0, , When a metallic conductor is rotated in normal magnetic, field about its one ends, the induced emf across the ends, e =, , 5., , B d, , R, , i0 1 e, , Mutual inductance : The induced emf in the second coil due to, change in current in first coil, is called mutual induction. If M is the, mutual inductance between the coils, then, or, , 13., , N2 21 = Mi1,, N1 12 = Mi2., , Mutual induction between two circular coils of radii R1 and, R2 with turns N1 and N2 is

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534, , ELECTRICITY & MAGNETISM, 2, 0 N1 N 2 R2, 2 R1, , M =, 14., , L1L2, , Combination of inductors :, (i), (ii), , In series :, In parallel :, , L, , 1, L, , = L1 + L2, =, , 1, L1, , 1, L2, , AC generator : If i0 be the maximum current (current amplitude),, then, i = i0sin t,, , If L1 and L2 are the self inductances of two coils, then mutual, induction between them, M =, , 15., , 16., , where, 17., , i0 =, , NBA, R, , Transformer : For the transformer with the turns Np and Ns in, the primary and secondary coils, the ratio of output and input, potentials, , Vs, VP, , =, , Ns, NP, , For the ideal transformer, VP iP = Vs is.

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ELECTROMAGNETIC INDUCTION, , E xercise 8. 1, , MCQ Type 1, , Only one option correct, 1., The graph gives the magnitude B(t) of a uniform magnetic field that, exists throughout a conducting loop, perpendicular to the plane of, the loop. In which region, the magnitude of induced emf will be, greatest :, , 5., , 535, , A circular loop of resistance R and enclosing an area A is subjected, to a magnetic field B0e, , bt, , perpendicular to its plane, B0 and b, being constants, and t is the time. What is the induced current at t, =0 ?, (a), , AB0b, R, , (b), , B0bR, A, , AB0 R, B0bR, (d), b, 2A, The magnetic field through a loop of area 1m2 and of resistance, , (c), , 6., , 2., , 3., , 4., , 10 , changes with time as shown in figure. The induced current, in the loop at t = 1s, , (a) I, (b) II, (c) III, (d) IV, The figure shows three circuits with identical batteries, inductor,, and registers. In which circuit, the current through the battery is, greatest just after closing of the switch, , (a) 1, (b) 2, (c) 3, (d) 1, 3, Figure shows three circuits with identical batteries, inductors, and, resistors. In which circuits the time taken to reach 50% of the, maximum current is greatest :, , (a) 0.1 A, (b) 0.2 A, (c) 0.4 A, (d) 1 A, In figure, a circular loop of wire 10 cm in diameter is placed with, , 7., , its normal N at an angle, , magnetic field B of magnetic 0.50 T. The loop is then rotated such, that N rotates in a cone about the field direction at the constant, rate of 100 rev/min; the angle remains unchanged during the, process. The induced emf in the loop :, , (a) 1, (b) 2, (c) 3, (d) 2, 3, If the circular conductor undergoes thermal expansion while it is in, a uniform magnetic field, a current will be induced in clockwise, around it. If B is the magnitude of the magnetic field, then its, direction is :, , (a) 0 V, (b) 10 V, (c) 100 V, (d) none, Transformers are used, (a) in AC circuits only, (b) in DC circuit only, (c) in both AC and Dc circuits, (d) neither in AC nor in DC circuits., A generator supplies 100 V to the primary-coil of a transformer of, 50 turns. If the secondary coil has 500 turns, then the secondary, voltage is, (a) 100 V, (b) 500 V, (c) 550 V, (d) 1000 V, , 8., , 9., (a), (c), , Bi, Bj, , (b), , Bj, , (d), , B kˆ, , = 30° with the direction of a uniform, , Answer Key, , 1, , (b), , 2, , (c), , 3, , (c), , 4, , (d), , Sol. from page 561, , 6, , (a), , 7, , (a), , 8, , (a), , 9, , (d), , 5, , (a)

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536, 10., , 11., , ELECTRICITY & MAGNETISM, , Consider the situation shown in figure. If the switch is closed and, after some time it is opened again, the closed loop will show, , (a), , a clockwise current, , (b), , an anticlockwise current, , (c), , an anticlockwise current and then clockwise, , (d), , a clockwise current and then an anticlock wise current., , 16., , 17., , In the circuit shown,, indicated a uniform magnetic field, which, is directed into the page and decreasing in magnitude at the rate of, 150 T/s. The reading of ammeter is :, 18., , 12., , (a), , 0.15 A, , (b), , 0.35 A, , (c), , 0.50 A, , (d), , 0.65 A, , A metal rod is moved with a constant velocity v in a magnetic field., A potential difference appears across the two ends, (a), , if v ||, , (b), , if v || B, , (c), , if, , (d), , none of these, , || B, , A rod of length rotates with a uniform angular velocity about, its perpendicular bisector. A uniform magnetic field B exists parallel, to the axis of rotation. The potential difference between two ends, of the rod is, (a), , 1, B, 2, , (c), , 2B, , 2, , (b), , B, , (d), , zero, , A magnet NS is suspended from a spring and while it oscillates, the, magnet moves in and out of the coil C. The coil is connected to a, galvanometer G. Then as the magnet oscillates,, , A circular wire loop of radius r is placed in a region of magnetic, field B such that the plane of the loop makes an angle with the, direction of B . Consider the following in this regard :, 1., , Change in B with time, , 2., , Change in r with time, , 3., , B being non-uniform in space, , 4., , Change in with time, , The conditions for an induced emf in the loop would include, , 13., , 14., , 15., , (a), , 1 and 4, , (b), , 1, 2 and 4, , (c), , 1 and 3, , (d), , 2, 3 and 4, , 19., , A series would dc motor has a total resistance of 1.5 ohm. When, connected across a 115 volt and running at a certain speed it draws, a current of 10 A. The back emf in the motor is, , (a), , G shows deflection to the left and right with constant, amplitude, , (b), , G shows deflection on one side, , (c), , G shows no deflection, , (d), , G shows deflection to the left and right but the amplitude, steadily decreases., , An infinitely long cylinder is kept parallel to an uniform magnetic, field B directed along positive z axis. The direction of induced, current as seen from the z axis will be, (a), , clockwise of the +ve z axis., anticlockwise of the +ve z axis., , (a), , 100 V, , (b), , 115 V, , (b), , (c), , 15 V, , (d), , 1.5 V, , (c), , zero emf., , (d), , along the magnetic field., , Which of the following can produce the maximum induced emf in, an inductor ?, (a), , 50 A, DC, , (b), , 50 A, 50 Hz AC, , (c), , 50 A, 500 Hz AC, , (d), , 100 A, DC., , 20., , An electric potential difference will be induced between the ends, of the conductor as shown in the diagram, when the conductor, moves in the direction, , Two circular loops of equal radii are placed coaxially at some, separation. The first is cut and a battery is inserted in between to, drive a current in it. The current changes slightly because of the, variation in resistance with temperature. During this period, the, two loops, , (a), , P, , (b), , Q, , (a), , attract each other, , (b), , repel each other, , (c), , L, , (d), , M, , (c), , do not interact, , (d), , none of these, , Answer Key, , 10, , (d), , 11, , (d), , 12, , (b), , 13, , (a), , 14, , (c), , Sol. from page 561, , 16, , (d), , 17, , (d), , 18, , (d), , 19, , (c ), , 20, , (d), , 15, , (a)

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ELECTROMAGNETIC INDUCTION, 21., , A conducting square loop of side L and resistance R moves in its, plane with a uniform velocity v perpendicular to one of its sides., A magnetic induction B constant in time and space, pointing, perpendicular and into the plane of the loop exists everywhere., The current induced in the loop is, , (a), , Blv, clockwise, R, , (b), , 2Blv, anticlockwise, (d) zero, R, The inductance of a closed-packed coil of 400 turns is 8 mH. A, current of 5 mA is passed through it. The magnetic flux through, each turn of the coil is, , 23., , (a), , 1, 4, , 0 Wb, , (b), , (c), , 1, 3, , 0, , Wb, , (d), , A thin semicircular conducting ring of radius R is falling with its, plane vertical in a horizontal magnetic induction B. At the position, MNQ, the speed of the ring is V and the potential difference, developed across the ring is, , (a) zero, (b) Bv R2 / 2 and M is at higher potential, (c), RBV and Q is at higher potential, (d) 2RBV and Q is at higher potential, A circular loop of radius R carrying current I lies in xy- plane with, its centre at origin. The total magnetic flux through xz-plane is, (a) directly proportional to I, (b) directly proportional to R, (c) directly proportional to R2, (d) zero, Two identical circular loops of metal wire are lying on a table, without touching each other. Loop-A carries a current which, increases with time. In response, the loop-B, (a) remains stationary, (b) is attracted by the loop-A, (c) is repelled by the loop-A, (d) rotates about its CM, with CM fixed (CM is the centre of, mass), A wire of length 1 m is moving at a speed of 2ms–1 perpendicular to, its length and a homogeneous magnetic field of 0.5 T. The ends of, , Blv, anticlockwise, R, , (c), , 22., , 25., , 1, 2, , 0, , Wb, , 537, , 26., , 27., , 0.4 0 Wb, , The adjoining figure shows two bulbs B1 and B2 resistor R and an, inductor L. When the switch S is turned off, , 28., , the wire are joined to a circuit of resistance 6 . The rate at which, work is being done to keep the wire moving at constant speed is, , 24., , (a) both B1 and B2 die out promptly, (b) both B1 and B2 die out with some delay, (c) B1 dies out promptly but B2 with some delay, (d) B2 dies out promptly but B1 with some delay, An electron moves along the line AB, which lies in the same plane, as a circular loop of conducting wires as shown in the diagram., What will be the direction of current induced if any, in the loop, , (a), (b), (c), (d), , (a), , 1, W, 12, , (b), , 1, W, 6, , 1, W, (d) 1 W, 3, Two circular coils can be arranged in any of the three situations, shown in the figure. Their mutual inductance will be, , (c), , 29., , (a), (b), (c), (d), , no current will be induced, the current will be clockwise, the current will be anticlockwise, the current will change direction as the electron passes by, , maximum in situation (A), maximum in situation (B), maximum in situation (C), the same in all situations, , Answer Key, , 21, , (d), , 22, , (a), , 23, , (c), , 24, , (d), , Sol. from page 561, , 26, , (d), , 27, , (c), , 28, , (b), , 29, , (a), , 25, , (d)

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538, 30., , ELECTRICITY & MAGNETISM, , As shown in the figure, P and Q are two coaxial conducting loops, separated by some distance. When the switch S is closed, a, clockwise current IP flows in P (as seen by E) and an induced, current I Q1 flows in Q. The switch remains closed for a long time., When S is opened, a current I Q2 flows in Q. Then the directions, of I Q1 and I Q2 (as seen by E) are, , 31., , 32., , 33., , (a) respectively clockwise and anticlockwise, (b) both clockwise, (c) both anticlockwise, (d) respectively anticlockwise and clockwise, A short-circuited coil is placed in a time-varying magnetic field., Electrical power is dissipated due to the current induced in the, coil. If the number of turns were to be quadrupled and the wire, radius halved, the electrical power dissipated would be, (a) halved, (b) the same, (c) doubled, (d) quadrupled, , (c), 36., , 37., , A coil of inductance 8.4 mH and resistance 6 is connected to a, 12 V battery. The current in the coil is 1.0 A at approximately the, time, (a) 500 sec, (b) 20 sec, (c) 35 milli sec, (d) 1 milli sec, A conducting wire frame is placed in a magnetic field which is, directed into the paper. The magnetic field is increasing at a constant, rate. The directions of induced current in wires AB and CD are, , 38., , 34., , 35., , (a) B to A and D to C, (b) A to B and C to D, (c) A to B and D to C, (d) B to A and C to D, A conducting ring of radius 1 meter is placed in an uniform magnetic, field B of 0.01 telsa oscillating with frequency 100 Hz with its, plane at right angles to B. What will be the induced electric field, (a), volt / m, (b) 2 volt / m, (c) 10 volt / m, (d) 62 volt / m, The variation of induced emf (E) with time (t) in a coil if a short bar, magnet is moved along its axis with a constant velocity is best, represented as, , (a), , (d), , An alternating current of frequency 200 rad/sec and peak value 1A, as shown in the figure, is applied to the primary of a transformer., If the coefficient of mutual induction between the primary and the, secondary is 1.5 H, the voltage induced in the secondary will be, (a) 300 V, (b) 191 V, (c) 220 V, (d) 471 V, A horizontal loop abcd is moved across the pole pieces of a magnet, as shown in fig. with a constant speed v. When the edge ab of the, loop enters the pole pieces at time t = 0 sec, which one of the, following graphs represents correctly the induced emf in the coil?, , (a), , (b), , (c), , (d), , The current i in an induction coil varies with time t according to the, graph shown in figure., , Which of the following graphs shows the induced emf (E) in the, coil with time?, , (b), , (a), , (b), , (c), , (d), , Answer Key, , 30, , (d), , 31, , (b), , 32, , (d), , 33, , (a), , Sol. from page 561, , 35, , (a), , 36, , (b), , 37, , (d), , 38, , (c), , 34, , (b)

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539, , ELECTROMAGNETIC INDUCTION, 39., , A magnet is made to oscillate with a particular frequency, passing, through a coil as shown in figure. The time variation of the, magnitude of e.m.f. generated across the coil during one cycle is, , 42., , A short magnet is allowed to fall along the axis of a horizontal, metallic ring. Starting from rest, the distance fallen by the magnet, in one second may be :, (a) 4.0 m, (b) 5.0 m, (c) 6.0 m, (d) 7.0 m, When a metallic plate swings between the poles of a magnet, (a) no effect on the plate, (b) eddy currents are set up inside the plate and the direction of, the current is along the motion of the plate, (c) eddy currents are set up inside the plate and the direction of, the current oppose the motion of the plate, (d) eddy currents are set up inside the plate, Pure inductance of 3.0 H is connected as shown below. The equivalent inductance of the circuit is, , 43., , 44., (a), , (b), , (c), , (d), , (a) 1 H, (b) 2 H, (c) 3 H, (d) 9 H, If the switch in the following circuit is turned off, then, , 45., 40., , In the given circuit containing an inductor and resistors, the time, constant of the circuit is :, 2H, , 6, , 41., , 4, , 1, s, 4, , (b), , (a), , (b), , K 4 K t4, 3KC 2t 2, t ,, and, 4R, L 4, 2, , (c), , K 3 K t4, t ,, and 3CKt 2, R, L 4, , (d), , K 3 K t3, t ,, and CKt 3, R, L 3, , L, , (a), , 1, s, 2, , the bulb B1 will go out immediately whereas B2 after, sometimes, (b) the bulb B2 will go out immediately whereas B1 after sometime, (c) both B1 and B2 will go out immediately, (d) both B1 and B2 will go out after sometime, The figure shows certain wire segments joined together to form a, coplanar loop. The loop is placed in a perpendicular magnetic field, in the direction going into the plane of the figure. The magnitude of, the field increases with time. I1 and I2 are the currents in the, segments ab and cd. Then,, , 46., , 1, 1, s, s, (c), (d), 3, 5, A voltage wave shape of the form V(t) = Kt3 is applied to a circuit, element at time t = 0. The current through a resistor (R), an inductor, (L) and a capacitor (C) respectively are :, , K 3 K t3, 3CK t 4, and, t ,, R, L 3, 4, , B2, R, , 12, , (a), , B1, , d, , c, a, , (a), (b), (c), (d), , b, , I1 > I2, I1 < I2, I1 is in the direction ba and I2 is in the direction cd, I1 is in the direction ab and I2 is in the direction dc, , Answer Key, , 39, , (a), , 40, , (a), , 41, , (c), , Sol. from page 561, , 44, , (a), , 45, , (d), , 46, , (d), , 42, , (a), , 43, , (c)

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540, , ELECTRICITY & MAGNETISM, , Level -2, Only one option correct, 1., , 2., , 4., , The figure shows three situations in which identical circular, conducting loops are in uniform magnetic fields that are either, increasing or decreasing in magnitude at identical rates. In each, the, dashes line coincides with a diameter. In which situation the, magnitude of the induced current is maximum :, , (a), , I, , (b), , I, II, , (c), , II, III, , (d), , I, III, , A square loop with 2.0 m sides is perpendicular to a uniform, magnetic field, with half the area of the loop in the field is shown, in figure. The loop contains a 20.0 V battery with negligible interval, resistance. If the magnitude of the field varies with time according, to B = 0.042–0.87 t, with B in tesla and t in second. The net emf of, the circuit is :, , 5., , The switch in the circuit of figure has been closed for a very long, time when it is then thrown to b. The resulting current through the, inductor is indicated in the figure for four sets of values for the, resistance R and inductance L : (1) R0 and L0 (2) 2 R0 and L0, (3), R0 and 2L0. (4) 2R0, 2L0. Select the correct matching :, , (a), , A-1, B-2, C-3, D-5, , (b), , A-4, B-3, C-2, D-1, , (c), , A-2, B-4, C-1, D-3, , (d), , none, , A coil having n turns and resistance R, , galvanometer of resistance 4R . This combination is moved in, time t second from a magnetic flux 1 weber to 2 weber. The, induced current in the circuit is, 2, , (a), , n, , 1, , (b), , 5 Rnt, , 2, , (c), , 3., , 20.0 V, , (b), , 18.26 V, , (c), , 21.74 V, , (d), , none, , 6., , Figure shows a circular region in which an increasing uniform, magnetic field is directed out of the page, as well as a concentric, Ed, , circular path along which, , is to be evaluated. The table, , gives the initial magnitude, and the time interval for the increase, in, three situations. The situation according to the greatest magnitude, of the electric field induced along the path :, , 1, , n, , (d), , Rnt, , 2, , 1, , 5 Rt, , ., , (a), , is connected with a, , 2, , 1, , Rt, , One conducting U tube can slide inside another as shown in figure,, maintaining electrical contacts between the tubes. The magnetic, field B is perpendicular to the plane of the figure. If each tube, moves towards the other at a constant speed v then the emf induced, in the circuit in terms of B, and v where is the width of each, tube, will be, , Situation Initial field Increase Time, A, B1, B1, t1, B, 2 B,, B1 / 2, t1, t1, C, B1 / 4, B1, 2, , 7., , (a), , A, , (b), , B, , (c), , C, , (d), , A and B, , (a), , zero, , (b), , 2 Blv, , (c), , Blv, , (d), , –B, , A small square loop of wire of side is placed inside a large square, loop of wire of side L (L > ). The loop are coplanar and their, centre coincide. The mutual inductance of the system is, proportional to, (a), (c), , /L, L/, , Answer Key, , 1, , (d), , 2, , (c), , 3, , (c), , Sol. from page 562, , 5, , (b), , 6, , (b), , 7, , (b), , (b), , 2, , (d), , L2, , /L, /, , 4, , (c)

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ELECTROMAGNETIC INDUCTION, 8., , A uniform but time-varying magnetic field B(t) exists in a circular, region of radius a and is directed into the plane of the paper, as, shown. The magnitude of the induced electric field at point P at a, distance r from the centre of the circular region, , 12., , 9., , 10., , is zero, , (b), , decreases as, , (c), , increases as r, , (d), , decreases as, , 1, r, , A variable magnetic field of induction B B0e t is established, inside the coil. If the key (K) is closed, the electrical power, developed right after closing the switch is equal to, , 14., , B02, , (b), , 2 4, , r R, , B010r 3, R, B02, , (a), (c), , (d), , X, A, Y, A2, , (b), (d), , 1, s, n2, , (b), , 2, s, n2, , (c), , 3, s, n2, , (d), , 4, s, n2, , A rectangular loop is being pulled at a constant speed v, through a, region of certain thickness d, in which a uniform magnetic field B is, set up. The graph between position x of the right hand edge of the, loop and the induced emf E will be, , (a), , (b), , (c), , (d), , A flexible wire bent in the form of a circle is placed in a uniform, magnetic field perpendicular to the plane of the coil. The radius of, the coil changes as shown in figure. The graph of induced emf in, the coil is represented by, , 2 4, , r, 5, R, A highly conducting ring of radius R is perpendicular to and, concentric with the axis of a long solenoid as shown in fig. The ring, has a narrow gap of width d in its circumference. The solenoid has, cross sectional area A and a uniform internal field of magnitude B0., Now beginning at t = 0, the solenoid current is steadily increased, to so that the field magnitude at any time t is given by B(t) = B0 +, t where > 0. Assuming that no charge can flow across the gap,, the end of ring which has excess of positive charge and the magnitude, of induced e.m.f. in the ring are respectively, (c), , 11., , B02 r 2, R, , (a), , 1, , r2, A coil of wire having finite inductance and resistance has a, conducting ring placed coaxially within it. The coil is connected to, a battery at time t = 0, so that a time-dependent current I1(t) starts, flowing through the coil. If I2(t) is the current induced in the ring, and B(t) is the magnetic field at the axis of the coil due to I1(t), then, as a function of time (t > 0), the product I2(t) B(t), (a) increases with time, (b) decreases with time, (c) does not vary with time, (d) passes through a maximum, Shown in the figure is a circular loop of radius r and resistance R., , (a), , 3, th of its steady state, 4, , value in 4s. The time constant of this circuit is, , 13., (a), , The current in a LR circuit builds up to, , 541, , (a), , (b), , (c), , (d), , X, R2, Y, R2, , Answer Key, , 8, , (b), , 9, , (d), , 10, , (d), , Sol. from page 562, , 12, , (b), , 13, , (b), , 14, , (b), , 11, , (a)

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542, 15., , ELECTRICITY & MAGNETISM, , A square loop of side 5 cm enters a magnetic field with 1 cms–1., The front edge enters the magnetic field at t = 0 then which graph, best depicts emf, , v = v0 iˆ . The emf induced in the loop is :, (a) zero, (b) v0B0 d, v 0 B0 d 2, a, a2, A circuit contains an inductance L, a resistance R and a battery of, emf . The circuit is switched off at t = 0. The charge flown, through the battery in one time constant ( ) is :, (c), , 18., , (a), (a), (c), 19., , (b), , 20., , (c), , 21., , v 0 B0 d 3, , (d), , 2, R.e, , ., R e, , (b), (d), , 2 Re, , zero, , Two coils, one primary of 500 turns and one secondary of 25, turns, are wound on an iron ring of mean diameter 20 cm and crosssectional area 12 cm2. If the permeability of iron is 800, the mutual, inductance is :, (a) 0.48 H, (b) 2.4 H, (c) 0.12 H, (d) 0.24 H, The current through an inductor of 1H is given by i = 3t sin t. The, voltage across the inductor of 1 H is :, (a) 3 sin t + 3 cos t, (b) 3 cos t + t sin t, (c) 3 sin t + 3t cos t, (d) 3t cos t + sin t, An L-shaped conductor rod is moving in transverse magnetic field, as shown in the figure. Potential difference between ends of the, rod is maximum if the rod is moving with velocity, , A, , y, , (d), , 3m, 16., , A sliding wire of length 0.25 m and having a resistance of 0.5, moves along conducting guiding rails AB and CD with a uniform, speed of 4 m/s. A magnetic field of 0.5 T exists normal to the plane, of ABCD directed into the page. The guides are short -circuited, with resistances of 4 and 2 as shown. The current through the, sliding wire is :, x, , A, , x, , x, , x, , 0.25 m, , x, , x, , 2m, (a), , x, , 2, x, , 4iˆ 6 ˆj m/s, , (b), , 4iˆ 6 ˆj m/s, , (d), (c) 3iˆ 2 ˆj m/s, 13 iˆ m/s, A rectangular coil has a long straight wire passing through its, centroid perpendicular to its plane as shown. If current through, the wire varies as i = i0sin t, induced current in the coil will be, (Given R = Resistance of the coil), , x, , D, , N, x, , B, , 22., , x, , B, , 4m/s, , x, , C, x, , M, 0.5, , x, , 4, , x, , x, , x, , x, , x, , x, , x, , a, , 0.5 T, x, , (a), (c), 17., , x, , x, , 0.27 A, 1.0 A, , (b), (d), , 0.37 A, 0.72 A, , b, , x ˆ, k. A, a, square loop of edge-length d is placed with its edges along the x, and y-axes. The loop is moved with a constant velocity, The magnetic field in a region is given by B, , (a), , B0 1, , (c), , i0 sin t, R, , (b), , a cos t, bR, , (d), , a sin t, bR, , zero, , Answer Key, , 15, , (c), , 16, , (a), , 17, , (d), , 18, , (b), , Sol. from page 562, , 19, , (d), , 20, , (c), , 21, , (c), , 22, , (d)

