Page 1 :
https://t.me/NEET_StudyMaterial, Chapter, , 19, , Moving Charges and Magnetism, Solutions, SECTION - A, Objective Type Questions, 1., , Numerically 1 gauss = x tesla, then x is, (1) 10–4, , (2), , 104, , (3), , 108, , (4), , 10–8, , Sol. Answer (1), 1 gauss = 10–4 T, 2., , An electron having a charge e moves with a velocity v in positive x direction. A magnetic field acts on it in, positive y direction. The force on the electron acts in (where outward direction is taken as positive z-axis)., (1) Negative direction of y-axis, , (2), , Positive direction of y-axis, , (3) Positive direction of z-axis, , (4), , Negative direction of z-axis, , Sol. Answer (4), ,  , F  – e v  B , , y, B, , So using right hand thumb rule, , F will be in –z direction., 3., , v, , x, , If a proton enters perpendicularly a magnetic field with velocity v , then time period of revolution is T. If proton, enters with velocity 2v, then time period will be, (1) T, , (2), , 2T, , (3), , 3T, , (4), , 4T, , Sol. Answer (1), , T , 4., , 2m, ⇒ independent of v ., qB, , A proton, a deutron and an -particle accelerated through the same potential difference enter a region of uniform, magnetic field, moving at right angles to it. What is the ratio of their kinetic energy?, (1) 1 : 1 : 2, , (2), , 2:2:1, , (3), , 1:2:1, , (4), , 2:1:1, , Sol. Answer (1), K.E. = qV, , V is same for all, , K.E.  q, So K.E. p : K.E d : K.E  = 1 : 1 : 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 2 :
2, 5., , Moving Charges and Magnetism, , Solution of Assignment, , A particle of mass m carrying charge q is accelerated by a potential difference V. It enters perpendicularly, q, in a region of uniform magnetic field B and executes circular arc of radius R, then, equals, m, (1), , 2V, 2, , B R, , (2), , 2, , V, 2B R, , (3), , VB, 2R, , (4), , mV, BR, , Sol. Answer (1), r, , 2mk, , qB, , 2mqV, qB, , 2V, B, , m, ⇒, q, , ⇒ r , , 6., , q, 2V, , m R 2B 2, , The net charge in a current carrying wire is zero still magnetic field exerts a force on it, because a magnetic, field exerts force on, (1) Stationary charge, , (2), , Moving charge, , (3) A positive charge only, , (4), , A negative charge only, , Sol. Answer (2), ,  , F  q V  B , Moving charge are only electrons and magnetic field exert force on moving charges only., 7., , If a charged particle enters perpendicularly in the uniform magnetic field then, (1) Energy remains constant but momentum changes, (2) Energy and momentum both remains constant, (3) Momentum remains constant but energy changes, (4) Neither energy nor momentum remains constant, , Sol. Answer (1), ,  , Work done by magnetic force will always be zero as F  V, so using work energy theorem, Wall = K, W=0,  k = constant, , P change as direction of velocity changes., 8., , The motion of a charged particle can be used to distinguish between a magnetic field and electric field in a, certain region by firing the charge, (1) Perpendicular to the field, , (2), , Parallel to the field, , (3) From opposite directions, , (4), , With different speeds, , Sol. Answer (3), , , To distinguish between E and B, , B, , E, q, , q, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 3 :
Solution of Assignment, , 9., , Moving Charges and Magnetism, , 3, , The correct expression for Lorentz force is, (1) q [E  (B  v )], , (2), , q [E  (v  B )], , q (v  B ), , (3), , (4), , qE, , Sol. Answer (2), ,   , F  q ⎣⎡E  (v  B )⎦⎤, 10. A charged particle moves in a gravity free space without change in velocity. Which of the following is not, possible in that space?, (1) E = 0, B = 0, , (2), , E  0, B = 0, , E = 0, B  0, , (3), , (4), , E  0, B  0, , Sol. Answer (2), , , , , , If E  0 and B  0 , E will definitely deflect / increase / decrease velocity of charge. If both E and B are, non zero, they may balance each other., 11. A wire is bent in the form of an equilateral triangle of side 100 cm and carries a current of 2 A. It is placed, in a magnetic field of induction 2.0 T directed perpendicular into the plane of paper. The direction and, magnitude of magnetic force acting on each side of the triangle will be, ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, ×, , ×, ×, , ×, ×, , ×, ×, , B, , (1) 2 N, normal to the side towards centre of the triangle, (2) 2 N, normal to the side away from the centre of the triangle, (3) 4 N, normal to the side towards centre of the triangle, (4) 4 N, normal to the side away from the centre of the triangle, Sol. Answer (3), ,  , F  i i  B, , F, ,  2(1)(2), , F, , 4N, , F, , Normal to side towards centre of triangle., 12. The oscillator frequency of a cyclotron is 10 MHz. What should be the operating magnetic field for accelerating, proton?, (1) 0.156 T, , (2), , 0.256 T, , (3), , 0.356 T, , (4), , 0.656 T, , Sol. Answer (4), f , , qB, 2m, , 2m ⎞, ⎛, ⎜⎝∵ T  qB ⎟⎠, , 1.6  10 –19  B, 6,  10  10  2, 1.6  10 –27 , , B = 0.656 T, 13. A negative charge is coming towards the observer. The direction of the magnetic field produced by it will be, (as seen by observer), (1) Clockwise, , (2), , Anti-clockwise, , (3) In the direction of motion of charge, , (4), , In the direction opposite to the motion of charge, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 4 :
4, , Moving Charges and Magnetism, , Solution of Assignment, , Sol. Answer (1), , Observer, –q, , , B will be clockwise as seen by observer., 14. When equal current is passed through two coils, equal magnetic field is produced at their centres. If the ratio, of number of turns in the coils is 8 : 15, then the ratio of their radii will be, (1) 1 : 1, , (2), , 15 : 8, , (3), , 8 : 15, , (4), , 1:2, , Sol. Answer (3), B1  B 2,  0 n1i  0 n 2 i, , 2r1, 2r 2, n1, r, 8, ,  1, n 2 15 r 2, , 15. The radius of a circular current carrying coil is R. At what distance from the centre of the coil on its axis,, the intensity of magnetic field will be, , (1) 2R, , (2), , 1, 2 2, , times that at the centre?, , 3R, 2, , (3), , R, , (4), , R, 2, , Sol. Answer (3), ,  0 iR 2, , 2R 2  x 2 , , 32, , , , 1  0i, 2 2 2R, , 2 2R 3   R 2  x 2 , , 2, , 2, , 2 3, , 32, , R2  R2  x2, , x 2  2R 2 – R 2  R 2 ⇒ x  R, 16. The figure below shows a current loop having two circular arcs joined by two radial lines. The magnetic field, at O is, , i, a, b, , , O, , (1), , 0i , (b  a ), 2 a b, , (2), , 0i , (b  a ), 4 a b, , (3), , Zero, , (4), , 0i , (b  a ), 3 a b, , Sol. Answer (2), ,   ,  0i ⎛  ⎞  0i ⎛  ⎞,  i ⎛ 1 1⎞, B  B1 – B 2 , ⎜⎝ ⎟⎠ –, ⎜⎝ ⎟⎠  0 ⎜ – ⎟, 2a 2, 2b 2, 4 ⎝ a b ⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 5 :
Solution of Assignment, , Moving Charges and Magnetism, , 5, , 17. Which one of the following graphs shows the variation of magnetic induction B with distance r from a long, wire carrying a current?, , B, , B, , B, , B, , (2), , (1), , (3), , (4), , r, , r, , r, , r, , Sol. Answer (3), B, ,  0i, 2r, , B, , 1, r, , B, r, , 18. Two straight long conductors AOB and COD are perpendicular to each other and carry currents i1 and i2. The, magnitude of the magnetic induction at a point P at a distance d from the point O in a direction perpendicular, to the plane ABCD is, C, , A, , (1), , 0, (i1  i 2 ), 2 d, , (2), , D, , 0, (i1  i 2 ), 4 d, , Sol. Answer (3),   ,  B  B1  B 2, , B, , O, , (3), ,  0 2 2 1/ 2, ( i1  i 2 ), 2 d, , (4), ,  0 ⎡ i1 i 2 ⎤, ⎢, ⎥, 2 d ⎣ i 1  i 2 ⎦, , i1, ,   i,  i, B  0 1 iˆ – 0 2 jˆ, 2d, 2d, , i2, , , 0, i 2  i 22, B , 2d 1, 19. A current of i ampere is flowing in an equilateral triangle of side a. The magnetic induction at the centroid, will be, (1), , 0 i, 3 3 a, , (2), , 3 0 i, 2a, , (3), , 5 2 0 i, 3 a, , (4), , Sol. Answer (4), O is centroid and using the OAD distance OD =, , C, , a, 2 3, , O, , By all the three sides AB, BC and CA, direction of magnetic, field produced will be same and inward to the plane of paper, So B total, ,  0i, ⎡, ⎤, 9 0 i,  sin60  sin60 ⎥, ⎢, =, 3, ⎛ a ⎞, ⎢ 4 ⎜, ⎥, 2a, ⎟, ⎢⎣ ⎝ 2 3 ⎠, ⎥⎦, , 9 0 i, 2a, , 60° 60°, A, , a2 D, , B, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 6 :
6, , Moving Charges and Magnetism, , Solution of Assignment, , 20. A square frame of side l carries a current i. The magnetic field at its centre is B. The same current is passed, through a circular coil having the same perimeter as the square. The field at the centre of the circular coil, B, is B. The ratio of, is, B, 8 2, , (1), , , , (2), , 2, , 8 2, , , (3), , 3, , (4), , l, 2, , l, , ⎡  0i, ⎤,  sin 45  sin 45 ⎥, ⎢, B4, ⎛l⎞, ⎢ 4 ⎜ ⎟, ⎥, ⎣ ⎝ 2⎠, ⎦, , 4 2, 2, , C, , D, , Sol. Answer (1), , OD , , 8 2, , , O, 45° 45°, A, , B, ,  0i, l, For circle, B2 2, , 2 r  4 l, , r , , r, , 2l, , , So B ' , ,  0i, , 2r, ,  B' , ,  0i , 4l, ,  So, ,  0i, ⎛ 2l ⎞, 2⎜ ⎟, ⎝ ⎠, , B 8 2,  2, B', , , 21. The magnetic induction at the point O, if the wire carries a current i, is, y, i, R, , i, z, 0i, 2R, , (1), , (2), , 0i, 2 R, , x, , O, i, (3), ,  0 i (  2  4)1/ 2, 4 R, , , Due to straight wires B1 at O, , i, ,  0i,  0i, ,  – jˆ, – jˆ  , 4R, 4R, ,  0i ˆ, j, 2R, Due to semicircle, ,  0 i (  2  4), 4 R, , y, , Sol. Answer (3), , B1 , , (4), , R, , –, , O, , R, , x, , i, , ,  i, B 2  – 0 kˆ, 4R, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 7 :
Solution of Assignment, , Moving Charges and Magnetism, , 7, , 2, 2,   , ⎛  i ⎞, ⎛ i⎞, Net B  B1  B 2  ⎜⎝ 0 ⎟⎠  ⎜⎝ 0 ⎟⎠, 2 R, 4R, , , ,  0i, 2R, , , ,  0i, 2  4, 2R  2 , , 1 1, , 2 4, , 22. The magnetic field intensity at the point O of a loop with current i, whose shape is illustrated below is, , b, i, b, , a, O, , (1), , 0i, 4, , ⎡ 3, 2⎤, ⎢ , ⎥, b ⎥⎦, ⎣⎢ 2a, , (2), , 0i ⎡ 2, ⎤,  b⎥, 2 ⎢a, 4 ⎣, ⎦, , (3), , Sol. Answer (1), , B due to square part :, B due to side OA and OC will be zero at point O, , B due to side AB and BC will be equal so, , 0i, 2, , ⎡ 1 1⎤, ⎢a  b ⎥, ⎣, ⎦, , A, i, D, , ,  0i, ⎡ i, ⎤, B1  2 ⎢ 0  sin 45  0 ⎥ , , ⎣ 4b, ⎦, 2 2b, , O, , (4), , b, 45°, 45°, , 0i, 4, , ⎡ 1 1⎤, ⎢a  b ⎥, ⎣, ⎦, , B, b, , C, , , B due to circular part,   i, B2  0, 2a, , ⎡ ⎛ 3 ⎞ ⎤, ⎢ ⎜⎝ 2 ⎟⎠ ⎥  3 0 i, ⎢, ⎥, 8a, ⎣ 2 ⎦, , , ,   , 1 ⎤  0i, ⎡3, Bnet  B1  B 2   0 i ⎢ , ⎥ , ⎣ 8a 2 2b ⎦ 4, , ⎡ 3, 2⎤, , ⎢, ⎥, b ⎦, ⎣ 2a, , 23. If an electron revolves around a nucleus in a circular orbit of radius R with frequency n, then the magnetic, field produced at the centre of the nucleus will be, (1), , 0 e n, 2R, , (2), ,  0 en, 4 R, , (3), , 4  0 e n, R, , (4), , 4  0 e, Rn, , Sol. Answer (1), Current (i) =, , q, ⎛ 1⎞,  q ⎜ ⎟  qn, ⎝T ⎠, T, ,   i   ne , So B  0  0, 2R, 2R, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 8 :
8, , Moving Charges and Magnetism, , Solution of Assignment, , 24. A thin disc of radius R and mass M has charge q uniformly distributed on it. It rotates with angular velocity, . The ratio of magnetic moment and angular momentum for the disc is, (1), , q, 2M, , R, 2M, , (2), , q2, 2M, , (3), , (4), , 2M, q, , Sol. Answer (1), , M  q, Standard result , 2M, L, , 25. The number of turns per unit length of a long solenoid is 10. If its average radius is 5 cm and it carries a, current of 10 A, then the ratio of flux densities obtained at the centre and at the end on the axis will be, (1) 1 : 2, , (2), , Sol. Answer (2), , B centre   0 ni, , 2:1, , (3), , 1:1, , (4), , 1:4, , n = 10, , ,  ni, B end  0, 2, ,  0 ni, B centre, 2, , , , ⎛  0 ni ⎞ 1, B end, ⎜⎝, ⎟, 2 ⎠, 1, 26. In a toroid the number of turns per unit length is 1000 and current through it is, ampere. The magnetic, 4, 2, field produced inside (in weber/m ) will be, , (1) 10–2, , (2), , 10–3, , 10–4, , (3), , (4), , 10–7, , Sol. Answer (3), n = 1000, , l, , 1, A, 4, , B   0 ni  4  10 –7 1000 , , 1, 4, , B = 10–4 (T), 27. A conducting circular loop of radius r carries a constant current i. It is placed in uniform magnetic field B,, such that B is perpendicular to the plane of the loop. The net magnetic force acting on the loop is, (1) irB, , (2), , 2 riB, , (3), , Zero, , (4), , riB, , Sol. Answer (3), Net force on a closed loop in uniform magnetic field will always be zero., 28. A circular loop of radius R is kept in a uniform magnetic field pointing perpendicular into the plane of paper., When a current I flows in the loop, the tension produced in the loop is, , (1) BIR, , (2), , BIR, 2, , B×, , ×, , ×I ×, , ×, , ×, , ×, , ×, R, × ×, , ×, , ×, , ×, , ×, , ×, , ×, , I, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , (3), , 2BIR, , (4), , Zero, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 9 :
Solution of Assignment, , Moving Charges and Magnetism, , 9, , Sol. Answer (1), , l(dl)B, ,  l  dl  B  2T sin , , T cos, , l  dl  B  2T , , T cos,  , , T T sin, , l  R  2 B   2T , , T sin, , T, , T = BlR, 29. If two straight current carrying wires are kept perpendicular to each other almost touching, then the wires, (1) Attract each other, , (2), , Repel each other, , (3) Remain stationary, , (4), , Become parallel to each other, , Sol. Answer (4), , A, , C, , D, B, , Due to wire AB, wire CD will experience torque due to which it will become parallel to AB., 30. Connecting wires carrying currents in opposite directions are twisted together in using electrical appliances, in order to reduce, (1) Electrical effect, , (2), , Magnetic effect, , (3), , Seebeck effect, , (4), , Peltier effect, , Sol. Answer (2), 31. 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 the wire,, the loop will, , i, , i, (1) Rotate about an axis parallel to the wire, , (2), , Move away from the wire, , (3) Move towards the wire, , (4), , Remain stationary, , Sol. Answer (2), , A, i, , B, F1, , D, , F2, i, , C, , Force on AB and CD will be zero and for AD and BC, F1 > F2,  Loop moves away from wire., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 10 :
10, , Moving Charges and Magnetism, , Solution of Assignment, , 32. A square loop of side l is kept in a uniform magnetic field B such that its plane makes an angle  with B ., The loop carries a current i. The torque experienced by the loop in this position is, (1) B i l2, , (2), , B i l2 sin , , (3), , B i l2 cos, , (4), , Zero, , Sol. Answer (3),  ,   M B, , M  iA  il 2, 2,    il sin  90 –   = B i l2 cos, , 33. Choose the correct statement, (1) It is possible for a current loop to stay without rotating in a uniform magnetic field, (2) If a uniform magnetic field exists in a cubical region and zero outside then it is not possible to project, a charged particle from outside into the field so that it describes a complete circle in the field, (3) A moving charged particle can be accelerated by a magnetic field, (4) All of these, Sol. Answer (4), 34. The effective radius of a circular coil is R and number of turns is N. The current through it is i ampere. The, work done in rotating the coil by angle of 180° in an external magnetic field B will be (initially plane of coil, is perpendicular to magnetic field), (1) NiR2B, , (2), , 2NiR2B, , (3), , (2 N i B ), , (4), , R2, , Zero, , Sol. Answer (2), , W  U  U f – U i  –Ni R 2 cos180 –  –Ni  R 2  cos0, ,  2NiR 2B, 35. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is doubled., Calculate, by what factor does the voltage sensitivity change?, (1) Becomes, , 3, times, 5, , (3) No change, , 2, times, 5, , (2), , Becomes, , (4), , Decreases by a factor of, , 7, 5, , Sol. Answer (1), , S i1 , , NiAB NAB, , ki, k, , Si2 , , 1.2NAB, k, , Sv1 , , NAB S i 1, , kR, R, , Sv 2 , , Si2, 2R, , , , 1.2  NAB , k  2R , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 11 :
Solution of Assignment, , Moving Charges and Magnetism, , 11, , SECTION - B, Objective Type Questions, 1., , A charged particle enters a uniform magnetic field perpendicular to it. The magnetic field, (1) Increases the speed of the particle, (2) Decreases the kinetic energy of the particle, (3) Changes the direction of motion of the particle, (4) Both (1) & (3), , Sol. Answer (3), ,  , Magnetic force F  V,  No work is done by magnetic field so speed and kinetic energy cannot be changed by magnetic field but, it can deflect the particle, 2., , A proton and an alpha particle enter the same magnetic field which is perpendicular to their velocity. If they have, same kinetic energy then ratio of radii of their circular path is, (1) 1 : 1, , (2), , 1:2, , (3), , 2:1, , (4), , 1:4, , Sol. Answer (1), r , , 2mk, qB, , rp, so r , , , 3., , here k is same, so r , mp, m, , , , q, , qp, , m, q, , 1 2,  1: 1, 4 1, , In which of the following situations, the magnetic field can accelerate a charge particle at rest?, I., , When the magnetic field is uniform with respect to time as well as position, , II., , When the magnetic field is time varying but uniform w.r.t. position, , III. When the magnetic field is time independent but position dependent, (1) I, II & III, , (2), , III only, , (3), , II only, , (4), , None of these, , Sol. Answer (3), When magnetic field is time varying, an induced electric field is produced which can accelerate charge., 4., , , A particle of charge –q and mass m enters a uniform magnetic field B at A with speed v1 at an angle  and, leaves the field at C with speed v2 at an angle  as shown. Then, v1, ×, ×, ×, , v2, , ×, , ×, , ×, , A ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , C ×, , ×, , ×, , (1)  = , (3) Particle remains in the field for time t =, , 2m(    ), qB, , (2), , v1 = v2, , (4), , All of these, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 12 :
12, , Moving Charges and Magnetism, , Solution of Assignment, , , , Sol. Answer (4), =, , v 1  v 2 ∵ Fm  v , , , , 2  –  m, T, qB, 5., , A proton moving with a constant velocity, passes through a region of space without change in its velocity. If E, & B represent the electric and magnetic fields respectively, this region may have, (1) E = 0, B  0, , (2), , E  0, B = 0, , (3) E & B both parallel, , (4), , E & B inclined at 45° angle, , Sol. Answer (1), 6., , A particle of charge per unit mass  is released from origin with a velocity v  v 0 iˆ in a uniform magnetic field, , B  B0 kˆ . If the particle passes through (0, y, 0) then y is equal to, 2v 0, (1)  B , 0, , (2), , v0, B0 , , (3), , (4), , , , v0, B0 , , (0, y, 0), , Sol. Answer (3), , q, , m, , 2R, , ⎛ mv ⎞ 2v 0, y  2R  2 ⎜, , ⎝ qB ⎟⎠ B 0, , 7., , 2v 0, B0 , , x, , A proton is accelerating in a cyclotron where the applied magnetic field is 2 T. If the potential gap is effectively, 100 kV then how much revolutions the proton has to make between the “dees” to acquire a kinetic energy of 20, MeV?, (1) 100, , (2), , 150, , (3), , 200, , (4), , 300, , Sol. Answer (1), Energy increased in each revolution = 2  100  103 eV, = 2  105 eV, Now for energy E = 2  107 eV, Number of revolution =, , 8., , 2  10 7 eV,  100, 2  10 5 eV, , 3, A current i ampere flows in a circular arc of wire which subtends an angle, radian at its center, whose radius is, 2, R. The magnetic field B at its center is, , (1), , 0i, R, , (2), , 3 0 i, 2R, , (3), , 3 0 i, 4R, , (4), , 3 0 i, 8R, , Sol. Answer (4), B, ,  0 i ⎛ 3  ⎞ 3 0 i, , 2R ⎜⎝ 2  2  ⎟⎠, 8R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 13 :
Solution of Assignment, , 9., , Moving Charges and Magnetism, , 13, , At what distance on the axis, from the centre of a circular current carrying coil of radius r, the magnetic field, becomes 1/8th of the magnetic field at centre?, (1), , 2r, , (2), , 23 / 2 r, , , ,  r 2  x 2  3 2  8r 3, , , ,  r 2  x 2    8r 3  2 3, , , , , r 2  x 2  4r, 3r2 = x2, , , , xr 3, , 3r, , (3), , (4), , 3 2r, , (4), , ⎛ 0 ⎞ 2I , ⎜, ⎟, ⎝ 4 ⎠ R, , Sol. Answer (3), ,  0 lr 2, 1  0i, , 32, 8 2r, 2r 2  x 2 , , 2, , 10. Magnetic field at P due to given structure is, , I, R, , I, I, R, R, , ⎛ 0 ⎞ I , (1) ⎜, ⎟, ⎝ 4 ⎠ 2R, , (2), , , P, , 0 6I , 4 5R, , (3), , ⎛ 0 ⎞ 5I , ⎜, ⎟, ⎝ 4 ⎠ 6R, , Sol. Answer (3), , ,  I,  0I ,  0I , Bp  0 , –, 4R 4  3R  4  2R , ,  I ⎡ 1, B p  0 ⎢1  –, 4R ⎣ 3, , 1 ⎤ 5  0I , , 2 ⎥⎦ 6 4R, , 11. A straight wire of finite length carrying current I subtends an angle of 60° at point P as shown. The magnetic field, at P is, P, , x, , 60°, , x, , I, (1), ,  0I, , (2), , 2 3 x, ,  0I, 2x, , (3), , 3 0I, 2x, , (4), ,  0I, 3 3 x, , Sol. Answer (1), , B, B, , , ,  0I,  sin30  sin30, 4x cos30,  0I, ⎛, ⎞, 4 x ⎜ 3 ⎟, ⎝ 2 ⎠, , ⎛1, ⎜⎝ , 2, , 1⎞, ⎟, 2⎠, , x, , 30° 30°, ,  0i, 2 3x, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 14 :
14, , Moving Charges and Magnetism, , Solution of Assignment, , 12. Magnetic field at the centre O due to the given structure is, , I, I, , R, , R, , I, , 0 I ⎡ 3, , 4R ⎢⎣ 2, , (1), , 1⎤, ,  ⎥⎦, , (2), , 0 I ⎡, 1⎤, 3 ⎥, ⎢, 2R ⎣, ⎦, , O, , (3), , 0 I ⎡ 3, , 4R ⎢⎣ 2, , 1⎤, ,  ⎥⎦, , (4), , 0 I ⎡, 2⎤, 3 ⎥, ⎢, 4R ⎣, ⎦, , Sol. Answer (3), B = Bdue to circular arc + Bdue to straight wires, , ,  0 i ⎛ 3 ⎞  0 i, , 2R ⎜⎝ 2.  2  ⎟⎠ 4R, ,  B, ,  0i ⎛ 3 1 ⎞, ⎜  ⎟, 4R ⎝ 2  ⎠, , 13. A current i flows in a thin wire in the shape of a regular polygon with n sides. The magnetic induction at the centre, of the polygon when n  is (R is the radius of its Circumcircle), (1), , 0n i, , tan, 2 R, 6, , (2), , 0n i, , tan, 2 R, n, , (3), , 0 i, 2R, , (4), , Zero, , Sol. Answer (3), For regular polygon having n sides where n   will be almost a circle, So B , ,  0i, 2R, , 14. Two long straight wires are placed along x-axis and y-axis. They carry current I1 and I2 respectively. The, equation of locus of zero magnetic induction in the magnetic field produced by them is, , (1) y = x, , (2), , ⎛I ⎞, y  ⎜ 2 ⎟x, ⎝ I1 ⎠, , (3), , ⎛I ⎞, y  ⎜ 1 ⎟x, ⎝ I2 ⎠, , (4), , y = (I1I2)x, , Sol. Answer (3), , x, , (B = 0), A, , l2, , y, l1, , On a general point A magnetic field will be zero when, y, ,  0l 2  0l 1, , 2x 2y, , l1, x, l2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 15 :
Solution of Assignment, , Moving Charges and Magnetism, , 15, , 15. Surface charge density on a ring of radius a and width d is  as shown in the figure. It rotates with frequency, f about its own axis. Assume that the charge is only on outer surface. The magnetic field induction at centre, is(Assume that d << a), d, , (1) 0fd, ,  0f  d, , (2), , (3), , 2  0 f  d, , 2, f d, 20, , (4), , Sol. Answer (1), Surface charge density = , Total charge on the ring (q) =   2a  d, ⇒i , , q,    2a  df, T, ,   l,   2adf , B 0  0,   0 df, 2a, 2a, ,  , , 16. If B1, B2 and B3 are the magnetic field due to I1, I2 and I3, then in Ampere’s circuital law, , , , , , ∫ B ·dl, , ,   0I, B is, , I1, I3, , × I2, ,  , , (1) B  B1 – B2, , (2), ,   , , B  B1  B2  B3, , (3), ,  , , , B  B1 – B2  B3, ,  , B  B3, , (4), , Sol. Answer (2), In ampere circuital law, on amperian loop B is due to all the current elements either inside or outside to the, amperian loop,   , , B  B1  B2  B3, 17. A charge Q moves parallel to a very long straight wire carrying a current I as shown. The force on the charge is, Y, I, , (1) Opposite to OX, , (2), , , +Q P, , O, , Along OX, , (3), , X, , Opposite to OY, , (4), , Along OY, , Sol. Answer (1), ,  , F  q V  B , Using right hand thumb rule F will be opposite to OX., , v, l, , F, , Y, ,  O, , Y, X, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 16 :
