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tf MOVING CHARGES AND MAGNETISM, s, , s, , Oersted’s experiments, Hans Christian Oersted noticed that a current carrying wire caused a noticeable deflection in a nearby magnetic, compass needle. On conducting a few experiments, he found that- The compass needle aligns tangentially to an imaginary circle at the center of which is the current carrying wire, and whose plane is perpendicular to the wire., - Reversing the direction of current, reverses the direction of orientation of the compass needle., - The deflection increases on bring the wire closer to the needle or increasing the current., - Iron filings sprinkled around the wire, arrange themselves in concentric circles with the wire as the center., Conclusion- Moving charges or currents produce a magnetic field in the surrounding space., Note: Important conventionCurrent emerging out of the plane is depicted by a dot, , | going into the plane is depicted by a cross, , ra, , Magnetic field (B), , Like static charges produce electrostatic field, moving charges produce both electrostatic and magnetic field., It is a vector quantity| SI Unit- Tesla (T) or Webber meter-2 (Wb m-2)|cgs unit- Gauss(G) 1G = 10-4 T, , an, g, , Principle of superposition of magnetic field, Total magnetic field at point due to multiple sources is the vector sum of all magnetic field of each individual source., , Magnetic force (PYQ 2020, 2019, 2018, 2017, 2016, 2014, 2013, 2011), , -, , It depends on q, v, and B. Force on a negative charge is opposite to that on a positive charge, The magnetic force q (v × B) includes a vector product of velocity and magnetic field. So, force will always be, perpendicular to v and B. Direction of force is given by cross product rule or Flemings’ left-hand rule, Since Fb is perpendicular to v, the work done by magnetic force is always zero., i.e. in a magnetic field, speed of particle remains constant, only the direction of velocity can change., Magnetic force becomes zero when v = 0 i.e. only moving charges can experience magnetic force., Magnetic force depends on the frame of reference., Magnetic force becomes zero when v is parallel to B i.e. when particle is moving along or opposite to the field, it, experiences no magnetic force., , ni, , -, , lJ, , 1. Magnetic Force on a moving, charged particle, The magnetic force Fb on charge q, moving with velocity v in a magnetic field B is given as-, , A, , :, , ••, , Case 1: v is perpendicular to B, If a charged particle moving with velocity v enters a, Magnetic field B perpendicularly, the force acts such, That the particle moves in a circular path (see fig) with, The same speed v., Let radius of circle be r. The magnetic force that the, Particle experiences provides the centripetal force, Required to move in the circle. Therefore, we can write-, , ', , T, , q, ••, , I, , Be, , →, , B, , ', , FI, , •, , q, , B, , E, , Anil Jangra, , (Where p is momentum, KE is kinetic energy and V is potential difference)

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Time period of the particle is independent of speed and radius of the particle i.e. if a charged particle of mass m, and charge q enters a magnetic field with different velocities, in each case its time period remains same., , Note: Use the following table to quickly solve the questions asking you to compare Time period and radius of various, charged particles in magnetic fields!, Proton, Charge +e, Mass, m, Important PYQs, , Electron, -e, m/1840, , Alpha particle, +2e, 4m, , Deuteron, +e, 2m, , ¥¥¥÷i7, ., , ¥ Ques: An alpha particle is accelerated through a potential difference of 10kV and moves along the x-axis. It enters in, ., , a region of uniform magnetic field B= 2 × 10-3 T acting along the Y-axis. Find the radius of its path. (mass of alpha, particle = 6.4 × 10-27 kg) (PYQ 2020) [3M], Ans:, , Ques: An alpha particle and a proton of the same kinetic energy are allowed are in turn allowed to pass through a, , ¥ magnetic field B, acting normal to the direction of the motion of the particles. Calculate the ratio of radii of the, ., , circular paths described by them. (PYQ 2019) [1M], Ans:, P, , I, , ¥ Ques: a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the, ., , presence of magnetic field B, b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going, into the plane of the paper as shown in the fig. Trace their trajectories and justify your answer. (PYQ 2016), Ans: a) F = q (v × B), b) The neutron will not be deflected as it is neutral in nature. The electron and the alpha particle will go around, circular trajectories (since they enter perpendicular to the field). However, they will be deflected in opposite, directions as they are oppositely charged, however they both wont be able to complete their circular, trajectories as they will exit the field., , Anil Jangra

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-, , Ques: A proton and deuteron having equal momenta enter a region of uniform magnetic field at right angle to the, direction of the field. Depict their trajectories in the field. (PYQ 2013) [2M], Ans: They will both move in circular trajectories., , p, , D, , l, , Case 2: v is inclined to B at an angle θ, If a charge q enters a magnetic field at an angle other than 90°, It will not move in a perfect circular path. Instead it will move in, A helical path. This can be explained by resolving the velocity into, Components. The component parallel to field will experience no, Force and hence continue moving in the same straight path. The, Component perpendicular to the field will experience a force due, To which the charge will move in a circle. The net result of these two, Will result in a helical trajectory of the charge., , USino, , →, , →, , B, , V, , )O, , an, g, , ra, , V cos O, , lJ, , To calculate the radius, we will take the perpendicular component only hence-, , A, , Important PYQ, ., , ¥E÷Ff, , ni, , Pitch- it is the forward distance moved in one complete revolution. Only the parallel component will contribute to, the pitch hence-, , f :/, ", , ", , ", ., , ", , ÷, ., , Pitch, , Ques: Two protons of equal kinetic energies enter a region of uniform magnetic field. The first proton enters normal, to the field direction while the other enters at angle of 30° to the field direction. Name the trajectories followed by, them. (PYQ 2018) [1M], Ans: The first will go around in a circular trajectory and the other will go in a helical path., , Motion in combined electromagnetic field/ Lorentz force (PYQ 2014, 2011), Consider a charge q moving with velocity v in a region with both electric E and magnetic field B. the charge will, experience both electrostatic and magnetic force. This was first given by H.A. Lorentz therefore it is called Lorentz, force., , Anil Jangra