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ELECTROMAGNETIC INDUCTION, 23., , The circuit shown here is in its steady state. Time constant of the, circuit is, , 3 mH, , 12, i, , 26., , i, , A nonconducting ring (of mass m, radius r, having charge Q) is, placed on a rough horizontal surface (in a cylindrical region with, transverse magnetic field). The field is increasing with time at the, rate R and coefficient of friction between the surface and the ring is, . For ring to remain in equilibrium should be greater than, , B, , 6, , 6, , 24., , Top view, , (a) 0.1 ms, (b) 0.2 ms, (c) 0.3 ms, (d) information is insufficient of determine time constant, Consider cylindrical region of the magnetic field shown in the, figure. Region I and II have fields directed perpendicularly, outward and inward respectively. Fields are varying with time as, , r2, , r, , 27., , r1, , (a), , Qr 2 R 2, 2mg, , (b), , QrR, 2mg, , (c), , Qr 2 R, 2 mg, , (d), , QrR 2, 2mg, , Consider a pair of smooth metallic rails joined at one of the ends., Rails are parallel and are inclined at 30° with horizontal. A jumper, of mass m, length and resistance R slides down the rails with, constant speed v. Magnetic field in the region is vertical. Strength, of magnetic field is, , Region I : B = 3B0t, Region II : B = B0t, , B, , Jumper, , r1, , 30°, , r2 such that there is no net induced electric field in the region, , 25., , 543, , outside the magnetic field is, (a) 0.5, (b) 0.6, (c) 0.8, (d) 0.2, Two conducting rings are moving on a smooth conducting, horizontal surface as shown. There is a transverse uniform magnetic, field of strength B in the region. Potential difference between highest, points of the rings is zero in this case. Had the rings been moving, in opposite direction with same speeds, potential difference, between the highest points would be, , (a), , (c), , 28., , B, , v, , mgR, 3v, , 2mgR, , (b), , 2, , 3mgR, , 3v 2, 3mgR, , (d), , 2, , v 2, A conducting rod of mass m and is placed on a smooth horizontal, surface in a region where transverse uniform magnetic field B exists, in the region. At t = 0, constant force F starts acting on the rod at, its mid point as shown. Potential difference between ends of the, rod, VP – VQ at any time t is, , v, , 3v, , F, , B, , 30°, , r, , Q, , Smooth conducting surface, (a), (b), (c), (d), , P, , 3 Brv, 6 Brv, 12 Brv, zero, as the surface on which rings are moving is conducting, and hence are equipotential, , Answer Key, , 23, , (b), , 24, , (a), , Sol. from page 562, , 27, , (b), , 28, , (a), , (a), , BF t, 2m, , (b), , BF t, 4m, , (c), , 5 BF t, 8m, , (d), , 7 BF t, 8m, , 25, , (c), , 26, , (b)

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544, 29., , ELECTRICITY & MAGNETISM, , Consider a region of cylindrical magnetic field, changing with time, at the rate x. A triangular conducting loop PQR is placed in the field, such that mid point of side PQ coincides with axis of the magnetic, field region. PQ = 2 , PR = 2 . E.m.f induced in the sides PQ, QR,, PR of the loop are, (a) x 2, 0, x 2, (b), (c), , A conducting loop is being pulled with speed v from region I of, magnetic field to region II. If resistance of the loop is R, current, induced in the loop at the instant shown is, , v, , R, , P, , x 2 3x 2, 0,, ,, 2, 2, 0, x 2, x 2, , 33., , Q, , 2, , 3, x, 0, x 2 ,, 2, 2, A conducting wire PQ of length , mass m and resistance r is, sliding down a smooth, vertical, thick pair of rails (as shown) with, constant speed. Choose the correct alternative, if the circuit is in, steady state, i, R, mg, mgCR, ,q, (a) i, B, B, B, mg, mgCR, ,q, (b) i, Q, P, B, B, , (c), , 30., , 31., , mg, ,q, B, , i, , (c), , B0, , 2B0, , Region -1, , Region -2, , S, , zero, , (a), , B0 v, , clockwise, R, , (b), , B0 v, , anticlockwise, R, , (c), , 3B0 v, , clockwise, R, , 3B0 v, , anticlockwise, R, For the circuit shown here keys k1 and k3 are closed for 1 second., Key k2 is closed at the instant k1 and k3 are opened. Maximum, charge on the capacitor after key k2 is closed is, , (d), , 34., , C, , 2V, , +q –q, mgCR, (d) i zero, q, B, A rectangular loop PQRS, is pulled with constant speed into a, uniform transverse magnetic field by a force F (as shown). E.m.f., induced in side PS and potential difference between points P and, S respectively are (Resistance of the loop = r), , P, , k1, 0.5, , 2F, , k3, , Q, 2, , 4V, , F, , R, , S, , B, , 2, (a), , 32., , Fr, zero,, B, , (b), , 35., , zero, zero, , Fr, Fr Fr, ,, (d), 6B, 6B 6B, A rod OPQ is rotating with angular speed about point O. A, uniform magnetic field B exists perpendicular to the plane of the, rod. Potential difference between point P and Q on the rod is, , (c), , (a), (b), (c), (d), , k2, , 2H, , 4 1, , 1, C, e, , (b), , 4 2 1, , (c), , 8 1, , 1, C, e, , (d), , zero, , Potential difference across the capacitor is varying as V e, volt. Potential difference between points A and B, VA – VB equals, (a), , 2, 2, , mV, , (b), (c), , t, mV, 10e 2, , (d), , 10e 2, , 60°, , B, zero, , 5e 2, , t, mV, 5e 2, , P, , 2, , 1, C, e, , 1, 2, , t, , zero,, , B, , (a), , t, , 6e, , –t/2, , 4F, + –, V, , A, , 10 A, 5 mH, , mV, , 2, , B, , O, , 8, , B, , Q, , Answer Key, , 29, , (c), , 30, , (a), , 31, , (c), , Sol. from page 562, , 33, , (c), , 34, , (b), , 35, , (d), , 32, , (c)

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545, , ELECTROMAGNETIC INDUCTION, 36., , Two capacitors of capacitances 3C and 2C are connected in series, with an inductor of inductance L. Potential differences across the, 7, V0 . Initial current in the, 2, circuit is zero. Energy in capacitor with capacitance 3C when, current in the inductor is maximum is, , capacitors are VP – VQ = V0, VR – VQ =, , 39., , Q, , Va Vb, , 10V, , (d), , Va Vb, , zero, , A cylindrical region of radius 1 m has instantaneous homogenous, magnetic field of 5T and it is increasing at a rate of 2T/s. A regular, hexagonal loop ABCDEFA of side 1 m is being drawn in to the, region with a constant speed of 1 m/s as shown in the figure. What, is the magnitude of emf developed in the loop just after the shown, instant when the corner A of the hexagon is coinciding with the, centre of the circle ?, C, , 2C, , 3C, , P, , 37., , (c), , (a), , 3, CV02, 2, , (b), , 3CV02, , (c), , 6CV02, , (d), , zero, , 40., , 1, , 1 mF, , 1, , 2, , 1m, E, , Key for the circuit shown is closed at time t = 0. Currents through, the inductor and the capacitor are equal at time t (in ms) equal to, 4mH, , A 1 m/s, , D, , R, , L, , B, , (a), , 5 / 3V, , (c), , 5 3, , 1m, , F, , 2 /3 V, , (b), , 2 /3 V, , (d), , 5 3, , V, , In given figure, a wire loop has been bent so that it has three, segments: segment ab (a quarter circle), bc (a square corner), and, ca (straight). Here are three choices for a magnetic field through, the loop:, , y, b, a, , Key, , 38., , (a), , n2, , (b), , n4, , (c), , n8, , (d), , n, , z c, 1, 2, , Regarding the given circuit, the correct statement is, , 10V, , (a), (b), , a, , b, , 4, , 4H, , (1), , B1, , 3iˆ 7 ˆj 5tkˆ, , (3), , B3, , 2iˆ 5tjˆ 12kˆ, , (2) B2, , 5tiˆ 4 ˆj 15kˆ, , where B is in milliteslas and t is in seconds. If the induced current, , 2, , 0.5F, , x, , in the loop due to B1 , B2 and B3 are i1, i2 and i3 respectively, then, (a) i1 > i2 > i3, (c) i3 > i2 > i1, , (b) i2 > i1 > i3, (d) i1 = i2 = i3, , Va Vb is increasing with time, Va Vb is decreasing with time, , Answer Key, , 36, , (c), , Sol. from page 562, , 40, , (b), , 37, , (b), , 38, , (d), , 39, , (c)

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546, , ELECTRICITY & MAGNETISM, , E xercise 8. 2, , MCQ Type 2, Multiple correct options, 1., The figure shows four wire loops, with edge lengths of either or, 2 . All four loops will move through a region of uniform magnetic, , 6., , field B at the same constant velocity. In which loop the emf, induced is maximum :, , Two metallic rings A and B, identical in shape and size but having, different resistivities A and B, are kept on top of two identical, solenoids as shown in the figure. When current I is switched on in, both the solenoids in identical manner, the rings A and B jump to, heights h A and hB, respectively, with h A > hB. The possible, relation(s) between their resistivities and their masses mA and mB, is(are), B, , A, , 2., , (a) I, (b) II, (c) III, (d) IV, Figure shows a circuit with two resistors and an ideal inductor., , 7., (a), (b), (c), (d), , 3., , The current in R1 is zero just after closing the switch S., The current in R1 is maximum just after closing the switch S., The current in R2 is zero just after closing of the switch S., The currents in the resistors are maximum of their values a, long time after closing the switch S., L, C and R represent inductance, capacitance and resistance, respectively. Which of the following have dimensions of frequency, ?, (a), , L, C, , P, Q, R, , 1, LC, , (b), , 1, R, (d), RC, L, A conducting loop is placed in a uniform magnetic field with its, plane perpendicular to the field. An emf is induced in the loop if, (a) it is translated, (b) it is rotated about its axis, (c) it is rotated about axis through bisector, (d) it is expanded, Two different coils have self-inductance L1 = 8 mH, L2 = 2mH., The current in one coil is increased at a constant rate. The current, in the second coil is also increased at the same rate. At a certain, instant of time, the power given to the two coils is the same. At, that time the current, the induced voltage and the energy stored in, the first coil are i1, V1 and W1 respectively. Corresponding values, for the second coil at the same instant are i2, V2 and W2 respectively., Then, , (d), , 4., , ., 5., , (a), , i1, i2, , (c), , W2, W1, , 1, 4, 4, , (a), (b), A > B and mA = mB, A< B and mA = mB, (d), (c), A > B and mA > mB, A < B and mA < mB, Figure shown plane figure made of a conductor located in a magnetic, field along the inward normal to the plane of the figure. The magnetic, field starts diminishing. Then the induced current, , (b), , i1, i2, , 48, , (d), , V2, V1, , 1, 4, , 8., , (a) at point P is clockwise, (b) at point Q is anticlockwise, (c) at point Q is clockwise, (d) at point R is zero, Two circular coils P & Q are fixed coaxially & carry currents I1 and, I2 respectively, I2, , I1, , P, , (a), (b), (c), (d), , Q, , if I2 = 0 & P moves towards Q, a current in the same direction, as I1 is induced in Q, if I1 = 0 & Q moves towards P, a current in the opposite, direction to that of I2 is induced in P., when I1 0 and I2 0 are in the same direction then the two, coils tends to move apart., when I1 0 and I2 0 are in opposite direction then the coils, tends to move apart., , Answer Key, , 1, , (a, b), , 2, , (b, c, d), , 3, , (b, c, d), , 4, , (c, d), , Sol. from page 567, , 5, , (a, c, d), , 6, , (b, d), , 7, , (a, c, d), , 8, , (b, d)

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547, , ELECTROMAGNETIC INDUCTION, 9., , A semicircle conducting ring of radius R is placed in the xy plane, as, shown in the figure. A uniform magnetic field is set up along the xaxis. No net emf, will be induced in the ring if, , ,r, K1, , r, , y, , + Q1, , L, B, , + Q2 – Q2, , R, , (a), z, , (a) it moves along the x-axis, (b) it moves along the y-axis, (c) it moves along the z-axis, (d) it remains stationary, Current growth in two-L-R circuits (b) and (c) as shown in figure, (a). Let L1, L2 R1 and R2 be the corresponding values in two, circuits. Then, , Q2 = 2Q1, , V, , t, (a), , 11., , V, , S, , Bulb 2, , R1, , S, , Key, , (a) R1 > R2, (b) R1 = R2, (c) L1 > L2, (d) L1 < L2., The arrangement shown which is confined in a vertical plane has, two rails inclined at angle with horizontal. A horizontal rod of, length moves on the rails with constant speed v, in the region, with transverse field B. Choose the correct alternative(s). The rod, starts moving at time t = 0, A, , r, , C, , 14., r, , (a), (b), , 1 2, Bulb 2 dies as soon as key is switched into position 1, Time in which brightness of bulb 1 becomes half its maximum, brightness does not depend on t., , (c), , If t =, , , total heat produced in bulb 1 is, , C, , B, , Capacitor-1, Rail, , 12., , At t = 0, current in the circuit is, , Switch, Inductor, (a) Frequency of oscillation of charge on left plate of capacitor-, , 2 BVl sin, r, , 1, 2LC, (b) Frequency of oscillation of charge on left plate of capacitor1 is, , Blv sin, r, (c) At any time t (except at t = 0) point A is at higher potential, than point B, (d) At any time t (except at t = 0) point D is at lower potential, than C, Keys K1 and K2 are simultaneously closed at t = 0. At any time t, current through K1 is i and current in inductor is increasing at the, rate x. Current in the resistor is zero. Choose the correct alternative, , (b), , At t =, , Capacitor-2, , L, , Rail, , (a), , L 2, , 2 R22, (d) Ratio of maximum power consumption of bulbs depends on, time, Consider the circuit shown with respective specifications of, elements marked in the figure. Capacitor-1 is charged such that, charge on it is Q0 and it’s left plate is positively charged. While, capacitor-2 is uncharged. The switch is closed at t = 0, , C, , r, , D, , R2, , L, , (c), , (b), , Lx, 2, , (d), , R2, , L2, , R1, , L1, , ir, , (b), , Lx, , Bulb 1, (c), , K2, , Q1, Lx, C, Key is in position 2 for time t. Thereafter, it is in position 1., Resistances of the bulb and inductance of inductor are marked in, the figure choose the correct alternative, , (c), , 13., , ir, , i, (b), , – Q1, , C, , x, , 10., , C, , ,r, , , current in the circuit is, , 1 is, , 1, , 2, LC, , (c) Maximum current through the inductor is, (d) Maximum current through the inductor is, , Answer Key, , 9, , (a, b, c, d), , 10, , (b, d), , Sol. from page 567, , 13, , (a, b, c), , 14, , (a, c), , 11, , (a, b), , 12, , Q0, 2 LC, Q0, LC, , (a, c, d)

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548, 15., , ELECTRICITY & MAGNETISM, , A rod OA of length l is rotating (about end O) over a conducting, ring in crossed magnetic field B with constant angular velocity, as shown in figure, , R, , 16., , In the figure shown the key is switched on at, t = 0. Let I1 and I2 be the currents through inductors having self, inductances L1 & L2 at any time t respectively. The magnetic, energy stored in the inductors 1 and 2 be U1 and U2. Then U1/U2, at any instant of time is, , B, , (a) L1/ L2, , A, , O, 2R, , (b) L2/ L1, (c) I1/ I2, , (a) Current flowing through the rod is, , 3B l 2, 4R, 2, , (b) Magnetic force acting on the rod is, , 3B l, 4R, , 3B 2 l 3, 8R, , Sol. from page 567, , (a, b, c, d), , 1 2, , L2, K, R, , 3, , 3B 2 l 4, 8R, (d) Magnitude of external force that acts perpendicularly at the, end of the rod to maintain the constant angular speed is, , 15, , L1, , (d) I2/ I1, , (c) Torque due to magnetic force acting on the rod is, , Answer Key, , V, , 16, , (b, c)

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ELECTROMAGNETIC INDUCTION, , Statement Questions, Read, (a), (b), (c), (d), 1., , 2., , 3., , 549, , E xercise 8. 3, , the two statements carefully to mark the correct option out of the options given below:, If both the statements are true and the statement - 2 is the correct explanation of statement - 1., If both the statements are true but statement - 2 is not the correct explanation of the statement - 1., If statement - 1 true but statement - 2 is false., If statement - 1 is false but statement - 2 is true., , Statement - 1, Induced emf will always occur whenever there is change in magnetic, flux., Statement - 2, Current always induces whenever there is change in magnetic flux., Statement - 1, An emf can be induced by moving a conductor in a magnetic field., Statement - 2, An emf can be induced by changing the magnetic field., Statement - 1, Figure shows a closed conducting loop and two axes of rotation., The magnetic fields is acted towards right. The emf in the loop will, induce when the loop is rotated about axis-1, , 7., , 8., , 9., , Statement - 1, Only a change in magnetic flux will maintain an induced current in, the coil., Statement - 2, The presence of large magnetic flux through a coil maintain a current, in the coil of the circuit is continuous., Statement - 1, An artificial satellite with a metal surface is moving above the, earth in a circular orbit. A current will be induced in satellite if the, plane of the orbit is inclined to the plane of the equator., Statement - 2, The induced emf across the ends of the moving conductor is given, by e = Bv sin ., Statement - 1, Figure shows an emf e induced in a coil. It happens due to rightward, decreasing current., , Statement - 2, ., , 4., , 5., , 6., , In the coil self induced emf e, , Statement - 2, The emf in the loop will induce when the loop is rotated about the, axis-2., Statement - 1, Lenz s law violates the principle of conservation of energy., Statement - 2, Induced current always opposes the change in magnetic flux, responsible for its production., Statement - 1, An aeroplane flies along the meridian, the potential at the ends of, its wings will be non zero., Statement - 2, Whenever there is change in magnetic flux, emf induces., Statement - 1, Faraday s laws are consequence of conservation of energy., Statement - 2, In a purely resistive ac circuit, the current legs behind the emf in, phase., , 10., , 11., , 12., , L, , di, ., dt, , Statement - 1, Figure shows a metallic conductor moving in magnetic field. The, induced emf across its ends is zero., , Statement - 2, The induced emf across the ends of a conductor is given by e =, Bv sin ., Statement - 1, In the phenomenon of mutual induction, self induction of each of, the coil persists., Statement - 2, Self induction arises due to change in current in the coil itself. In, mutual induction current changes in both the individual coil., Statement - 1, A transformer cannot work in dc supply., Statement - 2, There is no change in flux due to dc., , Answer Key, , 1, , (c), , 2, , (b), , 3, , (d), , 4, , (d), , 5, , (a), , 6, , (c), , Sol. from page 568, , 7, , (c), , 8, , (a), , 9, , (a), , 10, , (a ), , 11, , (a), , 12, , (a)

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550, 13., , ELECTRICITY & MAGNETISM, , Statement - 1, Eddy currents are produced in any metallic conductor when, magnetic flux is changed around it., Statement - 2, , Induced emf in the ring, e, 18., , Electric potential determines the flow of charge., 14., , Statement - 2, d, ., dt, , Statement - 1, Consider the situation shown in figure. When switch is closed, a, clockwise current will flow in the loop., , Statement - 1, The quantity L/R possesses dimensions of time., Statement - 2, To reduce the rate of increases of current through a solenoid should, increase the time constant, , 15., , L, ., R, , Statement - 2, , Statement - 1, , Direction of the current in the loop is according to the Fleming s, left hand rule., , Acceleration of a magnet falling through a long solenoid decreases., 19., , Statement - 2, , A flat, railroad car with a wooden base moves to the right with a, constant velocity. A voltmeter V1 is connected across the rails of, the car and another voltmeter V2 is connected across the axle of the, car and in the car. The reading of V1 is V volt., , The induced current produced in a circuit always flow in such, direction that it opposes the change to the cause that produced it., 16., , Statement - 1, , Statement - 1, An inductor is connected to a battery through a switch. The emf, induced in the inductor is much larger when the switch is opened, as compared to the emf induced when the switch is closed., Statement - 2, Induced emf in an inductor, e, , 17., , L di, ., dt, , Statement - 2, The reading of V2 will be 2V volt., , Statement - 1, Figure shows a horizontal solenoid connected to a battery and a, switch. A copper ring is placed on a smooth surface, the axis of the, ring being horizontal. As the switch is closed, the ring will move, away from the solenoid., , Answer Key, , 13, , (b), , 14, , (b), , 15, , (a), , 16, , (a), , Sol. from page 568, , 17, , (a), , 18, , (a), , 19, , (c), , 20, , (c)

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ELECTROMAGNETIC INDUCTION, , Passage & Matrix, , 4., , E xercise 8. 4, , How much is the electromotive force induced in the circuit at the, moment shown ?, , Passage for Q. 1 to Q. 3, Two capacitors of capacitances 2F and 1F are connected in series with, an inductor of inductance L. Initially capacitors have charge such that, VB – VA = 4 volt and VC- – VD = 1 volt. Initial current in the circuit is, zero. Find :, , 551, , 5., , (a) 0.04 V, , (b), , 0.12 V, , (c) 0.06 V, , (d), , 0.4 V, , How does the current change with time ?, (a) it increases, (b) it remains constant, (c) it decreases, , A B, , C D, , 2F, , F, , (d) first increases then remains constant, 6., , How much power is required for sliding at the instant shown if, friction is negligible?, , L= 1H, , 1., , 2., , 3., , Maximum current that will flow in the circuit, (a), , 3A, , (b), , 6A, , (c), , 9A, , (d), , 12 A, , Potential difference across AB capacitor at that instant (maximum current), (a) 2 V, (b) 3 V, (c) 4 V, (d) 5 V, Frequency of oscillation, , (a), , 8, mW, 15, , (b), , 1.6 mW, , (c), , 4, mW, 3, , (d), , 1.6 W, , Passage for Q. 7 to Q. 9, A circular coil P of 1000 turns and radius 2 cm is placed coaxially at the, centre of another circular coil Q of 100 turns and radius 20 cm., 7., , The mutual induction between the coils is :, , (a), , 0.5 rad/s, , (b), , 0.66 rad/s, , (a), , 8.94 10, , 4, , H, , (b), , 6.08 10, , 4, , H, , (c), , 1.5 rad/s, , (d), , 2.5 rad/s, , (c), , 4.49 10, , 4, , H, , (d), , 3.94 10, , 4, , H, , 8., , Passage for Q. 4 to Q. 6, The mobile side of the triangular conducting frame made of uniform wire, as shown in the figure is slid symmetrically at a uniform speed of v = 0.1, m/s along the two other sides. The horizontal frame is in a vertical, homogeneous magnetic field with an induction of B = 0.4 T. At a certain, instant, the length of each side is l = 1 m and combined resistance of all, three sides is 1 ., , The induced emf in coil P when current in the coil Q decreases, from 5A to 3A in 0.04 s :, , 9., , (a), , 10.28 10, , 3, , V, , (b), , 15.75 10, , 3, , V, , (c), , 19.72 10, , 3, , V, , (d), , 21.72 10, , 3, , V, , Charge passing through coil P if its resistance is 8 ohm :, , B, , v, , 5, , (a), , 9.86 10, , (c), , 14.29 10, , C, , (b), , 12.48 10, , 5, , C, , 5, , (d), , 18.86 10, , 5, , C, , C, , Answer Key, , 1, , (b), , 2, , (b), , 3, , (c), , 4, , (a), , Sol. from page 568, , 6, , (b), , 7, , (d), , 8, , (c), , 9, , (a), , 5, , (b)

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552, , ELECTRICITY & MAGNETISM, , Passage for Q. 10 to Q. 12, A thin non-conducting ring of mass m radius a carrying a charge q rotates, freely about its own axis which is vertical. At the initial moment, the ring, was at rest and no magnetic field was present. At instant t = 0, a uniform, magnetic field is switched on which is vertically downwards and increases, with time according to the law B = B0t. Neglecting magnetism emf induced, due to rotational motion of the ring., , 10., , B0 y ˆ, k is into the paper in the + z direction. B0, a, and a are positive constants. A square loop EFGH of side a and mass m, and resistance R, in xy-plane, start falling under the influence of gravity., Note the directions of x and y axes in figure., A magnetic field B, , The magnitude of electric field at any point of the ring is :, (a), , a B0, 2, , E, , (b), , E, , aB0, , 13., , B0, (d), zero, 2, The angular acceleration of the ring and its direction of rotation as, seen from above, (c), , 11., , Passage for Q. 13 to Q. 15, , (a), , E, , B0 q, m, , (b), , B0 q, 2m, , (a), , 14., , 2B0q, B0 q, (d), m, m, Power developed by the forces acting on the ring as a function of, time is :, 2, , (a), , 2 2, , q B0 a t, 2m, 2, , (c), , 2, , (b), , 2 2, , q B0 a t, 4m, , 15., , none of these, (c), , 16., , F, , B02 av, 2R, , ˆj N, , (b), , B02 a 2v ˆ, j N, (d), R, The terminal speed of the loop is, (c), , (a), (d), , (b), , 2B0av, anticlockwise, R, , 2B0av, B0 av, clockwise, (d), anticlockwise, R, R, The total Lorentz force acting on the loop in, (a), , 2 2, , q B0 a t, m, , B0 av, anticlockwise, R, , (c), , (c), 12., , If v is the speed of the loop at any time, the induced current in the, loop is :, , Column I, , F, , mgR, , (b), , 2 2, , B0 a, , 2mgR, , (d), , B02a 2, , B0a 2v, R, , F, , ˆj N, , none of these, , mgR, B0a, mgR, 2 B02 a 2, , Column II, , (A) Dielectric ring uniformly charged, , (p), , Constant electrostatic field out of system, , (B) Dielectric ring uniformly charged, , (q), , Magnetic field strength, , (C) Constant current in ring i, , (r), , Electric field (induced), , (D) i = i0cos t, , (s), , Magnetic dipole moment, , rotating with angular velocity, , Answer Key, , 10, , (a), , 11, , (b), , 12, , (b), , 13, , (a), , Sol. from page 568, , 14, , (c), , 15, , (a), , 16, , A-p ; B -p, q, s ; C-q, s ; D- r

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ELECTROMAGNETIC INDUCTION, 17., , Consider a uniform transverse magnetic field confined to a cylindrical region (with axis passing through O) as shown, in the figure. Also shown is an imaginary rectangualr loop abcd, with midpoint of its side cd at O. Magnetic field in the, region is increasing at a constant rate. Match the entries in Column I to the all possible entries in Column II., A., B., C., D., , Column I, Potential difference between points a and c, Va Vc, Work done in carrying charge q from point c to a, Component of electrostatic field at point a along side ad,, assuming direction from a towards d to be positive, Component of induced electric field at point c and cb,, assuming direction from c towards b to be positive, , Column II, (p) Zero, (q) Positive, (r) Negative, (s) Not defined, (t) May have non-zero value, , 18., , Column I gives some incomplete statements. Column II gives some completing statement. Match appropriately., Column I, Column II, A. A rod rotates in a uniform transverse magnetic field as, (p) May be zero, shown, about hinge at O. Potential difference between, points A and B., B, , A, , O, , B., , A conducting loop is moved in a region of transverse, magnetic field, downward the plane of paper, as shown., Value of induced current i __________, , (q) Must be zero, , i, v, , C., , If a constant force F is acting on the wire, rate of charge q, , (r) is/may be non-negative, , dq, stored by the capacitor,, _________, dt, , +q, q, , D., , F, , A square loop is rotated about diagonal in a region of uniform (s) is/may be non-positive, magnetic field as shown. Value of i at an instant__________, , (t) May have non-zero value, , Answer Key, , 17, , A- p ; B- p, q, r, t ; C-p ; D-q, t, , Sol. from page 568, , 18, , A-p, r, s,t ; B-p, r, s, t; C- r, s, t; D -p, r, s, t, , a, d O, , 553, b, c

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554, , ELECTRICITY & MAGNETISM, , Subjective Integer Type, , Exercise 8.5, Solution from page 569, , 1., , The magnetic flux through the loop shown in figure increases, according to the relation, , 2, , B, , 7.0 t 6.0 t , where, , B, , 4., , is in, , milliweber and t in second., (a) What is the magnitude of emf induced in the loop when t =, 2.0 s ?, (b) What is the direction of the current through R ?, , 5., , 2., , Ans : 30 mA, In figure, the square loop of wire has sides of length 2.0 cm. A, magnetic field is directed out of the page; its magnitude is given by, B = 4.0 t2y, where B is in tesla, t in second, and y in metre., Determine the emf around the square at t = 2.5 s and give its, direction., , Ans : (a) 31 mV (b) right to left., Two concentric coplanar circular loops made of wire, with, resistance per unit length 10, , 3., , is, In figure a 120-turn coil of radius 1.8 cm and resistance 5.3, placed outside a solenoid. The current in the solenoid is 1.5 A and, it reduces to zero at a steady rate in 25 ms. What current appears, in the coil ? The number of turns per unit length of the solenoid is, 220 turns/cm and its diameter D= 3.2 cm., , / m have diameters 0.2 m and 2, , m. A time varying potential difference (4 + 2.5 t) volt is applied to, the larger loop. Calculate the current in the smaller loop., Ans : 1.25 A, Consider the situation shown in figure. The wires P1Q1 and P2Q2, are made to slide on the rails with the same speed 5 cm/s. Find the, electric current in the 19, resistor if :, (a) both the wires move towards right and, (b) if P1Q1 moves towards left but P2Q2 moves towards right., , 6., , 7., , Ans : 80 V, clockwise., The current in a coil of self-induction 2.0 henry is increasing, according to i = 2 sin t2 ampere. Find the amount of energy spent, during the period when the current changes from 0 to 2 ampere., Ans : 4 J, The current in an RL circuits drops from 1.0 A to 10 mA in the, first second following removal of the battery from the circuit. If L, is 10 H, find the resistance R in the circuit., Ans : 46, , Ans : (a) 0.1 mA, (b) zero.