16, , Moving Charges and Magnetism, , Solution of Assignment, , 18. A uniform conducting wire ABC has a mass of 10 g. A current of 2 A flows through it. The wire is kept in a, uniform magnetic field B = 2 T. The acceleration of the wire will be, B, , ×, , (1) Zero, , (2), , ×, , ×, , ×, , ×, 4 cm, , ×, , 5×cm, , ×, , ×, , ×, , ×, , ×, , × A ×, , ×, , C, ×, , 12 ms–2, , 1.2 × 10–3 ms–2, , (3), , (4), , 0.6 × 10–3 ms–2, , Sol. Answer (2), Force on wire ABC will be same as force on wire AC,, ,  , using F  i  l  B , ⎛ 3 ⎞,   sin90, F  2⎜, ⎝ 100 ⎟⎠ 2, , ×, , ×, , ×, , ×, 4 cm, , ×, , 5×cm, , ×, , ×, , ×, , ×, , ×, ,  12  10 –2 N, a, , B, , ×, , × A × 3 cm, ×, , F 12  10 –2, ,  12 m/s 2, 10 –2, m, , C, ×, , 19. Figure shows a conducting loop ADCA carrying current i and placed in a region of uniform magnetic field B0., The part ADC forms a semicircle of radius R. The magnitude of force on the semicircle part of the loop is equal, to, , ×, , ×D ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, (1) RiB0, , ×, (2), , A, , Zero, , B0, , C, , ×, (3), , ×, , ×, 2RiB0, , (4), , 2iRB0, , Sol. Answer (4), The force on the semicircle part ADC, will be same as force on wire CA and force on wire CA = i (2R)(B0), (using F = ilB), = 2iRB0, , ⎛ x ⎞, 20. A wire carrying a current i is placed in a magnetic field in the form of the curve y  a sin ⎜ ⎟ 0  x  2L ., ⎝ L ⎠, Force acting on the wire is, , × ×, × ×, × ×, × ×, ×O ×, × ×, (1), , iBL, , , (2), , iBL, , ×, ×, ×, ×, ×, ×, , ×, ×, ×, ×, ×, ×, , ×, ×, ×, ×, ×, ×, , × ×, × ×, × ×, ×2L ×, × ×, × ×, , (3), , 2iBL, , (4), , Zero, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 17 :
Solution of Assignment, , Moving Charges and Magnetism, , 17, , Sol. Answer (3), , , l  2L, Now, using F  ilB,  i  2L  B, , , x⎞ ˆ, ⎛, 21. The magnetic field existing in a region is given by B  B0 ⎜ 1  ⎟ k . A square loop of edge l and carrying a, l ⎠, ⎝, current i, is placed with its edge parallel to the x-y axes. Find the magnitude of the net magnetic force, experienced by the loop, (1), , 1, iB0 l, 2, , (2), , Zero, , (3), , iB0l, , (4), , 2iB0l, , Sol. Answer (3), , , x⎞, ⎛, B  B 0 ⎜ 1  ⎟ kˆ, ⎝, l⎠, at x = 0, B1  B 0 kˆ, , F2, , F1, , at x = l, B 2  2B0 kˆ, Fnet = F2 – F1  il  B 2 – B1 , ,  il  2B0 – B0 ,  ilB 0, , 22. Two protons A and B move parallel to the x-axis in opposite directions with equal speeds v. At the instant shown,, the ratio of magnetic force and electric force acting on the proton A is (c = speed of light in vacuum), , e, , v, y, , A, d, , x, , e, v, (1), , v, c, , (2), , v, , 2, , c, , 2, , B, (3), , vd 2, c, , (4), , 2v, c, , Sol. Answer (2), 23. If the planes of two identical concentric coils are perpendicular and the magnetic moment of each coil is M,, then the resultant magnetic moment of the two coils will be, (1) M, , (2), , 2M, , (3), , 3M, , (4), , 2M, , Sol. Answer (4), M net  M 2  M 2, , M 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 18 :
18, , Moving Charges and Magnetism, , Solution of Assignment, , 24. In a hydrogen atom, an electron of mass m and charge e revolves in an orbit of radius r making n revolutions per, second. If the mass of hydrogen nucleus is M, the magnetic moment associated with the orbital motion of, electron is, (1), , ner 2 m, M m, , (2), , ner2, , (3), , ner 2, m, , (4), , ner 2 m, M, , Sol. Answer (2), Magnetic moment = NiA, , i, , q,  en, T, , A  r 2, and N = 1,  Magnetic moment  m   1  en   r 2 , 25. The phosphor bronze strip is used in a moving coil galvanometer because, (1) It is torsional constant is small, , (2), , It is easily available, , (3) It is paramagnetic, , (4), , It is diamagnetic, , Sol. Answer (1), 26. A uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the, centre due to ABC portion of the wire will be (length of ABC = l1, length of ADC = l2), , B, i, , A, , R O, C, , D, , (1), , 0, il1l 2, 2R (l1  l 2 )2, , (2), , 0, , il 2, 2R (l1  l 2 ), 2, , (3), ,  0 i (l1  l 2 ), 2R l1l 2, , (4), , Zero, , Sol. Answer (1), Let current in part ABC is i1, and in part ADC is i2, i, , il 2, l1  l 2, , (As ABC and ADC part are in parallel connection), , ⎛ 2 ⎞, and subtended by ABC at centre O will be  ⎜, l , ⎝ l 1  l 2 ⎟⎠ 1, , so using B , B, ,  0i ⎛  ⎞, ⎜ ⎟, 2a ⎝ 2 ⎠, ,  0 ⎛ il 2 ⎞ 2  l 1 , 2R ⎜⎝ l 1  l 2 ⎟⎠  l 1  l 2  2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 19 :
Solution of Assignment, , Moving Charges and Magnetism, , 19, , SECTION - C, Previous Years Questions, 1., , A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform, magnetic field of strength 0.2 Weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined, at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will, be, [Re-AIPMT-2015], (1) 0.12 Nm, , (2), , 0.15 Nm, , (3), , 0.20 Nm, , (4), , 0.24 Nm, , Sol. Answer (3), Torque on coil,  = nIABsin60°, ,   50  2  (0.12  0.1)  0.2 , , 3, 2, ,  = 0.20 Nm, 2., , A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the, field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton, is 1 MeV, the energy acquired by the alpha particle will be, [Re-AIPMT-2015], (1) 1 MeV, , (2), , 4 MeV, , (3), , 0.5 MeV, , (4), , 1.5 MeV, , Sol. Answer (1), 2mE, qB, , r=, rp, r, , , , , 3., , , , mp q, p, , Ep, , m q, , E, , Ep, E, Ep, E, , 1, , 1, , A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long, and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is, [AIPMT-2015], , Z, R, , I, Y, , O, I, , I, X, ,  0 I, iˆ  2kˆ, (1) B , 4 R, , , , , , , 0 I, iˆ  2kˆ, (3) B  , 4 R, , , ,   I, B 0, iˆ  2kˆ, 4 R, , 0 I, iˆ  2kˆ, (4) B  , 4 R, , (2), , , , , , , , , , , , Sol. Answer (4), , I, I, 0I ˆ, ⎡ i  2kˆ ⎤, B  0 2kˆ  0  iˆ , ⎦, 4R, 4R, 4R ⎣, , , , , ,  , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 20 :
20, 4., , Moving Charges and Magnetism, , Solution of Assignment, , An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced, at the centre has magnitude, [AIPMT-2015], , 0 ne, 2r, Sol. Answer (1), , (2), , (1), , 0 ne, 2r, , (3), , Zero, , (4), , 0 n 2 e, r, , 0 i, 2r, i=e×n, B=, , B=, 5., , 0 e  n, 2r, , Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other, such that O is their common point for the two. The wires carry I1 and I2 currents, respectively. Point 'P' is lying, at distance d from O along a direction perpendicular to the plane containing the wires. The magnetic field at the, point P will be, [AIPMT-2014], 0 ⎛ I1 ⎞, (1) 2d ⎜ I ⎟, ⎝ 2⎠, , (2), , 0, I1  I2 , 2d, , (3), , 0 2 2, I1  I2, 2d, , , , , , (4), , 0 2 2, I1  I2, 2d, , , , , , 1/2, , Sol. Answer (4), Since the wires are perpendicular, their magnetic fields also are perpendicular. So the resultant field will be, pythagoras of both the fields., 6., , A current loop in a magnetic field, , [NEET-2013], , (1) Can be in equilibrium in one orientation, (2) Can be in equilibrium in two orientations, both the equilibrium states are unstable, (3) Can be in equilibrium in two orientations, one stable while the other is unstable, (4) Experiences a torque whether the field is uniform or non-uniform in all orientations, Sol. Answer (3), 7., , 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 v0 it moves with an initial acceleration 3a0 toward west. The electric and, magnetic fields in the room are:, [NEET-2013], (1), , ma0, 2ma0, west,, down, e, ev 0, , (2), , ma0, 3ma0, east,, up, e, ev 0, , (3), , ma0, 3ma0, east,, down, e, ev 0, , (4), , ma0, 2ma0, west,, up, e, ev 0, , Sol. Answer (1), Case-I, , Case-II, , a0, +e, , ,  eE, west, Now, a 0 , m, ,  ma, 0, E, west, e, , 3a 0, , v0, +e, , Now, FB  m  2a 0   ev 0B, , B, , 2ma 0, ev 0, , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 21 :
Solution of Assignment, , 8., , Moving Charges and Magnetism, , 21, , Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents, flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre will be, [AIPMT (Prelims)-2012], (1), , 0 I, 2R, , 0I, R, , (2), , 50I, 2R, , (3), , 30I, 2R, , (4), , Sol. Answer (3), 9., , An alternating electric field, of frequency , is applied across the dees (radius = R) of a cyclotron that is being, used to accelerate protons (mass = m). The operating magnetic field (B) used in the cyclotron and the kinetic, energy (K) of the proton beam, produced by it, are given by, [AIPMT (Prelims)-2012], (1) B , , 2m, and K = 2m22R2, e, , (2), , B, , m, and K = m2R2, e, , (3) B , , m, and K = 2m22R2, e, , (4), , B, , 2m, and K = m2R2, e, , Sol. Answer (1), , R, , mV, qB, , here q  e ⇒ v , , frequency  , ,  B, , eBR, m, , 1, 1, , T ⎛ 2m ⎞, ⎜⎝, ⎟, eB ⎠, , 2m, e, 2, , k, , 1, 1 ⎛ eBR ⎞, e 2B 2R 2, mv 2  m ⎜, ⎟ , 2, 2 ⎝ m ⎠, 2m, , k, , e 2R 2 ⎡ 2m ⎤,  2m 2  2R 2, 2m ⎢⎣ e ⎥⎦, , 2, , 10. A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What, should be the energy of an -particle to describe a circle of same radius in the same field?, [AIPMT (Mains)-2012], (1) 2 MeV, , (2), , 1 MeV, , (3), , 0.5 MeV, , (4), , 4 MeV, , Sol. Answer (2), 11. A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic, , field acting along AB. If the magnetic force on the arm BC is F , the force on the arm AC is, [AIPMT (Prelims)-2011], , A, , C, , B, (1), , , 2F, , (2), , , – 2F, , (3), , , –F, , (4), , , F, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 22 :
22, , Moving Charges and Magnetism, , Solution of Assignment, , Sol. Answer (3), Net force on the loop = 0, , , Force on wire AB is zero because it is along B field. Hence Foce on AC = –(Force on BC) = –F, A, i, , B, , C, , 12. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an, electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron, [AIPMT (Prelims)-2011], (1) Will turn towards left of direction of motion, , (2), , Will turn towards right of direction of motion, , (3) Speed will decrease, , (4), , Speed will increase, , Sol. Answer (3), 13. Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency, f Hz. The magnitude of magnetic induction at the center of the ring is, [AIPMT (Mains)-2011], (1), , 0 q, 2fR, , (2), , 0 qf, 2R, , (3), , 0 qf, 2R, , (4), , 0 q, 2fR, , Sol. Answer (3), B=, , 0 qf, 2R, , 14. A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying, a steady current I1 at a distance d from the conductor as shown in figure.The loop will experience, , I1, d, , [AIPMT (Mains)-2011], , I, , I, (1) A net torque acting downward normal to the horizontal plane, (2) A net attractive force towards the conductor, (3) A net repulsive force away from the conductor, (4) A net torque acting upward perpendicular to the horizontal plane, Sol. Answer (2), , l1, F1, , F2, Fnet  F1 – F2 = F1  F2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 23 :
Solution of Assignment, , Moving Charges and Magnetism, , 23, , 15. A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the, force on one arm of the loop is F, the net force on the remaining three arms of the loop is:, [AIPMT (Prelims)-2010], , , (1) 3 F, , (2), , , F, , , 3F, , (3), , (4), , , F, , Sol. Answer (2), 16. A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted, such that the beam is not deflected. The specific charge of the cathode rays is given by (Where V is the potential, difference between cathode and anode), [AIPMT (Prelims)-2010], , (1), , B2, 2VE 2, , (2), , 2VB 2, E2, , 2VE 2, , (3), , B2, , (4), , E2, 2VB 2, , Sol. Answer (4), 17. A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the, other in x-z plane. If the current in the loop is i. The resultant magnetic field due to the two semicircular parts, at their common centre is, [AIPMT (Mains)-2010], , (1), , 0 i, 2 2R, , (2), , 0i, 2R, , 0 i, 4R, , (3), , (4), , 0 i, 2R, , Sol. Answer (1), 18. A particle having a mass of 10–2 kg carries a charge of 5 × 10–8C. The particle is given an initial horizontal, , , velocity of 105 ms–1 in the presence of electric field E and magnetic field B . To keep the particle moving in a, horizontal direction, it is necessary that, , , , (a) B should be perpendicular to the direction of velocity and E should be along the direction of velocity., , , , (b) Both B and E should be along the direction of velocity., , , (c) Both B and E are mutually perpendicular and perpendicular to the direction of velocity., , , (d) B should be along the direction of velocity and E should be perpendicular to the direction of velocity., Which one of the following pairs of statements is possible?, (1) (a) and (c), , (2), , (c) and (d), , (3), , [AIPMT (Mains)-2010], (b) and (c), , (4), , (b) and (d), , Sol. Answer (3), ,  , Fm  q V  B , , , Fe  qE, If Fm  Fe , particle will continuously move in horizontal direction, , , If B is in the direction of velocity Fm = 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 24 :
24, , Moving Charges and Magnetism, , Solution of Assignment, , 19. A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic, , field with a constant velocity V  viˆ . The magnetic field is directed along the negative z-axis direction. The, induced emf, during the passage of these loops, out of the field region, will not remain constant for, [AIPMT (Prelims)-2009], (1) The circular and the elliptical loops, , (2), , Only the elliptical loop, , (3) Any of the four loops, , (4), , The rectangular, circular and elliptical loops, , Sol. Answer (1), 20. The magnetic force acting on a charged particle of charge –2 C in a magnetic field of 2 T acting in, y-direction, when the particle velocity (2iˆ  3 ˆj ) × 106 ms–1, is:, (1) 4 N in z-direction, , (2), , 8 N in y-direction, , (3) 8 N in z-direction, , (4), , 8 N in –z direction, , [AIPMT (Prelims)-2009], , Sol. Answer (4), , B  2 Tjˆ, , q  –2 C, , , V   2iˆ  3 jˆ   10 6, ,  , F  –2 C V  B, , F  –2  10, , –6, , ⎡ iˆ ˆj kˆ ⎤, ⎢, ⎥ 6, ˆ, ⎢ 2 3 0 ⎥ 10  –8k, ⎢⎣ 0 2 0 ⎥⎦, , 21. Under the influence of a uniform magnetic field, a charged particle moves with constant speed V in a circle of, radius R. The time period of rotation of the particle:, , [AIPMT (Prelims)-2009], , (1) Depends on R and not on V, , (2), , Is independent of both V and R, , (3) Depends on both V and R, , (4), , Depends on V and not on R, , Sol. Answer (2), , T, , 2m, qB, , , 22. A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of induction B ., After 3 second the kinetic energy of the particle will be, (1) 4T, , (2), , 3T, , [AIPMT (Prelims)-2008], (3), , 2T, , (4), , T, , Sol. Answer (4), , , B does not change kinetic energy., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 25 :
Solution of Assignment, , Moving Charges and Magnetism, , 25, , 23. A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments, 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, [AIPMT (Prelims)-2008], , Q, P, , F3, , F1, S, , R, F2, , (1) F3 – F1 + F2, , (2), , F3 – F1 – F2, , (3), , (F3  F1)2  F22, , (4), , (F3  F1)2  F22, , Sol. Answer (3), Net force on loop will be zero in uniform magnetic field, So, force on QP will balance other forces, , , FQP , ,  F3 – F1  2  F22, , 24. A charged paritcle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic, moment  is given by, [AIPMT (Prelims)-2007], (1) q v R, , (2), , qvR, 2, , (3), , qvR2, , (4), , qvR 2, 2, , Sol. Answer (2), 25. A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields.If the electric, field is switched off, and the same magnetic field is maintained, the electrons move [AIPMT (Prelims)-2007], (1) Along a straight line, , (2), , In an elliptical orbit, , (3) In a circular orbit, , (4), , Along a parabolic path, , Sol. Answer (3), 26. In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric, potential V and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are, ⎛ charge on theion ⎞, ⎟ will be proportional to, kept constant, the ratio ⎜, ⎝ mass of theion ⎠, , (1) R, , (2), , 1, R, , (3), , 1, R2, , [AIPMT (Prelims)-2007], , (4), , R2, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 26 :
26, , Moving Charges and Magnetism, , Solution of Assignment, , , , 27. When a charged particle moving with velocity v is subjected to a magnetic field of induction B , the force on it, is non-zero. This implies that :, [AIPMT (Prelims)-2006], , , (1) Angle between v and B is necessarily 90°, , , (2) Angle between v and B can have any value other than 90°, , , (3) Angle between v and B can have any value other than zero and 180°, , , (4) Angle between v and B is either zero or 180°, Sol. Answer (3), ,  , F  q v  B   qvB sin , F  0 , sin   0 ,    0, 180°, , 28. Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil., What is the ratio of potential difference applied across them so that the magnetic field at their centres is the, same?, [AIPMT (Prelims)-2006], (1) 3, , (2), , 4, , (3), , 6, , (4), , 2, , Sol. Answer (2), 29. A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent, , magnet such that B is in plane of the coil. If due to a current i in the triangle a torque  acts on it, the side l, of the triangle is :, [AIPMT (Prelims)-2005], 1/2, , (1), , 2 ⎛  ⎞, ⎜ ⎟, 3 ⎝ Bi ⎠, , (2), , 1/2, , 2 ⎛  ⎞, ⎜ ⎟, 3 ⎝ Bi ⎠, , (3), , ⎛  ⎞, 2⎜, ⎟, ⎝ 3Bi ⎠, , (4), , 1 , 3 Bi, , Sol. Answer (3), , 30. 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 :, [AIPMT (Prelims)-2005], , Y, , I, , (1) Opposite to OX, , (2), , Q, P, , v, , O, , Along OX, , (3), , X, , Opposite to OY, , (4), , Along OY, , Sol. Answer (4), , i, , , , ,  , F  q V  B , , , using right hand thumb, F will be in the direction OY., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 27 :
Solution of Assignment, , Moving Charges and Magnetism, , 27, , 31. An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the, circle. The radius of the circle is proportional to, [AIPMT (Prelims)-2005], , (1), , B, v, , (2), , v, B, , (3), , v, B, , (4), , B, v, , Sol. Answer (3), B, ,  B, , r, ,  ,  0 q v  r , 4, r3, ,  0 qv, 4 r 2, , v, B, , 32. A beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength, 20 Vm–1 and 0.5 T at right angles to the direction of motion of the electrons. What is the velocity of the, electrons?, (1) 8 ms–1, , (2), , 5.5 ms–1, , (3), , 20 ms–1, , (4), , 40 ms–1, , Sol. Answer (4), qVB  qE, ,  V , , E 20, ,  40 ms –1, B 0.5, , 33. A 10 eV electron is circulating in a plane at right angles to a uniform field of magnetic induction, 10–4 Wb/m2 (= 1.0 gauss), the orbital radius of electron is, (1) 11 cm, , (2), , 18 cm, , (3), , 12 cm, , (4), , 16 cm, , Sol. Answer (1), r , , 2mk, qB, , 34. A positively charged particle moving due East enters a region of uniform magnetic field directed vertically, upwards. This particle will, (1) Move in a circular path with a decreased speed, (2) Move in a circular path with a uniform speed, (3) Get deflected in vertically upward direction, (4) Move in circular path with an increased speed, Sol. Answer (2), , , In uniform B , if charge enters perpendicular to the magnetic field. It will execute circular motion with uniform, speed., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 28 :
28, , Moving Charges and Magnetism, , Solution of Assignment, , 35. An electron having mass m and kinetic energy E enters in uniform magnetic field B perpendicularly, then its, frequency will be, , eE, (1) qB`, , (2), , 2m, eB, , eB, 2m, , (3), , (4), , 2m, eBE, , Sol. Answer (3), f=, , eB, 2m, , 36. A positive charge particle with charge q is moving with speed v in a region of uniform magnetic field B at the instant, shown in figure. An external electric field is to be applied so that the charged particle follows a straight line path., The magnitude and direction of electric field required are respectively, , y, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , q, , B, , v, , ×, ×, ×, ×, x, , (1) qvB, +y axis, , (2), , qvB, –y axis, , (3), , vB, –y axis, , (4), , vB, , –x axis, q, , Sol. Answer (3), , Fm  qvBjˆ, for constant velocity, , , F B  –F E, , , qvBjˆ  – qE, ,  E  –vBjˆ, 37. The magnetic field of given length of wire for single turn coil at its centre is B then its value for two turns coil, for the same wire is, , (1), , B, 4, , (2), , B, 2, , (3), , 4B, , (4), , 2B, , Sol. Answer (3), , BN, ,  0i, 2R, , B  2N, ,  0i,  i,  4N 0  4B, 2R, 2 R, 2, ,  , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 29 :
Solution of Assignment, , Moving Charges and Magnetism, , 29, , 38. The magnetic field dB due to a small current element dl at a distance r and element carrying current i is, ,  0 2 ⎛⎜ dl  r ⎞⎟, i, (1) dB , 4 ⎜⎝ r ⎟⎠, , (2), , ,  0 ⎛⎜ dl  r ⎞⎟, dB , i, 4 ⎜⎝ r 3 ⎟⎠, , ,  0 ⎛⎜ dl  r ⎞⎟, dB , i, 4 ⎜⎝ r ⎟⎠, , (3), , (4), , dB , , ,  0 2 ⎛⎜ dl  r ⎞⎟, i, 4 ⎜⎝ r 2 ⎟⎠, , Sol. Answer (2),   i ⎛   ⎞, dB  0 ⎜ dl 3 r ⎟, 4 ⎝ r ⎠, , 39. Two equal electric currents are flowing perpendicular to each other as shown in the figure. Lines AB and CD, are perpendicular to each other and symmetrically placed with respect to the currents. Where do we expect, the resultant magnetic field to be zero?, I A, C, O, , B, , l, D, , (1) On CD, , (2), , On AB, , (3) On both OD and BO, , (4), , On both AB and CD, , Sol. Answer (2), On line AB, as in this region magnetic field will be in opposite direction due to both the wires., 40. To convert a galvanometer into a voltmeter one should connect a, (1) High resistance in series with galvanometer, , (2), , Low resistance in series with galvanometer, , (3) High resistance in parallel with galvanometer, , (4), , Low resistance in parallel with galvanometer, , Sol. Answer (1), 41. A galvanometer having a coil resistance of 60  shows full scale deflection when a current of 1.0 A passes through, it. It can be converted into an ammeter to read currents upto 5.0 A by, (1) Putting in parallel a resistance of 15 , , (2), , Putting in parallel a resistance of 240 , , (3) Putting in series a resistance of 15 , , (4), , Putting in series a resistance of 240 , , Sol. Answer (1), , Rs , , Rg, 60, ,  15 , i, 5, –1, –1, 1, ig, , 42. A straight wire of a diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter, carrying the same current. The strength of the magnetic field far away is, (1) One-quarter of the earlier value, , (2), , No change, , (3) Twice the earlier value, , (4), , One-half of the earlier value, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 30 :