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Important PYQs, , i¥÷7, ., , Ques: Write the expression, in a vector form, for the Lorentz magnetic force F due to a charge moving with velocity v, in a magnetic field B. What is the direction of force? (PYQ 2014), Ans: F = q (v × B). The magnetic force acts in a direction perpendicular to both v and B., -, , Ques: Write the expression, in a vector form, for the Lorentz magnetic force F due to a charge moving with velocity v, in a magnetic field B. Show that no work is done by this force on the charged particle (PYQ 2011), Ans: F = q (v × B). Since the force acts perpendicular to the direction of motion of the charged particle, no work is, done by this force., W = F.s = F.s cos90°= 0, , S, , Velocity selector (PYQ 2017), Consider the following situation where the electric and magnetic fields are perpendicular to each other and to the, velocity of the particle. From the figure we can see that→, , E, , ', , F, , E, , (If net F =0), ••••, , q, Tf, , •••, , v→, , Fri, , So, we can balance the value of E and B such that the two forces become equal and since they act in opposite, directions, they will cancel out and the charged particle moves undeflected., This condition can be used to select particles of a particular velocity out of a beam containing charges moving with, different speeds. Therefore, the crossed (perpendicular) E and B fields act as a velocity selector., Use: 1) used by JJ Thomson to calculate e/m ratio of electron, 2) used in a Mass spectrometer- device that separates charges, usually ions according to their charged, aaa to mass ratio, , s, , Cyclotron, , It is a device used to accelerate charged particles or ions to high energies, Construction, It consists of two semicircular disc-like metal containers D1 and D2 separated by a small distance which are called, dees as they look like the letter D. It also contains an oscillator which produces an oscillating electric field. The, cyclotron makes use of crossed electric and magnetic field to increases the energy of charged particles, , Anil Jangra

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ra, , Principle, A cyclotron works on the principle that the time taken by an ion to complete one revolution in its orbit is, independent of the speed or radius of its orbit., Working, Inside the dees, the particle is shielded as no electric field acts on it. The magnetic field acts on the charged particle, and makes it go in a circular trajectory. Every time the charged particle moves out of one of the dees, it is acted upon, by the electric field. The direction of electric field changes in tune with the circular motion of the charge such that it, is always accelerated by the electric field. Each time the charge is accelerated, its energy increases and hence its, radius increases as well. So, it moves in a spiral trajectory. The increase in kinetic energy of the particle is qV (V is, potential difference across the dees at that time) every time it crosses one of the dees. These ions are repeatedly, accelerated till they have enough energy to have a radius approximately that of the dees. They are then directed by, a magnetic field and leave the system through an exit slit., Calculations, When positively charged particles like protons are released in the center of the dees, they move in a semicircular, Path in one of the dees and then arrive in the gap between them in a time interval T/2 which is given by-, , lJ, , an, g, , This frequency is known as the cyclotron frequency. The frequency of the applied voltage is adjusted such that the, polarity of the dees is reversed in the same time that the particle takes to complete one half revolution i.e., frequency of applied voltage = cyclotron frequency. This condition is called resonance condition., We can also calculate kinetic energy of the ions-, , ni, , Uses1. Bombard nuclei with energetic particles to study nuclear reactions., 2. To implant ions in solids to modify their properties or synthesize new materials., 3. Used in hospitals to produce radioactive substances used in diagnosis and treatment., , A, , 2. Magnetic force on a current carrying conductor, Consider a rod of length l and uniform cross-sectional are A. Let the number of electrons per unit volume be n. So, total number of electrons will be N=nlA. Let drift velocity of electrons be vd. So, the total magnetic force on the, electrons moving inside the conductor will beF = (nlA)e vd × B, Now, from the previous chapter we know that ne vd is the current density j and neA vd is the current I. SoF = (ne vd) Al × B = jAl × B, F = Il × B, Where l is a vector of magnitude equal to the length of the conductor and direction as that of the current I, Note: Current is not a vector, we transferred the vector sign from j to I(current), S, , Biot- Savart Law/ Magnetic field due to a current element (PYQ 2020,2019,2018,2017,2012), Consider a current carrying conductor carrying current I. Consider an infinitesimal element dl of the conductor. The, magnetic field produced due this element at a point at a distance r from the wire is directly proportional to the, magnitude of current, the length dl and inversely proportional to the square of the distance r. Its direction is, perpendicular to the plane containing dl and r. Mathematically, we can write-, , q••, , Anil Jangra

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Where µ˳/4π is a constant of proportionality (the above expression holds when the medium is vacuum), •, , Permeability of free space (µ˳), Magnetic permeability is the ability of a substance to acquire magnetization in magnetic fields. It is a measure of the, extent to which magnetic field can penetrate matter., µ˳/4π = 10-7 Tm A-1, Relation between ε˳ and µ˳, (Where c is speed of light in vacuum), , Similarities between Coulomb’s Law and Biot- Savart Law, - Both are long range forces since both depend inversely to the square of the distance., - The principle of superposition applies to both the fields, - Both magnetic and electrostatic fields are linear in their sources (I dl and electric charge resp), DifferencesMagnetic force, Produced by a vector source I dl, Field is perpendicular to the plane containing the, displacement vector r and the current source dl, There is an angle dependence, s, , Coulombic force, Produced by a scalar source, Field is along displacement vector joining the source, and the point of interest, There is no angle dependence, , Magnetic field on the axis of a circular current loop (PYQ 2020, 2018, 2017,2012), Consider a current carrying loop of radius R and current I placed in y-z plane. We want to calculate the field at a, point P on the x axis at a distance x from the center of the loop. Consider a differential element dl of the loop. Using, Biot-savart law-, , For N turns of coil, O, , ( due to symmetry), , a, , J, , Field at center, , Anil Jangra