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ELECTROMAGNETIC INDUCTION, , Subjective, , 555, , E xercise 8. 6, Solution from page 570, , 1., , Ans : e, 2., , 4., , 5., , 0ni0 A, , dB, dt, , 0.02, , cos t ., , T, . Find the current, s, , induced in the frame. Resistivity of copper = 1.7×10–8 W-m., Ans : 9.3×10–2 A., A conducting circular loop having a radius of 5.0 cm, is placed, perpendicular to a magnetic field of 0.50 T. It is removed from the, field in 0.50 s. Find the average emf produced in the loop during, this time., Ans : 7.8×10–3 V., (a) The magnetic field in a region varies as shown in figure., Calculate the average induced emf in a conducting loop of, area 2.0×10–3 m2 placed perpendicular to the field in each of, the 10 ms intervals shown., (b) In which intervals is the emf not constant ? Neglect the, behaviour near the ends of 10 ms intervals., , 1, R, , B, , 0, , B, , t, , and is independent, , of the way B has changed., If B(t) = B(0) in a particular case, we have q(t) = 0. Is the, induced current necessarily zero throughout the interval from, 0 to t ?, Ans : (b) no., A conducting loop of area 5.0 cm2 is placed in a magnetic field, which varies sinusoidally with time as B = B0 sin t where B0 =, 0.20 T and =300 s–1. The normal to the coil makes an angle of, 60° with the field. Find (a) the maximum emf induced in the coil,, (b), , 6., , (b) the emf induced at t, , t, 7., , Ans : (a)–2.0 mV, –4.0 mV, 4.0 mV, 2.0 mV; (b) 10 ms to 20, ms and 20 ms to 30 ms., In figure let the flux through the loop be B(0) at time t=0. Then let, the magnetic field B vary in continuous but unspecified way, in, both magnitude and direction, so that at time t the flux is represented, by B(t)., , Show that the net charge q (t) that has passed through resistor, R in time t is q t, , A uniform magnetic field B exists in a direction perpendicular to, the plane of a square frame made of copper wire. The wire has a, diameter of 2 mm and a total length of 40 cm. The magnetic field, changes with time at a steady rate, , 3., , (a), , A small loop of area A is inside of, and has its axis in the same, direction as, a long solenoid of n turns per unit length and current, i. If i = i0 sin t, find the emf induced in the loop., , 600, , s., , 900, , s and (c) the emf induced at, , Ans : (a) 0.015 V (b) 7.5×10–3 V (c) zero., , Figure shows a horizontal magnetic field which is uniform above, the dotted line and is zero below it. A long, rectangular, conducting, loop of width L, and mass m and resistance R is placed partly, above and partly below the dotted line with the lower edge parallel, to it. With what velocity should it be pushed downwards so that, it may continue to fall without any acceleration ?, , Ans : V, , 8., , 9., , mgR, , B 2 L2, An air plane, with a 20 m wing spread is plying at 250 m/s straight, south parallel to earth s surface. The earth s magnetic field has a, horizontal component of 2×10–5 Wb/m2 and angle of dip is 60°., Calculate the induced emf between the plane tips.Ans : 0.173 V., Figure shows a square loop of side 5 cm being moved towards, right at a constant speed of 1 cm/s. The front edge enters the 20, cm wide magnetic field at t = 0. Find the emf induced in the loop, at (a) t = 2s, (b) t = 10 s, (c) t = 22 s and (d) t = 30 s., , Ans :(a) 3×10–4 V, (b) zero, (c) 3×10–4 V and (d) zero.

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556, 10., , ELECTRICITY & MAGNETISM, , Figure shows a circular wheel of radius 10.0 cm whose upper half,, shown dark in the figure, is made of iron and the lower half of, wood. The two junctions are joined by an iron rod. A uniform, magnetic field B of magnitude 2.00×10–4 T exists in the space, above the central line as suggested by the figure. The wheel is set, into pure rolling on the horizontal surface. If it takes 2.00 second, for the iron part to come down and the wooden part to go up, find, the average emf induced during this period., 14., , 11., , Ans : 1.57×10–6 V., A pair of parallel horizontal conducting rails of negligible resistance, shorted at one end is fixed on a table. The distance between the, rails is L. A conducting massless rod of resistance R can slide on, the rails frictionlessly. The rod is tied to a massless string passes, over a pulley fixed to the edge of the table. A mass m tied to the, other end of the string hangs vertically. A constant magnetic field, B exists perpendicular to the table. If the system is released from, rest, calculate, (a) the terminal velocity achieved by the rod and, (b) the acceleration of the mass at the instant when the velocity, of the rod is half the terminal velocity., Ans : (a) VT, , 12., , A current i, , 3.36 1 2t, , mg R, B 2 L2, , 15., , (b) g/2., , Ans : (a) =BA cos t (b) e = BA sin t, A flexible circular loop 10 cm in diameter lies in a magnetic field, 1.2 T, directed into the plane of the diagram in figure. The loop is, pulled at the points indicated by the arrows, forming a loop to, zero area in 0.2 s., (a) Find the average induced emf in the circuit., (b) What is the direction of the current in R ?, , Ans : (a) 0.0471 V (b) From a to b., The cube is figure, 1 m on a side, is in uniform magnetic field of 0.2, T, directed along the positive y-axis. Wires A, C, and D move in, the direction indicated, each with a speed of 0.5 m/s., (a) What is the motional emf in each wire ?, (b) What is the potential difference between the terminals of, each ?, , 10 2 A increases at a steady rate in a, , long straight wire. A small circular loop of radius 10 –3 m has its, plane parallel to the wire and is placed at a distance of 1 m from, the wire. The resistance of the loop is 8.4 10, , 13., , 4, , . Find the, , magnitude and the direction of the induced current in the loop., Ans : 5.0×10–11, anticlockwise., The rectangular loop is figure, of area A and resistance R, rotates at, uniform angular velocity about the y-axis. The loop lies in a, uniform magnetic field B in the direction of the x-axis. Sketch the, following graph :, (a) the flux through the loop as function of time (let t = 0 in the, position shown);, (b) the induced emf in the loop., , 16., , Ans : (a) 0, 0.0707, 0.141 V (b) 0, 0.0707, 0.141 V, The magnetic field within a long, straight solenoid of circular cross, , dB, ., dt, What is the rate of change of flux through a circle of radius r1, inside the solenoid, normal to the axis of the solenoid, and, with centre of the solenoid ?, , section and radius R is increasing at a rate, (a), , (b), , (c), , Find the induced electric field En inside the solenoid, at a, distance r1 from its axis. Show that the direction of this field, in a diagram., What is the induced electric field outside the solenoid at a, distance r2 from the axis ?, , (d), , What is the induced emf in a circular turn of radius, , (e), (f), , of radius R ?, of radius 2 R ?, , R, ?, 2

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557, , ELECTROMAGNETIC INDUCTION, 2, Ans :(a) r1, , (d), 17., , R2, 4, , dB, dt, , dB, dt, , r1, 2, , (b), , (e) R, , 2, , dB, dt, , (c), , R2, 2r2, , (e), , dB, dt, , (f), , (g) Find the electrostatic field Ee in each side., (h) Find the potential differences Vac, Vce, Veg, and Vga. What, should be the sum of these potential differences ?, , dB, 2 dB, (f) R, dt, dt, , The magnetic field B at all points within the shaded circle of figure, is 0.5 T. It is directed into the plane of the diagram and is decreasing, at the rate of 0.1 T/s., , 20., (a), (b), , (c), (d), , 18., , Find the induced field En in each side of the square loop., Find the induced emf in each side, , What is the shape of the field lines of the induced En -field, within the shaded circle ?, What are the magnitude and direction of this field at any, point of the circular conducting ring of radius 10 cm, and, what is the emf in the ring ?, , Ans : (a) zero (b) 4 mV (c) 2 mA (d) zero (e) 5.00 mV/m (f) 1.00, mV (g) zero (h) zero, A wire is bent into three circular segments, each of radius r = 10, cm, as shown in figure. Each segment is a quadrant of a circle, ab, lying in the xy-plane, bc lying in the yz-plane, and ca lying in the, zx-plane. (a) If a uniform magnetic field B points in the positive, x-direction, what is the magnitude of the emf developed in the, wire when B increases at the rate of 5.0 mT/s ? (b) What is the, direction of the current in segment bc ?, , ?, What is the current in the ring, if its resistance is 2, What is the potential difference between points a and b of, the ring ?, , Ans :(a) Circles, clockwise (b) 0.005 V/m, 3.14 mV (c) 1.57 mA, (d) zero, A square conducting loop, 20 cm on a side, is placed in the magnetic, field 0.5 T, directed into the plane of the diagram and decreasing at, the rate of 0.1 T/s., , 21., , Ans : (a) 39.25 V, (b) from c to b., A rectangular coil of N turns and of length a and width b is rotated, at frequency f in a uniform magnetic field B, as indicated in, figure. The coil is connected to co-rotating cylinders, against which, metal brushes slide to make contact., , (a), , Prove that the component of En along the loop has the, same value at every point of the loop and is equal to that at, the ring of previous problem., , ?, What is the current in the loop, if its resistance is 2, What is the potential difference between points a and b ?, Ans : (b) 2.00mA (c) zero, A square conducting loop, 20 cm on a side, is placed in the magnetic, field 0.5 T, directed into the plane of the diagram with the side ac, along the diameter and b at the centre of the field and decreasing at, the rate of 0.1 T/s., (a) What is the induced emf in side ac ?, (b) What is the induced emf in the loop ?, (b), (c), , 19., , (c), (d), , What is the current in the loop, if its resistance is 2, ?, What is the potential difference between points a and c ?, Which is the higher potential ?, , (a), , Show that the emf induced in the coil is given (as a function, of time t) by e, , (b), , e0 sin 2 ft ,, , where e0 2 f NabB. This is the principle of the, commercial alternating-current generator., Design a loop that will produce an emf with e0 = 150 V when, rotated at 60.0 rev/s in a uniform magnetic field of 0.500 T., Ans : (b) Nab, , 5, 2, , m2

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558, 22., , ELECTRICITY & MAGNETISM, , A wire loop enclosing a semi-circle of radius a is located on the, boundary of a uniform magnetic field of induction B (figure). At, the moment t = 0 the loop is set into rotation with a constant, angular acceleration about an axis O coinciding with a line of, , 25., , Two long parallel wires of zero resistance are connected to each, other by a battery of 1.0 V. The separation between the wires is, 0.5 m. A metallic bar, which is perpendicular to the wires and of, resistance 10, , vector B on the boundary. Find the emf induced in the loop as a, function of time t. Draw the approximate plot of this function., The arrow in the figure shows the emf direction taken to be, positive., , , moves on these wires when a magnetic field of, , 0.02 T is acting perpendicular to the plane containing the bar and, the wires. Find the steady state velocity of the bar. If the mass of, bar is 0.002 kg, find its velocity as a function of time, Ans : V, 26., , 1, n, 1 Ba 2 t , where n = 1, 2, ....... is the number of, 2, the half-revolution that the loop performs at the given moment t., The plot ei(t) is shown in figure where tn, , 0.005t, , ), , At a given instant the current and self-induced emf in an inductor, are directed as indicated in figure., (a), , Is the current increasing or decreasing ?, , (b), , The induced emf is 25 V and the rate of change of the current, is 25 kA/s; find the inductance., , Ans : ei, , 2 n, , 100(1 e, , Ans : (a) The current in the inductor decreasing. (b) 1 mH., 27., , A coil of inductance 1 H and resistance 10, , is connected to a, , resistanceless battery of emf 50 V at time t = 0. Calculate the ratio, of the rate at which magnetic energy is stored in the coil to the rate, at which energy is supplied by the battery at t = 0.1 s., , ., , Ans : 0.36, 28., , The current (in ampere) in an inductor is given by I = 5 + 16t,, where t is in second. The self-induced emf in it is 10 mV. Find, (a), , the self-inductance and, , (b), , the energy stored in the inductor and the power supplied to, it at t = 1s., Ans : (a) 6.25 × 10–4 H (b) 210 mW., , 29., 23., , Consider the situation shown in figure. The wire PQ has a negligible, resistance and is made to slide on the three rails with a constant, speed of 5 cm/s. Find the current in the 10, , A solenoid having an inductance of 6.30 H is connected in series, with a 1.20 k, , resistor when the, , (a), , If a 14.0 V battery is switched across the pair, how long will, it take for the current through the resistor to reach 80.0 % of, its final value ?, , (b), , What is the current through the resistor at time t = 1.0 L ?, Ans : (a) 8.45 ns (b) 7.37 mA., , switch S is thrown to (a) the middle rail (b) the bottom rail., , 30., , resistor.., , In figure = 100 V, R1= 10.0 , R2, , 20.0 , R3, , 30.0, , and L, , = 2.00 H. Find the values of i1 and i2, , 24., , Ans : (a)0.1 mA, (b) 0.2 mA., A very small circular loop of area 5×10–4 m2, resistance 2 ohm, and negligible inductance is initially coplanar and concentric, with a much larger fixed circular loop of radius 0.1 m. A, constant current of 1 ampere is passed in the bigger loop and, the smaller loop is rotated with angular velocity rad/s about a, diameter. Calculate (i) the flux linked with the smaller loop, (ii), induced emf and (iii) induced current in the small loop, as a, function of time., Ans : (i), , 10, , 9, , cos t (ii), , (iii) 1.57×10–9, , 10, , 9, , sin t V, , sin t ampere., , (a), , immediately after the closing of switch S,, , (b), , a long time later, (c) immediately after reopening of switch, S, and (d) a long time after the reopening., , Ans : (a) i1, (c) i1, , i2, , 0, i2, , 3.33 A; (b) i1, , 4.55 A, i2, , 1.82 (reversed); (d) i1, , 2.73 A;, , i2, , 0.

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ELECTROMAGNETIC INDUCTION, 31., , In the circuit shown in figure, switch S is closed at time t = 0., Thereafter, the constant current source, by varying its emf,, maintains a constant current i out of its upper terminal. (a) Derive, an expression for the current through the inductor as a function of, time (b) Show that the current through the resistor equals the, current through the inductor at time t, , L, R, , n 2., , Ans : (a) i 1 e, 32., , 35., , Rt / L, , An inductor of inductance L = 400 mH and resistors of resistances, , R1, , 2, , and R2, , 2, , are connected to a battery of emf, , = 12 V as shown in the figure. The internal resistance of the, battery is negligible. The switch S is closed at t = 0., What is the potential drop across L as a function of time ? After, the steady state is reached, the switch is opened. What is the, direction and the magnitude of current through R1 as a function of, time ?, , Ans : VL, , e, , R 2t / L, , 6e, , 10t, , ., , In a circuit shown A and B are two cells of same emf but different, internal resistances r1 and r2 (r1 > r2) respectively. Find the value, of R such that the potential difference across the terminals of cell, A is zero a long time after the key K is closed., , Ans : R, 34., , , i, , Ans : (a) VA – VB = –5V., (b), (i) i0 = 0.6 A, (ii) 1.386×10–3 s, 4.5×10–4 J., An inductor of inductance 2.0 mH is connected across a charged, capacitor of capacitance 5.0 F and the resulting LC circuit is set, oscillating at its natural frequency. Let Q denote the instantaneous, charge on the capacitor and I the current in the circuit. It is found, that the maximum value of charge Q is 200 C., (a), , dI, , When Q = 100 C, what is the value of dt ?, , (b), (c), (d), , 36., , 33., , 559, , When Q = 200 C, what is the value of I ?, Find the maximum value of I., When I is equal to one-half its maximum value, what is the, value of |Q |, Ans :(a) 104 A/s (b) I = 0 (c) Imax = 2A (d) Q = 173.2 C., A metal rod OA of mass m and length r is kept rotating with a, constant angular speed in a vertical plane about a horizontal axis, at the end O. The free and A is arranged to slide without friction, along a fixed conducting circular ring in the same plane as that of, rotation. A uniform and constant magnetic induction B is applied, perpendicular and into plane of rotation as shown in figure. An, inductor L and an external resistance R are connected through a, switch S between the point O and the point C on the ring so as to, form an electrical circuit. Neglect the resistance of the ring and the, rod. Initially the switch is open., (a) What is the induced emf across the terminals of the switch ?, (b) The switch S is closed at time t = 0., (i) Obtain an expression for the current as a function of time., (ii) In the steady state obtain the time-dependence of the torque, required to maintain the constant angular speed, given that, the rod OA was initially along the positive x-axis at time t =0., , 4, r1 r2, 3, , A circuit containing a two position switch S is shown in figure., (a) The switch S is in position 1. Find the potential difference, VA – VB and the rate of production in joule heat in R1., (b) If now the switch S is put in position 2 at t = 0, find, (i) Steady current in R4 and, (ii) the time when the current R4 is half the steady value. Also, calculate the energy stored in the inductor L at that time., , Ans : (a), , B r2, (b) (i) i, 2, (ii), , B2 r 4, 4R, , B r2, 1 e, 2R, , mgr, cos t., 2, , Rt / L

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560, 37., , ELECTRICITY & MAGNETISM, , Figure shows, in cross-section, two coaxial solenoids. Show that, the mutual inductance M for a length of this solenoid-solenoid, combination is given by M, , R12, , 39., , 0 n1n2 in which n1 and n2, , toroid-coil combination is M, , are the respective numbers of turns per unit length and R1 is the, radius of the inner solenoid. Why does M depend on R1 and not, on R2 ?, , 38., , Figure shows a coil of N2 turns wound as shown around part of a, toroid of N1 turns. The toroid s inner radius is a, its outer radius, is b, and its height is h. Show that the mutual inductance M for the, 0 N1N 2 h, , 2, , n, , b, ., a, , A rectangular loop of N close-packed turns is positioned near a, long straight wire as shown in figure. What is the mutual inductance, M for the loop-wire combination ?, 40., , Ans :, , 0, , 2, , N, , n 1, , b, a, , A transformer on a utility pole operate at VP = 8.5 kV on the, primary side and supplies electrical energy to a number of nearby, houres at Vs=120 V, both quantities being rms values. Assume an, ideal step-down transformer, a purely resistive load, and a power, factor of unity., (a) What is the turns ratio NP/Ns of the transformer ?, (b) The average rate of consumption in the houres served by the, transformer is 78 kW. What are the rms currents in the, primary and secondary of the transformer ?, (c) What is the resistive load Rs in the secondary circuit ? What, is the corresponding relative load RP in the primary ?, Ans : (a), , 71 (b) 9.2 A, 650 A (c) RP = 926, , Rs= 0.18, , .

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ELECTROMAGNETIC INDUCTION, , 561, , S olutions Exercise 8.1Level -1, 1., , (b), , |e|=, , dB, d, = A, dt, dt, dB, dt, , As the slope, 2., , (c), , (c), , (d), , , and i3, , R, , L, ,, R, , 2, , L, and, 2R, , L, R/2, , 3, , (a), , BA, , AB0e, , At t = 0 , | e |, , 2L, ., R, , (a), , Bkˆ ., , 2, 2, , |e|, R, , (a), (a), , 1V, , e, 1, 0.1 A ., R 10, There is no change in the flux and so e = 0., , Ns, Np, , 500, 100, 50, , 9., , (d), , Vs = V p, , 10., , (d), , According to Lenz's law, when switch is closed, the flux in, the loop increases out of plane of paper, so induced current, will be clockwise., , 11., , (d), , dB, d, |e| =, = A, dt, dt, , 1000 V ., , 12., 13., , (b), (a), , 6.5, 10, , (a), , 16., , (d), , (d), , 4i, di, 2i, 4 f i 4 500 50 105 A/s, (1/ f ), dt T / 2, Due to the heating effect of current, the resistance of loop, increases and so current in the first loop decreases, whose, flux in second loop decreases. To compensate this flux the, second loop will come towards first loop., For enduced emf across the ends of the rod, one component, , 18., , (d), , 19., , (c), , 20., , (d), , 21., , (d), , 22., , (a), , to and B ., Till loop remains completely inside the magnetic field, the, flux will not change and hence induced emf/current will be, zero., We know that, N = Li, , 25., , (d), , 8 10, , 3, , 5 10, 400, , 3, , 10, , 7, , Wb, , 0, Wb ., 4, Because of inductor in bulb B2, current decreases slowly., When electron approaches nearby the loop flux inside loop, will increase and when electron recedes from the loop the, flux inside loop decreases and so current change in direction., The induced emf, e = Bv × (PQ) sin 90°, = Bv × 2R,, = 2 RvB., , =, , (c), (d), , /2, , Due to the change in flux in the coil enduced current starts, flowing alternatively in it, which opposes the motion of the, magnet., As there is no change in flux in the cylinder and so induced, emf in it will be zero., The velocity component of conductor must be perpendicular, , Li, N, , 23., 24., , must be, , perpendicular to B ., The two ends of the rod are at the same potential and so, e = 0., , /2, , 0.65 A ., , If eb is the back emf in the motor, then, , di, is to be maximum. In case (c) it is ;, dt, , of the velocity must be perpendicular to ; also, , = 100 × 10–4 × 150 = 1.5 V., , For decreasing flux into the plane of paper, this emf must be, added to battery emf, so e net = 5 + 1.5 = 6.5 V, and, i, , di, ., dt, , AB0b, ., R, , i=, , 7., 8, , 15., , 17., , AB0be bt, , AB0b , and current , i, , dB, e= A, =1, dt, , eb =100 V., In inductor,, e= L, , bt, , d, dt, , 115 – eb, 1.5, , For maximum emf,, , As 3 is greatest, and so it takes greatest to reach any specific, value., By right hand screw rule the direction of magnetic field will, , Induced emf, | e |, , 6., , (c), , eb, R, , 10 =, , ., , be out of the plane of the conductor. i.e., B, 5., , 14., , Clearly i3 i2 i1 ., Time constant of the circuits are ;, 1, , 4., , 2R, , or, , is greatest in II, and so | e | is greatest in, , this region., Just after closing of the switch, inductor offers infinite, resistance and so,, i1 = 0, i2 =, , 3., , i=