30, , Moving Charges and Magnetism, , Solution of Assignment, , Sol. Answer (2), For wire 1, , d = 0.5 mm, r = 0.25 mm, l = 1A, , For wire 2, , d = 1 mm, r = 0.5 mm, l = 1A, , B, ,  0i, 2x, , as x   B = 0 (for both the cases), , 43. A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10–4 m2 carries a current of, 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a, uniform magnetic field 5 × 10–2 tesla making an angle of 30º with the axis of the solenoid. The torque on the, solenoid will be, (1) 3 × 10–3 N.m, , (2), , 1.5 × 10–3 N.m, , (3), , 1.5 × 10–2 N.m, , (4), , 3 × 10–2 N.m, , Sol. Answer (3), N = 2000, ,  = 30°, , A  1.5  10 –4 m 2, , B  5  10 –2 T, , l = 2A,  ,   M B, ,   NlAB sin30  1.5  10 –2 Nm, 44. In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the, resistance of ammeter will be, (1), , 1, G, 499, , (2), , 499, G, 500, , (3), , 1, G, 500, , (4), , 500, G, 499, , Sol. Answer (3), 0.2 G = 100 RA, ,  RA =, , G, 500, , SECTION - D, Assertion-Reason Type Questions, 1., , A : Magnetic field lines are always perpendicular to the current producing it., R : Magnetic field due to a straight wire varies in inverse square proportion with distance., , Sol. Answer (3), 2., , A : Stationary charges do not experience a magnetic force., R : Magnetic force is a central force., , Sol. Answer (3), 3., , A : Net magnetic force experienced by a current carrying loop in a uniform magnetic field is always zero., R : A current loop placed in a uniform magnetic field never experiences a torque., , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 31 :
Solution of Assignment, , 4., , Moving Charges and Magnetism, , 31, , A : The trajectory of a charge when it is projected perpendicular to an electric field is a parabola., R : A moving charge entering parallel to the magnetic field lines moves in a circular path., , Sol. Answer (3), 5., , A : Like currents repel and unlike currents attract each other (in conductor)., R : Magnetic force acts in the direction of current., , Sol. Answer (4), 6., , A : A magnetic dipole experiences maximum torque when it is placed normal to the magnetic field., R : The minimum potential energy of magnetic dipole is zero., , Sol. Answer (4), 7., , A : The relation between magnetic moment and angular momentum is true for every finite size body., R : Ratio of magnetic dipole moment and angular momentum is just dependent on specific charge of the body., , Sol. Answer (1), 8., , A : When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic, field is taken into account., R : Magnetic field lines always form closed loops., , Sol. Answer (2), 9., , ,   , A : In the expression for Lorentz force, F  q(v  B  E ) , if one switches to a frame with instantaneous, , velocity v ., R : There exists an appropriate electric field in the new frame., , Sol. Answer (1), 10. A : Ampere's circuital law' is not independent of the Biot-Savart's law., R : Ampere's Circuital law can be derived from the Biot-Savart law., Sol. Answer (1), 11. A : The work done by magnetic field on a moving charge is zero., R : The magnetic force acting on a moving charge has no component in the direction of velocity., Sol. Answer (1), 12. A : In any magnetic field region the line integral, , R : The magnetic field B in the expression, , ∫ B.dl along a closed loop is always zero., , ∫ B.dl, , is due to the currents enclosed only by the loop., , Sol. Answer (4), 13. A : The magnetic field always accelerates a moving charge if the moving charge cuts the field lines., R : When a moving charge cuts the magnetic field lines, the magnetic force on the charge is always non, zero., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 32 :
32, , Moving Charges and Magnetism, , Solution of Assignment, , 14. A : The magnetic moment of a current carrying planar loop does not depend on the shape of the loop., R : The magnetic moment is a vector quantity., Sol. Answer (2), 15. A : Magnetic field is produced by moving charge(s)., R : The magnetic field in the central region of a solenoid is uniform., Sol. Answer (2), 16. A : In the middle to high latitudes on a dark night an aurora or the curtain of light hangs down from the sky., This curtain is local, several hundred kilometer high, several thousand kilometer long but less than 1 km, thick., R : Electrons and protons trapped in the helical terrestrial magnetic field collide with atoms and molecules, of air, causing that air to emit light., Sol. Answer (1), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 33 :
Chapter, , 20, , Magnetism and Matter, Solutions, SECTION - A, Objective Type Questions, 1., , Suppose an isolated north pole is kept at the centre of a circular loop carrying a electric current i. The magnetic, field due to the north pole at a point on the periphery of the wire is B. The radius of the loop is a. The force, on the wire is, (1) Nearly 2aiB perpendicular to the plane of the wire, (2) 2aiB in the plane of the wire, (3) aiB along the axis of the wire, (4) Zero, , i, , Sol. Answer (1), , N, , dF  ldlB, ,  B, F, , Ftotal  i  2a  B, Perpendicular to the plane of the paper., 2., , If L be the length of a bar magnet then separation between the two poles is nearly, (1), , 9, L, 10, , (2), , 6, L, 7, , (3), , 1, L, 3, , (4), , 1, L, 2, , Sol. Answer (2), 3., , Two identical thin bar magnets each of length l and pole strength m are placed at right angle to each other, with north pole of one touching the south pole of the other. Magnetic moment of the system is, (1) 2 ml, , (2), , ml, , (3), , 2 ml, , (4), , ml, 2, , Sol. Answer (3), S, , N, S, , N, ,   ml 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 34 :
34, 4., , Magnetism and Matter, , Solution of Assignment, , A solenoid of length 10 cm and radius 1 cm contains 200 turns and carries a current of 10 A. The value of, pole strength of each pole is, (1) 2 Am, , (2), , 2 Am, , (3), , 4 Am, , (4), , 10 Am, , Sol. Answer (2), NIA = mL, 5., ,  m = NIA/L = 200  (10–2)2  10/0.1 = 2, , A long magnetic needle of length 2L, magnetic moment M and pole strength m units is broken into two at, the mid point. The magnetic moment and pole strength of each piece will be, (1) M/2, m/2, , (2), , M, m/2, , (3), , M/2, m, , (4), , M, m, , Sol. Answer (3), M = m(2l), Pole strength does not change, on breaking the magnet but magnetic moment will become half., 6., , A bar magnet of magnetic moment M is cut into two equal parts along its length. The magnetic moment of, either part is, (1) 2M, , (2), , M, , (3), , M/2, , (4), , Zero, , Sol. Answer (3), M = ml, , m, l, 2, M' = M/2, M' , , 7., , A steel wire of length l has a magnetic moment M. It is then bent into a semi-circular arc. The new magnetic, moment is, (1) M, , (2), , 2M, , , (3), , M, , , (4), , 2 M, , Sol. Answer (2), l = r, l, , M = ml, r , , 2r, , M' = m2r, M' m, , 8., , 2l, 2M,  M' , , , , Points A and B are situated perpendicular to the axis of a small bar magnet at large distances x and 3x from, its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to, (1) 2 : 9, , (2), , 1:9, , (3), , 27 : 1, , (4), , 9:1, , Sol. Answer (3), , BA , ,  0 2M, 4 x 3, , BB , ,  0 2M, 4  3 x  3, , BA, 1, , B B 27, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 35 :
(Set-2), , 9., , Magnetism and Matter, , 35, , Figure shows two small identical magnetic dipoles a and b of magnetic moments M each, placed at a, separation 2d, with their axes perpendicular to each other. The magnetic field at the point P mid way between, the dipoles is, , a, , P, , b, , 2d, (1), , 2 0 M, 4d, , (2), , 3, ,  0M, 4d, , (3), , 3, , Zero, , (4), , 5 0 M, , 4d 3, , Sol. Answer (4), , N, , S, , B2, d, , d, 2d, , B1 , ,  0 2M, 4 d 3, , B2 , , 0 M, 4 d 3, , Bnet  B12  B 22 , , S, N, , 0 M, 5, 4 d 3, , 10. Two short magnets of equal dipole moments M are fastened perpendicularly at their centres as shown below, (point P is in the plane of the magnets), , N, S, , N, d, , S, P, The magnetic field (magnitude) at P, distance d from the centre on the bisector of the right angle is, (1), , 0 M, 4 d 3, , (2), , 0 M 2, 4 d 3, , (3), ,  0 2M, 4 d 3, , (4), ,  0 2 2M, 4 d 3, , Sol. Answer (4), , N, S, , N, S, , M can be broken of both the magnets in terms of their components, one in direction of bisector and other, perpendicular to it, Parallel components will be added whereas perpendicular components will be cancel out, , ,  0 2 2M, 2 0 ⎛ 2M ⎞, So B net , ⎜⎝ 3 ⎟⎠ cos 45  4 d 3, 4 d, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 36 :
36, , Magnetism and Matter, , Solution of Assignment, , 11. Two small bar magnets are placed in a line at certain distance d apart. If the length of each magnet is negligible, compared to d, the force between them will be inversely proportional to, (1) d 2, , (2), , d, , (3), , d3, , (4), , d4, , Sol. Answer (4), , F, ,  0 M 1M 2, 4 d 4, , so F , , 1, d4, , 12. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque, needed to maintain the needle in this position will be, (1) W, , (2), , 3W, , (3), , 3W, 2, , (4), , 2W, , Sol. Answer (2), , U i  – MB cos0  – MB, , 60°, , U f  –MB cos60, , , , – MB, 2, , W  Uf – Ui  –, , MB, MB,  MB , ⇒ MB  2W, 2, 2, ,   MB sin 60, ,   2W, , 3, W 3, 2, , 13. Potential energy of a bar magnet of magnetic moment M placed in a magnetic field of induction B such that, it makes an angle  with the direction of B is (take  = 90° as datum), (1) –M B sin , , (2), , –M B cos , , (3), , M B (1 – cos ), , (4), , M B (1 + cos ), , Sol. Answer (2), U = –MB cos, 14. A magnetic dipole is placed at right angles to the direction of lines of force of magnetic induction B. If it is rotated, through an angle of 180o, then the work done is, (1) 2MB, , (2), , MB, , (3), , –2MB, , (4), , Zero, , Sol. Answer (4), Ui = – MB cos90°, Uf = – MB cos90°, W=0–0=0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 37 :
(Set-2), , Magnetism and Matter, , 37, , 15. Which of the following statements regarding magnetic lines of force is correct?, (1) Total magnetic flux linked with a closed surface is always zero, (2) They need not be perpendicular to the surface from where they start or where they meet, (3) They may or may not pass through a conductor, (4) All of these, Sol. Answer (4), 16. A magnetic needle of negligible breadth and thickness compared to its length, oscillates in a horizontal plane with, a period T. The period of oscillation of each part obtained on breaking the magnet into n equal parts perpendicular, to the length is, (1) T/n, , (2), , T, , (3), , Tn, , (4), , 1/T n, , Sol. Answer (1), T  2, T '  2, , T', , I, MB, I, 2, I, , M, n MB, n3 B, n, , T, n, , 17. The time period of a freely suspended magnet is 4 s. If it is broken in length into two equal parts and one, part is suspended in the same way, then its time period will be, (1) 2 s, , (2), , 4s, , (3), , 0.5 s, , (4), , 0.25 s, , Sol. Answer (1), , T', , T,  2s, 2, , 18. The magnetic moments of two bar magnets of same size are in the ratio 1 : 2. When they are placed one over the, other with their similar poles together, then their period of oscillation in a magnetic field is 3s. If one of the magnets, is reversed, then the period of oscillation in the same field will be, (1), , 3s, , (2), , 3 3s, , (3), , 3s, , (4), , 6s, , (3), , Both (1) & (2), , (4), , Neither (1) nor (2), , (3), , Mass, , (4), , Size, , Sol. Answer (2), , T, 3, , T, , 1, M, M1 – M 2, 3, ⇒, , M1  M 2, T, , 2 –1, ⇒ T 3 3, 21, , 19. Magnetostatic screening or shielding can be created by, (1) Super conductor, , (2), , Soft iron ring, , Sol. Answer (3), 20. Planets producing larger magnetic field have larger, (1) Rotational speed, , (2), , Density, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 38 :
38, , Magnetism and Matter, , Solution of Assignment, , 21. To shield an instrument from external magnetic field, it is placed inside a cabin made of, (1) Wood, , (2), , Plastic, , (3) Diamagnetic substances, , (4), , Iron, , Sol. Answer (4), 22. A dip circle lies initially in the magnetic meridian, it shows an angle of dip  at a place. The dip circle is rotated, tan , is, through an angle  in the horizontal plane and then it shows an angle of dip . Hence, tan , (1) cos , (2) 1/sin , (3) 1/tan , (4) 1/cos , Sol. Answer (4), tan  ' , , tan , cos , , tan  ', 1, , ., tan  cos , 23. If a dip circle is placed in a vertical plane at an angle of 30° to the magnetic meridian, the dip needle makes, an angle of 45° with the horizontal. The real dip at that place is, (1) tan–1 (3/2), , (2), , tan–1 (3), , (3), , tan–1 (3/2), , (4), , tan–1 (2/3), , Sol. Answer (1), tan 45 , , 1, tan  , , tan , cos30, , 2 tan , 3, 3, ⎛, ⎞, ⇒   tan –1 ⎜ 3 ⎟, ⎝ 2 ⎠, 2, , 24. A dip needle free to move in a vertical plane perpendicular to the magnetic meridian will remain, (1) Horizontal, , (2), , Vertical, , (3) At an angle of 60° to the vertical, , (4), , At an angle of 45° to the horizontal, , Sol. Answer (2), tan  A , , tan , cos90, ,  tan  A  ,   A  90, 25. If 1 and 2 are the angles of dip in two vertical planes at right angles to each other and  is the true angle, of dip then, (1) cot2 = cot21 + cot22, , (2), , tan2 = tan21 + tan22, , (3) cot = cot1 + cot2, , (4), , tan = tan1 + tan2, , Sol. Answer (1), 26. How many neutral points will be obtained when a bar magnet is kept with magnetic moment parallel to earth's, magnetic field?, (1) One, , (2), , Two, , (3), , Four, , (4), , Infinite, , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 39 :
(Set-2), , Magnetism and Matter, , 39, , 27. Time period of oscillations of a magnet of magnetic moment M and moment of inertia I in a vertical plane perpendicular, to the magnetic meridian at a place where earth's horizontal and vertical component of magnetic field are BH and, BV respectively is, (1) T  2, , I, M (B BH2 )1 2, 2, V, , (2), , T  2, , I, MBV, , (3), , T  2, , I, MBH, , (4), , Infinite, , Sol. Answer (2), , BV, BH, , T  2, , I, MBv, , 28. A magnet is suspended in such a way that it oscillates in the horizontal plane. 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°. Ratio of the total earth’s magnetic field at the two places is, (1) 3 3 : 8, , (2), , 16 : 9 3, , (3), , 4:9, , (4), , 2 3 :9, , (4), , Any angle, , Sol. Answer (2), 1, I,  2, MB1 cos30, 20, , 1, I,  2, 15, MB 2 cos 60, 2, , B2, ⎛ 3⎞, B, 16, ⎜⎝ ⎟⎠ ,  1 , 4, B1 3, B2 9 3, , 29. Instruments based on tangent law are most accurate when the deflection is, (1) 30°, , (2), , 45°, , (3), , 60°, , Sol. Answer (2), 30. The magnetic induction and magnetising field intensity in a sample of magnetic material are B and H, respectively. The magnetic susceptibility of the material is, B, (1)  H, 0, , (2), , B, 1,  0H, , (3), , B, 1, H, , (4), ,  0H, 1, B, , Sol. Answer (2), B   0 I  H , , B   0  x mH  H , Xm , , B, –1,  0H, , 31. A frog is levitated in a magnetic field produced by current in a vertical solenoid below the frog is, (1) Diamagnetic, , (2), , Paramagnetic, , (3), , Ferromagnetic, , (4), , Ferrimagnetic, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 40 :
40, , Magnetism and Matter, , Solution of Assignment, , 32. When a ferromagnetic substance is heated to a temperature above its Curie temperature, it, (1) Behaves like a paramagnetic substance, , (2), , Behaves like a diamagnetic substance, , (3) Remains ferromagnetic, , (4), , Is permanently magnetised, , Sol. Answer (1), 33. Select the incorrect statement, (1) In a diamagnetic substance net magnetic moment of each atom/molecule is zero, (2) In a paramagnetic substance net magnetic moment of each atom/molecule is non-zero, (3) In a ferromagnetic material net magnetic moment of each domain is zero, (4) In a ferromagnetic material net magnetic moment of each domain is non-zero, Sol. Answer (3), 34. Soft iron is used to manufacture electromagnets because their, (1) Magnetic permeability is high and retentivity and coercive force are small, (2) Retentivity is high, (3) Coercive force is high, (4) Area of the hysteresis curve is large, Sol. Answer (1), 35. Relative permeability of superconductors is, (1) 0, , (2), , 1, , (3), , –1, , (4), , 0.5, , Sol. Answer (1), , SECTION - B, Objective Type Questions, 1., , At a certain place, vertical component of earth’s magnetic field is 3 times the horizontal component of earth’s, magnetic field. If a magnetic needle is suspended freely in air then it will incline, (1) 30° below horizontal, , (2), , 60° below horizontal, , (3) 30° above horizontal, , (4), , 45° above horizontal, , Sol. Answer (2), , BV  3BH, BV,  3 = tan , BH, ,  = 60°, 2., , The figure illustrates how B, the flux density, inside a sample of ferromagnetic material varies with external, magnetic field B0. For the sample to be suitable for making a permanent magnet, B, P, , Q, R, , S, , B0, , O, T, , (1) OQ should be large, OR should be small, , (2), , OQ and OR both should be large, , (3) OQ should be small and OR should be large, , (4), , OQ and OR both should be small, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 41 :
(Set-2), , 3., , Magnetism and Matter, , 41, , The work done in turning a magnet of magnetic moment M by an angle of 90° from the magnetic meridian is n, times the corresponding work done to turn it through an angle of 60°. The value of n is, (1) 2, , (2), , 1, , (3), , 1, 3, , (4), , 1, 4, , Sol. Answer (1), U1  –MB cos0, U 2  –MB cos90, U 3  –MB cos 60, U 2 – U 1  MB, , U 3 – U1 , 4., , MB, ⇒ n2, 2, , Soft iron is used in many parts of electrical machines for, (1) Low hysteresis loss and low permeability, , (2), , Low hysteresis loss and high permeability, , (3) High hysteresis loss and low permeability, , (4), , High hysteresis loss and high permeability, , Sol. Answer (2), 5., , A magnetic needle oscillates in a horizontal plane with a period T at a place where the angle of dip is 60°. When, the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be, (1), , T, , (2), , 2, , T, , (3), , 2T, , (4), , 2T, , Sol. Answer (1), , 6., , T  2, , I, T', ⇒,  cos 60, MB cos 60, T, , T '  2, , 1, I, T', T, ⇒, , ⇒ T'=, MB, T, 2, 2, , Two different magnets are tied together and allowed to vibrate in a horizontal plane. When their like poles are, joined, time period of oscillation is 5 s and with unlike poles joined, time period of oscillation is 15 s. The ratio of, their magnetic moments is, (1) 5 : 4, , (2), , 1:3, , (3), , 3:1, , (4), , 2:5, , Sol. Answer (1), 5  2, 15  2, , I, M1  M 2, 1, I, ⇒, , M1 – M 2, 3, , M1 – M 2, M1  M 2, , M1 – M 2 1, , M1  M 2 9, M 1 10 5, , , M2, 8, 4, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 42 :
42, 7., , Magnetism and Matter, , Solution of Assignment, , The values of the apparent angles of dip in two planes at right angles to each other are 45° and 30° respectively., The true value of angle of dip at the place is, (1) cot–1 (1), , (2), , cot–1 (2), , (3), , cot–1 (3), , (4), , cot–1 (4), , Sol. Answer (2), cot 2   cot 2 45  cot 2 30, , cot2 = 1 + 3,   = cot–1(2), , cot = 2, 8., , A magnet is placed horizontally on ground with its north pole towards the geographic north pole of the earth. The, neutral point is obtained, (1) Along the axis of the magnet, (2) On the east-west line through the centre of the magnet, (3) In only east side of the magnet, (4) In only west side of the magnet, , Sol. Answer (2), 9., , The value of horizontal component of earth’s magnetic field at a place is 0.35 × 10–4 T. If the angle of dip is 60°, the, value of vertical component of earth’s magnetic field is nearly, (1) 0.1 × 10–4 T, , (2), , 0.2 × 10–4 T, , (3), , 0.4 × 10–4 T, , (4), , 0.61 × 10–4 T, , Sol. Answer (4), BV  B H tan , , BH   0.35  10 –4  3, B H  0.61  10 –4, , 10. The period of oscillation of a magnet of a vibration magnetometer is 2.45 s at one place and 4.9 s at the other., The ratio of the horizontal component of earth magnetic field at the two places is, (1) 1 : 4, , (2), , 1:2, , (3), , 2:1, , (3), , tan  , , (4), , 4:1, , (4), , tan2 + tan2 = 1, , Sol. Answer (4), I, MB H1, , 2.45  2, , 4.9  2, , I, MB H 2, , Dividing both the equations, 2.45, , 4.9, , BH 2, B H1, , ⇒, , B H1, BH 2, ,  4 :1, , 11. If  is latitude and  is dip at a place then, (1) tan =, , tan , 2, , (2), , tan  , , tan , 2, , 1, tan , , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 43 :
(Set-2), , Magnetism and Matter, , 43, , 12. A tangent galvanometer has 80 turns of wire. The internal and external diameters of the coil are, 19 cm and 21 cm respectively. The reduction factor of the galvanometer at a place where, H = 0.32 oersted will be (1 oersted = 80 A/m), (1) 0.0064, , (2), , 0.64, , (3), , 0.064, , (4), , None of these, , (3), , Small and positive, , (4), , Small and negative, , (1) By the specimen, , (2), , Per unit volume of the specimen, , (3) Per unit volume per cycle of the specimen, , (4), , Per cycle of the specimen, , Sol. Answer (3), , 2RB h 2  0.1  0.32  80,  0.064, Reduction factor =  N , 80, 0, 13. Magnetic susceptibility for a diamagnetic substance is, (1) Large and positive, , (2), , Large and negative, , Sol. Answer (4), 14. Area of B-H curve measures the energy loss, , Sol. Answer (3), 15. A wire of length l m carrying a current I ampere, is bent in the form of a circle. The magnetic moment is, (1), , 4I, , (2), , l2, , 2Il 2, , , (4), , Il 2, 4, , (3), , eh 2, 4me, , (4), , eh, 4me, , (3), , Il 2, , , Sol. Answer (4), l  2 r, , r , , l, 2, 2, , ll 2, ⎛ l ⎞, M  l ⎜ ⎟ ⇒ M , ⎝ 2 ⎠, 4, , 16. Bohr magneton is given by (symbols have their usual meanings), , (1), , 4me, eh, , 2, , (2), , 4me, eh, , Sol. Answer (4), , Bm , , eh, 4m, , 17. The unit of magnetic susceptibility is, (1) weber, , (2), , weber per metre, , (3), , henry, , (4), , Dimensionless, , (2), , 104 maxwell, , (3), , 106 maxwell, , (4), , 108 maxwell, , Sol. Answer (4), 18. One weber is equal to, (1) 102 maxwell, Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 44 :