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n, , n, , ', , ", H, *, , road, , -, , O, , •, , t÷÷3, , an, g, , Important PYQs, , ra, , Direction of magnetic field due to current carrying loop (Right hand thumb rule), The direction of magnetic field is given by the right-hand thumb rule ‘curl the palm of your right hand around the, circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the, magnetic field.’, , Ques: Two identical circular coils, P and Q each of radius R carry current 1A and √3A resp, are placed concentrically, and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic, field at the center of the coils. (PYQ 2017) [2M], , ⇐, , Ans:, , B, , WEA, , YO, , BQ, , GEA IA, , A, , ni, , lJ, , Bp, , -, , Ques: Two circular coils P and Q of radius R carrying current I, are placed perpendicularly such that they have a, common center as shown in the fig. find the magnitude and direction of the net magnetic field at the center of the, two coils (PYQ 2012) [2M], Ans:, Bp, , Fu, , Bo, , ,, , Anil Jangra

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S, , Field due to straight current carrying wire (using integration), Consider a straight current carrying wire carrying current I. We want to calculate the magnetic field due to the wire, at a point P a distance r from the wire. Consider a differential element dlg of the wire at a distance x from P, , \, , k, , dyt, y, , Y, , I, H E, o, , o ,, , r, , g, , ., , /, , ', , *, , In, , Using Biot-Savart Law, , Case 1: If point is at middle of wire-, , =, , C:L =p ), , From the triangle, , Case 2: at mid point of infinite wire-, , f f Ha, , 3, , ⇐ a =p 907, -, , Case3: at one end of finite wireC. F- o ), , Case4: at one end of infinite wire-, , 4114 HE, , (:p, , -, , o, , ,, , L, , -=qoy, , I, , =, , =, ←, , E, -, , =, , I, , Direction of magnetic field due to current carrying wire (Right Hand Rule), Hold the wire in your right hand such that the thumb points in the direction of the current, then the direction in, which the fingers curl gives the direction of magnetic field., , ^, , o•, , (, , I, , Anil Jangra

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Ampere’s Circuital Law (ACL) (PYQ 2015, 2014, 2011, 2010), , S, , The integration, of a magnetic field along a closed loop is equal to µ˳ times the total current passing through, the surface of the loop., , Sign conventionLet the fingers of your right hand be curled in the sense that the boundary is traversed in the loop integral. Then the, direction of the thumb gives the sense in which the current I is regarded as positive., Conditions where Ampere’s circuital law is helpful1. B is tangential to the loop and is a•anon zero constant, 2. B is normal to the loop, 3. B vanishes, , ÷, , Field due to infinite straight current carrying wire (using ACL), Consider an infinitely long straight conducting wire with current I. we want to calculate the magnetic field at a point, P at a distance r from the wire. Consider an Amperian loop of radius r with the wire at its center., , S, , From ACL,, , -, , The field at every point on a circle of radius r is same i.e. the magnetic field due to a long straight current, carrying wire has cylindrical symmetry, The direction of the field at any point on the circle is tangential to it. Thus, the field lines of constant magnitude, form concentric circles, Even though the wire is infinite, the field due to it at finite distance is not infinite. It only blows up when we, come very close to the wire., Direction of field is given by right hand rule, , * Right hand rule and Right hand Thumb rule are NOT the same, Note: - Ampere’s circuital law is to Biot-savart law what Gauss’s law is to Coulomb’s law. ACL and Gauss’ law relate a, physical quantity on the boundary (electric or magnetic field) to another physical quantity., - Ampere’s law holds for steady currents which do not fluctuate with time., , ?÷sE3, , ., , Important PYQ ¥, Ques: a long straight wire of circular cross section (radius a) carrying steady current I. the current I is uniformly, distributed across this cross section. Calculate the magnetic field in the region i) r < a and ii) r > a (PYQ 2010), , Anil Jangra

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Ans: 1) For r>a, , ÷, , 332 Graph, , For r<a, Let, Be the current per unit areaConsider amperian loop 1, , Solenoid (PYQ 2015, 2014, 2011), , an, g, , ra, , (A is area), , A, , ni, , lJ, , It is a piece of equipment which generates magnetic field. It consists of an insulated conducting wire which is wound, (like a helix) around a core. For an ideal solenoid- The windings are tightly placed, - The length of the solenoid is very large as compared to its radius, - Field outside the solenoid is negligibly small, - Field around the middle region of its length is uniform and constant, - Direction of field is given by right hand rule, - Field can be made stronger by inserting a soft iron core inside the solenoid (PYQ 2011), , Magnetic field inside a solenoid (using ACL), Consider the following solenoid with number of turns per unit length as n, current I. Consider a rectangular, Amperian loop ABCD (with AB=l), , •!a, , Be, , Anil Jangra

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°, , °, , agog, , €8, , goes, , [email protected], , ( since BC, DA are To B), (Since CD is outside solenoid and field is negligible), , Using ACL,, enc, , (Since current in each turn is I), , Important PYQs, ", ", , ¥E÷?7, , Ques:, , (PYQ 2014) [3M], , Ans: i) inside, Field due to inner solenoid(Towards le=), , DirecHon- towards le= if n > n and vice versa, ., , t.BA, , ., , ii) field outside the system will be negligible, , Field due to outer solenoid(Towards right), Net field-, , Ques: An observer to the left of a solenoid of N turns and crosse section area A observes that a steady current I flows, in the clockwise direction. Depict the field lines due to the solenoid specifying its polarity. (PYQ 2015), , Go, , OBSERVER, , s, , #, , #, , Toroid (PYQ 2015, 2011), The toroid is hollow circular ring on which a large number of turns of a wire are closely wound. It can be viewed as a, solenoid which has been bent in a circular shape to close on itself., Consider the following toroid of inner radius a and outer radius b with current I and total turns N, , Anil Jangra