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562, 26., , (d), , 27., , (c), , 28., , (b), , ELECTRICITY & MAGNETISM, The magnetic field of the loop will lie parallel to xy-plane and, so there is no flux in this plane., With the increase in current in loop A, flux in loop B will, increase, so to compensate this flux loop B will be repelled., The induced emf, e = Bv = 0.5 × 2 ×1 = 1V., 12 1, W., 6 6, Magnetic field of one coil passes through other coil in case A, is maximum, and so mutual induction is also maximum., , 36., , (b), , = 1.5, , 2, The rate of work done, = e / R, , 29., , (a), , 30., , (d), , 31., , (b), , Resistance of coil R, , r2, ( / 4), , half radius, R ', , r/2, , 2, , =, 37., , (b), , 38., , (c), , . With the quadrupled turns and, , 33., , 34., , t/, , L, R, , 8.4 10, 6, , 3, , 39., , (a), , 40., , (a), , 1.4 10 3 s, , (a), , 3, 12, or 1, (1 e t /1.4 10 ), 6, t = 1 × 10–3s., As the magnetic field increases, its flux also increases into, the page and so induced current in bigger loop will be, anticlockwise. i.e., from D to C in bigger loop and then from, B to A in smaller loop., , or, , 2, E×2 r= r, , E=, , 41., , (c), , (a), , Rtotal =, , di, ., dt, , 6 12, 6 12, , 2, 8, , L, Rtotal, , 4 8, , 1, s., 4, , V, R, , Kt 3, ,, R,, , t, , iL, , Kt 3, dt, L, 0, , Kt 4, ,, 4L, , dq d (CV ) d (CKt 3 ), 3CKt 2 ., dt, dt, dt, The distance falls by an object in 1 second under gravity, and iC =, , d, dt, , 42., , (a), , 1 2 1, gt, 10 12 5m, 2, 2, The acceleration of magnet will be less than g (a < g) and so, distance falls will be less than 5 m., , dB, dt, , y=, , r, 4 B0 f, 2, , 1 4 0.01 100, 2, = 2 volt/m, When magnet enters through the coil, the rate of change of, magnetic flux starts decreasing and when magnet emerges out, from the coil, the rate of change of magnetic flux again starts, increasing, so(a) is the correct option., , 43., 44., , (c), (a), , The given inductors are in parallel in the circuit, and so, L 3, 1 H., 3 3, The presence of inductor will cause slow in decay of current, in both the bulbs., The flux in outer loop will increases into the plane of paper, and so induced current will be anticlockwise in outer loop., i.e, from d to c. This current flows into the smaller loop from, a to b., , Leq =, , 45., , (d), , 46., , (d), , S olutions Exercise 8.1Level -2, 1., , 2., , (d), , (c), , In I and III case the rate of change of flux is equal. But in II, case, it is zero., Induced emf. | e | =, , dB, d, = A dt, dt, , 2 2, =, 2, , ve, , Given, V = Kt3,, iR =, , =, , 35., , 191 V, , Induced emf in the coil, e = – L, , C, , 2 B0, B, =, = 4B0 f, t, T /2, , Now | e | = E × 2 r =, , 1.5 I, T, , di, di, di, 0 , then, – ve and finally, dt, dt, dt, Accordingly e = 0 , e = +ve, and finally e = –ve., , R, and so power dissipated, , i, , (b), , 4, , Initialy,, , (d), , i0 (1 e, , ) ;, , 2I, T /2, , 6 1, 2 / 200, , remains same., 32., , i, t, , e= M, , d, (0.042 – 0.87t ), dt, , = 2 × 0.87 = 1.74 V, , 3., , (c), , The net induced emf = 20 + 1.74 = 21.74 V, Induced electric field is given by, r dB, E = 2 dt, , In situation A : E = rA ×, B : E = rB, , B1, t1, , B1, 2 t1

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ELECTROMAGNETIC INDUCTION, B1, t1 / 2, , C : E = rC, 4., , (c), , 10., , (d), , e, , Clearly induced electric field is greater in case C., The decay of current in RL-circuit is given by, i = i0, , e–t/, , R, , e, , =, , d, dt, , A, , r 2 B0e, , t, , r2, , Power developed, P, i, , R0, , i, , e, , 2 R0, , i, , R0, , tR0 / L0, , e, , t 2 R0, L0, , (a), , 12., , e, , t, , (b), , 2 R0, 2 L0, , Amplitude in case (1) and (3) are equal and in case (2) and (4), are equal., 5., , (b), , The change in flux = n(, , 2, , –, , 2 – 1), ., 5 RT, , total, , 13., , (b), , 14., , (b), , v, , or, , 3, i0, 4, , A, , ( r 2 B0 ) 2, R, , d ( B0, dt, , or, , e, , or, , e4 /, , 0 i, (2sin 45 ), 4 L/2, , i0 (1 e, , i0 (1 e, , 2 4, , r B0 2, /, R, , t), , 4/, , t/, , ), , ), , 1, 4, , 4/, , 4, ln 4, 2, s, ln 2, , Till front side of the loop moves into the field the emf induced, e = Bv across it. When rear side comes inthe field, the emf is, induced across it., The flux through the coil is given by, = BA cos 0°, = (B × r2), Now | e | =, , = Blarge Asmall, =, , dB, dt, , We know that, i, , or, , e, v, , (b), , n(, , The induced emf across the sides AB and CD are shown in, figure and so net emf of the device becomes, = 2Bv, e, , 7., , A, , 1), , / t, R, , Thus induced current, i =, (b), , d, dt, , 4, , Total resistance of the circuit = 4 R + R = 5R, , 6., , |e|=, , e2, R, , =A ., , tR0, 2 L0, , i, e, 2 R0, , i, , 11., , d, ( B0e t ), dt, , r 2 B0 ., , At t = 0, e, , tR, L, , dB, dt, , 563, , 2, , Initially, , dr, dt, , d, dt, , 2 Br, , dr, ., dt, , 0 , and so e = 0. After that, , dr, dt, , is of constant, , 2, , On comparing with, 8., , (b), , En 2 r = – R 2, , Mi , we get M, , total, , L, , dB, dt, , 1, ., r, Till current in coil changes, the induced current I2 will occur, in the ring and the magnetic field in the ring increases so, product I2B increases. When I1 becomes steady, I2 becomes, zero and so product I2B becomes zero., , value, also r increases, so e is also increases. Finally, 15., , (c), , 16., , (a), , En, , 9., , (d), , becomes zero and so e becomes zero., From t = 0 to 5s : Emf will induced across the front edge., From t = 5 to 15s flux changes will be zero. Finally from 15, to 20s, emf is induced across rear edge of the loop., e = Bv = 0.6 × 0.01 × 0.05 = 3 × 10–4V., The induced emf across the sliding wire, e = Bv = 0.5 × 4 × 0.25 = 0.5 V, The effective circuit is shown in figure., , I1, 4, , Coil, ring, , dr, dt, , i, 0.5, 0.5 V, , 2, , The equivalent resistance of the circuit

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ELECTROMAGNETIC INDUCTION, , 567, , S olutions Exercise 8.2, 1., 2., , 3., , (a, b), , In case I and II, the induced emf, e = Bv(2 )., In case III and IV, the induced emf, e = Bv ., (b, c, d) At t = 0, inductor offers infinite resistance, and so current, in R2 is zero and in R1 is maximum. After a long time, inductor offers zero resistance and so both the resistors, will get their maximum values., L, (b, c, d), , CR and, R, , (c, d), , 5., , (a, c, d) For P1 = P2, we have V1i1 V2i2, di, i1, dt, , L2, , 2, =, 8, , (b, d), , 2, 8, , 4, 1, , – + ir – + ir +, , or, , 1, 4, , 8., 9., 10., , 2, , 4, , – + ir – + ir +, B, , ., , 11., , v, , so it does not depend on mass, 2g, , (a,b), , V, or R1, R2, , A, , r, , D, , r, , 13., , i0, , 0, , ...(ii), , Q2, =0, C, , R2, , Energy stored in the inductor, 1 2, Li0, 2, , U, , 1, = 2L R, 2, , 2, , L 2, 2 R22, , When key is in position 1, this energy will convert into, heat energy through resistor., 14., , e = Bv sin, The induced emf across the rod,, e = Bv sin, , Also VA > VB ., , ...(i), , Q2, Q1, = 2( – ir) = 2 Lx = 2, C, C, Q2 = 2Q1, (a,b,c) The maximum current in the inductor, , B, , e, 2 Bv sin, current, i = r 2 =, r, , F, , or, , R2 . Also L1 < L2., , The equivalent resistance between B and D is =, , K2, , Q1, + Lx = 0, C, , Q1, = Lx, C, Now in close loop ABGFEA,, , of the ring., (a, c, d) When magnetic field decreases, the flux of which in loops, decreases downwards and so induced current in the loops, will be clockwise. And no current in the connecting wire., (b, d), Use Lenz's law to get the answer., (a, b, c, d) In all the given cases, there is no flux and change of flux, across the semicircular ring., (b, d), As steady current in both the cases in same and so, V, R1, , – 2Lx +, , Lx, , or, , A, , –Q1, , Q1, + Lx = 0, C, Q1, C, , ir, , 2, , or, , 2, 8, , C, , –Q2, , In close loop ABCDA, – ir = Lx, In close loop, ABGFCDA, , Current in first ring must be greater and so, Height attained h, , +Q2, , ,r, , +Q1, , C, , 2, , 7., , C, , E, , 1, ., 4, , G, , L, , D, , di, i2, dt, , V2, L2 (di / dt ), V1 = L1(di / dt ), , 1, L2i22, W2, 2, =, W1, 1 2, L1i1, 2, , 6., , i, , B, r, , K1, , i1, L2, =, i2, L1, , Now, , ,r, , i, , A, , LC have dimensions of time and so, , 4., , L1, , (a,c,d), , 1, have dimensions of frequency.., LC, , R 1, ,, and, L CR, , or, , 12., , (a, c), , f, , 1, LCeq, , 2, , 1, 2 LC, , di, 0., dt, Charge on two capacitors is same, , When i is max,, r, 2, , 1 2, Li0, 2, , i0, , Q0, 2, 2C, , 2, , Q0, 2 LC, , ., , 2, , Q0 2, 4C

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568, 15., , ELECTRICITY & MAGNETISM, I, , (a, b, c, d), , 3, 2R, , 2R, 3, 3, 2R, , B l, 2, , 3B 2 l 3, 4R, , Force to be applied at the end, , 3B l 2, 4R, , 1, B l2, 2, , 16., , 2, , (b, c), , R, , e2 or L1, V1, V2, , Now, , Magnetic force F, , 3B l, 4R, , l, , B, , 3B l, 4R, , =, , 3, , di1, dt, , e1i1, e2i2, , 2R, 2, , 3B 2 l 3, ., 8R, , As inductors are in parallel, so, , e1, , 2, , 3B 2 l 4, 8R, , l, 2, , L2, , di2, dt, , L1i1, , L2i2, , i1, ,, i2, , L2, ., L1, , S olutions Exercise 8.3, 1., , (c), , 2., , (b), , 3., , (d), , 4., 5., 6., , (d), (a), (c), , 7., 8., , (c), (a), , 9., 10., , (a), (a), , 11., 12., 13., 14., , (a), (a), (b), (b), , 15., 16., , (a), (a), , In purely resistive circuit, the current and emf are in the same, phase., , 17., , (a), , In case when orbit of satellite inclined with the plane of, equator there is a component of magnetic field, which is, perpendicular to the velocity vector., , 18., , (c), , 19., , (c), , 3., , (c), , Emf will always induces whenever, there is change in, magnetic flux. The current will induced only in closed loop., In both the cases, the magnetic flux will change, and so there, is an induced current., The magnetic flux changes only in case when loop is rotated, about axis-2., Lenz's is law is based on conservation of energy., , In the given case, there is no component of velocity,, perpendicular to the magnetic field and so e = Bv sin 0°., , Both the statements are independently correct., The quantity L/R is inductive time constant and so it, possesses the dimensions of time., The rate of change of current in case of opening of switch in, much greater than the closing of the switch, and so induced, emf is greater., When switch is closed , the magnetic flux through the ring, will increase and so ring will move away form the solenoil so, as to compensate this flux. This is according to Lenz's law., The direction of induced current in the loop is according to, Lenz's law., The reading of voltmeter V2 will be zero., , S olutions Exercise 8.4, Passage Q.1 to Q.3, 1., (b) The equivalent circuit is shown in figure. The equivalent, capacitance,, , 3V 2/3 F, C, , 2 1, 2 1, , 2, F, 3, , 1, LC, , 1, 1 2/3, , Bvx, R, , 5., , (b), , i, , 6., , (b), , Power P =, , L = 1H, , or, , 2., , (b), , 1, 1, 1 i02 =, 2, 2, , 2 2, 3, 3, , i0 = 6 A, P. d across the capacitor, = 3V., , 1.5 rad / s, , Passage of Q.4 to Q.6, 4., (a) e = Bv = 0.4 × 0.1 × 1 = 0.04 V, Bvx, x, A, , BvA, , e2, R, , 0.042, 1, , If i0 is the maximum current, then, 1, 1, Li0 2 = CV 2, 2, 2, , 3, 2, , (constant), , 16 10, , 4, , 1.6 10, , 3, , W, , Passage of Q.7 to Q.9, 7., , (d), , M, , 0 N1 N 2 A1, , 4, , 2 R2, = 3.94 × 10–4 H., , 10, , 7, , 1000 100, 2 0.20, , 0.022

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ELECTROMAGNETIC INDUCTION, , Passage of Q.13 to Q.15, 13. (a) For simplicity we can take the moment when side EF of he, loop is at y = 0, so, B1 = 0 ; e1 = 0, , P, 2cm, , B0 a, a, , and at y = a, B2, , 20cm, , B0 ; e2, , The induced emf e = e1 e2, , Q, 8., , (b), , e, , M, , di2, dt, , 3.94 10, , (5 – 3), 0.04, , e, 19.72 10, t =, R, 8, Passage of Q.10 to Q.12, , 9., , 10., , (a), , E .d, , =, , 11., , or, , 2, E 2 a = na, , F a, , (b), , 16., B0qa 2, 2, , B0qa 2 / 2, I, , B0q, ., 2m, , ma 2, , Angular velocity at any time, B0qt, 2m, Power developed, P =, , B0 qa 2, 2, , (a), , (a), , 17., 18., , v=, , B0 2va 2, R, , mgR, B02a 2, , ., , A - p ; A charged ring can produced electric field out of the centre., B - p, q, s ; A charged rotating ring can produce electric field out of, centre, magnetic and dipole moment., C- q, s ; Current carrying produces magnetic field at the centre., D- r ; Alternating current can produce induced electric field ., See example., A-p, r, s, t ; Depending on value of q, the potential difference, between A and B may be zero, positive or negative., B-p, r, s, t ; When loop enters into the magnetic field, the direction, of induced current is anticlockwise, when lop moves into the, magnetic field, induced current is zero. When loop comes out, of the field, the induced current will be clockwise., B 2C, , q 2 B02 a 2t, ., 4m, , B, , =, , 7.0t 6.0t 2, , |e|, , =, , d, dt, , D- p, r, s, t ; The induced current, i, zero, positive or negative., , =, , 7 12 2, , 31 mV. Ans., , The resistance of the loops, R1 =, , 2 r1 10 2, , 0.1 10, , =, , Flux in the smaller loop,, , R2, , =, , 2 r2, , =, , 62.8 ., , =, , B2 A1, , 2, , µ0, , =, 1 10, , The induced current,, , i1, , =, , V, R2, , r12, , 2r2, , Right to left., , and, , F, ., B 2C 2, , µ0i2 2, r1, 2r2, , µ0, , 6.28 ., , m, , i0 sin t , and so it may be, , 7 12t, , =, , 2, , Depending on the direction of force, the current may be, positive or negative., , =, (b), 2., , Induced emf,, , P B 2 .va 2, ., v, R, For ternminal speed (constant speed), net force on the loop, is zero, and so, , C-r, s, t ; The induced current, i, , B0 qt, 2m, , S olutions Exercise-8.5, Given,, , B0va, , anticlockwise., R, , e2 ( B0va )2, R, R, So Lorentz force acting on the loop, , or, , t, , 1., , B0va ., , Power generated, P, , mg =, , B0 a, q a, 2, , =, , B0va ., , F, , B0a, ., 2, , Eqa, , (c), , 9.86 10 5 C, , d ( B0t ), dt, , Angular acceleration,, 12., , 0.04, , 15., , dB, E 2 a = A dt, , (b), , 14., , d, dt, , or, , E=, , 19.72 10 3V, , 3, , q = it =, , (a), , e, R, , Thus induced current, i, , 4, , 569, , e, R1, , [4 2.5t ] 2, r1, R2, 2r2, [ d / dt ], R1

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570, , 3., , ELECTRICITY & MAGNETISM, , After substituting the value and simplifying we get, Ans., i = 1.25 A., (a) The emf induced across each of the wire, e = Bv = 1 × 0.05 × 0.04, = 2 mV, The total emf,, The current,, (b), , 4., , i =, , In this case,, , e net, , The induced emf,, , |e|, , = (1.8 10 2 )2 (4, , 5., , enet, R, , 2, 19 1, , = 0,, , i, , 0.1mA ., , A, , =, , ( R ) 2 µ0 n, , e, R, , i, t, , 1.5, , =, The total flux through the square, , At t = 2.5 s,, , |e|, , = 4 × 2.5 × (0.02)3, = 8 × 10–6 V, Ans., , 30 mA, , 0 =, , 2sin t 2, , t, , and, , 2 =, , 2sin 2 t, , t2, , 7., , We have,, or, , 10, , 0, , ,, , 2, , 2, , E =, , 1 2, Li, 2, , =, , 1, L(2sin t 2 )2, 2, , =, , 2L sin 2 t, , =, , 2 2sin 2 / 2 4 J ., , i, , =, , i0 e, , 3, , = 1 e, = 46, , Ans., , (4t 2 y ) dy, , ., , For, , 68.6 mV, , 3, , 3, , 4t, , =, , The energy spent,, , 25 10, , 3, , |e|, , t, , The magnetic flux through the strip of area ( dy ) ,, d, , 6., , d, dt, , 2t 2, 0, , The induced emf,, , B, t, , =, , =, , 1, , i = 0., , 10 7 ) 220 10 2, , Current,, , 2 2, 2 2, , = 2mV and r, , e net, , y2, 2, , 4t 2, , =, , R, , tR / L, , 1R /10, , 4t 2 ydy, , =, 0, , S olutions Exercise-8.6, 1., , The magnetic flux through the loop, B, , Induced emf,, 2., , |e|, , = BA, , 4., , =, , µ0 niA, , =, , µ0 n(i0 sin t ) A, , =, , d B, dt, , The resistance of the wire R, , =, , Ans., , (1.7 10 8 ) (0.40), 2, , =, , A, , dB, dt, , =, , 10, , 2, , The induced current, i =, , e, R, , |e|, , (b), , (10 3 ) 2, , = 2.16 10 3, The area of the loop = 0.10 × 0.10 = 10–2 m2., The induced emf,, , 5., , The magnetic flux,, The induced emf,, , B, , =, , |e|, , =, =, =, , B, B, , 0.50, , 0.01, , 3, , 6., , r2, r2, , B, t, , (0.05) 2, 0.50, , 7.85 10 2 V, , Ans., , 10 10, , (b), , If B (t ), , B (0) , then, , (a), , The magnetic flux,, , B, , i, , =, , =, , 4, , 2.16 10, , t, , (a), , dB, dt, , A, , 3, , = –2.0 mV., Do the other part similarly, From 10 ms to 20 ms and 20 ms to 30 ms, the rate of change, of magnetic field is not constant and so emf is not constant in, these intervals, The charge flows in any interval t ,, , 2 10 4 V, , (0.02), , 2 10, , BA, , d B, dt, , e =, , q, , = 9.3 × 10–2 A., 3., , The induced emf,, , 3, = –(2.0 10 ), , µ0 ni0 A cos t, r, , (a), , q, , R, , [, , f, , R, , B (0), , B (t ), , R, , 0,, , =, , BA cos 60, , =, , ( B0 sin t ) A, , =, , 0.20, (5 10 4 ) sin t, 2, , =, , 5 10, , 5, , sin t, , 1, 2

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571, , ELECTROMAGNETIC INDUCTION, The induced emf,, , |e|, , d B, dt, , =, , 5 10, , When iron part comes down, the final flux, , 5, , cos t, , The maximum, value of emf,, , e0, , |e|, , 5, , =, , 5 10, , =, , (5 10 5 ) 300, , = 0.015 V, (b), , The emf, at t, , 300cos 300, =, , (c), , At t, , 600, , i, , The force on the loop, , For no acceleration, , =, =, , The force on the rod,, , F =, , BiL, , BiL, , VB 2 L2 / R, , tan, , e, , v, , 12., , =, , 2 3 10, , =, , Bv v, , =, , (2 3 10 5 ) 250 20, , =, , B, , =, , vB 2 L2, R, , mgR, , 1m, , or, , 14., , Ans., , B 2 L2, , =, , ma, , (a), , Ans., , µ0i, A, 2 r, , B, , i, , 13., , r2, 2, , vB 2 L2, R, , a = g/2, The magnetic flux through the loop,, = BA, , |e|, , =, , d, dt, , |e|, , =, , d µ0i, A, dt 2 r, , =, , 5 10, , T, , = 0.173 V, Ans., (a) At t = 2s, the length of the loop inside magnetic field,, x = 1 × 2 = 2 cm. so emf will, induce due to right side of the loop,, e = Bv = 0.6 0.01 0.20, = 3 × 10–4 V, (b) At t = 10 s,, x = 1 10 10 cm. The entire loop, is inside magnetic field and so, e = 0, (c) At t = 22 s,, x = 1 22 22 cm. The emf is, induced due to left side of the loop, so, e = 3 × 10–4 V., (d) At t = 20 s, the loop will out of the field so, e = 0, The magnetic flux through the iron part, i, , =, , =, , = BH tan, = 2 × 10–5 × tan 60°, 5, , mg, 2, , Ans., , BV, BH, , =, , BvL, L, R, , B, , If a be the acceleration, then, mg, , 2 2, , BvL, R, , =, , =, , (b), , mgR / B 2 L2, , 1.57 10 6 V, , Thus for terminal velocity,, mg = F, , = 0, , B[VBL]L, R, , mg =, , BV, , 10., , e, R, , Ans., , VBL, R, , VB L / R, , We have,, , The induced emf,, , e, R, , =, , V =, , 9., , i =, , If V is the velocity of the loop, then induced emf, e = VBL ,, and induced current,, , 8., , 600, , t, , The induced current,, , (a), , 300cos 300, , Br 2, 2 t, , The induced emf across the rod, e = BvL, , 900, , ,, , 5, e = 5 10, , 7., , Ans., , 7.5 10 3 V, , =, , Ans., 11., , 900, 5, , e = 5 10, , = 0, , t, , A, anticlockwise, , 11, , The magnetic flux through the loop,, = BA cos, , BA cos t, , d, dt, , The induced emf,, , (a), , = BA sin t, The initial flux through the loop,, , and, , e, , =, , d ( BA cos t ), dt, , (b), , i, , = BA, , f, , = 0, , Ans., , The average induced emf,, |e|, , =, =, , f, , t, BA, t, , 1.2, , = 0.0471 V., (b), , From a to b., , i, , t, (0.05) 2, 0.2

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572, 15., , ELECTRICITY & MAGNETISM, , (a, b) For wire A : e v Bv sin, , 0, , For wire C : e Bv sin 45, 16., , (a), , For wire D : e Bv 2 sin 90 0.141V, Ans., The magnetic flux through the circle of radius r1,, =, =, , r12, , En .d, , =, , d, dt, , En 2 r1, , =, , r12, , En, , =, , –, , 2 r2, , =, , – R2, , We have, or, , (c), , For r2, , R : En, , 20., , B r12, , BA cos 0, , d, dt, , Thus, (b), , (b), , 0.0707V, , The induced emf,, , |e|, , a2, , =, , (0.20)2 0.1 4 10, , =, , |e|, dB, dt, , dB, dt, , =, , (0.1) 2, 4, , e, , Ans., 22., , Do other parts similarly, 17, , (b), , =, , r dB, 2 dt, , =, , 0.1, 2, , e =, =, , 18., , Ans., , d, dt, , =, , –, , NBA sin t, , =, , NB(ab) 2 f sin 2 ft, , =, , BA, , =, , Ba 2, 2, , The flux through the small area,, R, , En, , The induced emf,, , (5 10 3 ), , = 39.25 µV, The flux through the coil at any instant, = NBA cos t, Induced emf,, , R dB, 2r2 dt, , =, , r 2 dB, 4 dt, , d, dt, , 2, , En, , V, , r2, 4, , BA = B, , =, , 21., , 3, , Do the other parts accordingly., The magnetic flux through the structure, , dB, ., dt, , r1 dB, 2 dt, , dB, dt, , =, , (c), , The induced current, i =, , (d), , Zero., , (a), , Induced emf, , r2, , 0.1, , –0.005 V/m, , dB, dt, , 0.1, , 3.14, 2, , 2, , 0.1, , Induced emf,, , e, , 1.57 10, , 3, , A, , d, dt, , =, , =, , d =, dt, , =, , Ba 2, 2, , =, , t, , =, , Ba 2 t, 2, , Ans., , In time t,, En .dl, , a, , a, 2, , a, , 3.14 × 10–3 V, e, R, , B, , e, , Ba 2 d, 2, dt, , Time taken to complete half rotation, or, , a, , (b), , Induced emf,, , Induced current,, 19., , (a), , dB, dt, , =, t =, , En 4a, , =, , a2, , En, , =, , a dB, 4 dt, , =, , 0.20, 0.1, 4, , |e|, , i, , 0.005 V/m, , 23., , =, , d, dt, , =, , (0.20) 2 0.1 4 10 3 V, , =, , e, R, , a, , =, , 4 10, 2, , d, 2cos, , t, , (a), , e =, =, =, , The induced emf,, , dB, dt, , 2, , 3, , 2 10, , 3, , Thus current,, , A., , The induced emf in any side of the loop is given by, e, , In general, , (b), , The induced emf,, , dB, dt, , where d is the distance of the side from the centre of the field., Here d = 0,, e=0, , Thus current,, , i, , =, , =, , 1 2, t, 2, 2, , 2 n/, Bv, , 1 × 0.05 ×0.02, 1 × 10–3 V, e, R, , 1 10, 10, , 3, , = 0.1 mA, e = Bv, = 1 × 0.05 × 0.04, = 2 × 10–3 V, i, , =, =, , e, R, , 2 10, 10, , 0.2 mA, , Ans., , 3, , Ans.