44, , Magnetism and Matter, , Solution of Assignment, , 19. If the number of turns and radius of cross section of the coil of a tangent galvanometer are doubled, then the, reduction factor K will become, (1) K, , (2), , 2K, , (3), , 4K, , (4), , K, 4, , Sol. Answer (1), K, , 2RB H,  0N, , , , K , , R, 2R, K, and K ' , N, 2N, , 20. Two short bar magnets of magnetic moments 'M' each are arranged at the opposite corners of a square of, side 'd', such that their centres coincide with the corners and their axes are parallel to one side of the square., If the like poles are in the same direction, the magnetic induction at any of the other corners of the square, is, (1), , 0 M, 4 d 3, , (2), ,  0 2M, 4 d 3, , (3), , 0 M, 2 d 3, , (4), ,  0 2M, 2 d 3, , Sol. Answer (1), , 2, , S, N, , S, , 1, , N, , B1, B2, , Bnet  B 2 – B1, , , ,  0 ⎛ 2M ⎞  0 ⎛ M ⎞, ⎜, ⎟–, ⎜ ⎟, 4 ⎝ d 3 ⎠ 4 ⎝ d 3 ⎠, , , , 0 M, 4 d 3, , 21. If declination at a place is known to be 15° E. And a compass needle points as shown, then geographic north, is represented by the direction numbered., , 2, , 1, 15° 15°, , 3, (1) 1, , (2), , 2, , 75°, , 4, , 75°, (3), , 3, , (4), , 4, , Sol. Answer (2), Geographic north will be 15° west of the direction in which the magnetic needle is pointing., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 45 :
(Set-2), , Magnetism and Matter, , 45, , 22. The magnetic moment of the arrangement shown in the figure is, , N, , N, , 2M, S, , M, M, , N, (1) Zero, , (2), , S, , 2 2M, , (3), , S, , 2M, , (4), , M, , Sol. Answer (3), , N, M, , N, , 2, , M, , S, N, , M, , S, , S, , Resultant of these three dipole moments will be, , M net , , M 2  2  M 2  2, , = 2M, , 23. Two identical short bar magnets are placed at 120° as shown in the figure. The magnetic moment of each, magnet is M. Then the magnetic field at the point P on the angle bisector is given by, P, N, , d, , S, , (1), , 0 M, ., 4 d 3, , (2), , 0 2M, ., 4 d 3, , 120°, S, , N, , (3), , 0 2 2M, ., 4 d 3, , (4), , Zero, , Sol. Answer (2), Since two equal vectors M are inclined at 120°, their resultant will also be M and along its angular bisector., So point P is on axial line of resultant moment M., , 2 0 ⎛ M ⎞, ⎜ ⎟, 4 ⎝ d 3 ⎠, , M, 60°, , Bnet , , 60°, , 2 0M, 4d 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 46 :
46, , Magnetism and Matter, , Solution of Assignment, , 24. The work done in rotating a bar magnet of magnetic moment M from its unstable equilibrium position to its, stable equilibrium position in a uniform magnetic field B is, (1) 2MB, , (2), , MB, , (3), , –MB, , (4), , –2MB, , Sol. Answer (4), Ui = MB, Uf = –MB, U = –2MB, 25. The variation of magnetic susceptibility () with absolute temperature (T) for a ferromagnetic material is, , , , , , , , (2), , (1), , , , (3), , T, , T, , (4), , T, , T, , Sol. Answer (2), , SECTION - C, Previous Years Questions, 1., , Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic, , [AIPMT-2014], dipole moment m . Which configuration has highest net magnetic dipole moment ?, , N, a., , b., , S S, , N, , N, , N, , (1) (a), , (2), , N, , S, , S, , N, , 30°, , c., , S, S, , (b), , (3), , (c), , 60°, , d., , N, , S, S, , (4), , N, , (d), , Sol. Answer (3), Resultant dipole moment =, 2., , M 2  M 2  2M 2 cos , , A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc as shown in figure. The, new magnetic dipole moment will be, [NEET-2013], , 60º, , (1), , 3, M, , , (2), , 2, M, , , r, , (3), , M, 2, , (4), , M, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 47 :
(Set-2), , Magnetism and Matter, , 47, , Sol. Answer (1), , l, r, ,  l, , 3 r, , 60°, , r, l, 3, , or r , , r, , 60°, 60°, , 3l, , , r, , M = ml, M' = mr, , ⎛ 3l ⎞ 3M,  M⎜ ⎟ , ⎝ ⎠, , 3., , A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It, [AIPMT (Prelims)-2012), (1) Will stay in north-south direction only, , (2), , Will stay in east-west direction only, , (3) Will become rigid showing no movement, , (4), , Will stary in any position, , Sol. Answer (4), Because here earth's magnetic field has vertical component only., 4., , A magnetic needle suspended parallel to a magnetic field requires, torque needed to maintain the needle in this position will be, (1) 2 3 J, , (2), , 3J, , (3), , 3 J of work to turn it through 60°. The, [AIPMT (Mains)-2012], , 3J, , (4), , 3, J, 2, , Sol. Answer (2), , W , 3, , MB, 2, MB, ⇒ MB  2 3, 2, ,   MB sin 60, ,   MB, 2 3, 5., , 3, 2, 3, 3J, 2, , There are four light-weight-rod samples, A, B, C, D separately suspended by threads. A bar magnet is slowly, brought near each sample and the following observations are noted, i), , A is feebly repelled, , ii), , B is feebly attracted, , iii) C is strongly attracted, , iv), , D remains unaffected, , Which one of the following is true ?, , [AIPMT (Prelims)-2011], , (1) A is of a non-magnetic material, , (2), , B is of a paramagnetic material, , (3) C is of a diamagnetic material, , (4), , D is of a ferromagnetic material, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 48 :
48, 6., , Magnetism and Matter, , Solution of Assignment, , A short bar magnet of magnetic moment 0.4 JT–1 is placed in a uniform magnetic field of 0.16 T. The magnet, is in stable equilibrium when the potential energy is, [AIPMT (Mains)-2011], (1) – 0.082 J, , (2), , 0.064 J, , (3), , – 0.064 J, , (4), , Zero, , Sol. Answer (3),  , U  – M.B,  –  0.4  0.16, , = – 0.064 J, 7., , Electromagnets are made of soft iron because soft iron has, , [AIPMT (Prelims)-2010], , (1) High retentivity and low coercive force, , (2), , Low retentivity and high coercive force, , (3) High retentivity and high coercive force, , (4), , Low retentivity and low coercive force, , Sol. Answer (1), 8., , A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes, oscillations with a time period of 2 s in earth's horizontal magnetic field of 24 microtesla. When a horizontal, field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time, period of magnet will be:, [AIPMT (Prelims)-2010], (1) 4 s, , (2), , 1s, , (3), , 2s, , (4), , 3s, , Sol. Answer (1), 9., , A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10–4 m2 carries a current of 2.0 A. It, is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a, uniform magnetic field 5 × 10–2 tesla making an angle of 30° with the axis of the solenoid. The torque on the, solenoid will be, [AIPMT (Mains)-2010], (1) 3 × 10–3 Nm, , (2), , 1.5 × 10–3 Nm, , 1.5 × 10–2 Nm, , (3), , (4), , 3 × 10–2 Nm, , Sol. Answer (3), 10. The magnetic moment of a diamagnetic atom is, , [AIPMT (Mains)-2010], , (1) Much greater than one, , (2), , 1, , (3) Between zero and one, , (4), , Equal to zero, , Sol. Answer (4), 11. Two identical bar magnets are fixed with their centres at a distance d apart. A stationary charge Q is placed, at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure, S, , P, D, O, , N, , N, , S, , d, , The force on the charge Q is, , [AIPMT (Mains)-2010], , (1) Zero, , (2), , Directed along OP, , (3) Directed along PO, , (4), , Directed perpendicular to the plane of paper, , Sol. Answer (1), If charge is stationary then, ,  , Using Fm  q v  B , as v = 0, Fm = 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 49 :
(Set-2), , Magnetism and Matter, , 49, , 12. A thin ring of radius R meter has charge q coulomb uniformly spread on it. The ring rotates about its axis with, a constant frequency of f revolutions/s. The value of magnetic induction in Wb/m2 at the centre of the ring is, [AIPMT (Prelims)-2010], , 0 qf, (1) 2 R, , (2), , 0 q, 2 fR, , (3), , 0 q, 2fR, , (4), , 0 qf, 2R, , Sol. Answer (4), 13. A bar magnet having a magnetic moment of 2 × 104 JT–1 is free to rotate in a horizontal plane. A horizontal, magnetic field B = 6 × 10–4 T exists in the space. The work done in taking the magnet slowly from a direction, parallel to the field to a direction 60° from the field is, [AIPMT (Prelims)-2009], (1) 12 J, , (2), , 6J, , (3), , 2J, , (4), , 0.6 J, , Sol. Answer (2), 14. If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is, [AIPMT (Prelims)-2009], (1) Repelled by the north pole and attracted by the south pole, (2) Attracted by the north pole and repelled by the south pole, (3) Attracted by both the poles, (4) Repelled by both the poles, Sol. Answer (4), 15. Curie temperature is the temperature above which, , [AIPMT (Prelims)-2008], , (1) Ferromagnetic material becomes diamagnetic material, (2) Ferromagnetic material becomes paramagnetic material, (3) Paramagnetic material becomes diamagnetic material, (4) Paramagnetic material becomes ferromagnetic material, Sol. Answer (2), 16. Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie, temperature, then it will show, [AIPMT (Prelims)-2007], (1) Diamagnetism, , (2), , Paramagnetism, , (3) Anti ferromagnetism, , (4), , No magnetic property, , Sol. Answer (2), 17. Above Curie temperature, , [AIPMT (Prelims)-2006], , (1) A ferromagnetic substance becomes paramagnetic, (2) A paramagnetic substance becomes diamagnetic, (3) A diamagnetic substance becomes paramagnetic, (4) A paramagnetic substance becomes ferromagnetic, Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 50 :
50, , Magnetism and Matter, , Solution of Assignment, , 18. If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic, material are denoted by d, p and f respectively, then, [AIPMT (Prelims)-2005], (1) d  0 and f  0, , (2), , p = 0 and f  0, , (3), , d = 0 and p  0, , (4), , d  0 and p = 0, , Sol. Answer (3), 19. A bar magnet having a magnetic moment of 2 × 104 JT–1 is free to rotate in a horizontal plane. A horizontal, magnetic field B = 6 × 10–4 T exists in the space. The work done in taking the magnet slowly from a direction, parallel to the field to a direction 60° from the field is, (1) 2 J, , (2), , 0.6 J, , (3), , 12 J, , (4), , 6J, , Sol. Answer (4), W , , MB 2  10 4  6  10 –4, , 6J, 2, 2, , 20. A bar magnet of magnetic moment M , is placed in a magnetic field of induction B , The torque exerted on, it is, (1) M  B, , (2), ,  M B, , (3), , M B, , (4), , BM, , (3), , Resistance, , (4), , Charge, , Sol. Answer (1),  ,   M B, 21. Tangent galvanometer is used to measure, (1) Potential difference, , (2), , Current, , Sol. Answer (2), 22. The work done in turning a magnet of magnetic moment M by an angle of 90° from the magnetic meridian is n, times the corresponding work done to turn it through an angle of 60°. The value of n is, (1) 2, , (2), , 1, , (3), , 1, 3, , (4), , 1, 4, , Sol. Answer (1), W 090  nW 060, MB , , nMB, 2, , n=2, 23. Due to earth’s magnetic field, the charged cosmic rays particles, (1) Can never reach the pole, (2) Can never reach the equator, (3) Require greater kinetic energy to reach the equator than pole, (4) Require less kinetic energy to reach the equator than pole, Sol. Answer (2), They will move in helical path while trapped in earth's magnetic field and will eventually move towards poles., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 51 :
(Set-2), , Magnetism and Matter, , 51, , 24. For protecting a sensitive equipment from the external magnetic field, it should be, (1) Surrounded with fine copper sheet, (2) Placed inside an iron can, (3) Wrapped with insulation around it when passing current through it, (4) Placed inside an aluminium can, Sol. Answer (2), Because stationary magnetic field is zero inside soft ring., 25. Two bar magnets having same geometry with magnetic moments M and 2M, are firstly placed in such a way, that their similar poles are same side then its time period of oscillation is T1. Now the polarity of one of the, magnet is reversed then time period of oscillation is T2, then, (1) T1 < T2, , (2), , T1 = T2, , (3), , T1 > T2, , (4), , T2 = , , Sol. Answer (1), T1  2, , I, 3MB, , T 2  2, , I, MB, , T1, 1, , ⇒ T2  T1 3 ⇒ T2  T1, T2, 3, 26. According to Curie’s law, the magnetic susceptibility of a substance at an absolute temperature T is proportional, to, (3), , 1, T2, , (1) From stronger to the weaker parts of the field, , (2), , From weaker to the stronger parts of the field, , (3) Perpendicular to the field, , (4), , In none of the above directions, , (1), , 1, T, , (2), , T, , (4), , T2, , Sol. Answer (1), , m , 27., , 1, T, , A diamagnetic material in a magnetic field moves, , Sol. Answer (1), 28. If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic, material are denoted by d, p. and f, respectively, then, (1) d = 0 and p  0, , (2), , d  0 and p = 0, , (3) p = 0 and f  0, , (4), , d  0 and f  0, , Sol. Answer (1),  d  0,  p  0 ,  f  0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 52 :
52, , Magnetism and Matter, , Solution of Assignment, , 29. A current loop in a magnetic field, (1) Can be in equilibrium in one orientation, (2) Can be in equilibrium in two orientations, both the equilibrium states are unstable, (3) Can be in equilibrium in two orientations, one stable while the other is unstable, (4) Experience a torque whether the field is uniform or non-uniform in all orientations, Sol. Answer (3), 30. An iron nail near a bar magnet experiences, (1) Only torque, , (2), , Torque and force of attraction, , (3) Only force, , (4), , Torque and force of repulsion, , Sol. Answer (2), Field is non-uniform., , SECTION - D, Assertion-Reason Type Questions, 1., , A : If a bar magnet is cut into two equal halves then magnetic dipole moment of each part is half that of the, original magnet., R : Magnetic dipole moment is the product of pole strength and magnetic length., , Sol. Answer (1), 2., , A : A magnetized needle in a uniform magnetic field experiences a torque but no net force, however, an iron, nail near a bar magnet experiences a force of attraction as well as torque., R : Bar magnet creates non-uniform magnetic field., , Sol. Answer (1), 3., , A : Every magnetic configuration need not have a north pole and south pole., R : North pole, south pole exists only if the source of the field has net magnetic dipole moment., , Sol. Answer (1), 4., , A : If different ends of two identical looking iron bars are brought closer and they always attract each other, then one of the bars is not magnetized., R : Repulsion is the sure check of presence of magnetization of both the bars., , Sol. Answer (1), 5., , A : The magnetic field lines also represent the lines of force on a moving charged particle at every point., R : Force on a moving charge acts parallel to the magnetic field., , Sol. Answer (4), 6., , A : Magnetic field lines can be entirely confined within the core of a toroid., R : Magnetic field lines cannot be entirely confined within the core of a straight solenoid., , Sol. Answer (2), 7., , A : A bar magnet does not exert a torque on itself due to its own field., R : One element of a current-carrying non-straight wire exert a force on another element of the same wire., , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 53 :
(Set-2), , 8., , Magnetism and Matter, , 53, , A : A system can have magnetic moments even though its net charge is zero., R : Magnetic moment is created by charges in motion., , Sol. Answer (1), 9., , A : The earth's magnetic field not only varies from point to point in space, it also changes with time., R : The earth's core is known to contain iron yet geologists do not regard this as a source of the earth's, magnetism., , Sol. Answer (2), 10. A : The earth may have even reversed the direction of its field several times during its history of, 4 to 5 billion years., R : Earth's magnetic field gets weakly 'recorded' in certain rocks during solidification., Sol. Answer (4), 11. A : The earth's field departs from its dipole shape substantially at large distances (greater than about 30,000, km)., R : At large distances, the field gets modified due to the field of ions in motion in the ionosphere which is, sensitive to extra-terrestrial disturbances such as, the solar wind., Sol. Answer (1), 12. A : Paramagnetic sample displays greater magnetisation (for the same magnetizing field) when cooled., R : The tendency to disrupt the alignment of dipoles (with the magnetising field) arising from random thermal, motion is reduced at lower temperatures., Sol. Answer (1), 13. A : Diamagnetism is almost independent of temperature., R : The induced dipole moment in a diamagnetic sample is always opposite to the magnetising field irrespective, of what the internal motion of the atom is., Sol. Answer (1), 14. A : If a toroid uses bismuth for its core, then the field in the core will be slightly less than when the core is, empty., R : Bismuth is paramagnetic., Sol. Answer (3), 15. A : Permeability of a ferromagnetic material is independent of the magnetising field., R : Permeability is given by the area under a hysteresis loop., Sol. Answer (4), 16. A : The maximum possible magnetization of a paramagnetic sample is of the same order of magnitude as the, magnetization of a ferromagnet., R : Saturation of paramagnetic substances requires impractically high magnetising fields., Sol. Answer (1), 17. A : A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory., R : Magnetisation of a ferromagnet is not a single valued function of the magnetising field rather it depends, both on the field and also on history of magnetization., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 54 :
54, , Magnetism and Matter, , Solution of Assignment, , 18. A : Ceramics (specially treated barium iron oxides) also called ferrites are used for coating magnetic tapes, in a cassette player, or for building 'memory stores' in a modern computer., R : A certain region of space surrounded by soft iron rings is approximately shielded from magnetic fields., Sol. Answer (2), 19. A : Magnetic field lines are continuous and form closed loops., R : Magnetic monopoles do not exist., Sol. Answer (1), 20. A : Superconductors are perfect diamagnetic., R : Superconductors are perfect conductors., Sol. Answer (2), 21. A : Superconducting magnets are gainfully exploited in running magnetically levitated superfast trains., R : Superconductors are diamagnetic substances which get repelled by strong external magnetic field., Sol. Answer (1), 22. A : Diamagnetism is exhibited by all the substances., R : Diamagnetism is due to paired electrons., Sol. Answer (1), 23. A : Above curie-point a ferromagnetic substance behaves as a paramagnetic substance., R : Magnetic susceptibility of a diamagnetic substance increases with rise in temperature., Sol. Answer (3), 24. A : Earth's magnetism protects us from many of the harmful cosmic rays., R : Earth's magnetism is due to a large permanent magnet inside earth., Sol. Answer (3), 25. A : In a diamagnetic substance each atom has a non-zero dipole moment but due to thermal agitation the, individual dipoles remain randomly oriented and, therefore, the net dipole moment in any finite volume of the, substance remains zero., R : On increasing the temperature the magnetism inside a permanent magnet increases., Sol. Answer (4), 26. A : The magnetic field lines have a tendency to avoid entering the body of a frog., R : The body of a frog is diamagnetic in nature., Sol. Answer (1), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 55 :
Chapter, , 21, , Electromagnetic Induction, Solutions, SECTION - A, Objective Type Questions, 1., , Dimensional formula of magnetic flux is, (1) [M L2 T–2 A–1], , (3), , [M L2 T–3 A–1], , (1) Stationary uniform magnetic field, , (2), , Stationary nonuniform magnetic field, , (3) Time varying magnetic field, , (4), , Not possible, , (1) The number of turns in the coil, , (2), , The rate of change of magnetic flux, , (3) Time of rotation, , (4), , The resistance of the circuit, , (2), , [M L1 T–1 A–2], , (4), , [M L–2 T–2 A–2], , Sol. Answer (1),  = BA, and F = ilB  B , , F, il, , So dimensional formula will be ML2T–2A–1, 2., , An emf can be induced in stationary coil if it is kept in, , Sol. Answer (3), , , Time varying B will induce emf as, –, , 3., , d, dt, , The induced e.m.f. in a coil does not depend on, , Sol. Answer (4),  does not depend on resistance., 4., , Flux  (in weber) in a closed circuit of resistance 10 ohm varies with time t (in second) according to the, equation  = 6 t2 – 5 t + 1. What is the magnitude of the induced current at t = 0.25 s?, (1) 1.2A, , (2), , 0.2A, , (3), , 0.6A, , (4), , 0.8A, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 56 :
56, , Electromagnetic Induction, , Solution of Assignment, , Sol. Answer (2),   6t 2 – 5t  1, , –, , d,  –12t  5, dt, , , l , R, , 5., , ⎛ 1⎞, –12 ⎜ ⎟  5, ⎝ 4⎠,  0.2 A, 10, , A cylindrical magnet is kept along the axis of a circular coil. On rotating the magnet about its axis, the coil, will have induced in it, (1) No current, , (2), , A current, , (3) Only an e.m.f., , (4), , Both an e.m.f. and a current, , Sol. Answer (1), No change in flux, so no current will be induced., 6., , A magnet is brought near a coil in two ways (i) rapidly (ii) slowly. The induced charge will be, (1) More in case (i), , (2), , More in case (ii), , (3) Equal in both the cases, , (4), , More or less according to the radius of the coil, , Sol. Answer (3), Induced charged is independent of speed of magnet., 7., , A circular loop of flexible conducting material is kept in a magnetic field directed perpendicularly into its plane., By holding the loop at diametrically opposite points its is suddenly stretched outwards, then, (1) No current is induced in the loop, , (2), , Anti-clockwise current is induced, , (3) Clockwise current is induced, , (4), , Only e.m.f. is induced, , Sol. Answer (3), Flux will increase by stretching outwards so by Lenz's law clockwise current will be induced to oppose the, change., 8., , An aeroplane is flying horizontally with a velocity of 360 km/h. The distance between the tips of the wings of, aeroplane is 25 m. The vertical component of earth's magnetic field is 4 × 10–4 Wb/m2. The induced e.m.f. is, (1) 1 V, , (2), , 100 V, , (3), , 1 kV, , (4), , Zero, , Sol. Answer (1),  = Bvl, ,  4  10 –4  360 , , 5,  25, 18, , =1V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 57 :
Solution of Assignment, , 9., , 57, , Electromagnetic Induction, , The magnetic flux through a coil varies with time t as shown in the diagram. Which graph best represents, the variation of the e.m.f. E induced in the coil with time t?, , , t, , E, , E, t, , (1), , E, t, , (2), , E, t, , (3), , t, , (4), , Sol. Answer (3),  = Asint, , –, , d,  – A cos  t, dt, , E, t, , 10. A coil having 500 square loops each of side 10 cm is placed with its plane perpendicular to a magnetic field, which increases at a rate of 1.0 tesla/s. The induced e.m.f. (in volts) is, (1) 0.5, , (2), , 0.1, , (3), , 1.0, , (4), , 5.0, , (4), , Energy, , Sol. Answer (4),   NA, , dB, dt, ,   500  0.1 2 1  5 V, 11. The physical quantity, which is conserved on the basis of Lenz's Law is, (1) Charge, , (2), , Momentum, , (3), , Mass, , Sol. Answer (4), 12. A short bar magnet passes at a steady speed right through a long solenoid. A galvanometer is connected, across the solenoid. Which graph best represents the variation of the galvanometer deflection  with time t?, , G, , S, , , , (1), , N, , t, , (2), , N, , S, , , t, , (3), , , t, , (4), , t, , Sol. Answer (1), Using Lenz's law, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 58 :