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Amperian loops, Toroid, Field, , Consider the following 3 Amperian loops, enc, , ( no current is enclosed), , (Current coming out of the plane is, cancelled by that going into the plane), , an, g, , enc, , ra, , enc, , Force between two parallel currents (PYQ 2018, 2016, 2011), , A, , ni, , lJ, , Consider two long, parallel conductors A and B separated by a distance d. A carries a current IA and B carries a, current IB (in the same direction). The conductor A produces a field Ba at all points along B. the direction of the field, will be downwards (conductors placed horizontally). From cross product rule, we can see that the direction of force, will be towards the left. From Ampere’s circuital law-, , The force (Fba) on segment of length L of wire B can written asTill the real gangsta arrives, , Similarly, the wire A will also experience an equal force (Fab) but in the opposite direction (towards right). Using, Newton’s third law of motion-, , Note: Parallel currents attract, antiparallel currents repel (can be verified by cross product or Fleming’s left-hand, rule), , Anil Jangra

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Force per unit length (fba), let fba be the magnitude of force Fba per unit length L. so we can write-, , The Ampere, The above expression can be used to define the ampere (A). ‘An ampere is that value of steady current which when, flows in each of the two long, straight, parallel conductors of negligible cross section, placed one meter apart in, 24× 10-7N per meter of length.’, vacuum, produces on each of these conductors a force of •, Similarly, we can define the Si unit of chargeWhen a steady current of 1A flows through a conductor, the quantity of charge that flows through it in one second is, 1C, , S, , Note: An instrument called the current balance is used to measure the force acting on the two long, straight, parallel, conductors., Important PYQs ¥Ei?, ¥ Ques: Two long straight parallel conductors A and B carrying current IA and IB in the same direction are placed a, ., , distance d apart. If a third conductor C carrying current Ic in the opposite direction is placed symmetrically in b/w A, and B, calculate the net force on C. (PYQ 2018) [2M], A, , B, , c, , Ans: The conductor C will experience repulsive forces from both the, conductors A and B as shown in the figure., Fc B, , A, , B, , C, , FCA, , d, , dy, , z, , s, , Torque on current carrying loop and magnetic dipole (PYQ 2020, 2019, 2012, 2010), 1. Torque on rectangular current loop in magnetic field, A loop carrying steady current I in a uniform magnetic field experiences a torque. It does not experience a net force, (analogous to behavior of dipole in electric field). This can be shown asa. Field is in the plane of the loop, Consider a rectangular loop ABCD carrying current I, kept in a magnetic field as shownForce on AD and BC will be zero as field along their length (sinθ=0), Force on arm AB will be into the plane and that on CD will be out of, Of the plane. Therefore, net force on the loop will be zero as all forces, Cancel out. But net torque is not zero. Torque can be calculated as-, , (A is area of coil), , b. Field is at an angle with the loop, Let the magnetic field B be at angle θ with the perpendicular to the, Area of the loop. The forces on arms BC and DA are equal and opposite, So, they cancel each other out. Also, they are collinear and act along the, axis of the loop and hence produce no torque. The forces on AB and CD, , Anil Jangra

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are also equal and opposite with magnitude-, , But since they are not collinear, they produce a couple as in the previous case. The torque acting on the loop can, be written as-, , Note: When θ→0, the perpendicular distance b/w the forces also approaches zero, therefore the forces become, collinear and net torque on the loop also becomes zero., , 2. Magnetic dipole/ magnetic moment, A current carrying loop of any shape behaves like a magnetic dipole. Consider the following circular loop with, current I and area A and square loop with side L and current I -, , N, Back view, , Front view, , -, , e, , I, , an, g, , S, , ra, , e, , The face of the loop with clockwise current acts South Pole| anticlockwise current acts as North pole, We define the magnetic moment(m) of the loop as-, , lJ, , We can also write it in vector form with the help of cross product as-, , ni, , Where direction of area vector A is given by the right-hand thumb rule| SI unit- Am2, Direction of moment- The direction of magnetic moment of a dipole is from clockwise face (S) to, anticlockwise face (N) (which will always be the same as the direction of the area vector), , A, , The torque on a current carrying loop in magnetic field can also be expressed in terms of its magnetic moment, as-, , Note: 1. This is analogous to the torque experienced by an electric dipole kept in an electric field2. m parallel to B – stable equilibrium| m antiparallel to B- unstable equilibrium, 3. If a loop has N turns, then m = N I A, Important PYQs, , tines, , ¥ Ques: A rectangular wire loop of 4cm × 10cm carries a steady current of 2 A. A straight long wire carrying 5A is kept, near the loop as shown. If the loop and wire are coplanar then find,, i), The torque on the loop, ii), Magnitude and direction of the net force acting on the loop (PYQ 2012)[3M], ., , Ans:, , D, , C, , l, , K, , A, , B, , b, , Anil Jangra

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I)Torque:, , 2) Force, , •B/, , Similar PYQ, -, , S, , /, , I, , (Towards le=), , ¥i÷7, :, , Ques: A square loop carrying current I is placed at a distance L next to a, wire carrying current I1 in the same plane. Explain how the loop experiences, no torque but a net force. Write the expression of this force. (PYQ 2010) [3M], , Circular current loop as a magnetic dipole, Consider circular loop of radius R, carrying current I. the magnetic field on its axis at a distance x from the center-, , #, , Let area of the loop be A. Then, A= πr2. Substituting in 1,, , Also, m = IA. Therefore,, , This expression is very similar to the expression for electric field due to an electric field at axial position. It can be, seen if we substitute –, µ˳→1/ε˳ |, m→p, | B→ E, Similarly, the expression for field due to an electric dipole at appoint on the perpendicular bisector is-, , If we make the above-mentioned substitutions, we obtain the result for B at a point in the field of the loop, at a, distance x from the center (for x>>R)-, , Difference b/w electric and magnetic dipole, , Anil Jangra