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573, , ELECTROMAGNETIC INDUCTION, 24., , (i), , The flux through the smaller loop, =, , 2, , =, , 10 7 ) 1 (5 10 4 ), cos t, 2 0.1, , (4, , =, (ii), , The induced emf,, , |e|, , (iii) The induced current, i, 25., , =, =, , (a), , (b), , Bi, , |e|, , =, , L, , =, , The power supplied P, , d, dt, , 10, , 9, , sin t, 9, , 1.57 10, , (a), , sin t, , t/, , ), , 0.8 i0, , =, , i0 (1 e, , t/, , ) ;, , 30., , (a), , i, , 10, , (b), , and, , 2, , (c), , Be, R, , B 2v, R, , 2, , =, , dv, dt, , m, , i2, , 26., , (b), , We have, , e, , =, , L, , =, , e, ( di / dt ), , L, , 27., , The required ratio,, , Estored, ESupplied, , 0.005t, , ), , =, =, =, , = 0, i2, , S i1, , i3, , –1.82 A, , i2, , R 10, , i, , 25, 25 103, , 1 10, , 3, , A, , V02 (1 e, , V02, , V02, t/, , t/, , i1R L(di2 / dt ), , ), , … (i), , i1 i2, , di, dt, , = 0, , … (ii), , di1, dt, , 0,, , di2, dt, , Thus from equation (ii), we have, i1R L, , )2, 0.1 10, 1, , =, , By loop rule, we have, , Since, , V2, , = 0.36, , = 2.73 A and i3 1.82 A, , By junction rule, , V02 / R, , 1 e, , 4.55 A, , 0, , Ans., , V /R, , (1 e, , V, 20 30, 20 30, , =, , i1 i2, , constant, current, , di, dt, , 3.33 A, , The direction of currents may be as shown in figure., , 2, , =, , 100, 10 20, , i2, , (d), , After simplify and substituting the values, we get, = 100 (1 – e, , =, , (a), , i, , v, , ) = 7.37 mA Ans., , After opening of the switch,, i1, , 31., , /, , i0 (1 e, , 10, , ( Bv ), B, R, , By Newton's second law, we have, F F ' = ma, or, , =, , After long time, inductor offers zero resistance so, i1, , =, , L, R, , Immediately after closing of switch, inductor offers infinite, resistance, so, i1, , B 2v, R, , (5 16t ), , i0 (1 e, , i, , Bi, , 3, , 10 10, , After substituting the values and simplifying, we get, t = 8.45 ns., , B = 0.02 T, , =, , di / dt, , =, , (b), , As soon as bar starts moving it experiences a retarding force,, , F', , 10 10 3, d, (5 16t ), dt, , ( e), , i, , We have, or, , e, B, R, , e = 1V, , di, dt, , = 10 × 10–3 × (5 + 16), = 210 mW., Ans., 29., , e, R, , L, , = 6.25 × 10–4 H, = ei, =, , 10 9 cos t, , Suppose field acts into the page. The force on the bar due to, current i, F, , We have,, , µ0i, A cos t, 2 R2 1, , =, , 1, , 28., , B2 A, cos t, , di1, dt, , = 0, , … (iii), , After simplifying equations (i) and (iii), we get, , 2, , Ans., , and, , i1, , =, , ie, , i2, , =, , i i1, , Rt / L, , i 1 e, , Rt / L

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575, , ELECTROMAGNETIC INDUCTION, 35., , (a), , If, , (ii), , di, is the rate of change of current, then by Kirchoff's, dt, , Torque,, , = force × moment arm, , second law, q, , =, , magnetic, , weight, , =, , Fmagnetic, , Fweight r / 2, , =, , [ Bir mg cos t ]r / 2, , =, , B2 r 4, 4R, , =, , N1B2 A1, , =, , n1 ( µ0 n2i2 ) R12, , =, , µ0, , i, L, , Q, C, , L, , di, dt, di, dt, , or, , =, =, , (b), , 37., , = 0, , The flux across first solenoid, 2, , Q, LC, 100 10, 2 10, , 3, , 6, , 5 10, , 6, , M, , 104 A/s, , When Q 200µC , there is no energy stored in inductor,, , 38., , (c), , (d), , Given, , =, , imax, , =, , i, , =, , 2C, , LC, , imax, 2, , i, x, , = 2A., , dx, , 1A, , From conservation of energy, we can write, 1 2, Li, 2, , Q2, 2C, , =, , µ0 i, ., 2 x, , =, , B, , 2, Qmax, , Qmax, , The flux through the area ( dx) ,, , 2, Qmax, 2C, , =, , d, , NB ( dx ), , After substituting the value, and simplifying, we get, =, , Q, , 36., , (a), , (b), , b, , Ans., , 173.2 µC, , total, , a, , =, , B r2, 2, , =, , N, , The effective circuit is shown in figure., =, , i, , B r, 2, , 2, , S, R, , On comparing with,, , L, , di, If i is the current and, is the increase in current in the, dt, , circuit then, iR, , Ldi, dt, , i, , or, , i, , total, , M, , 39., , µ0, 2, , b, , a, , i, dx, x, , =, , µ0 N i, | n x |(aa, 2, , =, , µ0 N i, a b, n, 2, a, , =, , Mi , we get, , =, , µ0 N, b, .n 1, 2, a, , =, , µ0 N1i, ., 2 r, , The total flux through the area (hdr ), =, =, , R, , (1 e, , Rt / L, , B r2, (1 e, 2R, , b, , ), total, Rt / L, , ), , Ans., , b), , Ans., , The magnetic field of the toroid a distance r from the centre,, B, , = 0, , After simplifying, we get, , (i), , NB dx, , =, , The induced emf across the whole rod, e, , R12 n1n2, , Magnetic field due to current carrying wire at a distance, x from, wire, , so i 0 ., 1 2, Limax, 2, , mgr cos t, .Ans., 2, , N 2 B(2 rdr ), , =, a

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576, , ELECTRICITY & MAGNETISM, b, , =, , N2, a, , M, , 40., , (a), , Np, Ns, , µ0 N1i, (hdr ), 2 r, , =, , µ0 N1N 2 h, b, n, i, 2, a, , =, , µ0 N1N 2 h, b, n, 2, a, , =, =, , and, , (c), and, , Vp, Vs, 8500, 120, , (b), , 71, , 78000, 8500, , ip, , =, , P, VP, , is, , =, , P, Vs, , RP, , =, , VP, 8500, iP = 9.2, , Rs, , =, , VP, is, , 78000, 120, , 120, 650, , 9.2 A, , 650 A, , 926, , 0.18, , Ans.

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578, , ELECTRICITY & MAGNETISM, , 9.1 ALTERNATING CURRENT (AC), Definition : The current which we use commonly in our houses is alternating, current. Alternating current or potential is one whose direction changes periodically,, often in a sinusoidal manner. Figure shows two alternating currents; one sinusoidal and, other square/rectangular., Skin effect : A direct current flows uniformly throughout the cross-section of, the conductor, while an alternating current, flows mainly along the surface of the, conductor. This is know as skin effect., By increasing surface area, we can decrease the resistance for ac. It can be possible by, making a cable which made of number of thin wires having same amount of metal as in, the case of a single wire cable., , Why alternating current used in practice?, 1., , Fig. 9.1, , AC potential can very easily be changed by using transformer according to the, requirement., 2., We can minimise transmission losses (i2R) by providing cable made of a number of, thin wires., 3., AC is more readily adaptable in rotating machinery such as generators and motors., Alternating current and potential : Figure 9.3 (a) and (b) represent, alternating current and potential respectively. They can be represently by the equations, i = i0 sin t and V = V0 sin t., , where i0 is peak value of current, is angular frequency in rad/s., The circuit symbol for an ac source is, , where V0 is the peak of potential., Fig. 9.3, ., , Phasor diagram, , Fig. 9.2, , Analysis of alternating current circuit is facilitated by the use of vector diagrams similar, to those used in the study of SHM. In such diagrams, the instantaneous value of a, quantity that varies sinusoidally with time is represented by the projection onto a, horizontal or vertical axis of a vector of length corresponding to the amplitude of the, quantity and rotating anticlockwise with angular velocity w. In the context of ac-circuit, analysis, these rotating vectors are often called phasors, and diagram containing them, are called phasor diagrams., From the diagram, the projection on x-axis diameter gives, V = V0 cos t,, and projection on y-axis gives, V = V0 sin t., , 9.2 AVERAGE, , AND ROOT MEAN SQUARE (RMS) VALUES, , Average value :, , The average value xav of any quantity x(t) that varies with time over a time interval t1 to, t2, is defined as, t2, , x t dt, Fig. 9.4, , xav or x, , =, , 1, , t1, t2, , dt, t1, , t2 t1, , t2, , x t dt, t1

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AC AND EM-WAVES, Let us now calculate average value of sinusoidally varying quantity, i = i0 sin t., (i) Average over half cycle :, T /2, , sin t dt, , i0, i av, , 0, , =, , T, 2, T /2, 0, , =, , =, , =, =, (ii), , 2, T, , sin, , i0, , t dt, , Fig. 9.5, , T, 2, T, 2, , 2 t, cos, T, 2, T, , 2i0, T, i0, , cos, , 0, , cos 0, , 2i0, , Average over complete cycle :, T, , i av, , = i0, 0, , sin t dt, T, , T, , =, , i0, 2, t dt, sin, T, T, 0, , =, , (iii) Average of half rectified wave :, , i av, , i0, T, , =, , 1, T, , 2 t, T, , 2, T, , i0, =, 2, = 0, , =, (iv), , cos, , cos 2, , T, , 0, , cos 0, , T, 2, , sin t dt, 0, , i0, , Fig. 9.6, , Average of full rectified wave :, i av, , =, , =, , 1, T, , T, , sin t dt, 0, , 2i0, , Fig. 9.7, , 579

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AC AND EM-WAVES, , 581, , Where average and RMS values are used ?, The usual moving-coil galvanometer, however, has very large moment of inertia to follow, the instantaneous values of current. It average out the fluctuating torque on its coil, and, its deflection is proportional to the average current., RMS values of voltage and current are useful in finding power or energy loss in ac, circuits., Most ac meters are calibrated to read not the maximum value of the current or voltage,, but the rms value of current or voltage., , rms value, ., average value, , Form factor R f, , Ex. 1, , An alternating current is given by the equation, i = i1 sin t + i2 cos t. Find its rms value., , Sol., , i2 =, =, i2 =, , or, and, , i, , 2, , =, =, , i rms =, , i1 sin t, , i2 cos t, , =, , =, , 2i1 i2 sin t cos t, , i12 sin 2 t i22 cos 2 t i1i2 sin 2 t, 1, 2, , i12, , 1, 2, , i22, , i1i2, , 0, , e2 1, , i0, e, , i2, , 2, , Ans., , 2, , i12, , i2, , i22, , Ans., , 2, , = 2 × 50 = 100 rad, V = V0 sin t = 311 sin 100 t., , -, , e0, , Find the average and effective values of the saw-tooth, wave form as shown in fig. 9.9., , Vrms = 220 V, or V0 = 2 220 311V, , i = i0 e, , 2, , Ex. 3, , i22, , Note:, , Ex. 2, , e, , 1, i0 2, 1 2, 2, e, , i rms =, , In our country the power supply for domestic purpose is 220 V, and 50 Hz. That is,, and f = 50 Hz, or, , 2, , 2, , i12 sin 2 t i22 cos 2 t, , i12, , i0 2, , The current in a discharging LR circuit is given by, , Sol., , Fig. 9.9, Since voltage increases linearly, so, Vav =, , 0 50, 2, , 25 V, , Ans., , Let V is the instantaneous value of voltage between 0 to 2s, then, V =, , 50, t, 2, , 25t, , The rms or effective value of the voltage is, T, , t, , where is the time constant of the circuit. Calculate, , the rms current for the period t = 0 to t = ., , Sol., , 2, Vrms, , =, , 1, V 2dt, T 0, , =, , 1, 2, , t, , Given,, , i = i0e, 2t, 2, , i 2 = i0 e, and, , i2, , =, , i0, , 2, , 2, , 25t dt, 0, 2, , 2t, , 2, , e, , =, , 625 2, t dt, 2 0, , =, , 625 t 3, 2 3, , dt, , 0, , 2t, , =, , i0, , 2, , e, , Vrms = 28.9 V, , 2, 0, , 2, , 833.3 V, 0, , Ans.

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ELECTRICITY & MAGNETISM, , 582, , Ex. 4, , Determine the rms value of a semicircular wave which, has a maximum value I0., Sol. The equation of a semicircular wave is, , =, , or, , 2, irms, , =, , 1, 2I0, , I0, , or, , 2, , y dx, , I0, , I 02, , x 2 dx, , I0, , 1, I 02 x, 2I0, , x3, 3, , =, , 1, 2I0, , I 03, 3, , =, , 2 I 02, 3, , =, , Fig. 9.10, x2 + y2 = I02, y2 = I02 – x2, , 1, 2I0, , I 03, , 2I 02, 3, , Irms =, , I0, I0, , I 03, , I 03, 3, , 2, I 0 Ans., 3, , I0, , Circuit elements :, In circuits, we have three circuit-elements : resistor, capacitor and inductor. All these, three offer resistance for ac. The resistance of resistor is real while the resistance offers, by capacitor and inductor is virtual., , Resistance :, The opposition offered by the resistor is called resistance. It is independent of frequency, of ac., , Capacitive reactance (XC) :, The opposition offered by the capacitor in an ac circuit is called capacitive reactance. It, is a virtual resistance. It is given by, Its unit is ohm., For dc,, , XC, , = 0, XC =, , =, , 1, C, , 1, ., 2 fC, , ., , Inductive reactance (XL) :, It is the resistance offered by the inductor in an ac circuit. It is also a virtual resistance., Its unit is ohm. It is given by, Fig. 9.11, , XL, For dc,, , =, , L., , = 0, XL = 0, , Impedance (Z) :, The total resistance of ac circuit is called impedance., , Admittance (A) :, Reciprocal of impedance is known as admittance; i.e., A, , 9.3 THREE, 1., , 1, . It s unit is mho., Z, , SIMPLE CIRCUITS, , A resistive circuit : Suppose that an alternating potential difference V = V0 sin t, is applied to resistance R, then current in resistor, , or, , V, R, , V0, sin t, R, , iR, , =, , iR, , = i0 sin t,, , ...(1)

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AC AND EM-WAVES, V0, , is the amplitude of the current., R, Current and voltage both vary with sin t, hence they are in same phase., A capacitive circuit : If a capacitor of capacitance C is connected across alternating, source, the charge at any instant,, , where i0, , 2., , q, , Current, , iC, , = CV0, = CV0 sin t, =, , dq, dt, , V0, sin, 1, C, , =, , or, , iC, , where, 3., , CV0 cos t, , = i0 sin, , t, , t, , 2, , V0, sin, XC, , t, , 2, , ...(2), , 2, , 1, = XC, is called capacitive reactance., C, , An inductive circuit : Suppose that an alternating potential V = V0 sin t is applied, across a pure inductor of self inductance L. From the definition of inductance, we, can write, L, , diL, dt, , VL, , =, , or, , V0sin t, , =, , LdiL, dt, , or, , di L, , =, , V0, sin t dt, L, , or, , iL, , =, , =, , diL, , Fig. 9.12, , V0, sin t dt, L, , V0, cos t c, L, , The value of the constant over complete cycle will be zero, XL = L, is called, inductive reactance., , or, , V0, sin, XL, , iL, , =, , iL, , = i0 sin, , t, , t, , 2, , 2, , ...(3), , Fig. 9.13, , 583

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584, , ELECTRICITY & MAGNETISM, , Fig. 9.14, , Phase and amplitude relations for alternating currents and voltage, Circuit element Symbol Impedance Phase of the current Phase angle, Resistor, , R, , R, , In phase with VR, , Capacitor, , C, , XC, , Lead VC by, , Inductor, , L, , XL, , Lags VL by, , Amplitude relation, VR, , 0, , 2, , 2, , 2, , 2, , iR R, , VC, , iC X C, , VL, , iL X L, , 9.4 RC-CIRCUIT, Consider a circuit with resistor R and capacitor C, connected in series with an alternating, source of potential V. If VR and VC are the p.d. across resistor and capacitor respectively,, then, , VR 2 VC 2, , (i), , Supply voltage, , V, , =, , (ii), , Impedance, , Z, , =, , (iii), , Current, , i, , = i0sin( t + ), , where, (iv) Peak current, , tan, i0, , V, i, , R2, , iR, XC2, , =, , XC, R, , 1, CR, , =, , V0, Z, , 2, , 2, , 2, , iX C, , 2, , 1, C, , R2, , V0, R, , XC2, , 9.5 LR-CIRCUIT, Fig. 9.15, , Fig. 9.16, , Consider a circuit with a resistor R and an inductor L connected in series with an alternating, source of potential V. If VR and VL are the p.d. across resistor and inductor respectively,, then, (i), , Supply voltage, , V, , (ii), , Impedance, , Z, , (iii), , Current, , i, , where, , tan, , =, , VR 2 VL 2, , V, R 2 X L2, i, = i0 sin( t– ), , =, , =, , XL, R, , L, R, , iR, , 2, , R2, , iX L, , 2, , L, , 2

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AC AND EM-WAVES, (iv) Peak current, , i0, , =, , V0, Z, , V0, R2, , X L2, , 9.6 LC-CIRCUIT, Consider a circuit with inductor L and capacitor C connected in series with an alternating, potential V. If VL and VC are the p.d. across inductor and capacitor respectively, then, , Fig. 9.16, Fig. 9.17, (i), , Supply voltage, , V, , = VL ~ VC, , (ii), , Impedance, , Z, , =, , (iii), , Current, , i, , = i0 sin, , (iv) Peak current, , i0, , 9.7 SERIES, , =, , V, i, , XL, , i XL, , XC, , XC, , L, , t, , V0, Z, , 1, C, , 2, , V0, XL, , V0, XC, , 1, C, , L, , RLC-CIRCUIT, , Consider a RLC- series circuit shown in fig. 9.18. If VR, VL and VC are the p.d. across the, resistor, inductor and capacitor respectively, then, (i), , (ii), , Supply voltage, , V, , Impedance, , Z, , =, , VR 2, , =, , iR, , =, , =, (iii), , Current, , where, , 9.8 RLC,, , i, , tan, , 2, , V, i, , iX L, , R2, , R2, , = i0 sin, =, , 2, , VL VC, , XL, , XL, , L, , 2, , XC, , 1, C, , 2, , 2, , t, XC, , R, , iX C, , 1, C, , L, R, , RESONANCE CIRCUIT, , Resonance will occur when oscillating quantities, here i and V are in same phase, i.e., = 0, XL – XC = 0, or, , L, , =, , 1, C, , Fig. 9.18, , 585

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ELECTRICITY & MAGNETISM, , 586, , or, , =, , 1, LC, , 0, , where 0 is called resonance frequency., At resonance,, Zmin = R., Current :, , i, , V, Z, , =, , V, , =, R2, , at, , = 0, i = 0, , at, , =, , i, 0, 0, , at, , =, , ,i, , L, , Fig. 9.19, 2, , 1, C, , V, R, , 0, , Remember :, 0,, , Fig. 9.20. Variation of current i with ., L, , 1, C, , 0,, , = 0, current in phase, with potential, , 1, C, , L, , current leads potential, , 0,, , L, , 1, C, , current lags potential, , Fig. 9.21, , Ex. 5, , A 60 Hz voltage of 230 V effective value is impressed on, an inductance of 0.265 H., (i), , Write the time equation for the voltage and the resulting, current. Let the zero axis of the voltage wave be at t = 0., , (ii), , Find the maximum energy stored in the inductance., , Sol., , V0 =, f =, , 2 f, , 2, , 2, , 230 V ;, , =, (ii), , Energy stored,, , E0 =, , 60 Hz,, , =, , 60 120, XL =, , (i), , 2 Vrms, , i =, , 2.3 2 sin 120 t, , 2, , 1, Li0 2, 2, 1, 2, , 0.265, , 2.3 2, , = 1.4 J, , L 120, , 0.265 100, , The time equation for voltage, V =, , 230 2 sin120 t., , i0 =, , V0, XL, , 230 2, 100, , 2.3 2 A, , Ans., , 2.3 2 cos120 t., , 2, , Ans., , Ex. 6, , The voltage applied to a purely inductive coil of self, inductance 15.9 mH is given by the equation, V = 100 sin 314t + 75 sin 942 t + 50 sin 1570 t., Find the equation of the resulting current wave., , Sol., , The standard equation of the voltage can be written as :, V = V01 sin, , 1t, , V02 sin, , 2t, , V03 sin, , 3t

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AC AND EM-WAVES, On comparing this with the given equation, we have, 1, , X L1, , 1L, , 2, , X L2, , 2L, , 3L, , 3, , 5, 942, , 3, , rad, ,, s, , 3, , and, , 15, 1570, , =, , 1570 15.9 10, , 9.9 POWER, , 3, , =, , 942 15.9 10, , and, , X L3, , =, , 314 15.9 10, , Hence :, , rad, 314, ,, s, , Thus, rad, ,, s, , i = i01 sin, = 20sin 314t, , i01 =, , V01, X L1, , 100, 5, , 20 A,, , i02 =, , V02, X L2, , 75, 15, , 5 A,, , i03 =, , V03, X L3, , 50, 25, , 2 A., , 1t, , 2, , i02 sin, , 2t, , i03 sin, , 5sin 942t, , 2, , 587, , 3t, , 2sin 1570t, , 2, Ans., , 25, , IN AC CIRCUITS, , Consider a circuit in which there is a phase difference between i and V. Current i can be, resolved into two mutually perpendicular components; icos along the direction of, applied voltage V, and i sin which is perpendicular to V., The power consumed by the circuit is given by the product of V and that component of, the current i which is in phase with V., , P, , So, and, , true, , [Ptrue]av, , = V i cos, = Vrms, =, , irms, , V0 i0, 2, , 2, , cos, , cos, , Fig.9.22, , V0i0, cos, 2, , where cos is called power factor., This power consumed is called true power or active power., cos, where, , cos, , =, , R, Z, Fig. 9.23, , 1., , Note:, In an ac circuit the product of rms voltage and rms current gives apparent power, , Pav app, (i), , (ii), , =, , Vrms × irms, , When circuit has resistor only,, = 0; cos = 1, and, Pav = Vrms irms, When circuit has inductor or capacitor only,, =, , 2, , , cos = 0 and Pav = 0, , Current in such a circuit is called wattless current., , L, , C, , Constant power, , Thermal power, , Electric, generator, , R, Fig. 9.24

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588, , ELECTRICITY & MAGNETISM, Active and reactive components of circuit current, Active component is that which is in phase with the applied voltage V, i.e., i cos . It is, also known as wattfull component. Reactive component is that component which is, perpendicular to V, i.e., i sin . It is also known as wattless or idle component., , Instantaneous power and average power, Consider the most general case, in which current and voltage differ in phase by an angle, . The instantaneous power, Pinst = V i, Fig. 9.25, , =, , Pinst, , V sin t, , i0 sin, , t, , =, , V0i0 sin t sin t cos, , =, , V0i0 sin 2 t cos, , =, , 2, , V0 i0 sin, , sin t cos t sin, , 1, cos, 2, , Pav, , =, , V0 i0, , or, , Pav, , =, , Vrms irms cos ., , 1, T, 1, T, , sin 2 t sin, 2, , t cos, , or, , cos t sin, , T, 0, T, 0, , sin 2 t dt, sin 2 t dt, , V0 i0, cos, 2, , 0, , .....(1), , Half power points and band width, We know that, , Pav, , =, , V0 i0, cos, 2, , =, , V0, 2, , =, , For RLC series circuit,, , Z, , =, , Pav, , =, , V0, Z, , R, Z, , V0 2 R, 2Z 2, R2, , L, , 0, , 1, LC, , Pav, , At, , L, , 2, , 1, C, , V0 2, 2R, P0 R 2, , =, , R, At, , ,, , = P0, , =, , Pav, , 2, , V02 R, 2 R2, , At, , 1, C, , 2, , L, , = 0, Pav = 0, =, , Pav 0, , 1, C, , 2, , .....(2), , 1, ,, 2, 0

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AC AND EM-WAVES, For,, , =, , P0, 2, , =, , 1, C, , L, , or, , Pav, , P0, , we have, 2, P0 R 2, R2, , 2, , 1, C, , L, , 2, , = R2, , 1, R, =, C, The above equation will have two values of , difference of which gives band width., L, , or, , L, , or, , 1, =+R, C, R, L, , 2, , =, , 1, =0, LC, , R, L, , R, L, , 1, = –R, C, , L, , R, L, , 2, , 2, , 1, LC, , 4, , R, L, , =, , 2, , 1, =0, LC, 2, , R, L, , 4, , 1, LC, , 2, , always be positive, therefore, , 2, , =, , R, L, , R, L, 2, , 2, , 4, LC, , R, L, , 1=, , R, L, 2, , 2, , 4, LC, , R, = 2 – 1=, L, , B and width, , The current corresponds to half power, irms =, , i0, 2, , 0.707 i0, , Quality factor or Q-factor, The characteristics of a series resonant circuit is determined by the Q-factor of the, circuit. It can be defined as, Q-factor =, =, , 2, , Resonance frequency, band width, , Q-factor =, , 0, , 1, LC, R, L, , =, , or, , maximum energy stored, energy dissipated per cycle, , 1, R, , L, C, , ....(3), , Acceptor circuit, If the frequency of ac supply is varied, the series RLC circuit passes a maximum current, and power output for = 0. This is the procedure by which a radio or television, receiving set may be tuned to receive the signal from a desired station which is sending, signals at a particular frequency. The circuit is thus known as an acceptor circuit., , Fig. 9.27, , 589