58, , Electromagnetic Induction, , Solution of Assignment, , 13. A metallic ring with a cut is held horizontally and a magnet is allowed to fall vertically through the ring, then, the acceleration of this magnet is, (1) Equal to g, , (2), , More than g, , (3) Less than g, , (4), , Sometimes less and sometimes more than g, , Sol. Answer (1), 14. A bar magnet is made to fall through a long vertical copper tube. The speed (v) of the magnet as a function, of time (t) is best represented by, v, d a, , c, , b, t, (1) a, , (2), , b, , (3), , c, , (4), , d, , Sol. Answer (4), 15. A loop of irregular shape made of flexible conducting wire carrying clockwise current is placed in uniform inward, magnetic field, such that its plane is perpendicular to the field. Then the loop, (1) Experiences force, , (2), , Develops induced current for a short time, , (3) Changes to circular loop, , (4), , All of these, , Sol. Answer (4), 16. A loop of irregular shape of conducting wire PQRS (as shown in figure) placed in a uniform magnetic field, perpendicular to the plane of the paper changes into a circular shape. The direction of induced current will be, , (1) Clockwise, , (2), , Anti-clockwise, , (3), , No current, , (4), , None of these, , Sol. Answer (2), Using Lenz's law, inward flux is increasing. So to oppose this change current will be anticlockwise., 17. When a conducting wire XY is moved towards the right, a current flows in the anti-clockwise direction. Direction, of magnetic field at point O is, , X, O, Y, (1) Parallel to motion of wire, , (2), , Along XY, , (3) Perpendicular outside the paper, , (4), , Perpendicular inside the paper, , Sol. Answer (3), Using Lenz's law, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 59 :
Solution of Assignment, , Electromagnetic Induction, , 59, , 18. A copper rod of length l is rotated about one end perpendicular to the uniform magnetic field B with constant, angular velocity . The induced e.m.f. between its two ends is, (1) Bl 2, , (2), , 3, B l 2, 2, , (3), , 1, B l 2, 2, , (4), , 2 Bl 2, , Sol. Answer (3), =, , 1, B l 2, 2, , 19. A coil having number of turns N and area A is rotated in a uniform magnetic field B with angular velocity , about its diameter. Maximum e.m.f. induced in it is given by, , (1) NAB, , (2), , NAB, , , (3), , NA, B, , (4), , B, NA, , Sol. Answer (1),   NBA cos   NBA cos t, –, , d, dt, ,   NBA sin t, ,  max  NBA, , 20. A flat coil of 500 turns, each of area 50 cm2, rotates in a uniform magnetic field of 0.14 Wb/m2 about an axis normal, to the field at an angular speed of 150 rad/s. The coil has a resistance of 5 . The induced e.m.f. is applied to an, external resistance of 10 . The peak current through the resistance is, (1) 1.5 A, , (2), , 2.5 A, , (3), , 3.5 A, , (4), , 4.5 A, , Sol. Answer (3), N = 500, A = 50 cm2, B = 0.14 Wb/m2,  = 150 rad/s,, , R=5, , max = NBA, , l max , ,  max,  3.5 A, R, , 21. Eddy currents are induced when, (1) A metal block is kept in a changing magnetic field, (2) A metal block is kept in a uniform magnetic field, (3) A coil is kept in a uniform magnetic field, (4) Current is passed in a coil, Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 60 :
60, , Electromagnetic Induction, , Solution of Assignment, , 22. A motor starter, (1) Is a variable resistor, , (2), , Offsets the back emf variations, , (3) Helps start a DC motor, , (4), , All of these, , Sol. Answer (4), 23. A simple electric motor has an armature resistance of 1  and runs from a d.c. source of 12 V. It draws a current, of 2 A when unloaded. When a certain load is connected to it, its speed reduces by 10% of its initial value. The, current drawn by the loaded motor is, (1) 3 A, , (2), , 6A, , (3), , 2A, , (4), , 1A, , Sol. Answer (1), 24. Which of the following is possible application of an RC circuit?, (1) Windshield wipers, (2) Flashing red lights on roadway construction sites, (3) Heart pacemakers, (4) All of these, Sol. Answer (4), 25. The current passing through a choke coil of self inductance 5 H is decreasing at the rate of 2 A/s. The e.m.f., developed across the coil is, (1) 10 V, , (2), , – 10 V, , (3), , – 2.5 V, , (4), , 2.5 V, , Sol. Answer (1), , L = 5 H,, , , , di,  2 A/s, dt, , Ldi,  5  2  10 V, dt, , 26. When the number of turns in a solenoid are doubled without any change in the length of the solenoid, its self, inductance becomes, (1) Half, , (2), , Double, , (3), , Four times, , (4), , Eight times, , Sol. Answer (3), L  n2, , So if n' = 2n, L' = 4L, 27. A coil of resistance 20 ohms and inductance 5 H has been connected to a 100 volt battery. The energy stored, in the coil is, (1) 31.25 J, , (2), , 62.5 J, , (3), , 125 J, , (4), , 250 J, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 61 :
Solution of Assignment, , Electromagnetic Induction, , 61, , Sol. Answer (2), l, , 100, 5A, 20, , , , 1 2 1, Ll  5  5 2  62.5 J, 2, 2, , 28. The magnetic energy stored in a long solenoid of area of cross-section A in a small region of length L is, B 2 AL, (1), 2 02, , (2), , AL, 2 0, , (3), , 1,  0 B 2 AL, 2, , B 2 AL, 2 0, , (4), , Sol. Answer (4), Energy, B2, B 2  AL , , ⇒ Energy , Volume 2 0, 2 0, , 29. An inductor is connected to a direct voltage source through a switch. Now, (1) Very large emf is induced in inductor when switch is closed, (2) Larger emf is induced when switch is opened, (3) Large emf is induced whether switch is closed or opened, (4) No emf is induced whether switch is closed or opened, Sol. Answer (2), 30. A long solenoid has self inductance L. If its length is doubled keeping total number of turns constant then its, new self inductance will be, , (1), , L, 2, , (2), , 2L, , (3), , L, , L, 4, , (4), , Sol. Answer (1), L   0 n 2 Al, 2, , ⎛N⎞, L   0 ⎜ ⎟ Al, ⎝ l ⎠, L, ,  0N 2, 1, A⇒L, l, l, , ⇒ L' , , L, 2, , 31. If L and R denote inductance and resistance respectively, then the dimension of L/R is, (1) [M0L0T–1], , (2), , [M0L0T1], , (3), , [M0L0T2], , (4), , [MLT2], , Sol. Answer (2), , , , L, represents time constant hence will have dimension of time., R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 62 :
62, , Electromagnetic Induction, , Solution of Assignment, , 32. A solenoid has 2000 turns wound over a length of 0.3 m. The area of its cross section is 1.2 × 10–3 m2. Around its, central section a coil of 300 turns is wound. If an initial current of 2 A is reversed in 0.25 s, the e.m.f. induced in the, coil is equal to, (1) 6 × 10–4 V, , (2), , 4.8 × 10–2 V, , (3), , 2.4 × 10–2 V, , (4), , 48 kV, , Sol. Answer (2), ,   NBA ⇒  , , d, dB ⎛, i ⎞, ⇒   NA,  ⎜ NA 0 n ⎟, ⎝, dt, dt, t ⎠, , i, 4, , , So   4.8  10 –2 V, t 0.25, , 33. With the decrease of current in the primary coil from 2 A to zero in 0.01s, the e.m.f. generated in the secondary, coil is 1000 V. The mutual inductance of the two coil is, (1) 1.25 H, , (2), , 2.50 H, , (3), , 5.00 H, , (4), , 10.00 H, , Sol. Answer (3),  = Mi, d  Mdi, , dt, dt, 1000 , , M, , M  2, 0.01, , 1, 10  5.00 H, 2, , 34. Two coaxial coils are very close to each other and their mutual inductance is 5 mH. If a current 50 sin 500t, is passed in one of the coils then the peak value of induced e.m.f. in the secondary coil will be, (1) 5000 V, , (2), , 500 V, , (3), , 150 V, , (4), , 125 V, , Sol. Answer (4),  = Mi, d  Mdi, , dt, dt,   5  10 –3  50   500  cos t, , , , max, , = 125 V, , 35. The coefficient of self induction of two inductor coils are 20 mH and 40 mH respectively. If the coils are connected, in series so as to support each other and the resultant inductance is 80 mH then the value of mutual inductance, between the coils will be, (1) 5 mH, , (2), , 10 mH, , (3), , 20 mH, , (4), , 40 mH, , Sol. Answer (2), L1 = 20 mH, L2 = 40 mH, L  L1  L 2  2 M, , 80 = 20 + 40 + 2 M, M = 10 mH, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 63 :
Solution of Assignment, , Electromagnetic Induction, , 63, , SECTION - B, Objective Type Questions, 1., , A conducting rod AB of length l is projected on a frictionless frame PSRQ with velocity v0 at any instant. The, velocity of the rod after time t is, , (1) v = v0, , (2), , v > v0, , (3), , v < v0, , (4), , None of these, , Sol. Answer (3), The rod will experience magnetic force in the direction opposite to initial force. So velocity will decrease with, time., 2., , In the given circuit, bulb will become suddenly bright, if, , L, , Bulb, , V, , S, , (1) Switch is closed or opened, , (2), , Switch is closed, , (3) Switch is opened, , (4), , None of these, , Sol. Answer (3), Sudden decrease in current due to circuit breaking (opening) will be compensated by sudden induced current, flow (as per Lenz's law) and hence sudden brightness., 3., , What will be the direction of current in the coil A as the switch S is closed?, , B, A, , S, V, , (1) Clockwise, , (2), , Anticlockwise, , (3) Anticlockwise and then clockwise, , (4), , Clockwise and then anticlockwise, , Sol. Answer (1), Using Lenz's law, current in A will be clockwise., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 64 :
64, 4., , Electromagnetic Induction, , Solution of Assignment, , The armature of a generator of resistance 1  is rotated at its rated speed and produces 125 V without load and, 115 V with full load. The current in the armature coil is, (1) 240 A, , (2), , 10 A, , (3), , 1A, , (4), , 2A, , Sol. Answer (2), 125  115,  10 A, 1, , l, , 5., , A copper disc of radius 0.1 m is rotated about its centre with 10 rev/s in a uniform magnetic field of 0.1T with its, plane perpendicular to field. The emf induced across the radius of disc is, , volt, 10, , (1), , (2), , , volt, 100, , (3), , , volt, 1000, , (4), , Zero, , Sol. Answer (2), , , , 1, , BR 2 , R  0.1 m, B  0.1 T, W  20, 2, s, , , , 1,  0.1  0.1 2  20 , 2, , , , 6., , , volt, 100, , , A frame CDEF is placed in a region where a magnetic field B is present. A rod of length one metre moves with, constant velocity 20 m/s and strength of magnetic field is one tesla. The power spent in the process is (take R =, 0.2  and all other wires and rod have zero resistance), P, D, B, , R, E, , (1) 1 kW, , (2), , 2 kW, , C, v, , Q, , (3), , F, , 3 kW, , (4), , 4 kW, , Sol. Answer (2), P = F.V , , , , B 2 l 2v 2, R, , 11 2  20  2, 0.2, ,  2 kW, , 7., , An ideal solenoid of cross-sectional area 10–4 m2 has 500 turns per metre. At the centre of this solenoid, another, coil of 100 turns is wrapped closely around it. If the current in the coil changes from 0 to 2 A in 3.14 ms, the emf, developed in the second coil is, (1) 1 mV, , (2), , 2 mV, , (3), , 3 mV, , (4), , 4 mV, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 66 :
66, , Electromagnetic Induction, , Solution of Assignment, , 10. Two coils A and B are wound on the same iron core as shown in figure. The number of turns in the coil A and B are, NA and NB respectively. Identify the correct statement, A, B, , IA, , IB, , (1) Both the coils have same magnitude of magnetic flux, , A NA, (2) The magnetic flux linked are in the ratio   N, B, B, EA ⎛ NA ⎞, ⎟, ⎜, (3) The induced emf across each coil are in the ratio, E B ⎜⎝ N B ⎟⎠, , 2, , (4) Both the coils have same magnitude of induced emf, Sol. Answer (2), A NA, , B NB, , 11. A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle, 2. The earth’s magnetic field component in the direction perpendicular to swing is B. The maximum potential, difference induced across the pendulum is, , , , L, , h, , ⎛⎞, (1) 2BL sin ⎜ ⎟ . gL, ⎝2⎠, , (2), , ⎛⎞, BL sin ⎜ ⎟ (gl ), ⎝2⎠, , (3), , ⎛⎞, BL sin ⎜ ⎟ .(gL )3/2, ⎝2⎠, , (4), , ⎛⎞, BL sin ⎜ ⎟ .(gL)2, ⎝2⎠, , Sol. Answer (2), Using conservation of energy, mgl 1– cos  , , 1 2, I, 2, , mgl 1– cos   , , 1 2 2, mI , 2, , ⎛, ⎛ ⎞ ⎞, 2gl ⎜ 2sin 2 ⎜ ⎟ ⎟  l 2  2, ⎝ 2⎠ ⎠, ⎝, 2, , , g, , sin, l, 2, , 1, 1 ⎛ g, ⎞ 2, B l 2   ⎜ 2, sin ⎟ l  B, 2, 2 ⎝, l, 2⎠, ,   Bl sin, , , gl, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 67 :
Solution of Assignment, , Electromagnetic Induction, , 67, , 12. The switch shown in the circuit is closed at t = 0. The current drawn from the battery by the circuit at t = 0, and t =  are in the ratio, , L, , E, R, (1) 2 : 1, , (2), , C, , R, , S, , 1:2, , R, (3), , 1:1, , (4), , 1:4, , Sol. Answer (3), At t = 0, L offers infinite resistance, At t , C offers infinite resistance, So, in both the cases, l , , , 2R, , Hence ratio is 1 : 1., 13. A small square loop of wire of side l is placed inside a large square loop of wire of side L (>>l). The loops, are coplanar and their centres coincide. The mutual inductance of the system is, , l, , (1), , 2 2 0 L, l, , (2), , 2 2 0 L2, l, , L, , (3), , 2 2 0 l, L, , (4), , 2 2 0 l 2, L, , Sol. Answer (4), , M, , , i, ,  due to outer loop in the smaller loop will be due to magnetic field Vector sum of all the four sides of bigger, square., So M , , 2 2 0 l 2 i, L  i , , , 14. A uniform magnetic field exists in the region given by B  3iˆ  4 jˆ  5kˆ . A rod of length 5 m placed along yaxis is moved along x-axis with constant speed 1 ms–1. Then induced e.m.f. in the rod is, (1) Zero, , (2), , 25 V, , (3), , 5V, , (4), , 10 V, , Sol. Answer (2),   ,   v  B  .l, ,  ⎡⎣1iˆ   3iˆ  4 ˆj  5kˆ  ⎤⎦ .5 ˆj,  ⎡⎣ 4kˆ – 5 ˆj ⎤⎦ .5 ˆj, , = 25 V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 68 :
68, , Electromagnetic Induction, , Solution of Assignment, , 15. A copper rod AB of length l, pivoted at one end A, rotates at constant angular velocity , at right angles to, a uniform magnetic field of induction B. The emf, developed between the mid point C of the rod and end B is, , , , , A, , (2), , 3, Bl 2, 4, , , , , , B l 2, 8, , (1), , B, , C, , (3), , Bl 2, 4, , (4), , 3, B l 2, 8, , Sol. Answer (4), , , , l, , ∫ Bxdx, l, , 2, , ⎛ x2⎞,  B ⎜ ⎟, ⎝ 2⎠, , l, l, , 2, , , , , , B 2 l 2, l –, 4, 2, , , , 3, B l 2, 8, , , , 16. Radius of a circular loop placed in a perpendicular uniform magnetic field is increasing at a constant rate of, r0 ms–1. If at any instant radius of the loop is r, then emf induced in the loop at that instant will be, (2), , (1) –2Brr0, , –2Br, , (3), , –Br0r, , (4), , –2Br0r, , Sol. Answer (4), , dr,  r 0 ⇒   Br 2, dt, , , –d , dr,  – B  2r , dt, dt, ,   –2Brr 0, , 17., , 1, , 2, , 3, , The figure shows three circuits with identical batteries, inductors and resistances. Rank the circuits according, to the currents through the battery just after the switch is closed, greatest first, (1) I2 > I3 > I1, , (2), , I2 > I1 > I3, , (3), , I1 > I2 > I3, , (4), , I1 > I3 > I2, , Sol. Answer (1), l 1  0, l 2 , , , , , l3 , R, 2R, ,  l2 > l3 > l1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 69 :
Solution of Assignment, , Electromagnetic Induction, , 69, , 18. In an inductor, the current I varies with time t as I = 5A + 16 (A/s) t. If induced emf in the inductor is 5 mV,, the self inductance of the inductor is, (1) 3.75 × 10–3 H, , (2), , 3.75 × 10–4 H, , (3), , 3.125 × 10–3 H, , (4), , 3.125 × 10–4 H, , Sol. Answer (4), , l  5  16t ⇒, , di,  16, dt, ,   5  10 –3 V,  , , Ldi,  L 16  5  10 –3, dt, , So, L  3.125  10 –4 H, 19. A magnet is moved in the direction indicated by an arrow between two coils AB and CD as shown in figure. The, direction of induced current in the straight wire is, , A, , C, , B, N, , (1) A to B and C to D, , (2), , D, , S, , B to A and C to D, , (3), , A to B and D to C, , (4), , B to A and D to C, , Sol. Answer (3), Using Lenz's law, 20. Two coils of self inductance L1 and L2 are placed near each other so that the total flux in one coil is completely, linked with the other. Their mutual inductance (M) will be given by, (1) M = L1 L2, , (2), , M  L1 L2, , (3), , M  L1 L2, , (4), , M  L1 L2, , Sol. Answer (2), M  L1L2, 21. The network shown in figure is a part of a complete circuit. If at a certain instant, the current i is 4 A and is, increasing at a rate of 103 A/s. Then VB – VA will be, A, , (1) –11 V, , (2), , 11 V, , i, 1, , 12 V, , 5 mH, , (3), , –21 V, , B, , (4), , 21 V, , Sol. Answer (3), di,  10 3, dt, , So, V A –  4  1 – 12 – 5  10 –3 10 3   V B,  VB – V A  –21 V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 70 :
70, , Electromagnetic Induction, , Solution of Assignment, , 22. The magnetic flux through a stationary loop with resistance R varies during interval of time T as,  = at (T – t). The heat generated during this time neglecting the inductance of loop will be, , a 2T 3, 3R, , (1), , (2), , a 2T 2, 3R, , (3), , a 2T, 3R, , (4), , a 3T 2, 3R, , Sol. Answer (1),   at T – t   aTt – at 2, , d,  –  aT – 2at , dt, , –, , P–, ,  2  aT – 2at  2, , R, R, , T, , T, , 0, , 0, , H  ∫ Pdt  ∫, , , ,  aT – 2at  2, R, , a 2T 3, 3R, , 23. 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 electric power, dissipated would be, (1) Halved, , (2), , The same, , (3), , Doubled, , (4), , Quadrupled, , Sol. Answer (2), ,   nBr 2, , dB, dt, 2, , ⎛ r ⎞ dB,  '  4nB ⎜ ⎟, ⎝ 2 ⎠ dt, ,  '  nBr 2, , P' , ,   ' 2, R, , dB, , dt, , , , 2, R, , 24. A small square loop of wire of side l is placed inside a large circular loops of radius r. The loop are coplanar, and their centre coincide. The mutual inductance of the system is proportional to, (1), , l2, r, , (2), , l2, r2, , (3), , r, l2, , (4), , r2, l, , Sol. Answer (1), Using   Mi, M, ,  ⎛  0i ⎞ l 2 l 2, ⎜, ⎟. , i ⎝ 2r ⎠ i, r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 71 :
Solution of Assignment, , Electromagnetic Induction, , 71, , SECTION - C, Previous Years Questions, 1., , An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron., What will be the direction of current, if any, induced in the coil?, [Re-AIPMT-2015], a, , b, , d, c, electron, , X, , Y, , (1) No current induced, (2) abcd, (3) adcb, (4) The current will reverse its direction as the electron goes past the coil, Sol. Answer (4), When electron goes in straight line electric flux first increases then decreases., 2., , A conducting square frame of side a and a long straight wire carrying current I are located in the same plane, as shown in the figure. The frame moves to the right with a constant velocity V. The emf induced in the frame, will be proportional to, [AIPMT-2015], , x, I, V, a, , (1), , 1, (2 x  a )(2 x  a ), , (2), , 1, x2, , 1, (2 x  a )2, , (3), , (4), , 1, (2 x  a )2, , Sol. Answer (1), eABCD = eAB + eCD, , , x, ,  0I,  0I, av , av, a⎞, a⎞, ⎛, ⎛, 2 ⎜ x  ⎟, 2 ⎜ x  ⎟, 2⎠, 2⎠, ⎝, ⎝, , , , 0Iav ⎡ 2, 2 ⎤, , 2 ⎢⎣ 2 x  a 2 x  a ⎥⎦, , , , 0Iav, , , , , 20Ia 2v, (2 x  a )( x  a ), , ⎡ 2x  a  2x  a ⎤, ⎢ (2 x  a )(2 x  a ) ⎥, ⎣, ⎦, , x–, , a, 2, , A, +, , D, +, V, , B– a, 2, x+, , a, 2, , V, a, 2, , –, C, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 72 :
72, 3., , Electromagnetic Induction, , Solution of Assignment, , A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic, field B, as shown in figure. The potential difference developed across the ring when its speed is v is, , ×, , ×, , ×, , ×, , ×, , ×, , P, , ×, Q, r, , ×, ×, , [AIPMT-2014], , ×, B, , R, , ×, ×, , (1) Zero, , (2), , Bvr2/2 and P is at higher potential, , (3) rBv and R is at higher potential, , (4), , 2rBv and R is at higher potential, , Sol. Answer (4),  = BLeffv (Leff = Diameter), = B 2Rv, 4., , A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decrease, when, [NEET-2013], (1) Number of turns in the coil is reduced, (2) A capacitance of reactance XC = XL is included in the same circuit, (3) An iron rod is inserted in the coil, (4) Frequency of the AC source is decreased, , Sol. Answer (3), 5., , A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is, [NEET-2013], (1) Twice per revolution, , (2), , Four times per revolution, , (3) Six times per revolution, , (4), , Once per revolution, , Sol. Answer (1),   BA cos t, , d,  BA sin t, dt, , , 6., , A coil of resistance 400  is placed in a magnetic field. if the magnetic flux  (Wb) linked with the coil varies, with times t (s) as  = 50t2 + 4. The current in the coil at t = 2 s is, [AIPMT (Prelims)-2012], (1) 2 A, , (2), , 1A, , (3), , 0.5 A, , (4), , 0.1 A, , Sol. Answer (3),   50t 2  4, ,  , , d,  100t, dt, , , ,  100  2  200 V, , t 2, , l, ,  200, ,  0.5 A, R 400, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 73 :
Solution of Assignment, , 7., , Electromagnetic Induction, , 73, , The current (I) in the inductance is varying with time according to the plot shown in figure., , I, T/2, t, , T, , Which one of the following is the correct variation of voltage with time in the coil?, v, , v, , v, , (1), , T, , T/2, , (2), , v, t, , t, , (3), , T/2, , t T, , T/2, , [AIPMT (Prelims)-2012], , T, , (4), T/2, , t, , T, , Sol. Answer (2), Use  , for 0 to, , Ldi, dt, , T, ,, 2, , and from, 8., , di, is constant and positive, dt, , di, T, to T,, is constant and negative., dt, 2, , In a coil of resistance of 10 , the induced current developed by changing magnetic flux through it,is shown in, figure as a function of time. The magnitude of change in flux through the coil in Weber is [AIPMT (Mains)-2012], , i (A), 4, , 0, (1) 8, , (2), , 2, , 0.1, (3), , t(s), 6, , (4), , 4, , Sol. Answer (2), , ⎛  ⎞, l⎜, ⎝ t .R ⎟⎠,   l tR, or d    ldt  R, Total flux   R ∫ ldt, , ∫ idt, , represents area under i – t curve, , ⎡1, ⎤, So   R ⎢  0.1  4 ⎥  2, ⎣2, ⎦, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 74 :
74, 9., , Electromagnetic Induction, , Solution of Assignment, , The current i in a coil varies with time as shown in the figure. The variation of induced emf with time would be, [AIPMT (Prelims)-2011], i, , 0, , T/4, , T/2, , 3T/4, , emf, , (1), , 0, , emf, T/2 3T/4 T, , t, , T/4, , (2), , 0, , 0, , T/4, , t, , T/2 3T/4 T, , emf, , emf, , (3), , t, , T, , T/4 T/2 3T/4 T, , t, , (4), , 0, , T/4 T/2 3T/4 T, , t, , Sol. Answer (2), , , –Ldi, dt, , di, is slope of l – t curve., dt, , 10. A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to, the loop. The radius of the loop is made to shrink at a constant rate of 1 mms–1. The induced emf when the, radius is 2 cm is, [AIPMT (Prelims)-2010], (1) 2V, , (2), ,  V, , (3), , , V, 2, , (4), , 2V, , Sol. Answer (2), 11. A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the, magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius, is 2 cm is, [AIPMT (Prelims)-2009], (1) 4.8 V, , (2), , 0.8 V, , (3), , 1.6 V, , (4), , 3.2 V, , Sol. Answer (4), , 12. A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction, , , axis makes an angle of 60° with B . The magnetic flux linked with the disc is, (1) 0.01 Wb, , (2), , 0.02 Wb, , (3), , 0.06 Wb, , 1 ⎛ Wb ⎞, in such a way that its,  ⎜⎝ m2 ⎟⎠, [AIPMT (Prelims)-2008], (4), , 0.08 Wb, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 75 :