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An electric dipole is built up of two elementary units- charges/ electric monopoles but in magnetism, the magnetic, dipole (any current carrying loop) is the most elementary element i.e. magnetic monopoles do not exist., Note: - We seen that a current carrying loop i) produces a magnetic field and behaves like a magnetic dipole at large, distances ii) experiences a torque in external magnetic field like a magnetic needle. This led Ampere to believe that, all magnetism is due to circulating currents, which is partly true., - Electrons and protons also carry an intrinsic magnetic moment, which is not due to any circulating current., S, , Magnetic dipole moment of a revolving electron, Consider the situation shown in the fig. Due to the revolving electron,, A current is setup in the opposite direction which is given byI= e/T, Where T is the time period of the electron., , Substituting value of T, we get-, , egg, , ra, , We know -, , (Where L is angular momentum of electron), , an, g, , (since current flows in opposite direction), , A, , ni, , lJ, , The ratio µ/L is called the gyromagnetic ratio and is a constant. Its value is 8.8×1010 C/kg for an electron., From Bohr’s model of an atom we know that angular momentum of an electron is quantized i.e.L= nh/2π, Substituting this in the previous expression-, , This value is called the Bohr magneton, , -, , s, , Note: - Any charge in uniform circular motion will have a magnetic moment associated with it which is called the, orbital magnetic moment, The electron has another intrinsic magnetic moment, called the spin magnetic moment which has a value of 1, B.M (Bohr magneton), It is not that the electron is spinning. This is because it is an elementary particle and doesn’t have an axis to spin, about, like our earth. But it does have an intrinsic magnetic moment., , Moving coil galvanometer (PYQ 2020, 2018, 2016, 2015, 2011, 2010), , Anil Jangra

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It is a device which is sensitive to small current, ConstructionIt consists of a coil of many turns which is free to rotate about a fixed axis, in a uniform radial magnetic field. It also, consists a soft iron core which makes the field radial and also increases the strength of the magnetic field., Reasons for a radial field- (PYQ 2020,2019), Cylindrical magnets are used to generate radial fields so that the magnetic moment of the coil is always, perpendicular to the field., Principle- (PYQ 2016, 2015, 2011, 2010), A current carrying loop in a uniform magnetic field experiences a torque, WorkingWhen a current flows through the coil, it experiences a torque of magnitude-, , Since field is radial (i.e. perpendicular to the area vector) we take sinθ=1, The galvanometer is fitted with a spring which provides a torque in the opposite direction which balances the, magnetic torque resulting in a steady angular deflection. Let the spring have angle of twist φ and torsional constant, K-, , Note- Torsional constant is restoring torque per unit angular displacement., Current sensitivity of a galvanometer (PYQ 2020,2010), It is defined as the deflection per unit current We can increase the current sensitivity of the galvanometer by increasing the number of turns N of the, coil, Voltage sensitivity of galvanometer (2011), It is defined as the deflection per unit voltage-, , We cannot increase voltage sensitivity by increasing the number of turns because if the number of turns increase,, the resistance also increases by the same amount, hence voltage sensitivity remains the same. For e.g.-, , S, , Galvanometer as an Ammeter (PYQ 2020, 2011, 2010), The galvanometer cannot be used to measure the current in a circuit as it is becausei), It gives full scale deflection for currents of the order of µA, ii), For measuring current, it will have to be connected in series and since it has such a high resistance RG , it, will change the value of current in the circuit, To convert the Galvanometer into an ammeter –, We attach a very small resistance rs (called shunt resistance) in parallel with the galvanometer so that most of, the current passes through the shunt. Since both are in parallel, the resistance of the combination isAG, G, , G, , G, , •, , ⑧, , Anil Jangra

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S, , Galvanometer as a Voltmeter (PYQ 2015), Galvanometer can also be converted to a voltmeter to measure the potential difference across a section of the, circuit for which it must be connected in parallel to the circuit and draw very less current. To achieve this, we attach, a large resistance R in series with the galvanometer•o•, , G, , GG, , Important PYQs, , Ques: An ammeter of resistance 0.80Ω can measure current up to 1.0 A, 1. What must be the value of the shunt resistance to enable the ammeter to measure current up to 5A?, 2. What is the combined resistance of the ammeter and the shunt (PYQ 2013) [2M], The current should split as shown. Since both are in parallel, , Rg, , -, , Mr, , ra, , IA, , bA, , R, , g, , an, g, , JA, , ✓, , 4A, , ni, , lJ, , Rs, , A, , a, , iE÷Ef, , Anil Jangra

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tf Magnetism and Matter, , s, , Some commonly known ideas about magnetism, , s, , i., ii., , iii., iv., v., , The earth behaves like a magnetic with the magnetic field pointing approximately from the geographical south, to north., A bar magnet when suspended freely points in the North-South direction. The tip which points to the, geographical north is called the north pole of the magnet and that which points to the geographical south is, called the south pole of the magnet., Like poles of the magnet repel and unlike poles attract., Magnetic monopoles do not exist. If we cut a magnet in half, we end up with two smaller magnets with both, north and south pole., It is possible to make magnets out of iron and its alloys, , Bar magnet, , s, , A bar magnet has two poles. One pole is designated as the North pole and the other as the south pole. When iron, filings are sprinkled around a bar magnet, they are arranged in a pattern similar to the one seen around a current, carrying solenoid., , Magnetic field lines (PYQ 2019), , s, , -, , Magnetic field lines for a bar magnet or a current carrying solenoid form closed loops, this is because magnetic, monopoles do not exist., Tangent to a field line at a point gives the direction of the magnetic field at that point., The greater the density of field lines in a region, the greater is the magnitude of the magnetic field in that region., Magnetic field lines do not intersect. This is because, at the point of intersection there will be two tangents which, means that the magnetic field at that point will have two directions, which is not possible., , (i) Bar magnet, , (ii) solenoid, , Note: 1. For a bar magnet (or a current carrying solenoid) the direction of magnetic field is from the North pole to, the South pole outside the magnetic and from the south pole to the north pole inside the magnet., 2. Unlike in electrostatics, the magnetic field lines do not indicate the direction of force on a moving charge., , s, , Bar magnet as an equivalent solenoid, Ampere hypothesized that all magnetic phenomenon are due to circulating currents. The similarity between, magnetic field lines produced due to a bar magnet and a current carrying solenoid suggests that a bar magnet may, be thought of as a large number circulating atomic currents in analogy with a solenoid., , Anil Jangra, , Ann, , Titian, , .