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590, , ELECTRICITY & MAGNETISM, Choke coil, Choke coil is a device having high inductance and negligible resistance. It is used to, control ac in the circuit as rheostat is used for dc. Ideal choke coil has no power loss. But, practical choke coil has some power loss because of resistance of inducting coil. Practical, choke coil is in LR circuit as discussed earlier., , Parallel LCR circuit, Consider parallel LCR parallel circuit as shown in fig. 9.29., Fig. 9.28, , We have,, , iR =, , V0, R, , iL =, , V0, XL, , iC =, , V0, XC, , From phasor diagram, i =, and, , tan, , =, , iR 2, iC, , iC, , iL, , 2, , iL, iR, , Admittance (A) :, , or, Fig. 9.29, Current, , At,, , Fig. 9.30, , Power, At,, , V0, Z, , =, , V0, R, , 1, Z, , =, , 1, R, , i =, , Fig. 9.31, , or, , V0, XC, , V0, XL, , C~, , 1, L, , 2, , V, , 1, R, , 2, , C, , 2, , 2, , 1, L, , 2, , = 0, i =, =, , i, 0, min, , =, , ,i=, , P =, , V2, Z, , V2, , V, R, , 1, R, , = 0, P =, =, , Resonance :, , V, Z, , 2, , 0,, , Pmin, , V2, R, , =, ,P=, = 0, iC = iL gives imin = iR, 1, C =, L, , 2, , 2, , C, , L

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AC AND EM-WAVES, 1, LC, , =, , or, , f =, , 1, 2, , 0, , 1, ., LC, , Use of operator j in AC circuits, In complicated ac circuits, the use of the operator j is of very great importance. In, Cartesian form, a phasor A can be written as,, A = a+jb, where a is the x-component and b is the y-component of phasor A., The value of, , j =, , From the geometry, , A =, , j2, , 1,, a2, , 1, j 3, , j , and j 4, , 1, , b2, , Fig. 9.32, , b, and, tan =, a, By using j operator we can write XL and XC in the following manner., XL = j L, , and, , XC =, , =, , or, , XC =, , j, , 1, C, , j, , j, , 1, C, , j, , Fig. 9.33, , 1, j C, , Parallel circuit (Rejector circuit), Let us consider an ac source is connected across an inductance L in parallel with C. The, resistance in series with the inductance is R. Let the instantaneous value of potential be, V and the corresponding current be i, the current through the inductance be iL and, through the capacitance be iC. These currents iL and iC will be almost in opposite phase, if R is very small. The total current,, i = iL + iC, or, , V, Z, , =, , or, , 1, Z, , =, , 1, Z, , =, , or, , A, , V, R j L, , V, 1, j C, , 1, j L, , R, , R, , j C, , R, , j L, , j L, , R, , j L, , j C, , Fig. 9.34, R, , =, , R, , R, , =, , j L, , 2, , 2 2, , j C, , L, , j, , CR 2, R2, , 3 2, , LC, , 2 2, , L, , L, , 591

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ELECTRICITY & MAGNETISM, , 592, , The magnitude of the admittance, R2, , 1, Z, , A, , CR 2, , =, , 3 2, , LC, , R2, , L, , 2, , 2 2, , L, , For resonance, i will be in phase with V, if j component of A becomes zero. i.e.,, CR2 + L2 C– L = 0, or, , =, , 0, , =, , R2, , 1, LC, , L2, , At resonance, Z will be maximum. It will be, Z =, , R2, , 2 2, , L, , , is called dynamic resistance, , R, R2, , L2, , =, , 1, LC, , R2, L2, , L, CR, , R, , At resonance the peak current from supply is, known as make up current, which is, , or, , For the values of L, C, R and, , i =, , V0, Z, , i =, , CRV0, L, , V0, L, CR, , 1, LC, , satisfying the equation, , R2, L2, , , the reactive, , component of A is real. At such a maximum impedance, the current in the circuit is, minimum. Thus the parallel circuit does not allow this frequency from the source to pass, through the circuit. Due to this reason the circuit with such a frequency is known as a, rejector circuit., , Ex. 7, , (a) In a circuit containing a capacitor and an AC source,, the current is zero at the instant the source voltage is, maximum. It is consistent with ohm s law ?, , (b), , Can you have an AC series circuit in which there is a phase, difference of 180° between potential and current., , (c), , Is rms value of alternating current can equal to peak value of, current ?, , Sol., (a), , or, , No. (b) No., , XL =, L =, C =, , Sol., (a), , For resonance, , XC, , 1, C, , 1, 2, , 2, , Ex. 8, , A series circuit contains a resistance of 4 , an, inductance of 0.5 H and a variable capacitor across a 100 V, 50 Hz, supply. Find, (a) the capacitance for getting resonance,, (b) p.d. across inductance and capacitance, (c) the Q-factor of the series circuit., , L, , 2 f, , 2, , L, , 1, , =, (c)Yes, in a square wave., , 1, , 50, , 2, , 20.3 10, , 6, , 0.5, Ans., , (b), , 100, 25 A, 4, P.d. across inductor or capacitor, V L = VC = iXL = 25 (2 × 50 × 0.5), = 3925 V, Ans., , Current at resonance =, , F

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AC AND EM-WAVES, (c), , 0L, R, , Q-factor =, =, =, , 2, , i =, , 50, 4, , 39.25, , Ans., , An ac source of angular frequency is fied across a, resistor R and a capacitor C is series. The current registered is i. If, now the frequency of the source is changed to /3 (but maintaining, the same voltage), the current in the circuit is found to be halved., Calculate the ratio of reactance to resistance at the original, frequency., At angular frequency , the current in CR-circuit is given by, , Vrms, , irms =, , 1, C, , R2, , 2, , Sol., , Z =, , or, , ...(i), , R2, , or, R2, , L, , 100 L, , 2, , 2, , R2, , =, , The ratio, , Ex. 10, , 1/ C, R, , R, Z, R, 88, , L = 42.2, For power factor to be 1, the resistor remain effective, cos = 1, , 2, , which gives, , L =, , 1, C, , 1, , 3, 5, , =, , ...(1), , From (1), we get, , 5, C, , C =, , Ans., , L, , In inductor-coil, a capacitor an AC source of rms voltage, , 24V are connected in series. When the frequency of the source is, varied, a maximum rms current of 6.0 A is observed. If the inductor, coil is connected to a battery of emf 12 V and internal resistance 4.0, , what will be the current ?, , Sol., The maximum current will occur at resonance in series circuit, which is, , =, , 2, , 1, 50, , =, = 75 F, Ans., So capacitor of capacitance 75 F should be connected in series, in the circuit., , Ex. 12 A box contains L, C and R. When 250 dc is applied to the, terminals of the box, a current of 1.0 A flows in the circuit. When an, ac source of 250 V rms at 2250 rad/s is connected, a current of 1.25, A rms flows. It is observed that the current rises with frequency and, becomes maximum at 4500 rad/s. Find the values of L, C and R., Draw the circuit diagram., When dc is used in the circuit, only resistor remain effective, , V 250, 250, ...(i), i, 1, When dc is used, there is current in the circuit it means capacitor must, not be in series. It may be in parallel to resistor together with inductor., R =, , or, , V, Z, , Z =, , V, i, , 24, 6, , 4, , The impedance of the circuit = 4 . Now if this coil is connected to a, battery of 12 V and internal resistance 4, become, Z = 4 + 4=8, , 42.2, , 7.5 × 10–5 F, , Sol., Fig. 9.35, i =, , 88, , R = 70.4, , 2, , From equations (i) and (ii), we get, 3R2 =, , 2202, 550, , = 882, , 0.8 =, , ...(ii), , V2, P, , = 882, , cos, , or, 1, C, 3, , Ans., , The impedance of the circuit, , and, , Vrms, , =, , 1.5 A, , A circuit draws a power of 550 watt from a source of 220, volt, 50 hertz. The power factor of the circuit is 0.8 and the current, lags in phase behind the potential difference. To make the power, factor to the circuit as 1.0, what capacitance will have to be connected, with it ?, , When frequency becomes /3,, , irms, 2, , 12, 8, , Ex. 11, , 0.5, , Ex. 9, , Sol., , V', Z', , 593, , , the impedance of circuit, Fig. 9.36

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ELECTRICITY & MAGNETISM, , 594, , When ac source is used in the circuit, all the circuit element remain, effective., Z =, , or, , 1, R2, , 1, 1, C, , L, , 2, , =, , V, i, , 250, 1.25, , 1, , Z2, , 2002, , ...(ii), , With increase of frequency, current increases and becomes maximum at, = 4500 rad/s. It means Z must be minimum. It will when, , L, , 1, C, , =, , or, , 2, , or, , LC =, , =, , 0, , 1, LC, , 1, , 1, , 2, , 4500, , L =, , 4, H, 81, , C =, , 1, , and, , (b), , We have,, , for, , =, , 0, cos = 1, , for, , =, , 90°, cos = 0, , Ex. 14, , A circuit with R = 70, , 2, , (i), , Find the impedance of the circuit., , (ii), , What is the rms value of total current ?, , (iv), , Find the total power input to the circuit and power factor., , (v), , How will the circuit behave if, , =, , L, , 117, , P0 =, , V02, 2R, , 1, ?, LC, , Fig. 9.38, The resultant impedance of L and C in parallel, , 1, Z1, , =, , 1, XL, , 1, XC, , 1, , 1, 1, j C, , j L, , 1, , =, , 2, , 60, , 2, , 60 10, , =, , F, , 30 2, 2 5, , 90 W, , 2, , L, , 1, C, , j L, , j C, , 1, , 2, , LC, j L, j L, , Ans., , Z1 =, , 1, The total impedance of the circuit, , 2, , R, , Z1, , Z =, , 2, , Pav will be minimum, when C is zero that is Pav = 0 Ans., P0 = 90 W, and Pm = 0, Ans., Phase angles :, , |Z| =, (i), , Fig. 9.37, , 1, , 3, , P0 R 2, , Pav =, , in series with a parallel, , combination of L = 1.5 H and C = 30 F is driven by a 230 V supply, of angular frequency 300 rad/s., , =, , 1, , R, , (c), (d), , 90°, , 1, LC, , =, , C =, , =, , Ans., , In series RLC circuit maximum power is dissipated at, resonance., , or, , Pav,, , Power factor :, , Sol., , F, , Sol.(a), , 0, , = 0, , ...(iii), , 2, , Ex. 13 In RLC series circuit, assume R = 500 , L = 60.0 mH,, f = 60.0 Hz, and V0 = 30.0 V. For what values of the capacitance, would the average rate at which energy is dissipated in the resistor, be (a) a maximum and (b) a minimum ? (c) what are these maximum, and minimum energy dissipation rates ? What are (d) the, corresponding phase angles and (e) the corresponding power factor, ?, , At resonance, , minm, , (iii) What are the current amplitudes in the L and C arms of the, circuit ?, , After solving equations (i), (ii) and (iii), we get, , and, , At, (e), , 200, , 1, , At resonance,, , Given, R = 70, , R2, , LC, , 70 2, , =, , 163.3, , 2, , 1, , 1, , 2, , LC, , 2, , L, , , L = 1.5 H, C = 30 ×, , Z =, , j L, , R, , LC, , 10–6, , F and, , = 300 rad/s., 2, , 300 1.5, 1 300 2 1.5 30 10, Ans., , 6

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AC AND EM-WAVES, (ii), (iii), , irms =, , Vrms, Z, , 230, 163.3, , 1.41 A, , Sol. If ZA and ZB are the impedances of the branches A and, B respectively, then their combined impedance, , Ans., , If iL and iC are the currents in inductor and capacitor respectively,, then, iL + iC = irms, ...(i), and, XL iL = XC iC, ...(ii), On solving above equations, we have, , and, , Again, , ZAB =, Here, , 6 8j 4 3j, , iL =, , ZAB =, , iC =, , XL, irms, X L XC, , =, , iL =, , irms, , 1, , j L, , irms, 2, , 1, , LC, , j C, , 1.41, =, , 1 300, , 2, , 1.5 30 10, 2, , j L, , iC =, , 0.46 A, , 6, , =, =, (iv), , Power factor, , (v), , When, , cos, , =, , 300 2 1.5 30 10, , R, Z, , 70, 163.3, , 1, ; or, LC, , =, , Z =, , 10 5 j, , 10 5 j, , 10 5 j, , 48 14 j, 10 5 j, 550 100 j, 125, , i0 =, , 7.4, , 2, , 7.8, , 2, , V0, Z, , 1.41, 6, , 1, , =, , Ans., , =, , 100, 7.4 7.8 j, , 1.87 A, , j L, Now, , 6, , 4 3j, , 48 14 j, , |Z| =, , LC 1, , Total current, 3002 1.5 30 10, , 6 8j, , = 4.4 + 0.8 j, The impedance of the branch C,, ZC = 3 + 7 j, Thus total impedance of the circuit, Z = ZAB + ZC, = (4.4 + 0.8j) + (3 + 7 j), = 7.4 + 7.8 j, , LC irms, , 2, , 1, j C, , j L, , Z AZB, Z A ZB, , ZA = (6 – 8j) and ZB = 4 + 3j, , XC, irms, X L XC, , 1, j C, , 1, , 2, , LC, , 0.43, 2, , LC 1, , j L, 1 1, , =, , and so, i, , 0, , Ans., In the circuit shown in fig. 9.39, the voltage applied is, 100 V. Find the current in each branch, voltage drop across each, branch, the power factor and the power supplied by the line to the, circuit., , 100 7.4 – 7.8 j, (7.4 7.8 j )(7.4 – 7.8 j ), , 740 780 j, 54.76 60.84, , 740 780 j, 115.6, , = 6.4 – 6.74 j, Current lags behind the applied voltage by, , Ex. 15, , tan, , =, , 6.74, or, 6.4, , ; 45, , The actual value of the current, iC =, , 6.4, , 2, , 6.74, , 2, , = 9.3 A, Voltage drop across branch C, = VC = IC ZC, = (6.4 – 6.74j) × (3 + 7j), = 66.38 + 24.58 j, or, , Fig. 9.39, , 595, , VC, , =, , 66.382, , 24.582, , Voltage drop across AB,, VAB = 100 – VC, = 100 – 70.8 = 29.2 V., , 70.8 V

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ELECTRICITY & MAGNETISM, , 596, , From the figure, , Fig. 9.40, cos, Thus power, , =, , 6.4, 9.3, , Sol., , Fig. 9.41, The impedance of the circuit is given by, , 0.69, , = V0 i0 cos, = 100 × 9.3 × 0.69, = 641.7 W, , where, , Z =, , R2, , XL =, , L, , Z =, , 1., , Voltage amplitude, , When a value is given for alternating current or voltage, it, is ordinarily the rms value., The power rating of an element used in ac circuits refers to, its average value not rms., Though in a phasor diagram, voltage and current are, represented by vectors, but these quantities are not really, vectors themselves. They are scalar quanties., Only resistor is the dissipative element and so no power, losses associated with pure inductances or capacitances., A transformer (step-up) changes a low-voltage into a high, voltage. The current is reduced by the same proportion., This does not violate the law of conservation of energy., , 2., 3., , 4., 5., , Current amplitude, , V0 =, , 50 35 10, , 112 112, , 3, , 11 2, , 2 Vrms, , =, , 2, , i0 =, , V0, Z, , 220 V ., , 2, , 220, , 11 2, , = 20 A, If is the phase between V and i, then, tan, , =, =, , L, R, 4, , 11, 1, 11, , rad ., , Thus we can write, V =, , Ex. 16, , 2, , = 11, , Ans., , Note:, , X L2, , 220 2 sin t, , In L R series circuit, a sinusoidal V =V0 sin t is applied., , and, It is given that L = 35 mH, R = 11, , , Vrms = 220 V,, , 2, , i =, , = 50Hz and, , 20 sin, , t, , 4, , 22, . Find the amplitude of current is steady state and obtain, 7, the phase difference between the current and the voltage. Also plot, the variation of current for one cycle on the given graph., =, , 9.10 ELECTROMAGNETIC, , WAVES, , The idea of electromagnetic waves was proposed by Maxwell in 1864. The wave equation, for light propagating in x-direction in vacuum may be written as, E = E0 sin(kx – t), where E is the sinusoidally varying electric field., The electric field E is in the yz- plane, that is perpendicular to direction of propagation of, wave., There is also a sinusoidally varying magnetic field associated with the electric field when, light propagates. This magnetic field is perpendicular to both electric field and direction, of propagation of wave. The magnetic field, B = B0 sin(kx – t)

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AC AND EM-WAVES, Such a combination of mutually perpendicular electric and magnetic fields can represent, an electromagnetic wave in vacuum., , More about displacement current, According to Kirchhoff s junction rule, for a conducting circuit, in the steady state, the, total current into any given portion must be equal to the current out of that portion. But, this rule is not obeyed for a capacitor that is being charged. In fig. 9.43 there is a, conduction current into the left plate but there is no conduction current out of this plate;, similarly, there is conduction current out of the right plate, but not into it., According to Maxwell, as the capacitor charges, the conduction current increases the, charge on each plate, and this in turn increases the electric field between the plate., Because of the changing electric field there is a current between the plates, is called, displacement current (id): A current without movement of electrons. For any charge, density , electric field between the plates of capacitor, , Fig. 9.43, , q, , E =, , A, , 0, , or, , And displacement current, or, , q =, , 0, , EA, , =, , 0, , E, , id =, , dq, dt, , id =, , 0, , 0, , d, , 0, , E, , dt, d E, dt, , Fig. 9.44, , Note: The current due to the flow of charges is often called conduction current., Generalised Ampere's law :, It was James Clerk Maxwell who generalised Ampere s law. According to him, if there, exists an electric current as well as a changing electric field, then Ampere's law can be, written as:, , or, , B d, , =, , 0, , ic id, , B d, , =, , 0, , ic, , =, , 0 ic, , =, , 0, , 0, , 0, , d E, dt, 0, , d E, dt, , In vacuum, ic = 0 therefore, B d, , 0, , d E, dt, , Maxwell's equations :, Maxwell discovered that all the basic principles of electromagnetism can be formulated, in terms of four fundamental equations called Maxwell s equations. They are, (1) Gauss s law for electricity, E dA =, , qin, , ...(1), , 0, , (2), , Gauss s law for magnetism, B dA = 0, , (3), , Faraday s law of induction, , ...(2), , 597

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598, , ELECTRICITY & MAGNETISM, , (4), , Ampere s law, , E d, , =, , d B, dt, , B d, , =, , 0, , ...(3), , ic, , 0, , d E, dt, , ...(4), , Faraday's law, d B, ., dt, Consider a rectangular path abcd in the xy- plane as shown in fig. 9.44. The electric, , E d, , =, , field is parallel to y-axis and magnetic field is parallel to z-axis. The circulation of E over, the close path abcd,, E d, , E dE h Eh, , =, , = hdE, And the magnetic flux through rectangle,, B = B (hdx), , ...(i), , d B, dB, = hdx, dt, dt, Now from equations (1), (i) and (ii), we have, , Differentiating, we get, , or, Fig. 9.45, or, , d E0 sin kx, , hdx, , dE, dx, , =, , dB, dt, , =, , d, B0 sin kx, dt, , t, , t, , =, , B0 cos kx, , or, , E0 =, , or, , E 0 = cB 0., , where, , k, , dB, dt, , hdE =, , dx, , E0 k cos kx, , ...(ii), , k, , t, , t, , B0, , ...(5), , = c, speed of the light wave., , d E, … (2), dt, Consider the rectangular path efgh in the xz- plane as shown in fig. 9.45. The circulation, , Ampere's law, , B d, , =, , B d, , = Bh – (B + dB), , 0, , 0, , of B is, = –h dB, The magnetic flux through the rectangle is, E = E (h dx), On differentation gives, , ...(i), , d E, dE, = h dx, dt, dt, Now from equations (2), (i) and (ii), we have, , –h dB =, Fig. 9.46, , 0, , 0, , h dx, , ...(ii), , dE, dt

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600, , ELECTRICITY & MAGNETISM, 1, , E 0 = cB0 and c, , uav =, , 1, 2, , 0, , E02, , Intensity of wave, , 0, , B0 2, ., 2 0, , 0, , ....(9), , The energy crossing per unit area per unit time perpendicular to the direction of, propagation is called the intensity of wave. From above figure we have, dx = cdt, Energy contained, U = uav (dV) = uav (A dx), = uav Acdt, The intensity, , I =, , U, Adt, , or, , I =, , 1, 2, , Momentum :, , 0, , uav c, , E0 2 c., , ....(10), , The electromagnetic wave also carries momentum with it. The momentum carried by the, portion of wave having energy U is gives by, P =, , 9.12 ELECTROMAGNETIC, , U, ., c, SPECTRUM, , Maxwell predicted the existence of electromagnetic waves. The only familiar, electromagnetic waves were visible light waves. By the end of nineteenth century, Xrays and gamma rays had also been discovered. We know now were electromagnetic, waves include infrared waves, visible light waves, X-rays, gamma-rays, radio waves,, microwaves, and ultraviolet waves. Their classification according to frequency is called, the electromagnetic spectrum.

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AC AND EM-WAVES, , Ex. 17, , A plane electromagnetic wave of frequency 25 MHz, travels in free space along the x-direction. At a particular point in, , Sol. Bulb radiates energy uniformly in all direction. At a distance of 3, m from the bulb, the surface area of the envelop, , the space and time, E = 6.3j V/m. What is B at this point ?, , Sol. The magnitude of the magnetic field is given by, B =, , E, c, , 6.3, , 2.1 10, , 3 108, , A =, , T., I =, , Thus, , Ex. 18, , B =, , i, indicates that Aˆ, 8, , 2.1 10, , =, , k., , t–, , x, C, , 1, 2, , Thus, , Sol., , or, , The maximum value of electric field is 300 V/m. Therefore the maximum, electric force, Fe = qE0 = 1.6 × 10–19 × 300, = 4.8 × 10–17 N, The maximum magnetic field, B0 =, , 3 108, The maximum magnetic force on electron, Fm = qvB 0, =, , 1.6 10, , 19, , 10, , 8, , =, , 3.2 10, , 0, , Power, area, , 2.5, 100, , 100, 113, , 0.022 W/m2., , Erms 2c = 0.011, , Erms =, , =, , 8.85 10, , 12, , 3 108, , 2.9 V/m, , The peak value of electric field, , T, , E0 =, =, , 2.0 107, , 0.011 2, 0 c, , 0.11 2, =, , 10, , 8, , Now, 18, , 113 m 2 ., , I, = 0.011., 2, , . An electron is, , constrained to move along the y-direction with a speed of 2.0 × 107, m/s. Find the maximum electric force and the maximum magnetic, force on the electron., , 300, , 2, , As half of the intensity is provided by the electric field and half by the, magnetic field, so, , A light beam travelling in the x-direction is described, , E0, c, , 3, , Ans., , k T., , by the electric field E y = (300 V/m) sin, , 4, , The intensity of radiation at this distance is, 8, , The vector E × B should be along x-direction, and so, , j ( Aˆ ) =, , 601, , Brms =, , 2 Erms, , 2, , 2.9, , 4.1 V/m, , Erms, c, , Ans., , 2.9, 3 108, , N., , Ex. 19 Calculate the electric and magnetic fields produced by, the radiation coming from a 100 watt bulb at a distance of 3 m., Assume that the efficiency of the bulb is 2.5% and it is a point, source., , and, , =, , 9.6 10, , B0 =, , 2 Brms, , 9, , T,, 1.4 10, , 8, , T., , Ans.