Solution of Assignment, , Electromagnetic Induction, , 75, , 13. A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with, [AIPMT (Prelims)-2008], each turn of the solenoid is 4 × 10–3 Wb. The self-inductance of the solenoid is, (1) 4.0 H, , (2), , 2.5 H, , (3), , 2.0 H, , (4), , 1.0 H, , Sol. Answer (4), N = 500, l = 2, L, , N  500  4  10 –3 , , i, 2, , 14. What is the value of inductance L for which the current is a maximum in a series LCR circuit with C = 10 F, [AIPMT (Prelims)-2007], and  = 1000 s–1?, (1) 10 mH, , (2), , 100 mH, , (3) 1 mH, , (4), , Cannot be calculated unless R is known, , Sol. Answer (2), 15. Two coils of self-inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is, completely linked with the other. The mutual inductance between these coils is, [AIPMT (Prelims)-2006], (1) 10 mH, , (2), , 6 mH, , (3), , 4 mH, , (4), , 16 mH, , Sol. Answer (3), L1 = 2mH, L2 = 8 mH, M  L1L 2  16  4 mH, 16. A metal ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis, of the ring. The acceleration of the falling magnet is, (1) More than g, , (2), , Equal to g, , (3), , Less than g, , (4), , Either (1) or (3), , Sol. Answer (3), , S, N, , due to Lenz's law a < g, 17. The magnetic flux through a circuit of resistance R changes by an amount  in a time t. Then the total, quantity of electric charge Q that passes any point in the circuit during the time t is represented by, (1) Q , , 1 , ., R t, , (2), , Q, , , R, , (3), , Q, , , t, , (4), , Q  R., , , t, , Sol. Answer (2), l, , , Q, , ⇒, , tR, t, tR, , ⇒ Q , , , R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 76 :
76, , Electromagnetic Induction, , Solution of Assignment, , 18. As a result of change in the magnetic flux linked to the closed loop as shown in the figure, an e.m.f V volt, is induced in the loop. The work done (in joule) in taking a charge Q coulomb once along the loop is, , (1) QV, , (2), , 2QV, , (3), , QV, 2, , (4), , Zero, , Sol. Answer (1), W = QV, (By the defination of e.m.f), 19. A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform, , magnetic field with a constant velocity V  viˆ. The magnetic field is directed along the negative z-axis direction., The induced emf, during the passage of these loops, out of the field region, will not remain constant for, (1) Any of the four loops, , (2), , The rectangular, circular and elliptical loops, , (3) The circular and the elliptical loops, , (4), , Only the elliptical loop, , Sol. Answer (3), For circular and elliptical loop, area coming out from the field per unit time is not constant, i.e.,, , dA,  constant, dt, , 20. A conductor of 3 m length is moving perpendicular to its length as well as a magnetic field of 10–3 T with a speed, of 102 m/s, then the force required to move it with this constant speed is, (1) 0. 3 N, , (2), , 0.9 N, , (3), , 0, , (4), , 3 × 10–3 N, , Sol. Answer (3), Motional emf will induce ( = Bvl). But loop is not closed so no current will flow, hence no magnetic field force, will act upon it., 21. In a circular conducting coil, when current increases from 2 A to 18 A in 0.05 s, the induced emf is 20 V. The self, inductance of the coil is, (1) 62.5 mH, , (2), , 6.25 mH, , (3), , 50 mH, , (4), , 0, , Sol. Answer (1), , , , Ldi L 18 – 2, ,  20, 0.05, dt, , So L = 62.5 10–3 H, 22. In the circuit given in figure, 1 and 2 are ammeters. Just after key K is pressed to complete the circuit, the, reading will be, R1, , 1, L, , R2, , 2, , K, , (1) Zero in 1, maximum in 2, , (2), , Maximum in both 1 and 2, , (3) Zero in both 1 and 2, , (4), , Maximum in 1, zero in 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 77 :
Solution of Assignment, , Electromagnetic Induction, , 77, , Sol. Answer (4), At t = 0, C offers zero resistance, L offers infinite resistance, So reading of Ammeter (1)  max, Ammeter (2)  Zero, 23. For a coil having L = 2 mH, current flowing through it is I = t2 e–t then the time at which emf become zero, (1) 2 s, , (2), , 1s, , (3), , 4s, , (4), , 3s, , Sol. Answer (1), , , Ldi, dt, , When, , di, becomes zero,  will becom zero., dt, , di,  – t 2  e – t   e – t  2t   0, dt, t=2, 24. When the number of turns and the length of a solenoid both are doubled, its self inductance becomes, (1) Four times, , (2), , Doubled, , (3), , Halved, , (4), , Unchanged, , Sol. Answer (2), L   0 n 2 Al, , L  0, ,  0N 2 A, N2, Al, , l2, l, , So, on doubling N and l, L' , ,  0  2N  2 A 2 0N 2 A, , 2l, l, ,  L' = 2L, 25. The time constant of L-R circuit is doubled if, (1) Both L and R become two times, (2) L becomes four times and R becomes two times, (3) L becomes two times and R becomes four times, (4) L becomes two times and R becomes eight times, Sol. Answer (2), , , L, R, ,  '  2 , , 4L, 2R, , 26. Two neighbouring coils A and B have a mutual inductance of 20 mH. The current flowing through A is given by, i, = 3t2 – 4t + 6. The induced emf at t = 2 s is, (1) 160 mV, , (2), , 200 mV, , (3), , 260 mV, , (4), , 300 mV, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 78 :
78, , Electromagnetic Induction, , Solution of Assignment, , Sol. Answer (1), M = 20 mH, , l  3t 2 – 4t  6, di,  6t – 4, dt, , , , Mdi, dt, , t 2, ,  20  10 –3  6  2 – 4 = 160 mV, , 27. The self inductance L of a solenoid depends on the number of turns per unit length ‘n’ as, (1) L  n, , (2), , L  n2, , (3), , L  n–1, , (4), , L  n–2, , Sol. Answer (2), 28. Two coils have a mutual inductance 0.005 H. The current changes in the first coil according, to equation / = /0 sin t, where /0 = 10 A and  = 100  radian per second. The maximum value of e.m.f. (in, volt) in the second coil is, (1) , , (2), , 5, , (3), , 2, , (4), , 4, , Sol. Answer (2), M = 0.005 H, l  I 0 sin t  10 sin t,   1 rad/s, , , , Mdi,   0.005  10 100  cos t    0.005 1000  = 5, dt, , 29. A transformer has 500 primary turns and 10 secondary turns. If the secondary has resistive load of 15 , the, currents in the primary and secondary respectively, are, (1) 0.16 A, 3.2 × 10–3 A, , (2), , 3.2 × 10–3 A, 0.16 A, , (3) 0.16 A, 0.16 A, , (4), , 3.2 × 10–3 A, 3.2 × 10–3 A, , Sol. Answer (2), Use, , i P nP, , i S nS, , 30. A magnet is made to oscillate with a particular frequency through a coil as shown in figure. The time variation, of magnitude of emf generated across the coil during one cycle is, , S, N, , V, , e, , e, (1), , t, , (2), , e, , e, t, , (3), , t, , (4), , t, , Sol. Answer (3), emf induced will be zero when magnet will be inside coil. Use Lenz's law., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 79 :
Solution of Assignment, , Electromagnetic Induction, , 79, , 31. Two coils have self inductance L1 = 4 mH and L2 = 1 mH respectively. The currents in the two coils are, increased at the same rate. At a certain instant of time both coils are given the same power. If I1 and I2 are, I1, the currents in the two coils, at that instant of time respectively, then the value of I is, 2, (1), , 1, 8, , (2), , 1, 4, , (3), , 1, 2, , (4), , 1, , Sol. Answer (2), L1 = 4 mH,, , L2 = 1 mH, , di 1 di 2, , dt, dt, , l1  2, , l 2 1, , 1 l 2 ⎞, ⎛, ⎜⎝∵ P  i ⇒   l ⎟⎠, 2, 1, , di 2, l 1 L 2 dt, l, L, 1, , ⇒ 1  2 , di, l2, l 2 L1 4, L1 1, dt, , SECTION - D, Assertion-Reason Type Questions, 1., , A : Total induced emf in a loop is not confined to any particular point but it is distributed around the loop in, direct proportion to the resistances of its parts., R : In general when there is no change in magnetic flux, no induced emf is produced., , Sol. Answer (2), 2., , A : The induced current flows so as to oppose the cause producing it., R : Lenz's law is based on energy conservation., , Sol. Answer (1), 3., , A : Faraday's law is an experimental law., R : Time varying magnetic field cannot generate induced emf., , Sol. Answer (3), 4., , A : Unlike electrostatic field the lines of induced field form closed loop., R : Electrostatic field is conservative unlike induced fields., , Sol. Answer (1), 5., , A : The mutual induction between the two coils infinitely apart is zero., R : If the mutual induction between the two coils is zero, it means that their self inductances are also zero., , Sol. Answer (3), 6., , A : An inductor is called the inertia of an electric circuit., R : An inductor tends to keep the flux constant., , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 80 :
80, 7., , Electromagnetic Induction, , Solution of Assignment, , A : At any instant, if the current through an inductor is zero, then the induced emf will also be zero., R : In one time constant, the current flows to 37 percent of its maximum value in a series LR circuit., , Sol. Answer (4), 8., , A : There may be an induced emf in a loop without induced current., R : Induced current depends on the resistance of the loop as well., , Sol. Answer (2), 9., , A : When the magnetic flux through a loop is maximum, induced emf is maximum., R : When the magnetic flux through a loop is minimum, induced emf is minimum., , Sol. Answer (4), 10. A : When a conducting loop is kept stationary in a non-uniform magnetic field an emf is induced., R : As per Faraday's law, whenever flux changes, an emf is induced., Sol. Answer (4), 11. A : When an electric motor is started, a variable resistance (that decreases with time) is used in series. This, resistance is known as motor starter., R : The back-emf in the beginning, when motor starts, is very small., Sol. Answer (1), 12. A : When a bar magnet is dropped into a vertical long hollow metallic tube, the magnet ultimately moves with zero, acceleration., R : The magnet falling into metallic tube causes the eddy currents in the metal tube, so the motion of the magnet, is damped., Sol. Answer (1), 13. A : The power output of a practical transformer is always smaller than the power input., R : A transformer works on the principle of mutual induction., Sol. Answer (2), 14. A : Electrical power through transmission lines is transmitted at high voltage., R : At high voltage theft of power is checked., Sol. Answer (3), 15. A : The electric field induced due to changing magnetic field is non-conservative., R : The line integral of the electric field induced due to changing magnetic field along a closed loop is always zero., Sol. Answer (3), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 81 :
https://t.me/NEET_StudyMaterial, , Chapter, , 22, , Alternating Current, Solutions, SECTION - A, Objective Type Questions, 1., , Hot wire ammeters are used for measuring, (1) Both AC and DC., , (2), , Only AC, , (3), , Only DC, , (4), , Neither AC nor DC, , (3), , Mean value, , (4), , Mean square value, , Sol. Answer (1), 2., , In alternating current circuits, the a.c. meters measure, (1) r.m.s. value, , (2), , Peak value, , Sol. Answer (1), 3., , The mean value of current for half cycle for a current variation shown by the graph is, , i, , O, , (1), , i0, 2, , (2), , i0, , i0, , T, 2, , 3T, 2, , T, , (3), , t, , i0, , (4), , 3, , i0, 3, , Sol. Answer (1), T, , I mean , , 2, , ∫ Idt, 0, , T, , From 0 to, , So I mean, , , , 2, , i0, 2i, T, .t  0 t, graph is straight line so the function (I) will be =, T, T, 2, 2, , 2, , T, ,  , , T, , 2, , ∫, 0, , T, , 2 ⎛ 2⎞ ⎛ t 2 ⎞ 2, 2i 0, tdt  ⎜⎝ ⎟⎠ i 0 ⎜ ⎟, ⎝ 2⎠0, T T, T, , 4 ⎛ 1⎞ ⎛ T 2, ⎞, i, i ⎜ ⎟⎜, – 0⎟  0, ⎠, T 2 0 ⎝ 2⎠ ⎝ 4, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 82 :
82, 4., , Alternating Current, , Solution of Assignment, , A 110 V d.c. heater is used on an a.c. source, such that the heat produced is same as it produces when, connected to 110 V dc in same time-intervals. What would be the r.m.s. value of the alternating voltage?, (1) 110 V, , (2), , 220 V, , (3), , 330 V, , (4), , 440 V, , Sol. Answer (1), Given that HAC = HDC, l 2 rmsRt  l 2Rt, , V 2 rms V 2, , R, R, , Vrms = V, so Vrms = 110 V, 5., , The peak value of an alternating e.m.f. E = E0 sin t is 10 volt and its frequency is 50 Hz. At a time t , the instantaneous value of the e.m.f. is, (1) 1 volt, , (2), , 5 3 volt, , (3), , 5 volt, , (4), , 1, ,, 600 s, , 10 volt, , Sol. Answer (3), E = E0 sint,,  = 50 Hz, t , , E0 = 10 V, 1, 600, , E  10sin  2  50 , ,  10 sin, 6., , 1, 600, , , ⎛ 1⎞,  10 ⎜ ⎟  5 volt, ⎝ 2⎠, 6, , The time required for a 50 Hz sinusoidal alternating current to change its value from zero to the r.m.s. value, (1) 1.5 × 10–2 s, , (2), , 2.5 × 10–3 s, , (3), , 10–1 s, , (4), , 10–6 s, , Sol. Answer (2), I  I 0 sin t, , I0, , I rms , , So, , sin, , 2, , I0, 2, ,  I 0 sin t, , ,  sin  2   50 t, 4, , ,  2  50 t, 4, , t, , 1, s, 400, , t  0.25  10 –2 ⇒ 2.5  10 –3 s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 83 :
Solution of Assignment, , 7., , Alternating Current, , 83, , There is no resistance in the inductive circuit. Kirchhoff 's voltage law for the circuit is, , V = V0 sin  t, , (1) V  L, , di, 0, dt, , (2), , V L, , di, dt, , L, , (3), , V  L2, , di, 0, dt, , (4), , None of these, , Sol. Answer (2), , +, +, , V = V0sin t, , –, , –, Using Kirchhoff's voltage law, –Ldi, V  0, dt, , ⇒V , 8., , Ldi, dt, , A sinusoidal supply of frequency 10 Hz and r.m.s. voltage 12 V is connected to a 2.1 F capacitor. What is, r.m.s. value of current?, (1) 5.5 mA, , (2), , 20 mA, , (3), , 26 mA, , (4), , 1.6 mA, , Sol. Answer (4), f = 10 Hz, Vrms = 12 V, C = 2.1 F, I rm s , , V rm s, XC ,, , XC , , 1, C, ,  = 2f, Putting all values, I rms  1.6 mA, , 9., , An inductive circuit contains a resistance of 10 ohms and an inductance of 2 henry. If an alternating voltage of 120, V and frequency 60 Hz is applied to this circuit, the current in the circuit would be nearly, (1) 0.32 A, , (2), , 0.80 A, , (3), , 0.48 A, , (4), , 0.16 A, , Sol. Answer (4), R = 10 , L = 10 H, Vrms = 120 V, f = 60 Hz, I rms , , Vrms, Z, , Z  R 2   2L2  10  2   2  60  2 2, ⇒ Z  753.6 , , I rms , , 120,  0.16 A, 753.6, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 84 :
84, , Alternating Current, , Solution of Assignment, , 10. A coil and a bulb are connected in series with a 12 volt direct current source. A soft iron core is now inserted, in the coil. Then, (1) The intensity of the bulb remains the same, , (2), , (3) The intensity of the bulb increases, , (4), , The intensity of the bulb decreases, Nothing can be said, , Sol. Answer (1), Intensity of bulb remains the same because source is DC, so steady state current will be independent of the, inductance of the inductor for DC circuit,, i steady , , E source, R bulb, , 11. When 100 volt d.c. is applied across a solenoid, a current of 1.0 A flows in it. When 100 volt a.c. is applied, across the same coil, the current drops to 0.5 A. If the frequency of a.c. source is 50 Hz the impedance and, inductance of the solenoid is, (1) 200 ohm and 0.55 henry, , (2), , 100 ohm and 0.86 henry, , (3) 200 ohm and 1.0 henry, , (4), , 100 ohm and 0.93 henry, , Sol. Answer (1), When there is direct current,  = 0, XL = 0, so only R is there, So, V = IR, 100 = 1(R)  R = 100 , When, f  50 Hz,, , Z, , X L2  R 2, , X L2  100 2, 100   0.5 2, 40000  X L2  10000, Impedance z  200  X L2  30,000, , V  IZ, , 200 , , X L  100 3 , , Also XL = L, 100 3  2fL, , L, , 100 3, 1.73, ,  0.55 H, 2  3.14  50 3.14, , 12. In an LCR series circuit R = 10 , XL = 8  and XC = 6  The total impedance of the circuit is, (1) 10.2 , , (2), , 17.2 , , (3), , 10 , , (4), , None of these, , Sol. Answer (1), R = 10 , XL = 8  and XC = 6 , Z  R 2   XL – XC , , 2, ,  10  2  2 2, ,  104  10.2 , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 85 :
Solution of Assignment, , Alternating Current, , 85, , 13. In a series RLC circuit, potential differences across R, L and C are 30 V, 60 V and 100 V respectively as shown in, figure. The e.m.f. of source (in volts) is, , (1) 190, , (2), , 30 V, , 60 V, , 100 V, , R, , L, , C, , 70, , (3), , 50, , (4), , 40, , Sol. Answer (3), emf  Vrms  VR2  VL – VC , , 2, ,   30  2  100 – 60  2,   30  2   40  2  50 V, , 14. In a series RLC circuit, the r.m.s. voltage across the resistor and the inductor are respectively, 400 V and 700 V. If the equation for the applied voltage is   500 2 sin t , then the peak voltage across, the capacitor is, , R, , L, , C, ,  = 500 2sint, (1) 1200 V, , (2), , 1200 2 V, , (3), , 400 V, , (4), , 400 2 V, , Sol. Answer (4),   500 2 sin t, VR = 400 V, VL = 700 V,  rms , , 0, 2, , , , 500 2, = 500 V, 2, ,  rms  VR2  VL – VC , , 2, ,  500 2   400 2  VL – VC , , 2, , 250000 – 160000  V L – VC , 90000  V L – VC , , 2, , 2, , VL – VC  300, VC  700  300, , VC  400 V, , V0  Vrms 2  400 2 V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 86 :
86, , Alternating Current, , Solution of Assignment, , 15. In the following circuit the emf of source is E0 = 200 volt, R = 20 , L = 0.1 henry, C = 10.6 farad and, frequency is variable then the current at frequency f = 0 and f =  is, C, R, L, , (1) Zero, 10 A, , (2), , 10 A, zero, , (3), , 10 A, 10 A, , (4), , Zero, zero, , Sol. Answer (4), E0 = 200 volt, R = 20 , L = 0.1 henry,C = 10.6 F, When f = 0, XL = L = 2fL, XL = 0, When f =  XL = , XC = 0, In both case at least one component has value infinite. So in both cases current will be zero as they are, connected in series., 16. In series LCR circuit, the phase difference between voltage across L and voltage across C is, (1) Zero, , (2), , , , (3), , , 2, , (4), , 2, , Sol. Answer (2), IN LCR circuit  angle between VL and VC = 180° (rad)., 17. With increase in frequency of an a.c. supply, the impedance of an LCR series circuit, (1) Remains constant, (2) Decreases, (3) Increases, (4) Decreases at first, becomes minimum and then increases, Sol. Answer (4), , X L  2fL,, , XC , , 1, 2fC, , When f is increased, XL will increase, , XC will decrease, So XL – XC will increase, hence z will increase, 18. Mark the incorrect statement, (1) In any AC circuit, the applied instantaneous voltage equals to the algebraic sum of the instantaneous, voltages across the series elements of the circuit., (2) For circuits used for transporting electric power, a low power factor implies large power loss in transmission, (3) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit, (4) The use of a capacitor is avoided in the circuit of an induction coil, Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 87 :
Solution of Assignment, , Alternating Current, , 87, , 19. The reading of ammeter in the circuit is, , A, XC = 2 , , V, XL = 2 , , (1) 2 A, , (2), , 3A, , 110 V, R = 55 , (3), , Zero, , (4), , 1A, , Sol. Answer (1), , A, , XC = 2 , , V, , 110 V, , XL = 2 , , R = 55 , , Z  R 2   XL – XC , , 2, ,   55  2  55 , , V=IZ, , 110, I ⇒I 2 A, 55, 20. In an LCR circuit, the resonating frequency is 500 kHz. If the value of L is doubled and value of C is decreased, 1, to, times of its initial values, then the new resonating frequency in kHz will be, 8, (1) 250, , (2), , 500, , (3), , 1000, , (4), , 2000, , Sol. Answer (3), f1 = 500  103 Hz, L '  2L, C ' , , f , , 1, 2 LC, , f1, , f2, , L2C 2, L1C1, , 1, C, 8, , ⇒f , , , , 1, CL, 1, ⎝8 ⎠, LC, ,  2L  ⎛⎜ C ⎞⎟, , f1 1,  ⇒ f 2  1000 kHz, f2 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 88 :
88, , Alternating Current, , Solution of Assignment, , 21. In an LCR circuit L = 8.0 henry, C = 0.5 F and R = 100 ohm are in series. The resonance angular frequency, is, (1) 500 rad/s, , (2), , 600 rad/s, , (3), , 800 rad/s, , (4), , 1000 rad/s, , Sol. Answer (1), L = 8.0 henry, C = 0.5 F and R = 100  in series, , , , 1, , LC, , 1, 8   0.5  10 –6, 1, , , , 4  10 –6, , , , 10 3, 2 = 500 rad/s, , 22. In series LCR circuit voltage leads the current when (Given that 0 = resonant angular frequency), (1)  < 0, , (2), ,  = 0, , (3), ,  > 0, , (4), , None of these, , (4), , One, , (4), , Not defined, , Sol. Answer (3), X L  L,, , XC , , 1, C, , If voltage leads  X L  X C , for this condition,  > 0, So it will be an inductive circuit with voltage leading current by 90°., 23. Power factor of an ideal choke coil (i.e., R = 0) is, (1) Near about zero, , (2), , Zero, , (3), , Near about one, , Sol. Answer (2), cos  , , R, 0, Z, , 24. At resonance, the value of the power factor in an LCR series circuit is, (1) Zero, , (2), , 1, , (3), , 1, 2, , Sol. Answer (2), At resonance, Z = R, , cos  , , R,  1 = Power factor., Z, , 25. In an a.c. circuit, the instantaneous values of e.m.f. and current are E = 200 sin 314 t (volt) and, i = sin (314 t + /3) A. The average power consumed in watts is, (1) 100, , (2), , 200, , (3), , 50, , (4), , 25, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 89 :
Solution of Assignment, , Alternating Current, , 89, , Sol. Answer (3), P  E rmsI rms cos , , E, , I ⎛ 1⎞, ⎜ ⎟, 2 2 ⎝ 2⎠, , , ,  200  1 , , , , , ⎛, ⎞, ⎜⎝∵ cos   here⎟⎠, 3, , 1 1, , 2 2, , https://t.me/NEET_StudyMaterial, , 100, W, 2, , = 50 W, 26. An a.c. of frequency f is flowing in a circuit containing only an ideal choke coil of inductance L. If V0 and i0, represent peak values of the voltage and the current respectively, the average power given by the source to, the choke coil is equal to, , (1), , 1, iV, 2 0 0, , (2), , 1 2, i (2fL), 2 0, , (3), , Zero, , (4), , 1, V (2fL), 2 0, , Sol. Answer (3), For an ideal choke coil cos = 0, so, P = ErmsIrmscos, P=0, 27. When a voltage V = V0 cos t is applied across a resistor of resistance R, the average power dissipated per, cycle in the resistor is given by, (1), , V0, , (2), , 2R, , V0, , (3), , 2R, , V02, 2R, , (4), , V02, 2R, , Sol. Answer (3), P = ErmsIrmscos, cos =, , R, Z, , cos = 1 for resistor, , Vrms , , P, , P, , V0, 2, , V0, , V0, , ⎛, , V, , V, , ⎞, , 1 ⎜∵ I rms  rms  0 ⎟, ⎝, R, R 2⎠, 2 R 2, ,, , V02, 2R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 90 :