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Consider a solenoid of length 2L and radius a carrying current I, having no of turns per unit length as n. Let us, calculate the field at a point P on its axis at a distance r from its center. Consider a differential element of thickness, dx a distance x from the center. It consists of n.dx turns. We know the expression for field due a to a circular element, at a point on its axis-, , l, , Z, , -, , l, , For, , l, l, , ra, , -, , ( m = NIA), :, , an, g, , r, , lJ, , This is also the magnetic field for a bar magnet a point far on its axis (obtained experimentally). Thus, a bar magnet, and a solenoid produce similar magnetic fields. The magnetic moment of a bar magnet is thus equal to that of an, equivalent solenoid which produces the same magnetic field., , 1., 2., 3., 4., , ni, , Magnetic pole strength/ Magnetic charge (qm), , It is called magnetic charge and is analogous to electric charge| SI unit- A m (ampere-meter), It depends on area of cross-section and intensity of magnetization, North pole has a magnetic charge +qm and south pole has -qm, Magnetic moment of a bar magnet of length 2L can be written as –, , A, , S, , 5. Consider a solenoid of length L, current I and n no of turns per unit length. Its magnetic moment can be, written asAlsoEqua+ng both-, , 6. The magnetic field strength due to qm at a distance can be written ass, , Dipole in a uniform magnetic field (PYQ 2013), Consider a magnetic needle of magnetic moment m and moment of inertia I kept in a magnetic field B making an, angle θ with the field. The needle experiences a torque which is given by-, , Ann, , Anil Jangra Titian, ., , .

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Where τ is restoring torque (-ve sign indicates that torque is restoring) which can be written as-, , For small angular displacements (θ→0), sinθ→θ therefore,, , -, , ①, , This represents simple harmonic motion-, , ②, , Comparing both equations, we get-, , Important PYQs, , ¥i÷¥7, , Ques: A small compass needle of magnetic moment m is free to turn about an axis perpendicular to the direction, magnetic field B. The moment of inertia of the needle is I. the needle is slightly displaced from the, equilibrium position and released. Prove that it executes SHM and hence find its Time period (PYQ 2013), [3M], Ans: (Exactly as given above), , :, , Potential energy of dipole in magnetic field, Potential energy of dipole in magnetic field can be calculated similar to that in electrostatic field. the potential, energy (U) is given by-, , Note: -The zero of potential energy is taken when the dipole is perpendicular to the field i.e. θ=90°, -Potential energy is minimum at θ=0° (stable equilibrium) | maximum at θ=180° (unstable equilibrium)., , Electrostatic analogue, The equation for magnetic field due to bar magnet of moment m can be obtained from the equation of field due to, an electric dipole of moment p by making the following replacements, E → B | p → m | 1/4πε˳ → µ˳/4π, , April, , Anil JangraThirst, ., , .

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The field at equatorial position (BE) of a bar magnet at a distance r (r>>L), where L is the length of the magnet-, , The field at axial position (BA) of a bar magnet at a distance r (r>>L), where L is the length of the magnet-, , assay, , ra, , Gauss’ law in magnetism states that the net flux through any closed surface is zero, , -, , ni, , Earth’s Magnetism, , lJ, , an, g, , This follows from the fact that magnetic field lines always form closed loops so for any given Gaussian surface, the, no of field lines entering the surface will be equal to the number of field lines exiting the surface. Gauss’ law for, magnetism is a reflection of the fact that magnetic monopoles do not exist., , The earth’s magnetic field is thought to arise due to electric currents produced by convection and rotation of, molten metallic fluids (nickel and iron) in the outer core of the earth. This is known as the dynamo effect., The magnetic field lines of the earth resemble that of a hypothetical magnetic dipole located at the center of the, earth. The axis of this dipole is at angle of 11.3° to the axis of rotation of the earth., The pole near the geographical north pole is called the north magnetic pole and that near the geographical south, pole is called the south magnetic pole, But the field lines enter the earth from the north magnetic pole and exit from the south magnetic pole. This, convention came around because magnetic north was the direction in which the north pole of a magnet pointed., Thus, in reality the north magnetic pole of earth behaves like the south pole of a bar magnet and vice-versa., , A, , :, , Gauss’ law in Magnetism (PYQ 2019), , Nm, v, , Nsm, Anil JangraAtkins, ., , .

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Let us define some terms1. Geographic meridian- At a given place, it is the plane containing the longitude and the Earth’s axis, 2. Magnetic meridian- At a given place it is a vertical plane containing a freely suspended magnet, 3. Declination (θ)- At a given place, it is the angle between the magnetic meridian and the geographic meridian., Declination is greater near high latitudes and smaller near the equator., For e.g. A declination of 10° west means magnetic north is 10° west of geographic north., Declina7on, , or, , Geographic meridian, , Magne7c, meridian, , Earth’s magnetic field-, , -, , At magnetic south, is vertically upwards, At magnetic north, is vertically downwards, At equator, is parallel to the surface, In southern hemisphere, is inclined above the horizontal, In northern hemisphere, is inclined below the horizontal, , Inclination/ Dip (𝛿) (PYQ 2020, 2013, 2012), , s, , It is the angle that a freely suspended magnet makes with the horizontal. The earth’s magnetic field BE at a point can, be resolved into a horizontal component HE and a vertical component ZE. The angle that BE makes with HE is the angle, of dip 𝛿, At poles,, At equator,, , Convention- Dip in the northern hemisphere is positive | southern hemisphere is negative, Declina7on, , a, HE, , 8, , Dip, , Geographic meridian, , Magne7c meridian, , F-, , BE, , Note: Dip needle- it is a compass pivoted to move in a vertical circle containing the magnetic field of the earth., Important PYQs, ., , i÷÷E?, , Ques: Earth’s magnetic field and the angle of dip at a point is 0.3G and 30°. Calculate the vertical component of, Earth’s magnetic field. (PYQ 2020) [1M], Ans:, , ¥ Ques: A compass needle free to turn in the vertical plane, orients itself vertically at a certain place on the earth., ., , Calculate (i) the horizontal component 0f earth’s magnetic field (ii) the angle of dip (PYQ 2013) [2M], Ans:, , Any, , Anil JangraThirst, ., , .