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ELECTRICITY & MAGNETISM, , 602, , Review of formulae & Important Points, 1., , 2., , Alternating current (AC) : Alternating current is one whose, direction changes periodically; i = i 0 sin t is the sinusoidal, alternating current., RMS value of AC : For sinusoidal AC, i = i0 sin t, the RMS value, irms =, Similarly, , 3., , electric field E and a magnetic field B with magnitudes which, depend on x and t :, , = 0.707 i0., , 2, , V0, = 0.707 V0., 2, , 1, C, is the capacitive reactance : the current here leads the potential, by 90°., (iii) Circuit having inductor only : VL = iXL, in which XL= L is, the inductive reactance; the current here lags the potential by, 90°., Impedance and admittance : The total resistance of AC circuit is, called impedance Z, and reciprocal of impedance is called, Circuit having capacitor only : Vc = iXc in which X c, , 10., , 6., , E dA, , and, , tan, , =, , R2, , L, , L, , 1, C, , =, , 1, C, , (ii), , R, 7., , 2, , E d, , =, , =, , 0, , d B, dt, , d E, dt, Energy flow : The rate per unit area at which energy is transported, via an electromagnetic wave is given by the Pointing vector, Bd, , 2, , 11., , ,, , =, , S, , iC, , 0, , E, , 0, , B, , =, 0, , 1, C, , Intensity of EM-wave : The time averaged rate per unit area at, which energy is transported, is called the intensity of wave :, , 1, 2, 0 E0 c., 2, The intensity of the waves at distance r from a point source of, power P is, I =, , 2, , Resonance : The current amplitude i0 in a series LRC circuit, driven by a sinusoidal external emf is a maximum i, , V0, R, , P, I =, , when, , the driving angular frequency equals the natural angular frequency, 0 (resonance frequency) of the circuit. Then XC = XL, = 0, and, the current is in phase with the potential., , 8., , ., , (iv) Ampere s law, , 12., , L, , q, , =, , Gauss s law for magnetism, B dA, , V0, , i =, , 0, , (iii) Faraday s law of induction, , R, current, , 0, , 0, , V = 220 2 sin 100 t., Series RL C circuit : For a series RLC circuit,, Z, , 1, , Maxwell s equations : Maxwell discovered that all the basic, principles of electromagnetism can be formulated in terms of four, fundamental equations, called Maxwell s equations. These are :, (i) Gauss s law for electricity, , 2 V, and, , 220, , E, =, B, , c =, , 1, ., Z, In India domestic supply is 220 V, 50 Hz. Thus, V0 =, , t, , t ., and, B = B0 sin kx, Electric field induces the magnetic field and vice-versa. The speed, of EM-waves is c, which can be written as, , admittance A. Thus A =, 5., , E0 sin kx, , E =, , For square wave AC, irms = i0., Three simple circuits :, (i) Circuit having resistor only : The alternating potential, difference across a resistor has amplitude VR = iR; the current, is in phase with the potential difference., (ii), , 4., , Vrms =, , i0, , 9., , cos, 1. For purely inductive or capacitive circuit cos = 0 and, so Pav = 0., EM-waves : An EM-wave consists of oscillating electric and, magnetic fields. An EM-wave travelling along an x-axis has an, , 1, ., 0 =, LC, Power in AC circuits : In a series RLC circuit, the average power, Pav equal to the production rate of thermal energy in the resistor :, Pav = Vrms irms cos ., , R, . Its value, Here cos is called power factor and is equal to, Z, , 13., , 4 r2, Radiation pressure : When a surface intercepts electromagnetic, radiation, a force is exerted on the surface. If the radiation is totally, absorbed by the surface, the force is, IA, c, where A is the area of the surface perpendicular to the path of the, radiation. If the radiation is totally reflected back along its original, path, the force is, F =, , F =, , 2IA, c

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AC AND EM-WAVES, , E xercise 9. 1, , MCQ Type 1, , Only one option correct, 1., A capacitor in an LC oscillator has a maximum potential difference, of 17 V and a maximum energy of 160 J. When the capacitor has, a potential difference of 5V and an energy of 10 J, what is the, energy stored in the magnetic field ?, , 2., , 3., , (a), , 10, , (c), , 160, , J, J, , (b), , 150, , J, , (d), , 170, , J, , If we increase the driving frequency in a circuit with a purely, resistive load, then amplitude VR, (a), , remains the same, , (b), , increase, , (c), , decrease, , (d), , none, , The figure shows a sine curve S(t) = sin t and three other, sinusoidal curves A(t), B(t) and C(t), each of the form sin ( t– )., Which curve is according to the most negative value of, , 4., , A, , (b), , 7., , B, , (c) C, (d) B and C, Figure shows three oscillating LC circuit with identical inductors, and capacitors. If t1, t2, t3 are the time taken by the circuits I, II,, III for fully discharge, then, 9., , the charge on the capacitor is zero after a time, , (2), , The charge on the capacitor is zero after time, , (3), , the energy stored in the capacitor maximum at, , (4), , the energy stored in the inductor is maximum at, , 50 , 100, , (a), , t1 > t2 > t3, , (b), , t1 < t2 < t3, , t1t2, t2 < t1 < t3, (d) t3, In an oscillating LC circuit with L = 50 mH and, C = 4.0 F, the current is initially a maximum. How long will it, take before the capacitor is fully discharged for the first time :, (a) 7 × 10–4 s, (b) 14 × 10–4 s, –4, (c) 28 × 10 s, (d) none, A charged capacitor and an inductor are connected in series at, time t = 0. Read the statements : (T is the time period of, oscillations), (c), , 5., , 6., , T, ., 2, , (1), , T, 4, T, 2, T, 4, , Out of these statements; correct statement(s) is/are, (a) 2, (b) 3, (c) 1, 2, (d) 2, 3, 4, If we increase the driving frequency in a circuit with a purely, capacitive load. Read the following statements :, (1) amplitude VC increases (2) amplitude VC decrease, (3) amplitude iC increase, (4) amplitude iC decreases, Out of these, the correct statement(s) is/are, (a) 1, (b) 1, 2, (c) 3, (d) 2, 3, 4, If we increase the deriving frequency in a circuit with a purely, inductive load. Read the following statements :, (1) amplitude VL remain constant, (2) amplitude VL increases, (3) amplitude iL increases, (4) amplitude iL decreases, Out of these, the correct statement (s) is / are, (a) 1, (b) 2, (c) 1, 2, 3, (d) 1, 4, Here are the capacitive reactance and inductive reactance,, respectively, for three sinusoidally driven RLC circuits : (1), , 8., (a), , 603, , 10., , (2) 100 , 50, , (3) 100 , 100 . Which is in, , resonance ?, (a) 1, (b) 2, (c) 3, (d) 1, 2, An alternating current emf device has a smaller resistance than that, of the resistive load, to increase the transfer of energy from the, device to the load, a transformer will be connected between two., Then, (a) NS should be greater than NP, (b) NS should be less than N/P, (c) NS = NP, (d) none, , Answer Key, , 1, , (b), , 2, , (a), , 3, , (a), , 4, , (c), , 5, , (a), , Sol. from page 613, , 6, , (d), , 7, , (c), , 8, , (d), , 9, , (c), , 10, , (a)

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ELECTRICITY & MAGNETISM, , 604, 11., , An AC source is rated 220 V, 50 Hz. The average voltage is, calculated in a time interval of 0.01s. It, (a) may be zero, (b) must be zero, (c), , 12., , 13., , 14., , is never zero, , (d), , is, , 220, V, 2, , An electromagnetic wave going through vacuum is described by E, = E0 sin(kx– t); B = B0 sin (kx– t). Then, (a) E0 k = B0, (b) E0 B0 = k, (c) E0 = B0k, (d) none of these., Alternating current can not be measured by dc ammeter because, (a) ac cannot pass through dc ammeter, (b) average value of complete cycle is zero, (c) ac is virtual, (d) ac changes its direction, If a current I given by I 0 sin, , t, , 2, , flows in an ac circuit across, , 19., , (a), (c), 15., , 16., , 17., , 18., , P, , V0 I 0, , (b), , 2, , P, , (d), , An alternating voltage V, , 20., , 200 2 sin 100t is connected to a 1, , microfarad capacitor through an ac ammeter. The reading of the, ammeter shall be, (a) 10 mA, (b) 20 mA, (c) 40 mA, (d) 80 mA, A 220 V, 50 Hz ac source is connected to an inductance of 0.2 H, and a resistance of 20 ohm in series. What is the current in the, circuit, (a) 10 A, (b) 5 A, (c) 33.3 A, (d) 3.33 A, In the circuit shown below, the ac source has voltage V = 20, cos( t) volt with = 2000 rad/s. The amplitude of the current will, be nearest to, , (b), , 3.3A, , (c), , 2 / 5A, , (d), , 5A, , L/R, , (d) R / R 2, , The power factor of an ac circuit having resistance (R) and, inductance (L) connected in series and an angular velocity is, , 22., , L, , 2 2, , L, , 1/ 2, , 1/ 2, , R2, , V 2R, , 2 2, , L, , (b), , V, , R, , 2, , (d), , 2 2, , L, , R2, , 2 2, , L, , R2, , 2 2, , L, , V, , 2, , In an LCR series ac circuit, the voltage across each of the, components, L, C and R is 50 V. The voltage across the LC, combination will be, 50 V, , (b), , 50 2V, (c) 100V, (d) 0 V (zero), A bulb and a capacitor are connected in series to a source of, alternating current. If its frequency is increased, while keeping the, voltage of the source constant, then, (a) bulb will give more intense light, (b) bulb will give less intense light, (c) bulb will give light of same intensity as before, (d) bulb will stop radiating light, An alternating e.m.f. of angular frequency is applied across an, inductance. The instantaneous power developed in the circuit has, an angular frequency, (a), , 23., , 2 2, , An inductor of inductance L and resistor of resistance R are joined, in series and connected by a source of frequency . Power, dissipated in the circuit is, , (a), , P=0, , 2A, , (c), , V, , 2V0 I 0, , (a), , (b) R / R 2, , (c), , 21., , V0 I 0, 2, , R/ L, , (a), , which an ac potential of V V0 sin t has been applied, then the, power consumption P in the circuit will be, , P, , (a), , (b), , 4, , 2, (c), (d) 2, The diagram shows a capacitor C and a resistor R connected in, series to an ac source. V1 and V2 are voltmeters and A is an ammeter., , Consider now the following statements, I., Readings in A and V2 are always in phase, II. Reading in V1 is ahead in phase with reading in V2., III. Readings in A and V1 are always in phase. Which of these, statements are/is correct, (a) I only, (b) II only, (c) I and II only, (d) II and III only, , Answer Key, , 11, , (a), , 12, , (a), , 13, , (b), , 14, , (d), , 15, , (b), , 16, , (d), , Sol. from page 613, , 18, , (b), , 19, , (b), , 20, , (d), , 21, , (a), , 22, , (d), , 23, , (d), , 17, , (a)

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AC AND EM-WAVES, 24., , At t < 0, the capacitor is charged and the switch is opened. At t =, 0 the switch is closed. The shortest time T at which the charge on, the capacitor will be zero is given by :, L, , 30., , 605, , The value of alternating emf E in the given circuit will be, , VR = 80V, , VC = 100V, , VL= 40 V, , Q0 C, , S, , E, 50 Hz, (a), , (c), 25., , (c), , 27., , 29., , LC, , (d), , LC, , 2, , 31., , LC, , Q, 2, , 32., , Q, 2, , (b), , (a), , 100 V, , (b), , 20 V, , (c), , 220 V, , (d), , 140 V, , The power factor of LCR circuit at resonance is, (a), , 0.707, , (b), , 1, , (c), , zero, , (d), , 0.5, , (a), , Q, 3, , Q, 3, , (d), , (c), 33., , (a), , ac cannot pass through dc ammeter, , (b), , average value of complete cycle is zero, , (c), , ac is virtual, , (d), , ac changes its direction, , 1, 1000, , The frequency for which a 5µF capacitor has a reactance of, ohm is given by, , Alternating current can not be measured by dc ammeter because, , 100, , MHz, , 1000, , (b), , 1, Hz, 1000, , (d), , Hz, , 1000 Hz, , Two identical incandescent light bulbs are connected as shown in, figure. When the circuit is an AC voltage source of frequency f,, which of the following observations will be correct?, , An alternating current is given by the equation, , i, , 28., , 2, , 3, 2, , In an oscillation of L-C circuit, the maximum charge on the capacitor, is Q. The charge on the capacitor, when the energy is stored equally, between the electric and magnetic field is, (a), , 26., , (b), , LC, , R, , i1 cos t i2 sin t . The r.m.s current is given by, , (a), , 1, (i1 i2 ), 2, , (b), , (c), , 1 2, (i1, 2, , (d), , i22 )1/2, , C, , 1, (i1 i2 )2, 2, 1 2, (i1, 2, , R, , L, Bulb b1, , i2 2 )1/2, , Bulb b2, , A choke coil has, , (a), , Both bulbs will glow alternatively, , (a), , high inductance and low resistance, , (b), , Both bulbs will glow with same brightness provided, , (b), , low inductance and high resistance, , (c), , high inductance and high resistance, , (d), , low inductance and low resistance, , f, , An alternating voltage is connected in series with a resistance R, and an inductance L. If the potential drop across the resistance, is 200 V and across the inductance is 150 V, then the applied, voltage is, (a), (c), , 350 V, , (b), , 500 V, , (d), , 34., , 1, 2, , (1/ LC ), , (c), , Bulb b1 will light up initially and goes off, bulb b2will be ON, constantly, , (d), , Bulb b1 will blink and bulb b2 will be ON constantly, , For high frequency, a capacitor offers, , 250 V, , (a), , more reactance, , (b), , less reactance, , 300 V, , (c), , zero reactance, , (d), , infinite reactance, , Answer Key, , 24, , (c), , 25, , (b), , 26, , (b), , 27, , (c), , 28, , (a), , Sol. from page 613, , 30, , (a), , 31, , (b), , 32, , (a), , 33, , (a ), , 34, , (b), , 29, , (b)

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ELECTRICITY & MAGNETISM, , 606, 35., , 36., , In a LCR circuit capacitance is changed from C to 2C. For the, resonant frequency to remain unchanged, the inductance should be, change from L to, , Which of the following plots shows the variation of voltage in the, coil, , V, , (a), , 4L, , (b), , 2L, , (c), , L/ 2, , (d), , L/ 4, , The current i in an inductance coil varies with time t according to, following graph., , (a), , V, , t, , (0, 0), , (b), , t, , (0, 0), , i, , V, , (c), t, , (0, 0), , Answer Key, , 35, , (c), , 36, , V, , t, , (0, 0), , (d), , t, , (0, 0), , (b), , Sol. from page 613, , Level -2, Only one option correct, 1., A parallel-plate capacitor with rectangular plates is being, discharged. A rectangular loop, centered on the plates and between, them, measures L by 2L : the plates measure 2L by 4L. The, fraction of the displacement current is encircled by the loop if that, current is uniform, (a), , 2., , 3., , 1, , (b), , 1, 2, , 1, 1, (c), (d), 8, 4, Charges on the capacitors in four oscillating LC circuits vary as, follows : (1) q = 2 cos 4t, (2) q = 4cos t, (3) q = 3 cos 4t, (4) q = 4, cos 2t, with q in coulomb and t in second. In which circuit(s), current amplitude is greatest :, (a) (1), (b) (2), (c) (3), (d) (4), The figure here gives the electric field of an electromagnetic wave, at a certain point and a certain instant. The wave is transporting, energy in the negative z-direction. The direction of the magnetic, field of the wave at that point and instant is :, , (a), (c), , + ve x-direction, +ve z-direction, , 4., , 5., , Which of the following plots may represent the reactance of a, series LC combination ?, (a), , I, , (b), , II, , (c), , III, , (d), , IV, , The voltage of an ac supply varies with time (t) as V = 120 sin 100, t cos 100 t. The maximum voltage and frequency respectively, are, (a), , 6., , (b) –ve x-direction, (d) –ve y-direction, , Answer Key, , 1, , (c), , 2, , (c), , Sol. from page 614, , 5, , (d), , 6, , (b), , 120 volts, 100 Hz, , (b), , (c) 60 volts, 200 Hz, Match the following :, Currents, , 120, volts, 100 Hz, 2, , (d) 60 volts, 100 Hz, r.m.s. values, , (1), , x0 sin t, , (i) x0, , (2), , x0 sin t cos t, , (ii), , (3), , x0 sin t, , (iii), , (a), (c), , 1. (i), 2. (ii), 3. (iii), 1. (i), 2. (iii), 3. (ii), , 3, , x0 cos t, , x0, 2, x0, 2 2, , (b) 1. (ii), 2. (iii), 3. (i), (d) None of these, , (a), , 4, , (d)

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AC AND EM-WAVES, 7., , 8., , An ac source of angular frequency is fed across a resistor r and, a capacitor C in series. The current registered is I. If now the, frequency of source is changed to /3 (but maintaining the same, voltage), the current in the circuit is found to be halved. Calculate, the ratio of reactance to resistance at the original frequency, (a), , 3, 5, , (b), , 2, 5, , (c), , 1, 5, , (d), , 4, 5, , (c), , 10., , When 100 volts dc is supplied across a solenoid, a current of 1.0, amperes flows in it. When 100 volts ac is applied across the same, coil, the current drops to 0.5 ampere. If the frequency of ac source, is 50 Hz, then the impedance and inductance of the solenoid are, , 12., , (a), , 200, , and 0.55 henry, , (b) 100, , and 0.86 henry, , (c), , 200, , and 1.0 henry, , (d) 100, , and 0.93 henry, , If the total charge stored in the LC circuit is Q0, then for t, (a), , The charge on the capacitor is Q, , Q0 cos, , (b), , The charge on the capacitor is Q, , Q0 cos, , I0, 2, , (c), , The charge on the capacitor is Q, , LC, , 2I 0, , (d), , The charge on the capacitor is Q, , The output current versus time curve of a rectifier is shown in the, figure. The average value of output current in this case is, (a) 0, (b), , 9., , 11., , (d) I0, When an ac source of e.m.f. e = V0 sin(100 t) is connected across a, circuit, the phase difference between the e.m.f. e and the current i, in the circuit is observed to be / 4, as shown in the diagram. If the, circuit consists possibly only of RC or LC in series, find the, relationship between the two elements, (a), , R 1k , C 10 F, , (b), , R 1k , C 1 F, , (c), , R 1k , L 10 H, , (d), , R 1k , L 1H, , 607, , 13., , 2, , t, LC, , 2, , t, LC, , 0, , d 2Q, dt 2, , 1, , d 2Q, , LC dt 2, , In the circuit shown in the figure, the ac source gives a voltage V, = 20 cos 200 t. Neglecting source resistance, the voltmeter and, ammeter reading will be, , 6, A, 5mH, , 4, , 50µF, , V, , The figure shows three circuits with identical batteries, inductors,, and resistors. Rank the circuits, in the decreasing order, according, to the current through the battery (i) just after the switch is closed, and (ii) a long time later, , 14., , (a), , 0V, 0.97 A, , (b) 1.68 V, 0.47 A, , (c), , 0V, 1.4 A, , (d) 5.64 V, 1.4 A, , If i1 = 3 sin t and i2 = 4 cos t, then i3 is, , i2, , i1, , i3, , (a) 5 sin ( t + 53º), 15., , (1), , (2), , (3), , (a), , (i) i2 > i3 > i1 (i1 = 0) (ii) i2 > i3 = i1, , (a), , A, , (b), , (i) i2 < i3 < i1 (i1, , 0) (ii) i2 > i3 > i1, , (b), , B, , (c), , (i) i2 = i3 = i1 (i1 = 0) (ii) i2 < i3 < i1, , (c), , C, , (d), , (i) i2 = i3 > i1 (i1, , (d), , D, , 0) (ii) i2 > i3 > i1, , (b) 5 sin ( t + 37º), , (c) 5 sin ( t + 45º), (d) 5 cos ( t + 53º), The figure shows graphs of the electric field magnitude E versus, time t for four uniform electric fields, all contained within identical, circular regions. Which of them is according to the magnitudes of, the magnetic field greatest :, , Answer Key, , 7, , (a), , 8, , (c), , 9, , (a), , 10, , (a), , Sol. from page 614, , 12, , (a), , 13, , (d), , 14, , (a), , 15, , (c), , 11, , (a)

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ELECTRICITY & MAGNETISM, , 608, , MCQ Type 2, Multiple correct options, 1., Figure shows a parallel plate capacitor and the current in the, connecting wires that is discharging the capacitor., , 4., , 5., , (a), (b), , The displacement current is leftward., The displacement current is rightward, , (c), , 2., , 3., , The electric field E is rightward, (d) The magnetic field at point P is into the page., If the inductance L in an oscillating LC circuit having a given, maximum charge Q is increased, then, (a) the current magnitude increases, (b) the maximum magnetic energy increases, (c) the maximum magnetic energy decreases, (d) current magnitude and maximum magnetic energy remain, constant., The reactance of a circuit is zero. It is possible that the circuit, contains, (a) an inductor only, (b) a capacitor only, (c) an inductor and a capacitor, (d) a resistor only., , 6., , E xercise 9. 2, , In an AC series circuit, the instantaneous current is zero when the, instantaneous voltage is maximum. Connected to the source may, be, (a) pure inductor, (b) pure capacitor, (c) pure resistor, (d) combination of an inductor and a capacitor, The magnetic field can be produced by, (a) a moving charge, (b) a changing electric field, (c) a non uniform field, (d) none of them, An alternating e.m.f. of frequency f, , 1, 2 LC, , is applied to a, , series LCR circuit. For this frequency of the applied e.m.f., , Answer Key, , 1, , (a, d), , 2, , (a, b), , Sol. from page 615, , 5, , (a, b), , 6, , (b, d), , (a), , The circuit is at resonance and its impedance is made up only, of a reactive part, , (b), , The current in the circuit is in phase with the applied e.m.f., and the voltage across R equals this applied emf, , (c), , The sum of the p.d.'s across the inductance and capacitance, equals the applied e.m.f. which is 180° ahead of phase of the, current in the circuit, , (d), , The quality factor of the circuit is L / R or 1 / CR and this, is a measure of the voltage magnification (produced by the, circuit at resonance) as well as the sharpness of resonance of, the circuit, , 3, , (c, d), , 4, , (a, b, d)

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AC AND EM-WAVES, , Statement Questions, Read, (a), (b), (c), (d), 1., , Statement - 1, An alternating current is given by i = 3 sin t + 4cos t. The. rms, , 7, A., 2, , 7., , 8., , Statement - 2, The rms current is, , 5, A., 2, , 9., , Statement - 1, A capacitor is connected to a direct current source. Its reactance is, infinite., Statement - 2, 10., Reactance of a capacitor is given by X c, , 3., , 4., , 5., , 1 ., C, , Statement - 1, In a circuit containing a capacitor and an AC source the current is, zero at the instant the source voltage is maximum. It is not consistent, with Ohm s law., Statement - 2, According to Ohm s law, V = iR., Statement - 1, When the frequency of the AC source in an LCR circuit equals the, resonant frequency, the reactance of the circuit is zero, and so, there is no current through the inductor or the capacitor., Statement - 2, The net current in the inductor and capacitor is zero., Statement - 1, The alternating current legs behind the emf by a phase angle of, , 2, , 6., , Exercise 9.3, , the two statements carefully to mark the correct option out of the options given below:, If both the statements are true and the statement - 2 is the correct explanation of statement - 1., If both the statements are true but statement - 2 is not the correct explanation of the statement - 1., If statement - 1 true but statement - 2 is false., If statement - 1 is false but statement - 2 is true., , current of it is, , 2., , 609, , 11., , (path-1 follows the edge of the plate). The value of, , Power in AC circuit is given by P, , V0i0, cos ., 2, , for, , Bd, , the path-1 is smaller., , , when AC flows through an inductor.., , Statement - 2, The inductive reactance increases as the frequency of AC source, increases., Statement - 1, The voltage and current in a series AC circuit are given by V = V0, sin t and i = i0cos t. The power dissipated in the circuit is zero., Statement - 2, , Statement - 1, The power is produced when a transformer steps up the voltage., Statement - 2, In an ideal transformer VI = constant., Statement - 1, Hot-wire ammeter can be used to measure DC., Statement - 2, DC produces heating effect., Statement - 1, Choke coil is preferred over a resistor to control the current in an, AC circuit., Statement - 2, Power factor of an ideal inductor is zero., Statement - 1, A bulb connected in series with a solenoid is connected to AC, source. If a soft iron core is introduced in the solenoid, the bulb, will glow brighter., Statement - 2, On introducing soft iron in the solenoid, the electrical inertia, increases., Statement - 1, The figure is a view of one plate of a parallel-plate capacitor from, within the capacitor. The dashed line show two integration paths, , 12., , Statement - 2, The value of magnetic field for the path-2 is smaller., Statement - 1, The displacement current in a parallel-plate capacitor of capacitance, C can be written as id, , C, , dV, , where V is the potential, dt, , difference between the plates., Statement - 2, The displacement current in free space is given by id, , 0, , d e, ., dt, , Answer Key, , 1, , (d), , 2, , (a), , 3, , (a), , 4, , (d), , 5, , (b), , 6, , (a), , Sol. from page 615, , 7, , (d), , 8, , (a), , 9, , (a), , 10, , (d ), , 11, , (d), , 12, , (a)

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ELECTRICITY & MAGNETISM, , 610, , Passage & Matrix, , 6., Passage for Q.1 to Q.3, Two coils are connected in series. With 2A dc through the circuit, the, p.ds. across the coils are 20 V and 30 V respectively. With 2 A ac at 40, Hz, the p.ds. across the coils are 140 V and 100 V respectively. The two, coils are connected in series with a 230 V, 50 Hz supply., 1., , 2., , 3., , The current in the coils is :, (a) 1 A, (b) 1.45 A, (c) 1.55 A, (d) 2.15 A, The power factor of the circuit is :, (a) 0.17, (b) 0.5, (c) 1, (d) none, The power dissipated in the circuit :, (a) 40 W, (b) 60 W, (c) 80 W, (d) 100 W, , Exercise 9.4, , The p.d. across the capacitor at the instant when the current is, one-half of its maximum value :, (a) 60.7 V, (b) 73.5 V, (c) 95.0 V, (d) 100 V, , Passage for Q.7 to Q.9, A parallel-plate capacitor has square plate 1.0 m on a side as shown in, figure. A current of 2.0 A charges the capacitor, producing a uniform, electric field E between the plates, with E perpendicular to the plates., , Passage for Q. 4 to Q. 6, 7., , A voltage V = 100 sin 314 t is applied to a circuit consisting of a 25, resistor and 80 F capacitor in series., 4., The instantaneous current is given by :, (a) i = 100 sin (314 t), (b) i = 2.13 sin (314 t), (c) i = 2.13 sin (314 t + ), (d) zero, 5., The power consumed, (a) 35.5 W, (b) 50 W, (c) 56.7 W, (d) 90 W, , 10., , 8., , 9., , The displacement current through the region between the plates, (a) zero, (b) 1.0 A, (c) 2.0 A, (d) 2.5 A, The value of dE/dt in the region, (a), , 1.1 1011 V / m, , (b) 2.1 1011 V / m, , (c), , 2.3 1011 V / m, , (d) 3.4 1011 V / m, , The value of, (a), (c), , B.d, , around the square dashed path, , zero, 0.75 T-m, , (b) 0.63 T-m, (d) none of these, , Consider the circuit shown in the figure. Currents in various branches of the circuit have been marked. Match the entries in Column I to, the entries in Column II., I4 L, Column I, Column II, A. I1 at t = 0, (p) 0, R, B., , I2 at t =, , C., , I3 at t = 0, , D., , I4 at t =, , (q), , E, R, , I3, R, , E, 2R, (s) I1 at t =, (t) I2 at t = 0, , I2, , L, , (r), , I1, R, E, , Key closed, at t = 0, , Answer Key, , 1, , (c), , 2, , (a), , 3, , (b), , 4, , (c), , Sol. from page 616, , 7, , (c), , 8, , (c), , 9, , (b), , 10, , A-r ; B-p, t ; C-r ; D-q, s, , 5, , (c), , 6, , (b)

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AC AND EM-WAVES, 11., , In an LCR sereis circuit connected to an ac source, the supply voltage is V, VL, , VC, , VR, , L, , C, , R, , Column I, , 12., , V0 sin 100 t, , . VL = 40 V, VR = 40V, Z = 5, , 6, , 611, and R =4, , Column II, , A., , Peak current (in A), , (p) 10 2, , B., , V0 (in volts), , C., D., , Effective value of applied voltage (in volts), XC (in ), , (q) 50 2, (r) 50, (s) 1, , In a series LCR circuit, the e.m.f. leads current. Now the driving frequency is decreased slightly., Column I, , Column II, , A., , Current amplitude, , (p) Increases, , B., , Phase constant, , (q) Decreases, , C., , Power developed in resistor, , (r) Remains same, , D., , Impedance, , (s) May increase or decrease, (t) Becomes maximum if resonace is achieved, , Answer Key, , 11, , A-p ; B-q, ; C-r ; D-s, , 12, , A-p, t ; B-q ; C-p, t ; D-q, , Sol. from page 616, , Subjective Integer Type, , E xercise 8. 5, Solution from page 617, , 1., , A 25.0 F, a 0.10 H inductor and 25.0, , resistor are connected, , in series with an ac source whose emf is given by e = 310 sin 314 t., (i) What is the frequency of emf ?, (ii) What is rms value of emf ?, (iii) What is reactance of circuit ?, (iv) What is impedance of the circuit ?, (v) What is the current in the circuit ?, (vi) What are the effective values of voltage across capacitor,, inductor and resistor, Ans : (i) 50 Hz (ii) 219 V (iii) 96, 2., , 3., , 4., , (iv) 99.2, , (v) 2.20 A, , (vi) VR = 55 V, VC = 280 V, VL = 69 V., An inductor of inductance 100 mH is connected in series with a, resistance, a variable capacitance and an AC source of frequency, 2.0 kHz. What should be the value of the capacitance so that, maximum current may be drawn into the circuit ?, Ans : 63 nF., A 60 Hz AC voltage of 160 V impressed across an LR-circuit, results in a current of 2 A. If the power dissipation is 200 W,, calculate the maximum value of the back emf arising in the, inductance.Ans : 125 V., A LC circuit ( inductance 0.01 H and capacitance 1 F) is, connected to an AC source of variable frequency. If the frequency, is varied from 1 kHz to 2 kHz, then show the consequent variation, of current by a rough sketch., , 5., , Ans : fr = 1592 Hz, A 100 V AC source of frequency 500 Hz is connected to LCR, circuit with L = 8.1 mH, C = 12.5 F and R = 10, , , all connected, , in series. Find the potential across the resistance., Ans : 100 V, 6., , An LCR series circuit with 100, , resistance is connected to an, , AC source of 200 V and angular frequency 300 rad/s. When only, the capacitance is removed, the current legs behind the voltage, by 60°. When only the inductance is removed, the current leads, the voltage by 60°. Calculate the current and the power dissipated, in the LCR circuit., Ans : 2A, 400 W.