90, , Alternating Current, , Solution of Assignment, , 28. In a series L-C circuit, if L = 10–3 H and C = 3×10–7 F is connected to a 100 V-50 Hz a.c. source, the, impedance of the circuit is, (1), , 10 5,  10 , 3, , (2), , 0.1 – 3 × 10–5, , 10 5, , , 3 10, , (3), , (4), , None of these, , Sol. Answer (3), L = 10–3 H and C = 3×10–7 F, f = 50 Hz, V = 100 V, Z, ,  XL – XC 2, ,   X L – XC   L –, , 1, C, , Putting , L & C, ⎛ 10 5, ⎞, Z⎜, –, ⎝ 3 10 ⎟⎠, , 29. Switch is in position A for long time. At time t = 0 it is shifted to position B. Find the maximum charge that, will accumulate on capacitor, E, R, , A, , C, (1) ( LC ), , E, R, , (2), , E, ( LC )R, , Y, (3), , B, X, ⎛ L ⎞E, ⎜⎜, ⎟⎟, ⎝ C ⎠R, , (4), , ⎛ C ⎞E, ⎜⎜, ⎟⎟, ⎝ L ⎠R, , Sol. Answer (1), lmax = 0qmax, lmax =, , 0 , , , R, , 1, LC, E, , q max , , 1, , R, , ⇒ q max  ( LC ), , LC, , E, R, , 30. In Q.29, which of the following graphs best represents the current flowing from X to Y?, , I, , I, t, , (1), , I, (3), , t, , (2), , I, t, , (4), , t, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 91 :
Solution of Assignment, , Alternating Current, , 91, , Sol. Answer (2), At first, current will start charging the capacitor and then it will start oscillating between maximum and minimum, value., or at t = 0,, , l = imax, , so, l = l0cost, Hence graph will be as per cost with time t., 31. In oscillating LC circuit, the total stored energy is U and maximum charge upon capacitor is Q. When the, charge upon the capacitor is, U, 2, , (1), , Q, , the energy stored in the inductor is, 2, , (2), , U, 4, , 4, U, 3, , (3), , (4), , 3U, 4, , Sol. Answer (4), Initial energy = total energy =, U' , , , Q2, U, 2C, , Q2, , Q⎞, ⎛, ⎜⎝ as Q'= ⎟⎠, 4  2C , 2, Q2, 8C, , Uinductor = Utotal – Ucapacitor, , , Q2 Q2, –, 2C 8C, , , , 3Q 2 3,  U, 8C, 4, , SECTION - B, Objective Type Questions, 1., , The virtual current of 4 A and 50 Hz flows in an AC circuit containing a coil. The power consumed in the coil is 240, W. If the virtual voltage across the coil is 100 V, then its resistance will be, (1), , 1, , 3, , (2), , 1, , 5, , (3), , 15 , , (4), , 1, , , , Sol. Answer (3), Vrms = IrmsZ, 100 = 4 Z, Z = 25 , P = VIcos, 240 = (100)(4)cos, , cos  , , 3 R, , 5 Z, , 3 R, , 5 25, R = 15 , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 92 :
92, 2., , Alternating Current, , Solution of Assignment, , For an AC circuit the potential difference and current are given by V  10 2 sin t (in V) and I  2 2 cos t (in A), respectively. The power dissipated in the instrument is, (1) 20 W, , (2), , 40 W, , (3), , 40 2 W, , (4), , Zero, , Sol. Answer (4), Here phase difference  , So, P  VrmsI rms cos, 3., , , 2, , , 0, 2, , If the power factor in an AC circuit changes from, , 1, 1, to then by what percent reactance will change (approximately),, 3, 9, , if resistance remains constant?, (1) Increase by 200%, , (2), , Decrease by 200%, , (3), , Increase by 100%, , (4), , Decrease by 100%, , Sol. Answer (1), Power factor cos  , R, R X, 2, , 2, , , , R, R   X ', 2, , 2, , 1, 3, , , , R, Z, , ...(1), , 1, 9, , ...(2), , Using (1) and (2), , X R 8, X '  R 80, X',  10, X, X '– X,   10 – 1, X, , so percentage change   10 – 1  100,  200%, 4., , A step-up transformer operates on 220 V line and supplies 2.2 A. The ratio of primary and secondary winding is, 11 : 50. The output voltage in the secondary is, (1) 220 V, , (2), , 100 V, , (3), , 1000 V, , (4), , 0V, , Sol. Answer (3), N P VP, , N S VS, , VS , , , N SVP, NP, , 50,  220  1000 V, 11, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 93 :
Solution of Assignment, , 5., , Alternating Current, , 93, , An alternating power supply of 220 V is applied across a series circuit of resistance 10 3  , capacitive reactance, 40  and inductive reactance 30 . The respective currents in the circuit for zero and infinite frequencies are, (1) 2 A,, , 1, A, 2, , (2), , 0 A, 10 A, , (3), , 10 A, 0 A, , (4), , 0 A, 0 A, , Sol. Answer (4), For f = 0, XL = 0 and f  , XC = 0, 6., , In L-C oscillation, maximum charge on capacitor can be Q. If at any instant, electric energy and magnetic energy, associated with circuit is equal, then charge on capacitor at that instant is, , (1), , Q, , (2), , 2, , Q, 2, , 3, , (3), , Q, 2, , (4), , 3Q, 2, , Sol. Answer (1), , Magnetic energy =, , Electric energy =, , 1 ⎛ Q 2 ⎞ Q ' , ⎜, ⎟, 2 ⎝ 2C ⎠, 2C, , Q' , , 7., , Total energy, 2, , Total energy, 2, , 2, , Q, 2, , A series LCR circuit consists of an inductor L a capacitor C and a resistor R connected across a source of emf , = 0 sin t. When  L , , 1, the current in the circuit is I0 and if angular frequency of the source is changed to, C, , , the current in the circuit becomes, , I0, 1, , then the value of  ' L , is, 2,  'C, , L, , C, , R, ,  = 0 sin t, , (1) R, , (2), , 3R, , (3), , 15 R, , (4), , Zero, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 94 :
94, , Alternating Current, , Solution of Assignment, , Sol. Answer (2), , I0 , , , , I0, , 2, , R, , E0, R, E0, 1 ⎞, ⎛, R  ⎜W ' L –, ⎟, ⎝, W 'C ⎠, 2, , 2, , ...(i), , E0, I0, , So in (i), , 1 ⎞, ⎛, 2, R 2  ⎜W ' L –, ⎟⎠  4R, ⎝, W 'C, 2, , 1 ⎞, ⎛, 2, ⎜⎝ W ' L –, ⎟  3R, W 'C ⎠, , W 'L –, 8., , 1, R 3, W 'C, , A direct current of 10 A is superimposed on an alternating current I = 40 cos t (A) flowing through a wire. The, effective value of the resulting current will be, (1) 10 2 A, , (2), , 20 2 A, , (3), , 20 3 A, , (4), , 30 A, , Sol. Answer (4), , ⎛ 40 ⎞, I net  10 2  ⎜, ⎝ 2 ⎟⎠, , 2, , I net  100  800,  900  30 A, , 9., , The sinusoidal potential difference V1 shown in figure applied across a resistor R produces heat at a rate W. What, is the rate of heat dissipation when the square wave potential difference V2 as shown in figure is applied across the, resistor?, , V2, V0, , V1, V0, t, –V0, (1), , W, 2, , (2), , W, , t, –V0, (3), , 2W, , (4), , 2W, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 95 :
Solution of Assignment, , Alternating Current, , 95, , Sol. Answer (4), Power (P) =, , 2, Vrms, R, , According to the problem, ⎛ V0 ⎞, ⎜⎝, ⎟, 2⎠, w, R, , 2, , for square wave Vrms = V0, W', , V02, R, , W',  2, W '  2W, W, , 10. In the circuit shown in figure, if both the bulbs B1 and B2 are identical then, , B1, , C = 600 µF, , B2, L = 12 mH, (1) Both bulbs have equal brightness, (2) B2 will be brighter than B1, (3) As the frequency is increased, brightness of B1 will increase and that of B2 will decrease, (4) As the frequency is decreased, the brightness of B1 will increase and that of B2 will decrease, Sol. Answer (3), As frequency is increased, XC will decrease and XL will increase, so brightness of B, will increase and that, of B2 will decrease., 11. In the following circuit rms current through inductor is 0.8 A, the rms current through capacitor is 0.4 A and, rms current through resistor is 0.3 A, the current delivered by a.c. source is, , L, R, C, , (1) 0.7 A, , (2), , 1.5 A, , (3), , 0.5 A, , (4), , Approximately 1 A, , Sol. Answer (3), I  I R2   I L – I C , , 2, ,   0.3  2   0.8 – 0.4  2,   0.3  2   0.4  2  0.5 A, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 96 :
96, , Alternating Current, , Solution of Assignment, , 12. The output current versus time curve for a rectifier is shown in the figure. The average value of output current, in this case is, I, I0, t, , (1) 0, , (2), , I0, 2, , (3), , 2I0, , , (4), , I0, , Sol. Answer (3), , I avg , , So I avg , , ∫ Idt, ∫ dt, 2I 0, , , 13. If reading of voltmeter V shown in the figure at resonance is 200 V, then the quality factor of the circuit is, V, , 50 V, , ~, , (1) 2, , (2), , 4, , (3), , 1, , (4), , 3, , Sol. Answer (2), , Q, , L X L, , R, R, , ⎛i⎞, ⎜⎝ ⎟⎠, i, , So Q can be written as Q , , VL 200, , 4, VR, 50, , 14. In the given A.C. circuit, the instantaneous current through inductor and capacitor are 0.8 A and 0.4 A, respectively. The instantaneous current through resistor is, L, R, C, , ~, (1) 1.2 A, , (2), , 0.6, , (3), , 0.4, , (4), , 0.8 A, , Sol. Answer (3), , l R  i L – iC, IR = 0.4, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 97 :
Solution of Assignment, , Alternating Current, , 97, , 15. A capacitor of capacitance C has initial charge Q0 and connected to an inductor of inductance L as shown., At t = 0 switch S is closed. The current through the inductor when energy in the capacitor is three times the, energy of inductor is, , + + + +, , Q0, L, , – – – – C, , (1), , Q0, , (2), , 2 LC, , Q0, , (3), , LC, , 2Q0, LC, , (4), , 4Q0, LC, , Sol. Answer (1), E total  E i  E c , , 4E i , , Q 02, 2C, , Q 02, 2C, , Ei , , Q 02 1 2,  Li, 8C 2, , i2 , , Q 02, 4LC, , i, , Q0, 2 LC, , 16. An inductor L and a capacitor C ac connected in the circuit as shown in the figure. The frequency of source, is equal to resonance frequency of the circuit. Which ammeter will read zero ampere?, , A1, A2, A3, , ~, , E = E0 sin t, (1) A1, , (2), , A3, , (3), , A2, , (4), , None of these, , Sol. Answer (2), Given that L , , 1, C, , So net current lL – lC = 0 ∵ l L  l C , So A3 reads zero ampere., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 98 :
98, , Alternating Current, , Solution of Assignment, , 17. Average energy loss due to inductance (L) of circuit is (symbols have their usual meaning), 2, (1) i rms, , (2), , 2, Vrms, XL, , (3), , i 02 X L, , (4), , Zero, , Sol. Answer (4), No energy will be lost as there is no resistor (inductor only stores energy)., 18. The virtual current of 4 A and 50 Hz flows in an ac circuit containing a coil. The power consumed in the coil is 240, W. If the virtual voltage across the coil is 100 V, then its resistance will be, (1), , 1, , 3, , (2), , 1, , 5, , (3), , 15 , , 1, , , , (4), , Sol. Answer (3), P  VI cos , 240  4 100 cos , 3, R,  cos  , 5, Z, , ...(i), , and 4Z = 100  Z = 25 , , 3 R, , ⇒ R  15 , 5 25, , So in (i), , 19. Different alternating voltages are given below. In which case the peak value and rms value are same?, , V0, , V, , (a) 0, –V0, , V0, , V0, (b), , t, , 0, –V0, , V, , V0, , 0, –V0, , t, , (1) (c) only, , (2), , (c), , (d) only, , V, t, V, , (d), , 0, –V0, , (3), , (a), (b) & (c), , t, (4), , (b) & (c), , Sol. Answer (2), For square pulse, Vrms = Vpeak, 20. Two cables of copper are of equal lengths. One of them has a single wire of area of cross-section A, while, other has 10 wires of cross-sectional area, , A, each. Give their suitability for transporting A.C. and D.C., 10, , (1) Only multiple strands for A.C, either for D.C., (2) Only multiple strands for A.C, only single strand for D.C., (3) Only single strand for D.C, either for A.C., (4) Only single strand for A.C, either for D.C., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 99 :
Solution of Assignment, , Alternating Current, , 99, , 21. An AC source is rated 222 V, 60 Hz. The average voltage is calculated in a time interval of 16.67 ms. It, (1) Must be zero, , (2), , May be zero, , (3), , Is never zero, , (4), , Is ( 111 2 ) V, , Sol. Answer (1), , T , , 2, 2, ,  16.67 ms, W 2  60, ,  for time interval 16.67 ms (which is time period) average volatge will always be zero., 22. In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of, a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be, (1) R, , (2), , 2R, , (3), , R, 4, , (4), , R, 2, , Sol. Answer (3), R, , V2, P, , R, R' , ,  220  2, 60, , 110 2, 60, , R, R,  4 ⇒ R' , R', 4, 23. An electric bulb of 100 W – 300 V is connected with an AC supply of 500 V and, inductance to save the electric bulb is, (1) 2 H, , (2), , 1, H, 2, , (3), , 4H, , (4), , 150, Hz . The required, , , 1, H, 4, , Sol. Answer (3), , 50, 0, , V, , 300, , V   300   500 2, 2, , 2, , V 2   400 2, V = 400, and l , , 100 1,  A, 300 3, , V  lX L, , ⎛ 1⎞, 400  ⎜ ⎟ X L, ⎝ 3⎠, X L  1200 , ,  2f  L  1200 , So L = 4 H, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 100 :
100, , Alternating Current, , Solution of Assignment, , 24. In a RLC circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the, inductance should be changed from L to, (1) 4 L, , (2), , 2L, , L, 2, , (3), , (4), , L, 4, , Sol. Answer (3), , 1, , W , , LC, , LC  L ' 2C, ⇒ L' , , L, L, ⇒ L' , 2, 2, , SECTION - C, Previous Years Questions, 1., , A series R-C circuit is connected to an alternating voltage source. Consider two situations, (a) When capacitor is air filled., (b) When capacitor is mica filled., Current through resistor is i and voltage across capacitor is V then, (1) Va = Vb, , (2), , Va < Vb, , (3), , Va > Vb, , [Re-AIPMT-2015], (4), , ia = ib, , Sol. Answer (3), , C, , R, , R 2 + XC2, , Z=, i=, , V, Z, , If C increases, XC decreases, current will increases hence voltage across resistance increases, so voltage, across capacitor decreases., 2., , Across a metallic conductor of non-uniform cross- section a constant potential difference is applied. The quantity, which remains constant along the conductor is, [AIPMT-2015], (1) Electric field, , (2), , Current density, , (3), , Current, , (4), , Drift velocity, , Sol. Answer (3), Across a metallic conductor of non-uniform cross section the rate of flow of charge through every cross section, is constant hence. Current is constant., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 101 :
Solution of Assignment, , 3., , Alternating Current, , 101, , A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary, coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are, [AIPMT-2014], (1) 300 V, 15 A, , (2), , 450 V, 15 A, , (3), , 450 V, 13.5 A, , (4), , 600 V, 15 A, , Sol. Answer (2), Power ouput = 3kW , , 90,  2.7 kW, 100, , Ib  6 A, , 4., , VS , , 2.7 kW,  450 V, 6A, , IP , , 3 kW,  15 A, 200 V, , In an electrical circuit R,L,C and an a.c. voltage source are all connected in series. When L is removed from the, , circuit, the phase difference between the voltage and the current in the circuit is . If instead, C is removed, 3, , from the circuit, the phase difference is again . The power factor of the circuit is [AIPMT (Prelims)-2012], 3, (1) 1, , (2), , 3, 2, , (3), , 1, 2, , (4), , 1, 2, , Sol. Answer (1), Magnitude of phase difference is constant  XL = XC, So cos  , 5., , R R,  1, Z R, , The instantaneous values of alternating current and voltages in a circuit are given as, i=, , e=, , 1, 2, , sin(100t) A, , ⎞, ⎛, sin ⎜ 100t  ⎟ V, 3⎠, 2, ⎝, , 1, , The average power in watts consumed in the circuit is, (1), , 1, 4, , (2), , 3, 4, , [AIPMT (Mains)-2012], (3), , 1, 2, , (4), , 1, 8, , Sol. Answer (4), P  VrmsI rms cos , I rms , Vrms , , l0, 2, V0, 2, , 1, , , ,  2  2 , , , ,  2  2 , , 1, , , , 1, 2, , , , 1, 2, , 1, , 1 1, 1 1, So, P  ⎛⎜ ⎞⎟ ⎛⎜ ⎞⎟ cos  ⎛⎜ ⎞⎟ ⎛⎜ ⎞⎟  W, ⎝ 2⎠ ⎝ 2⎠, 3 ⎝ 4⎠ ⎝ 2⎠ 8, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 102 :
102, 6., , Alternating Current, , Solution of Assignment, , In an ac circuit an alternating voltage e  200 2 sin100t volts is connected to a capacitor of capacity 1 F., The rms. value of the current in the circuit is, [AIPMT (Prelims)-2011], (1) 20 mA, , (2), , 10 mA, , (3), , 100 mA, , (4), , 200 mA, , Sol. Answer (1), , 7., , XC , , 10 6,  10 4, 100  1, , I rms , , Vrms, XC, , I rms , , 200, = 0.02 A = 20 mA, 10 4, , An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are, both equal to 3 , the phase difference between the applied voltage and the current in the circuit is, [AIPMT (Prelims)-2011], (1) Zero, , , 6, , (2), , , 4, , (3), , (4), , , 2, , Sol. Answer (3), cos  , , 8., , R, R  XL, 2, , 2, , ⇒ cos  , , 3, , 3 2 , So  = 45°, , The rms value of potential difference V shown in the figure is, , [AIPMT (Mains)-2011], , V, V0, O, (1), , V0, 2, , (2), , V0, 3, , t, , T/2 T, (3), , V0, , (4), , V0, 2, , Sol. Answer (4), , I rms , , ∫ I dt, ∫ dt, , Vrms , , ∫ V dt, ∫ dt, , 2, , 2, , from O , , So Vrms , , T, T, T , V = 0, , V = V0 and, 2, 2, , V02 T, , 2  V0, T, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 103 :
Solution of Assignment, , 9., , Alternating Current, , 103, , A coil has resistances 30  and inductive reactance 20  at 50 Hz frequency. If an ac source, of, 200 V, 100 Hz is connected across the coil, the current in the coil will be, [AIPMT (Mains)-2011], (1), , 20, 13, , A, , (2), , 2.0 A, , (3), , 4.0 A, , (4), , 8.0 A, , Sol. Answer (3), 20 = 100 L, 1, L, 5, , ⎛ 1⎞, X L  L  2 100  ⎜ ⎟  40 , ⎝ 5 ⎠, and R = 30 ,  Z  R 2  X L 2   30 2   40 2  50 , So l , , 200, 4A, 50, , 10. A 220 V input is supplied to a transformer. The output circuit draws a current of 2 A at 440 V. If the efficiency, of the transformer is 80%, the current drawn by the primary windings of the transformer is, [AIPMT (Prelims)-2010], (1) 5.0 A, , (2), , 3.6 A, , (3), , 2.8 A, , (4), , 2.5 A, , Sol. Answer (1), Pout,  0.8, Pinput, ,  given, , VoutI out,  0.8, VinputI input, , 2  440,  0.8,  220 I, I, , 4,  5.0 A, 0.8, , 11. In the given circuit the reading of voltmeter V1 and V2 are 300 volts each. The reading of the voltmeter V3 and, ammeter A are respectively:, [AIPMT (Prelims)-2010], L, C, R = 100 , V1, , V2, , V3, , A, ~, , 220V, 50 Hz, (1) 150 V, 2.2 A, , (2), , 220 V, 2.2 A, , (3), , 220 V, 2.0 A, , (4), , 100 V, 2.0 A, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 104 :
104, , Alternating Current, , Solution of Assignment, , Sol. Answer (2), Potential drop on L & C is same  Circuit is at resonance, So V3 = 220 V, , 220 220, ,  2.2 A, R, 100, , and l , , 12. Power dissipated in an LCR series circuit connected to an a.c source of emf  is, , 1 ⎞, ⎛, 2 R 2  ⎜ L ,  ⎟⎠, C, ⎝, (1), R, , (3), , 2, , (2), , 2R, 1 ⎞, ⎛, R  ⎜ L , C  ⎟⎠, ⎝, 2, , (4), , 2, , [AIPMT (Prelims)-2009], , 2, ⎡, 1 ⎞ ⎤, ⎛, ⎥,  2 ⎢ R 2  ⎜ L , C  ⎠⎟ ⎥, ⎝, ⎢⎣, ⎦, R, , 2R, 2, ⎡, 1 ⎞ ⎤, ⎛, ⎢ R 2  ⎜ L , ⎥, ⎟, C ⎠ ⎥, ⎝, ⎢⎣, ⎦, , Sol. Answer (4), ⎛V 2 ⎞, P  ⎜ rms ⎟ cos , ⎝ Z ⎠, , now in problem, , P, ,  2 R  2R, .  2 , Z Z, Z, ,  2R, 1 ⎞, R 2  ⎛⎜ L –, ⎟, C ⎠, ⎝, , 2, , 13. In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by, e = E0 sint, i = l0 sin(t – ), The average power in the circuit over one cycle of a.c. is, (1) E0I0, , (2), , E0 l 0, 2, , (3), , [AIPMT (Prelims)-2008], , E0 l 0, sin, 2, , (4), , E0 l 0, cos, 2, , Sol. Answer (4), Pavg =, , E 0I 0, cos , 2, , 14. The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux, linked with the primary coil is given by  = 0 + 4t, where  is in webers, t is time in seconds and 0 is a, constant, the output voltage across the secondary coil is, [AIPMT (Prelims)-2007], (1) 30 V, , (2), , 90 V, , (3), , 120 V, , (4), , 220 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 105 :
Solution of Assignment, , Alternating Current, , 105, , Sol. Answer (3), , , , d, 4, dt, , N P VP, , N S VS, 50 1500, , ⇒ V = 120 V., 4, V, 15. A transformer is used to light a 100 W and 110 V lamp from a 220 V (Mains). If the main current is 0.5 A, the, efficiency of the transformer is approximately, [AIPMT (Prelims)-2007], (1) 10%, , (2), , 30%, , (3), , 50%, , (4), , 90%, , Sol. Answer (4), 100 = 110 I, I, , 10, A, 11, , 16. A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in, series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be, [AIPMT (Prelims)-2006], (1), , f, 4, , (2), , 8f, , (3), , f, 2 2, , (4), , f, 2, , Sol. Answer (3), 17. A coil of inductive reactance 31  has a resistance of 8 . It is placed in series with a condenser of capacitative, reactance 25 . The combination is connected to an a.c. source of 110 V. The power factor of the circuit is, [AIPMT (Prelims)-2006], (1) 0.56, , (2), , 0.64, , (3), , 0.80, , (4), , 0.33, , Sol. Answer (3), , cos  , , R, Z, , Z  R 2   XL – XC , , 2, , XL = 31 , XC = 25 , So Z  8 2   31– 25  2  10, , So cos , , 8, 4,   0.8, 10 5, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 106 :
106, , Alternating Current, , Solution of Assignment, , 18. The core of a transformer is laminated because, , [AIPMT (Prelims)-2006], , (1) Energy losses due to eddy currents may be minimised, (2) The weight of the transformer may be reduced, (3) Rusting of the core may be prevented, (4) Ratio of voltage in primary and secondary may be increased, Sol. Answer (1), 19. In a circuit, L, C and R are connected in series with an alternating voltage source of frequency f. The current, leads the voltage by 45°. The value of C is, [AIPMT (Prelims)-2005], 1, (1) 2f  2fL  R , , (2), , 1, f  2fL  R , , (3), , 1, 2f  2fL  R , , (4), , 1, f  2fL  R , , Sol. Answer (1), cos 45 , , R, 1 ⎞, ⎛, R 2  ⎜ L –, ⎟, ⎝, C ⎠, , 2, , 2, , 1 ⎞, ⎛, 2, R 2  ⎜ L –, ⎟  2R, ⎝, C ⎠, , L –, , 1, R, C, , C , , 1, L – R, , C , , 1, 1, ,    L – R  2  f  2  fL – R , , 20. In an ac circuit an alternating voltage e  200 2 sin100t volts is connected to capacitor of capacity 1 F. The, r.m.s. value of the current in the circuit is, (1) 20 mA, , (2), , 10 mA, , (3), , 100 mA, , (4), , 200 mA, , (3), , Irms , , I0, , , (4), , Irms , , Sol. Answer (1), 21. In an A.C. circuit, Irms and /0 are related as, (1) Irms = I0, , (2), , Irms  2 I 0, , I0, 2, , Sol. Answer (4), 22. The electric current in a circuit is given by i = 3 sin t + 4 cos t. The rms current is, (1), , 5, 2, , (2), , 5, , (3), , 4, 2, , (4), , 3, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 107 :
Solution of Assignment, , Alternating Current, , 107, , Sol. Answer (1), l  3 sin t  4 cos t, , ⎞, ⎛,  3 sin t  4cos ⎜ t  ⎟, ⎝, 2⎠, l 0  l 12  l 22  2l 1l 2 cos, , , 2, , l1 is max of l1 = 3, and l2 = 4, l 0   3 2   4 2  2  3  4 cos, , , 2, , =5, , l rms , , l0, 2, , , , 5, 2, , 23. An L-C-R circuit is connected to a source of A.C. current. At resonance, the phase difference between the, applied voltage and the current in the circuit, is, (1) , , (2), , Zero, , (3), , 1,  RC, , (3), , , 4, , (4), , , 2, , (4), , L, R, , Sol. Answer (2),  = 0 at resonance, 24. The value of quality factor is, (1), , L, R, , (2), , 2, , LC, , Sol. Answer (1), Q, , L, R, , 25. A capacitor of capacity C has reactance X. If capacitance and frequency become double then reactance will, be, (1) 4X, , (2), , X, 2, , (3), , X, 4, , (4), , 2X, , Sol. Answer (3), , X , , , , 1, 1, , X', 2fC, 2  2f   2C , , X, X,  4, So X ' , X', 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 108 :