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Ques: The horizontal component of Earth’s magnetic field at a point is B and the angle of dip is 60°. What is the, vertical component? (PYQ 2012) [1M], Ans:, , S, , Time period of oscillation in earth’s magnetic field, For a compass in the horizontal plane-, , Neutral Points (NCERT back Ques 5.13, 5.14, 5.18), Neutral points are those points where field due toa magnet cancels out the horizontal component of Earth’s, magnetic field at that point, , HE, , s, , v, , lJ, , :B, , an, g, , w, , The field due to the magnet will be opposite to HE along the axial line, There will be two neutral points, , Case 2:, , ni, , -, , E, , w, v, , ra, , Case I:, , HE, , A, , S, , B, , E, , w, , S, , -, , :B, -, , Field due to the magnet will be opposite to HE in the equatorial plane, There will be infinite neutral points, , Any, , Anil JangraThirst, ., , .

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NCERT BACK EXERCISES, , Ans 5.13, , ¥:¥÷?, , We know that for a Bar magnet-, , 5.14 This is similar to case 2 men7oned above, From the previous part we know that-, , ①, , ATQ,, -, , ②, , Equa7ng 1,2, Now, ATQ, -, , ③, , Also, we know-, , ④, , From 3,4, (On the perpendicular bisector), , ( in the direc7on of Earth’s field), , co, , G. M, , M, , -, , M, , N, , •, , BE, =, , f, , ., , 18, , ., , 100, , W, , 2 -5 A, , E, , •, ., , w, •, , B, •, , •, , S, , Anti, , Anil Jangra, , #, , ', , 1, , RAKSHA, #, , c, , ,

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Ans: We know field due to an infinite current elementTherefore, the line of null points will be parallel to the, wire at a distance of 1.5 cm above the wire (using right, hand rule), , ATQ,, , To obtain null point we equate the two-, , S, , 3 Magnetic vectors, , ra, , 1. Magnetization vector (M or I) – It is a characteristic of material. The magnetic moment of various electrons, in a bulk material can add up vectorially and give a non-zero net magnetic moment. The magnetization of a, sample is defined as the net magnetic moment per unit volume| SI unit- A m-1, , 2. Magnetic field intensity (H)- it is a characteristic of field| SI unit- A m-1, , Relation between H, B, M, , The magnetic field in the interior of a solenoid is given by-, , lJ, , If the interior of the solenoid is filled with a material with non-zero magnetization, the field inside the solenoid will, be greater than B˳. the net field B may be expressed as-, , Also,, , ni, , Where Bm is the field contributed by the material core. It is found that this additional field is directly proportional to, M-, , A, , s, , an, g, , 3. Magnetic field/ Magnetic induction vector (B)- it is a characteristic of field| SI Unit –, Tesla, , Here we see that the net magnetic field inside the solenoid is due to factors. One, due to external factors like, current in the solenoid which is represented by H, and two, due to nature of magnetic material represented by M., mathematically, we can write-, , Where χm is a dimensionless quantity called magnetic susceptibility. So, we can write-, , Where, , Magnetic susceptibility (χm) (PYQ 2018), It is a measure of how magnetic materials respond to external magnetic field., For paramagnetic materials χm is small and positive| for diamagnetic materials χm is small and negative, , Amu, , Anil JangraTitian, ., , .

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Relative magnetic permeability (µr), It is a dimensionless quantity given by-, , It is the analog of dielectric constant in electrostatics. The magnetic permeability of a substance thus can be written, as-, , Calculating H due to various magnetic, configurations, , S, , 1. Due to solenoid Consider a solenoid with number of turns per unit length n and carrying current I. We know,, Also,, Equa7ng, , 2. Due to toroid, Consider a toroid of total turns N and current I. we know,, , Also,, Equa7ng, , Note: Even if the solenoid or toroid are filled with a magnetic material the value of B will change but H will remain, the same. Therefore, we can say that H is the property of the geometry and NOT the material., , Magnetizing current (Im), , s, , It is the additional amount of current that needs to be passed through the windings of the solenoid in absence of the, core which would give the value of B as in the presence of the core i.e.-, , NCERT EXAMPLE, ., , o, , ¥÷÷i3, , Ques: A solenoid has a core of material with relative permeability 400. The windings of the solenoid are insulated, from the core and carry a current of 2A. if the number of turns is 1000 per meter, calculate a) H b) M c) B d), magnetizing current. (NCERT e.g. 5.10), Ans:, , a), , c), , I, , b), , d), , Amn, , Anil JangraThirst, ., , .