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612, , ELECTRICITY & MAGNETISM, , Subjective, , E xercise 9. 6, Solution from page 618, , 1., , The electric current in a circuit is given by i, , i0, , t, , frequency and becomes maximum at 4500 rad/s. Find the values, of L, C and R. Draw the circuit diagram., , for some, , time. Calculate the rms current for the period t = 0 to t = ., Ans :, , 2., , 3., , 4., , 5., , Ans : L =, , i0, , 3, A current is made up of two components : a 5 A DC component, and a 60 cycle AC sinusoidal component of peak value 4 A., Write an expression for resultant current and calculate the average, current over a complete cycle and the effective value of current., Ans : i = 5+4 sin 377t; 5A; 5.75 A., A current of 4 A flows in a coil when connected to a 12 V DC, source. If the same coil is connected to a 12 V, 50 rad/s AC, source. a current of 2.4 A flows in the circuit. Determine the, inductance of the coil. Also find the power developed in the, circuit if a 2500 F capacitor is connected in series with the coil., Ans : 0.08 H, 17.28 W., A capacitor of capacitance 12.0 F is joined to an AC source of, frequency 200 Hz. The rms current in the circuit is 2.00 A. (a), Find the rms voltage across the capacitor (b) Find the average, energy stored in the electric field between the plates of the, capacitor., Ans : (a) 133 V, (b) 0.106 J., A coil has a resistance of 10, , 10., , 11., , and an inductance of 0.4 henry. It, , is connected to an AC source of 6.5 V,, , 30, , power consumed in the circuit., Ans :, , 6., , 7., , 8., , 1 F, R, , 250, , ,, , In the circuit shown in figure the, switch is kept in position for a, long time. It is then thrown to, position b. (a) calculate the, frequency of the resulting, oscillating current. (b) What is, the amplitude of the current, oscillations ?, Ans : (a) 275 Hz (b) 364 mA., In figure, R 15.0 , C = 4.70, F, and L = 25.0 mH. The, generator provides a sinusoidal, voltage of 75.0 V (rms) and, frequency f = 550 Hz. (a), Calculate the rms current (b), Find the rms voltages Vab, Vbc,, Vca, Vbd, Vad. (c) At what average, rate is energy dissipated by each, of the three circuit elements ?, , Hz. Find the average, , 5, W., 8, The energy in an oscillating LC circuit containing a 1.25 H inductor, is 5.70 J. The maximum charge on the capacitor is 175 C. Find, (a) the mass, (b) the spring constant, (c) the maximum, displacement, and (d) the maximum speed for a mechanical system, with the same period., Ans : (a) 1.25 kg (b) 372 N/m (c) 1.75×10–4 m (d) 3.02 mm/s, , 4, , H, C, 81, , Ans : (a) 2.59 A; (b) 38.8 V, 159 V, 224 V, 64.2 V, 75.0 V; (c) 100 W for R,, 0 for L and C., 12., In an oscillating series RLC circuit, show that the fraction of the, , A LCR circuit has L = 10 mH, R = 3 , and C = 1, F connected, in series to a source of 15 cos t volt. Calculate the current, amplitude and the average power dissipated per cycle at a, frequency that is 10% lower than the resonant frequency., Ans : 0.704 A, 0.744 W., An AC source is connected to two circuits as shown. Obtain the, current through the resistance R at resonance in both the circuits., , energy lost per cycle of oscillation, , U, , is given to a close, U, , 2 R, L, . The quantity, is often called the Q, L, R, of the circuit. A high-Q circuit has low resistance and a low, , approximation by, , 2, per cycle., Q, A light beam travelling in the x-direction is described by the, , fractional energy loss, 13., , x, . An electron is constrained, c, to move along the y-direction with a speed of 2.0×107 m/s. Find, the maximum electric force and the maximum magnetic force on, the electron., Ans : Fe = 4.8×10–17 N, Fb = 3.2×10–18 N., electric field E y, , 9., , Ans : (a) V/R (b) 0., A box contains L, C and R. When 250 DC is applied to the, terminals of the box, a current of 1.0 A flows in the circuit. When, an AC source of 250 V rms at 2250 rad/s is connected, a current, of 1.25 A rms flows. It is observed that the current rises with, , 14., , 300sin, , t, , W, . Find the amptitude, m2, of electric and magnetic fields in the beam., V, Ans : 4.3 108 , 1.44 T ., m, , A laser beam has intensity 2.5 1014

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AC AND EM-WAVES, , 613, , Solutions Exercise9.1 Level -1, 1., , 2, 3., 4., , (b), , (a), (a), (c), , U e U m = 160, Um = 160 – Ue, = 160 – 10 = 150 µJ., VR does not depend on the frequency of ac., Time period, T1, Clearly t2, , 5., , (a), , (d), , 7., , (c), , 2, , LC, , T3, 2, , 2, , 2, , 2 (50 10 3 ) 4 10, = 28 × 10–4 s, Time taken by capacitor to charge fulley,, , T, 4, , 2, , 2 LC ., , LC, , 4, , 7 10 s ., , 8., , (d), , 9., , (c), , 10., , (a), , To increase transfer of energy, Vs and hence N s should be, , (a), , 12., , (a), , greater than N p ., The average of AC may be zero, when it is calculated between, i0 to – i0 ., We know that, in EM-waves E0 = C B., , k, , , and so E0 k, , B0 ., , 13., , (b), , 14., , (d), , Power, P V I cos / 2 = 0, , 15., , (b), , XC, , 1, C, , 1, 100 10, , Vrms, XC, , Thus irms, , L (2, , XL, , (d), , Z, i, , 17., , (a), , (b), , 20., , (d), , X L2, , R 2 ( L )2, R, Z, , V2, V2, cos, Z, Z, V = 50 – 50 = 0, , R, R, , 2, , ., , L, , V 2R, , R, , P, , 2 2, , Z2, , ( R2, , ., , 2 2, , L), , 1, , so with increase in frequency of ac, X C, C, decreases and so current in the bulb wil increase., As X C, , 21., 22., 23., , (a), (d), (d), , 24., , (c), , Current in resistor and p.d. across it are in same phase., Also p.d. across resistor is /2 ahead of p.d. across capacitor., The time period of LC oscillations, T, , 2, , LC ., , X L2, , V, Z, , 220, 65.91, , R2, , Z, , L, , V0, Z, , 20, 10, , 200, , 25., , (b), , q2, 1 Q2, =, 2C, 2 2C, , 26., , (b), , 27., , (c), , 2, i 2 = (i1 cos t i2 sin t ), , or, , 1, i2 2, 2, , 65.91, , 3.33 A., , irms =, , i2, , R, Z, , cos, , 2000 50 10, , i22 ) / 2, , R, , 29., , (b), , V, , VR2 VL2, , 30., , (a), , V, , VR2 (VL VC ) 2, , 31., , (b), , At resonance LCR, Z = R,, , 32., , (a), , 2, , =, , We have, , 2, 6, , (i12, , . For low value of cos , R should, 2 2, R, L, be small and L should be large., , and so cos, , 1, , 0, , (a), , 2, , 3, , 1, 2, , 28., , XC =, =, , 2A, , Q, ., 2, , q=, , 2, i 2 = i1, , 20 2 62.82, , 1, C, , LC ., 2, If q is the required charge, then, , = i12 cos 2 t i22 sin 2 t 2i1i2 cos t sin t, , 20 10 3 A ., , 10 4, , T, ., 4, , So t, , 10 4, , 6, , 2000 5 10, , = 10, , The time at which charge on the capacitor will be zero is, , 50) 0.2 62.8, , R2, , 10 2, , i, , R2, , Z, , As X C, , Also, C, , 16., , 19., , 6, , 1, , so with the increase in frequency, X C, C, decreases and so iC increases., L , so with the increase in frequency, XL increases, As X L, and so current iL will decrease., For resonance, X C X L ., , 11., , (b), , Power factor, cos, , t1 t3 ., , Time period, T, , t, 6., , LC , T2, , 18., , 1, C, , 1, X CC, , 2002 1502, , R, Z, , 1., , 250V ., , 802 (40 –100)2, , 100V

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AC AND EM-WAVES, , 615, , Solutions Exercise 9.2, 1., , (a, d) According to conservation of charge, the displacement, current must be leftward., , id, , i, , 3., , Q2, , so with the increase in Q, current will increase, 2C, and U also increases., (c, d), 4. (a, b, d), 5. (a, b), , 6., , (b, d) For f, , 8., , (d), , 9., 10., , (a), (d), , 11., , (d), , 2., , (a, b) As U, , 1, 1, 2 f, ,, 0., 2 LC, LC, This is the condition of resonance of LCR circuit., At resonance, Z = R, and so V iZ iR ., , Solutions Exercise 9.3, 1., , (d), , irms =, , 2., , (a), , As X C, , 3., 4., , (a), (d), , 5., , (b), , 6., , (a), , 7., , (a), , i12 i22, 2, , 32, , 42, , 5, A, 2, , 2, , 1, , so for, C, , = 0, X C, , ., , Hot wire ammeter is based on heating effect; dc produces, heat., When soft iron rod is inserted in solenoid, its electrical inertia increases, but ultimate intensity of bulb remains same., B.d, , 0iin ; iin, , is same for both the paths and so, , The currents in capacitor and in inductor are opposite and so, net current is zero., In case of inductive circuit emf leads current by /2 rad, / 2), V = V0 sin t i = i0 cos t = i0 sin( t, , or, , , and cos, 0., 2, Transformer cannot produce power, but it transfer from primary to secondary., , 1, r, As r is greater for second path and so, path., , B.d, , Also, , = µ0iin, 0iin, , B 2 r =, B, , 12., , (a), , Solutions Exercise 9.4, Passage Q.1 to Q.3, , R1, , V, i, , 20, 10, 2, , V 140, i, 2, For first coil;, Z1, , or, , Passage Q.4 to Q.6, , 30, 15, 2, , and R2, , 70, , 100, 2, , and Z 2, , Zp, , R12, , 70 =, , 102 (2, , XC, , 50, , 2 2, L1, , 40)2, , L12, , 1, C, , or, , R22, , 50 =, , 152 (2, , 1., , 50 Hz , Z1, , (a), , R2, , X C2, , 4., , (c), , i0, , 5., , (c), , cos, , V, Z, , 252, V0, Z, , =, , 2 2, L2, , 6, , 402, , 100, 47.17, R, Z, , 40, , 47.17, , ., , 2.13 A, , 25, 47.17, , 0.53, , 10 2, , (2, , 230, 148.52, , 50 0.276)2, , 87, , 2.13, 0.53, 2, , 7., , (c), , The displacement current, i1, , 8., , (c), , We know that, id, , dE, dt, , 1.55 A, R, Z, , 100, 2, , = 56.7 W, 6., (b), Passage Q.7 to Q.9, , 40) 2 L22, , Z2, 152 (2, 50 0.19)2 61.5, Now Z = Z1 + Z2 = 148.52 , R = R1 + R2 = 25, , i, , 314 80 10, , P = Vrmsirms cos, , L2 = 0.19 H, At f, , 1, , Z, , L1 = 0.2764 H, For second coil ;, Z2, , 25, 148.52, , 2., , (a), , Power factor, cos, , 3., , (b), , Power dissipated, P = Vrmsirms cos, = 230 × 1.55 × 0.17 = 60 W, , B.d, , is same for both the paths., , =, , 0, , A, , id, 0, , 9., , (b), , B.d, , =, , 2A ., , 12, , 12, , dE, dt, 2, , A, , 8.86 10, , ; 2.3 103 V/m., , 0.17, , i, , 0iin, , = 4 × 10–7 × (2 × 0.52), = 0.63 × 10–6 T-m, , is smaller for second

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ELECTRICITY & MAGNETISM, , 616, 10., , A-r : At t = 0, inductor offers infinite resistance, and so circuit is, completed through I1 and I3 ., , 11., , E, ., 2R, B-p, t : At t =, , inductor will offer zero resistance, and so, E, I 2 I3 0 , I1 I 4, R, , Thus i1, , C-r : i3, , i3, , A - p : irms, B-q ;, C-p :, , Vrms, , 40, 10 A ; i0, 4, , 2irms, , iZ 10 5 50V ; V0, , 2 2 A, 2 Vrms, , 50 2 V, , D-s : Now V 2 VR2 (VL VC ) 2, or 502, , E, ., 2R, , 402 (40 – VC ) 2, , VC 10V ,, , E, R, , D - q, s : i4, , VR, R, , VC 10, 1, i, 10, See theory of the chapter., and X C, , 12., , Solutions Exercise-9.5, 1., , Given, C 25 10 6 F , L 0.10H , R 25, , and e 310sin 314t, , We know that,, , Z =, , R2, , or, , 80 =, , 50 2, , On comparing with , e e0 sin t , we have, = 314 ,, (i), , f =, , (ii), (iii), , erms =, XL, , 2, , The back emf,, , 219 V, , 4., , 1, C, , 1, , (iv) Impedance, Z =, Current,, , (vi), , i =, , X L : XC, , VR = iR, , 2.2 A, , 2.2 25 55, , 6., , X C , so resonance will occur and VR 100V ., , If first case, tan 60 =, 3 =, , In second case, tan 60, or, , 2 103 ) 2 100 10, , Vrms, Z =, i, , 160, 2, , 3, , Ans, , 3 =, , 300 L, 100, , Z =, , 80, , =, , 2, Vrms, R, P=, 2, , 802, , 1, 300C 100, , 1, C, , 2, , R2, , L, , 1002, , 300 0.58, , 2, , 1, 300 19.2 10, , 6, , = 100, , Z, , 1602 R, , 1, CR, , XC, R, , C = 19.2 µF, The impedance of the circuit is, , We know that,Power,, , R = 50, , L, R, , =, , (2 f )2 L, , L, , = 63 10 9 F, , 200 =, , XL, R, , L = 0.58 H, , 1, , or, , 25.4, , 500) (12.5 10 6 ), , Ans., , 1, , = (2, , 6, , Ans., , 1, (2, , or, , 1, , The impedance,, , 1, C, , L, , 2, , 1, 0.01 1 10, , 500) 8.1 25.4, , Ans, , XC = X L, , 3., , L (2, , As X L, , For the maximum current,, , C =, , XL, , and X C, , VL = iX L = 2.2 × 31.40 = 69, , 1, =, C, , 1, LC, , = 1592 Hz, 5., , VC = iX C = 2.2 × 127.38 = 280, , or, , Ans., , 1, , = 2, , 99.2, , 219, 99.2, , erms, Z, , 1, 2, , 127.38, , 96, , 252 962, , 2 62.5, , = 125 V., Resonance frequency,, fr =, , 314 (25 10 6 ), , The reactance, X, , 2., , VL = iX L, , L 314 0.10 31.40, , and X C, , (v), , 50 Hz, , 310, 2, , e0, , X L2, , X L = 62.5, , 314, 2, , 2, , X L2, , Current,, Power dissipated, , i =, , V, 200, =, Z, 100, , 2A ., , = Vrmsirms cos, = 200 × 2 × 1 = 400 W, , Ans.

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AC AND EM-WAVES, , 617, , Solutions Exercise-9.6, 1., , Given,, , (b), , t, , i = i0, , Energy stored,, U =, , 2, i2 = i ., , t2, 2, , 5., , P =, , i dt, , i2 =, , 0, , t 2 dt, , 3, , where, , 0, , =, , and, 2., , i02 t 3, 3, 3, i, , irms =, , Given,, , P =, , Ans., , 3, , 6., , We know that,, , = 5 4sin(2 60)t, = 5 4sin 377t, The average value of current, , or, , T, 2, i 2 = (5 4sin t ), , i2, , i2, , irms =, , 3., , As, , 6, , The reactance,, , (d), , 5.75 A, , 4., , (a), , F, , 1, , vmax = imax =, , 7., , =, , 1, LC, , 6, , Q, , 175 10, , LC, , 1.25 2.69 10, , 10–3, , = 3.02 ×, The resonance frequency,, , X L2, , 4, 50, , Thus, , 0.08H, , m/s, , 3, , Ans., 1, , 10 10, , = 0.90, XL =, , 1, 50 2500 10, , 8, , 6, , 3, , 1 10, , 6, , R, , 2, , (X L, , XC ), , 2, , 2, , 3, , Vrms R, , 12 2 3, , Z2, , 52, , 1, C, , (4 8), , 2, , and, , The impedance,, 17.28 W, , Ans., , 200) 12 10, , XC =, , Z =, , L = 9 103 (10 10 3 ), , 1, C, , 1, 9 103 10, , 6, , i0 =, , VC = irms X C = 2 × 66.35, , Ans., , R2, , ( X L ~ X C )2, , = 32 (111.11 90)2, = 21.32, The current amplitude,, , 6, , = 66.35, = 133 V, , = 9 103 rad / s, , = 111.11, , 1, (2, , r, , = 90, , The capacitive reactance, XC =, , 3, , The maximum speed vmax corresponds to the maximum current. Thus, , = 5, =, , (175 10 6 ) 2, 2C, , xmax = 175 × 10–6 m., , The total impendance,, , Power, , Q2, 2C, , = 104 rad/s, , 1, C, , Z =, , Ans., , (c), , 5, , XL, , XC =, , 0.625 W, , 262, , 25 16sin 2 t 40 sin 2 t, , XL = 4, L =, , =, , 6.52 10, , 1, , r, , and, , 0.4)2, , Force constantk= C, 2.69 10 3, = 372 N/m., The maximum displacement is corresponding to the maximum charge, and so, , 0 33, 33, , R2, , Z =, , 30, , (2, , (b), , V, 12, R = i = 4 = 3, 12, 2.4, , 10 2, , T, , The resistance of the coil,, , Impedance, Z =, , ( L)2, , The mass m is corresponding to the inductance so,, m = 1.25 kg, , = 5 A., , and, , 5.70 10, , 0, , 16, = 25, 2, , ,, , Z2, R2, , Ans., , (a), (5 4 sin 377t )dt, , 0, , 2, Vrms, R, , C = 2.69 10, , T, , idt, , iav =, , For rms value,, , U =, , i = 5 4sin t, , T, , (133)2, , = 26, , i0, , 2, , Z =, , i02, 3, , 0, , 6, , = 0.106 J, Average power consumed is given by, , 2, , i0 2, , 1, 1, CVC 2 =, 12 10, 2, 2, , Average power, Pav =, , V0, 15, =, Z, 21.32, V02 R, 2Z, , 2, , = 0.74 W, , 0.704 A, , 152 3, 2 21.322, , Ans.

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ELECTRICITY & MAGNETISM, , 618, 8., , In series circuit at resonance, , 2, = 15, , 9., , V, R, , i, , impedance Z = R,, , 250, V, =, 1, i1, , 2, , irms, , (b), , 250, , Vbc, , ( L, , XC, , V, Z, , L, , i12, , i22, , i2 =, , i2, , i12, , = 1.252 – 12, From equation (i), we have, , 1, C, , L, , =, , … (i), , 1, C, , =, , 0, , i2, , (c), , 4500 =, , After substituting, and (iii), we get, , (a), , ; 100 W, The power dissipated by inductor and capacitor will be zero., , 0.75 A, , 12., , If U0 be the initial energy stored in the capacitor, then energy at, any time t, Rt / L, , U (t ) = U 0e, , and U (t T ) = U 0e R(t, The fractional loss in energy is, , … (ii), , … (iii), , LC, , = (1 e, , 2250 rad/s and simplifying equations (ii), , Also, , Ans., , e, , The frequency of oscillations, , RT / L, , ; 1, , U, = 1, U, , 1, LC, , 1, (54 10 3 ) (6.20 10 6 ), , = 275 Hz, The maximum charge stored on the capacitor, Q = CV = 6.20 × 10–6 × 34, = 210.8 × 10–6 C, , 13., Ans., , 11., , (a), , i0 =, , Q2, =, LC, , = 0.364A, Impedance of the circuit, 2, Z = R, , L, , 1, C, , E0, , 300 V/m ,, , RT, L, , RT, L, , 1, , 2 R, L, , B0 = E0, C, , Ans., 300, , 1 10, , 3 108, , = 4.8 ×, The maximum magnetic force,, = 3.2 ×, The intensity is given by, , (210.8 10 6 )2, 6.20 10, , 1, I = 2, , 6, , Ans., , Rt / L, , RT, L, , 10–7, , 14., 3, , R (t T ) / L, , ), , 6, , N/A-m, , or, , 0, , 19, , N., , Fb = qvB (1.6 10, , 54 10, , e, , F0 = E0 q 300 1.6 10, , 1 2, Li0 = Q2 / 2C, 2, , e, , The maximum electric force,, , If i0 be the amplitude of current, then, , or, , RT / L, , Rt / L, , e, , RT, Assume that, is small compared to 1, (as resistance is small)., L, , =, , (b), , T)/ L, , U (t ) U (t T ), U, =, U (t ), U, , LC, , 1, = 2, , 15, 28.5, , = 75 2.6, , i1, , 1000, 3, , 1, f =, 2, , Ans., , 75 V, , P = Vrmsirms cos, , i, , 4, L = 81 H , C = 1µF, , 10., , 224 –159 65 V, , Power dissipated by resistor,, , 1, , or, , 6, , 550 25 10 3 ], , 2.6 [2, , 2, 2, Vab, Vbd, , Vad =, , 1, , Also,, , 550 4.70 10, , = 224 V, , 1, C, , L, , 2, , Vbd = Vcd Vbc, , 250, , 0.75 =, , Vab, , 1, , 2.6, , Vcd = irms X L, , 250, , i =, , The net current,, , Vad, , = 159 V, 1, ), C, , Current in LC- branch,, i2 =, , = irms X C, , 6, , Vbd, , Vab = irms R 2.6 15 38.8V, , The impedance of LC-branch,, Z = XL, , 550 4.70 10, , 75, ; 2.6A, 28.5, , Vrms, Z, , 2, , 1, , 3, , 550) 25 10, , = 28.5, , In parallel circuit at resonance, impedance, Z =, i 0,, When DC is applied, the resistor is effective, R =, , (2, , 10–18, , 19, , ) (2 107 ) (1 10 6 ), , N., , Ans., , E 2C, , 1, 2.5 1014 = 2 (8.86 10, , 12, , ) E02 (3 108 ), , E0 = 4.3 108 V/m, , 2, , and, , B0 =, , E0, C, , 1.44 T, , Ans.