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Magnetic Properties of materials (PYQ 2019, 2018, 2012), 1. Diamagnetism, -, , The substances which are weakly repelled by magnetic field are called diamagnetic substances, They move from a region of stronger field to a weaker field., The resultant magnetic moment of a diamagnetic atom is zero., E.g. Bismuth, Copper, Lead, Silicon, Nitrogen (STP), Water, NaCl, , Explanation for diamagnetismWhen an external field is applied, the electrons having orbital magnetic moment slow down and those having that in, the opposite direction speed up (this happens in accordance with Lenz’ law). Therefore, the substance develops a, net magnetic moment is the direction opposite to the external field and hence is repelled., Susceptibility of diamagnetic materials- It is small and negative. It is independent of temperature., Xm, , an, g, , ra, , I, , lJ, , Diamagne7c, , Superconductors- They are the most exotic diamagnetic materials. They are metals which when cooled to very, low temperatures exhibit perfect conductivity and perfect diamagnetism. Here, the magnetic field lines are, completely expelled out of the substance. For a superconductor-, , ni, , = -1, , The phenomenon of diamagnetism in super conductors is called Meissner effect. They are used to make, magnetically levitated superfast trains., , A, , S, , 2. Paramagnetism, -, , They are substances which weakly magnetized when kept in an external field. They are weakly attracted towards, the field, They move from a region of weak field to strong field, E.g. Aluminum, Sodium, Calcium, Oxygen (STP), Copper Chloride, , Explanation, The individual atoms/ions/molecules of a paramagnetic material have a permanent magnetic dipole moment. In the, absence of an external field, due to random thermal motion of the constituent atoms, the net magnetic moment of a, paramagnetic material is zero. But, in the presence of an external field B˳ and at low temperatures, the magnetic, moments of constituent atoms align in the direction of field and we get a net magnetic moment in the direction of, external field. The field lines get concentrated inside the material and the field inside the material gets enhanced., , Amn, , Anil JangraTitian, ., , .

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SusceptibilityFor paramagnetic materials, it is small and positive, , Curie’s Law, The susceptibility of a paramagnetic substance is inversely proportional to Absolute temperature i.e., x, , (Where C is called Curie’s constant), , T, , Note: As the field is increased or the temperature is lowered, the magnetization increases until it reaches its, saturation MS at which point the dipoles are perfectly aligned with the field. Beyond this point, curie’s law is no, longer valid., , 3. Ferromagnetism, -, , They are materials which get strongly magnetized when placed in an external magnetic field., They move from region of weak field to region of strong field i.e. they are strongly attracted, , ExplanationThe constituent atoms/ions/molecules of ferromagnetic substances possess a permanent dipole moment and they, align themselves in a common direction over a macroscopic volume called domain (a domain contains about 10 11, atoms). In absence of external field, the orientation of the domains is random and hence there is no net magnetic, moment. In the presence external field, the domains orient themselves in its direction. Hence, the field inside the, ferromagnet becomes stronger and the field lines inside become extremely dense, , External field, , -, , Domain, , Hard ferromagnets- When external field is removed, in some ferromagnetic substances, the magnetization persists., E.g. Alnico, an alloy of iron, aluminum, nickel, cobalt and copper; Lodestone. They are used to make permanent, magnets like compass needle, Soft ferromagnets- when external field is removed, the magnetization is also removed. E.g. Soft Iron, Susceptibility- for ferromagnetic substances, it is very large and positive, , April, , Anil Jangra Titian, ., , .

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Curie- Weiss Law, At temperatures above Curie Temperature (TC), ferromagnets become paramagnetic. The domain structures, disintegrate with increase in temperature. The susceptibility above the curie temperature is described as(This C is NOT curie’s constant), , X, , •, , -, , =, -, , I, , Important PYQs, , T, , ra, , tc, , :3, ta, , Ans: A- diamagnetic (µr<1), B- ferromagnetic (µr>>1), , an, g, , Ques: Two magnetic materials A and B have relative magnetic permeability as 0.96 and 500 resp. Identify A and B, (PYQ 2018) [1M], , lJ, , Ques: The magnetic susceptibility of a material is -2.6 × 10-5. Identify the material (PYQ 2012) [1M], Ans: Since χm < 0, it is diamagnetic, , ni, , Hysteresis curve, , Hysteresis means ‘lagging behind’. Let us study the relation between B and H. Consider a unmagnetized, ferromagnetic substance kept inside a solenoid., - As the current in the solenoid is increased, the value of B also rises and becomes saturated as shown in the curve, oa. At this point all the domains are aligned with external field., - Now, when H (or I) is decreased to zero, we see that B does not come down to 0, represented with curve ab. The, value of B at H=0 is called retentivity or remanence, - Next, the current in solenoid is increased in the opposite direction till the value of B becomes 0, represented by, curve bc. The value of H when B=0 is called coercivity., - Therefore, we conclude that for a given value of H, B is not unique but depends on the previous history of the, sample. This phenomenon is called hysteresis., - The area under B-H curve of hysteresis cycle gives loss of energy per unit volume during a cycle of magnetization, and demagnetization., , A, , s, , t.mu, , Anil JangraThirst, ., , .

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S, , Permanent Magnets and Electromagnets (PYQ 2017, 2013), 1. Permanent magnetsSubstances which at room temperature retain their ferromagnetic property for a long period of time are called, permanent magnets., Preparation- 1. Hammering an iron rod kept in the north-south direction, 2. Stroking a steel rod with one end of a bar magnet in the same sense repeatedly., 3. Placing a ferromagnet in the center of a solenoid and pass a current. The magnetic field of the solenoid, magnetizes the rod, Properties of materials used as permanent magnets1. High retentivity- so that magnet is strong, 2. High coercivity- so magnetization is not removed by stray fields, temp fluctuations or minor mechanical damage., 3. High permeability, e.g.- steel, alnico, Cobalt steel and ticonal (alnico + Titanium), , 3. Electromagnets, They are ferromagnetic materials which are made into magnets by passing a current through it using a coil, Properties of materials used to make electromagnets1. High permeability, 2. Low retentivity, e.g. soft iron, 3. The area under hysteresis curve should be small so that loss of energy in cycle of magnetization and, demagnetization is less e.g. in the case of transformer cores and telephone diaphragm, 4. high resistivity- to lower eddy current losses, Uses- electric bells, loudspeakers, telephone diaphragm, cranes., , Apna, , Anil JangraTinian, ., , .