108, , Alternating Current, , Solution of Assignment, , 26. For a series LCR circuit the power loss at resonance is, V, , 2, , (1) ⎡  L – 1 ⎤, ⎢⎣,  C ⎥⎦, , (2), , I2L, , (3), , I2R, , (4), , V 2, C, , Sol. Answer (3), Power loss = l2R, 27. A capacitor and a bulb are connected in series with a source of alternating emf. If a dielectric slab is inserted, between the plates of the capacitor, then, (1) The brightness of the bulb decreases, , (2), , The brightness of the bulb increases, , (3) The brightness of the bulb remains same, , (4), , The brightness of the bulb becomes zero, , Sol. Answer (2), i, , E, ⎛ 1 ⎞, R2  ⎜, ⎝ C ⎟⎠, , 2, , 1, will decrease, so i will increase. Hence the brightness of the bulb increases., C, 28. The primary winding of a transformer has 500 turns whereas its secondary has 5000 turns. The primary is, connected to an A.C. supply 20 V, 50 Hz. The secondary will have an output of, (1) 2 V, 50 Hz, , (2), , 2 V, 5 Hz, , (3), , 200 V, 50 Hz, , (4), , 200 V, 500 Hz, , Sol. Answer (3), N P VP, , N S VS, , 500 5000, , 20, V, V = 200 V, no change in frequency., 29. What is the value of inductance L for which the current is maximum in a series LCR circuit with, C = 10 F and  = 1000 s–1?, (1) 1 mH, , (2), , Cannot be calculated unless R is known, , (3) 10 mH, , (4), , 100 mH, , Sol. Answer (4), XL = XC for maximum current, WL , , 1, WC, , 1000L , , 10, 1000  10, , L  100 mH, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 109 :
Solution of Assignment, , Alternating Current, , 109, , 30. A step-up transformer operates on 220 V line and supplies 2.2 A. The ratio of primary and secondary winding is, 11 : 50. The output voltage in the secondary is, (1) 220 V, , (2), , 100 V, , (3), , 1000 V, , (4), , Zero, , Sol. Answer (3), N P VP, , NS VS, , 11 220, , ⇒ V  1000 V, 50, V, 31. A coil of self inductance L is connected in series with a bulb and an A.C. source. Brightness of the bulb decreases, when, (1) Number of turns in the coil is reduced, (2) A capacitance of reactance XC = XL is included in the same circuit, (3) An iron rod is inserted in the coil, (4) Frequency of the A.C. source is decreased, Sol. Answer (3), i, , E2, , ...(1), , R 2  WL  2, , An iron rod is inserted in the coil, L will increase hence i will decrease., 32. A condenser of capacity C is charged to a potential difference of V1. The plates of the condenser are then, connected to an ideal inductor of inductance L. The current through the inductor when the potential difference, across the condenser reduces to V2 is, , ⎛ C(V1  V2 )2 ⎞, (1) ⎜, ⎟⎠, L, ⎝, , 1/ 2, , (2), , C (V12  V22 ), L, , (3), , C (V12  V22 ), L, , (4), , ⎛ C(V12  V22 ) ⎞, ⎜⎝, ⎟⎠, L, , 1/ 2, , Sol. Answer (4), , SECTION - D, Assertion-Reason Type Questions, 1., , A : Direct current is more dangerous than Alternating current of same value., R : An electrocuted person sticks to direct current line. while alternating current repels the person from the, line., , Sol. Answer (4), 2., , A : AC can be transmitted over long distances at high voltage without much power loss., R : The average value of AC is defined over any half cycle., , Sol. Answer (3), 3., , A : An inductor and a capacitor are called low pass filter and high pass filter respectively., R : Reactance of an inductor is low for low frequency signals and that of a capacitor is high for high frequency, signals., , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 110 :
110, 4., , Alternating Current, , Solution of Assignment, , A : The chief characteristic of series resonant circuit is voltage magnification., R : At resonance the voltage drop across inductance (or capacitance) is Q times the applied voltage., , Sol. Answer (1), 5., , A : Wires of the transmission lines carrying A.C. are made of multiple strands., R : A.C. flows on surface of the conductor., , Sol. Answer (1), 6., , A : The ammeters and voltmeters used for measuring alternating current and voltages have non-uniform divisions, on their scales., R : The instruments used for measuring alternating current and voltage are based on heating effect of current., , Sol. Answer (1), 7., , A : A series resonant circuit is also known as an acceptor circuit., R : For large value of Ohmic resistance, the quality factor of a series resonant circuit is high., , Sol. Answer (3), 8., , A : For a practical choke coil the power factor is very small., R : In a practical choke coil the power dissipation reduces if frequency of the a.c. is increased., , Sol. Answer (2), 9., , A : If a current has both ac and dc components, then a dc ammeter used to measure this current will measure, the average value of the total current., R : The scale of a dc ammeter is uniformly divided., , Sol. Answer (2), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 111 :
Chapter, , 23, , Electromagnetic Waves, Solutions, SECTION - A, Objective Type Questions, 1., , According to modified Ampere's circuital law, (iD = displacement current), d ⎞, ⎛,   0 ⎜ iC   0 E ⎟, dt ⎠, ⎝, , (1), , ∫ B  dl, , (3), , ∫ B  dl  , , 0, , i, , (2), , ∫ B  dl, , (4), , ∫B d l  , ,  0 0, , 0, , d E, dt, , ⎞, ⎛ d E,  iD ⎟, ⎜ iC, dt, ⎠, ⎝, , Sol. Answer (1), According to modified Ampere's circuital law., , ∫ B  dl, 2., , d ⎞, ⎛,   0 ⎜ iC   0 E ⎟, dt ⎠, ⎝, , Displacement current is set up between the plates of a capacitor when the potential difference across the plates, is, (1) Maximum, , (2), , Zero, , (3), , Minimum, , (4), , Varying, , Sol. Answer (4), Displacement current, I d  A 0, , dE, dt, , So, the potential difference has to change with time, 3., , A parallel plate capacitor with circular plates of radius R is being charged as shown. At the instant shown,, R, the displacement current in the region between the plates enclosed between, and R is given by, 2, +, –, , i, , (1), , 3, i, 4, , (2), , 1, i, 4, , i, , (3), , 3i, , (4), , 4, i, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 112 :
112, , Electromagnetic Waves, , Solution of Assignment, , Sol. Answer (1), , Id   0A, , dE, dE,   0  R 2 , i, dt, dt, , ⎛, R 2 ⎞ dE, I ' d   0 ⎜ R 2 –, ⎟, ⎝, 4 ⎠ dt, 3, dE,  0 R 2, 4, dt, , , , I 'd , , 4., , 3, i, 4, , Figure shows a circular region of radius R in which uniform magnetic field B exists. The magnetic field is, dB, . The induced electric field at a distance r from the centre for r < R is, dt, , increasing at a rate, , B, R, , dB R 2, dt 2r, , (3), , dB, dt, , (1) Wavelength, , (2), , Frequency, , (3) Intensity, , (4), , Medium, in which it travels, , (1), , dB r, dt 2, , O, , (2), , Zero, , (4), , Sol. Answer (1), Rate of increase of B , , or, , E in  2r  –, , d, dt, , E  2r  – A, , dB, dt, , E  2r  – r 2, , , , 5., , E, , dB, dt, , dB, dt, , – r dB, 2 dt, , The speed of electromagnetic waves depends upon, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 113 :
Solution of Assignment, , Electromagnetic Waves, , 113, , Sol. Answer (4), I, Speed of wave  , , I, , , ,  0 r  0  r, , It depends upon the medium., 6., , If E and B represent the electric and magnetic field vectors of an electromagnetic wave, then the direction, of propagation of the electromagnetic wave is in the direction of, (1) E, , (2), , B, , (3), , E B, , (4), , B E, , Sol. Answer (3), ,  , Direction of propagation is given by E  B, , 7., , If an electromagnetic wave propagating through vacuum is described by E y = E 0 sin (kx –  t);, Bz = B0 sin (kx –  t), then, (1) E0 k = B0 , , E0 B0 =  k, , E 0 B0 , , , k, , (3), , E0  = B0 k, , (1) Electric field, , (2), , Magnetic field, , (3) Both (1) & (2), , (4), , Neither electric field nor magnetic field, , (2), , (4), , Sol. Answer (1), Speed of wave , , 8., , E0 , , B0 k, , kE 0  B 0, , Electromagnetic wave is deflected by, , Sol. Answer (4), Electromagnetic wave consists of unchanged particle called photons which are neither deflected by electric, field nor by magnetic field., 9., , Out of the following, choose the ray which does not travel with the velocity of light, (1) X-ray, , (2), , Microwave, , (3), , -rays, , (4), , -rays, , (4), , Amplitude, , (4), , Infra-red ray, , Sol. Answer (4), -rays do not travel with speed of light, as they are not em waves., 10. Red light differs from blue light in its, (1) Speed, , (2), , Frequency, , (3), , Intensity, , Sol. Answer (2), Frequency is different for each light as they have different wavelengths., 11. Which of the following has the largest wavelength?, (1) Radio wave, , (2), , X-ray, , (3), , Ultraviolet ray, , Sol. Answer (1), Radio waves : 10–2 m to 104 m., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 114 :
114, , Electromagnetic Waves, , Solution of Assignment, , 12. Velocity of electromagnetic waves in a medium is,  1/2, (1) ( 0 0 ), , (2), , ( 0  r 0  r ) 1/2, , (3), , 3×, , 108, , m/s, , (4), , ⎛ 0 r ⎞, ⎜⎝   ⎟⎠, 0 r, ,  1/2, , Sol. Answer (2), , 1, , , , Speed of light in medium , , 1,  0 r  0  r, ,    0 r  0  r , , –, , 1, 2, , 13. Which of the following is incorrect about a plane electromagnetic wave?, (1) The electric field and magnetic field have equal average values, (2) The electric energy and the magnetic energy have equal average values, (3) The electric field and magnetic field both oscillate in same phase, (4) The electric field and magnetic field oscillate in opposite phase, Sol. Answer (4), , , E and B are in the same phase but oscillate in different planes which are perpendicular to each other., 14. In a plane electromagnetic wave, which of the following has/ have zero average value in one complete cycle?, (a) Magnetic field, , (b), , Magnetic energy, , (c) Electric field, , (d), , Electric energy, , (1) (a), (c), , (2), , (b), (c), , (3), , (a), (d), , (4), , All of these, , Sol. Answer (1), , , Both E and B are sinusoidal, so over a complete cycle, its value will be zero., 15. An electromagnetic wave is propagating in vacuum along z-axis, the electric field component is given by Ex = E0, sin(kz – t), then magnetic component is, (1) Bx =, , E0, sin  kz – t , C, , (2), , By =, , B0, sin  kz – t , C, , (3) By =, , E0, sin  kz – t , C, , (4), , By = B0C sin(kz – t), , Sol. Answer (3), Direction of propagation  Z, E x  E 0 sin  kZ – t , , B  will be in the y-direction, ,  By  B 0 sin  kZ – t , E0,  B c, 0, B0 , , So, B y , , E0, c, E0, sin  kZ – t , c, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 115 :
Solution of Assignment, , Electromagnetic Waves, , 115, , 16. The speed of electromagnetic wave in a medium (whose dielectric constant is 2.25 and relative permeability, is 4) is equal to, (1) 0.5  108 m/s, , (2), , 0.25  108 m/s, , (3), , 0.75  108 m/s, , (4), , 1  108 m/s, , Sol. Answer (4), Er = 2.25, r = 4, , V, , , 1,  0 r  0  r, 1, 1, ,  0 0, 2.25  4, , 3  10 8, 1.5  2, V = 108 m/s, , , 17. The magnetic field in a plane electromagnetic wave is given by = 2 × 10–7sin(0.5 × 103x + 1.5 × 1011t). This, electromagnetic wave is, (1) Visible light, , (2), , Infrared, , (3), , Microwave, , (4), , Radiowave, , (3), , [M L–1 T–2, , (4), , [M L2 T–1, , 10–6 cm, , (4), , 10–7 cm, , Sol. Answer (4), B  2  10 –7 sin  0.5  10 –3 x  1.5  10 11t , , f , , c, , k  0.5  10 –3, , ,   1.5  10 11, V , ,  1.5  10 11, ,  3  10 14, k 0.5  10 –3, , ,  f, k, , , , k, , 2, , , , , 2, 2  22, ,  1.25  10 4 m, k, 7  0.5  10 –3, Radio waves, , 18. The dimensional formula of, , 1,  0 E 2 is, 2, , (E = electric field), (1) [M L T–1, , (2), , [M L2 T–2, , Sol. Answer (3), 1, E  2 –2 ,  0E 2  Energy density   ML 3T, 2, V, L , , = ML–1T –2 , 19. Which of the following represents an infra-red wavelength?, (1) 10–4 cm, , (2), , 10–5 cm, , (3), , Sol. Answer (1), 10–6 m or 10–4 cm  Infrared, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 116 :
116, , Electromagnetic Waves, , Solution of Assignment, , 20. Which of the following is not transported by electromagnetic waves?, (1) Energy, , (2), , Momentum, , (3), , Charge, , (4), , Information, , Sol. Answer (3), Only energy, momentum and information can be transferred with the help of em waves, not any matter., 21. Hertz experiment is used for, (1) Production of electromagnetic wave, , (2), , Detection of electromagnetic wave, , (3) Both (1) & (2), , (4), , None of these, , Sol. Answer (3), Hertz was the scientist who produced and detected the em waves., 22. Electromagnetic waves are produced due to, (1) A charge at rest, , (2), , A moving charge, , (3) An accelerated charge, , (4), , None of these, , Sol. Answer (3), Em waves are produced by accelerated charged particles., 23. Ozone layer blocks the radiation of wavelength, (1) Less than 4 × 10–7 m, , (2), , Between 4 × 10–7 m to 8 × 10–7 m, , (3) More than 8 × 10–7m, , (4), , None of these, , Sol. Answer (1), Radiation having wavelength less than 4  10–7 m are blocked by ozone layer, 24. Which of the following can be used in cancer treatment?, (1) X-rays, , (2), , UV-rays, , (3), , -rays, , (4), , Both (1) & (3), , (4), , D>C>B>A, , Sol. Answer (4), Both X-rays and -rays are used for the treatment of cancer., 25. The following can be arranged in decreasing order of wave number, A. AM radio, , B., , TV and FM radio, , C. Microwave, , D., , Short radio wave, , (3), , A>B>C>D, , (1) A > B > D > C, , (2), , C>D>B>A, , Sol. Answer (2), Decreasing order of wave number, Microwave > short radiowave > TV and FM radio > AM radio., , SECTION - B, Objective Type Questions, 1., , E, where the symbols have, c, their usual meanings and the energy in a given volume of space due to the electric field part is U, then the energy, due to the magnetic field part will be, , If the electric field and magnetic field of an electromagnetic wave are related as B =, , (1), , U, c, , (2), , U, c2, , (3), , U, 2, , (4), , U, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 117 :
Solution of Assignment, , Electromagnetic Waves, , 117, , Sol. Answer (4), , B, , E, C, , Energy due to E , , 1,  E2, 2 0, , B0 2, B, , Energy due to, 2 0, , E part , , U, 2, , Bpart , , U, 2, , E part  B part, , 2., , The direction of poynting vector represents, (1) The direction of electric field, (2) The direction of magnetic field, (3) The direction of propagation of EM wave, (4) The direction opposite to the propagation of EM wave, , Sol. Answer (3),  ,  E  B, S, 0,  , E  B ⇒ direction of wave propagation, 3., , A plane electromagnetic wave is incident on a plane surface of area A normally, and is perfectly reflected. If energy E, strikes the surface in time t then average pressure exerted on the surface is (c = speed of light), (1) Zero, , (2), , E, Atc, , (3), , 2E, Atc, , (4), , E, c, , Sol. Answer (3), P, , 2I, [Perfect reflection], c, , , , 4., , 2E, Atc, , P, , 2E, Atc, , 5% of the power of 100 W bulb is converted to visible radiation. Average intensity of visible radiation at a distance, of 10 m from the bulb is, (1), , 5, 2(10 ), , 2, , watt/m2, , (2), , 5, 4(10), , 2, , watt/m2, , (3), , 5, (10), , 2, , watt/m2, , (4), , 5, 8(10 )2, , watt/m2, , Sol. Answer (2), I, , E, P, , At A, , 5,  100, P, 5, ,  100, , watt/m 2, 2, 2, 4r, 400, 4 10, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 118 :
118, 5., , Electromagnetic Waves, , Solution of Assignment, , Which of the following physical quantities contained in a small volume oscillates at double the frequency of, passing electromagnetic wave?, (1) Electric field, , (2), , Magnetic field, , (3), , Magnetic energy, , (4), , All of these, , Sol. Answer (3), The electric and magnetic energy oscillate at double the frequency as compared to electric and magnetic field., 6., , A capacitor is connected across a battery which delivers a current of 1 A at an instant in the capacitor. Displacement, current through the capacitor at that instant is, (1) 1 A, , (2), , 0A, , (3), , 2A, , (4), , 1, A, 2, , Sol. Answer (2), I=1A, Id = 1 A, 7., , The magnetic field in a plane electromagnetic wave is given by, B = 3.01 × 10–7 sin (6.28 × 102x + 2.2 × 1010t) T., [where x in cm and t in second]. The wavelength of the given wave is, (1) 1 cm, , (2), , 628 cm, , (3), , 1.129 cm, , (4), , 314 cm, , Sol. Answer (1), , B  3.01 10 –7 sin  6.28  10 2 x  2.2  10 10 t  T, , k  6.28  10 2,   2.2  1010, , , 8., , k, , 2, , , , , 2,  10 –2 m, 6.28  10 2, , At a particular instant the current in the circuit given below is i. The displacement current between the plates of the, capacitor shown below is, , C, i, V, (1) Zero, , (2), , i, , (3), , i, 2, , i, 4, , (4), , Sol. Answer (2), Displacement current id = i, 9., , To establish an instantaneous displacement current of I ampere in the space between the plates of a parallel plate, dV, 1, farad, the value of, is, capacitor of, dt, 2, (1) 2I, , (2), , I, 2, , (3), , 1, 2I, , (4), , I, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 119 :
Solution of Assignment, , Electromagnetic Waves, , 119, , Sol. Answer (1), Id , , A 0 dE, d, d dt, , I, , 1 dE, d, 2 dt, , I, , 1 dV 1,  .d, 2 dt d, , dV,  2I, dt, , 10. A plane electromagnetic wave of frequency 28 MHz travels in free space along the positive x-direction. At a, particular point in space and time, electric field is 9.3 V/m along positive y-direction. The magnetic field (in T) at, that point is, (1) 3.1 × 10–8 along positive z-direction, , (2), , 3.1 × 10–8 along negative z-direction, , (3) 3.2 × 107 along positive z-direction, , (4), , 3.2 × 107 along negative z-direction, , Sol. Answer (1), f = 28106 Hz, E = 9.3 V/m   ĵ , Bc = E, B, , 9.3, 3  10 8, , B = 3.1 × 10–8 along positive z-direction, , SECTION - C, Previous Years Questions, 1., , The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong?, [Re-AIPMT-2015], (1) -rays, , (2), , X-rays, , (3), , Infra-red rays, , (4), , Ultraviolet rays, , Sol. Answer (2), E = 15 keV, , , , hc, 6.6  1019  3  108, , E 15  106  1.6  10–19, ,  = 0.8 Å  1 Å, 2., , A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface, is (C = velocity of light), [AIPMT-2015], (1), , E, C2, , (2), , E, C, , (3), , 2E, C, , (4), , 2E, C2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 120 :
120, , Electromagnetic Waves, , Solution of Assignment, , Sol. Answer (3), Surface is perfectly reflecting, So change in momentum i.e., momentum transferred is p  p –  – p , p  2 p, , So, p , , 3., , 2E, C, , Light with an energy flux of 25 × 104 Wm–2 falls on a perfectly reflecting surface at normal incidence. If the, surface area is 15 cm2, the average force exerted on the surface is, [AIPMT-2014], (1) 1.25 × 10–6 N, , (2), , 2.50 × 10–6 N, , (3), , 1.20 × 10–6 N, , (4), , 3.0 × 10–6 N, , Sol. Answer (2), , Fav , , 2 I A 2  25  10 4  15  10 4, , N, c, 3  10 8, , = 250 × 10–8 N = 2.5 × 10–6N, 4., , The condition under which a microwave oven heats up a food item containing water molecules most efficiently, is, [NEET-2013], (1) The frequency of the microwaves has no relation with natural frequency of water molecules, (2) Microwaves are heat waves, so always produce heating, (3) Infra-red waves produce heating in a microwave oven, (4) The frequency of the microwaves must match the resonant frequency of the water molecules, , Sol. Answer (4), 5., , , The electric field associated with an e. m. wave in vacuum is given by E  iˆ 40 cos (kz – 6 × 108t), where E,, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is [AIPMT (Prelims)-2012], (1) 6 m–1, , (2), , 3 m–1, , (3), , 2 m–1, , (4), , 0.5 m–1, , Sol. Answer (3), 6., , The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating, in vacuum is equal to, [AIPMT (Mains)-2012], (1) The speed of light in vacuum, (2) Reciprocal of speed of light in vacuum, (3) The ratio of magnetic permeability to the electric susceptibility of vacuum, (4) Unity, , Sol. Answer (2), , E0, C, B0, , , B0 1, , E0 C, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 121 :
Solution of Assignment, , 7., , Electromagnetic Waves, , 121, , The electric and the magnetic field, associated with an e.m. wave, propagating along the +z-axis, can be, represented by, [AIPMT (Prelims)-2011], (1), , , , ⎡E  E0 jˆ, B  B0 kˆ ⎤, ⎣, ⎦, , (2), , (3), , , , ⎡E  E0 kˆ , B  B0 iˆ⎤, ⎣, ⎦, , (4), , , , ⎡E  E0 iˆ, B  B0 ˆj ⎤, ⎣, ⎦, , , ⎡E  E0 jˆ, B  B0 iˆ⎤, ⎣, ⎦, , Sol. Answer (2), , , If wave is propagating in +z direction then E and B will be in x-y plane., ,  , Also, E  B = direction of propagation, , , E  E 0 iˆ, B  B0 jˆ, 8., , The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is, [AIPMT (Prelims)-2011], (1) Infrared, microwave, ultraviolet, gamma rays, , (2), , Microwave, infrared, ultraviolet, gamma rays, , (3) Gamma rays, ultraviolet, infrared, microwaves, , (4), , Microwaves, gamma rays, infrared, ultraviolet, , Sol. Answer (2), Maximum wavelength = microwaves, Minimum wavelength = -rays, 9., , Which of the following statement is false for the properties of electromagnetic waves?, [AIPMT (Prelims)-2010], (1) These waves do not require any material medium for propagation, (2) Both electric and magnetic field vectors attain the maxima and minima at the same place and same time, (3) The energy in electromagnetic wave is divided equally between electric and magnetic vectors, (4) Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of, propagation of wave, , Sol. Answer (4), 10. The electric field of an electromagnetic wave in free, , space is given by E  10cos 107 t  kx jˆ V/m, where t and x are in seconds and metres respectively. It can be, , , , , , inferred that, (a) The wavelength  is 188.4 m, (b) The wave number k is 0.33 rad/m, (c) The wave amplitude is 10 V/m, (d) The wave is propagating along +x direction, Which one of the following pairs of statements is correct ?, (1) (c) & (d), , (2), , (a) & (b), , (3), , [AIPMT (Mains)-2010], (b) & (c), , (4), , (a) & (c), , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 122 :
122, , Electromagnetic Waves, , Solution of Assignment, , 11. The electric field part of an electromagnetic wave in a medium is represented by, ⎡⎛, N, ⎛, 6 rad ⎞, 2 rad ⎞ ⎤, Ex = 0; E y =2.5 C cos ⎢ ⎜⎝ 2  10 s ⎟⎠ t  ⎜⎝   10 m ⎟⎠ x ⎥ ;, ⎣, ⎦, , Ez = 0. The wave is, , [AIPMT (Prelims)-2009], , (1) Moving along x-direction with frequency 106 Hz and wavelength 100 m, (2) Moving along x-direction with frequency 106 Hz and wavelength 200 m, (3) Moving along –x-direction with frequency 106 Hz and wavelength 200 m, (4) Moving along y-direction with frequency 2 × 106 Hz and wavelength 200 m, Sol. Answer (2), Ex = 0, E y  2.5, , N, as  2  10 6 t –   10 –2 x , C, , Ez = 0, f , , 2  10 6, w, ,  10 6 s –1, 2, 2, , , , 2, 2, ,  200 m ., ,   10 –2, , 12. The velocity of electromagnetic radiation in a medium of permittivity 0 and permeability 0 is given by, [AIPMT (Prelims)-2008], , (1), , 0, 0, , (2), , 0, 0, , (3), , 0 0, , (4), , 1,  0 0, , Sol. Answer (4), , C, , 1,  0 0, , 13. The electric and magnetic field of an electromagnetic wave are, , [AIPMT (Prelims)-2007], , (1) In opposite phase and perpendicular to each other (2), , In opposite phase and parallel to each other, , (3) In phase and perpendicular to each other, , In phase and parallel to each other, , (4), , Sol. Answer (3), 14. If v, x and m represent the wavelengths of visible light, X-rays and microwaves respectively, then :, [AIPMT (Prelims)-2005], (1) m > x > v, , (2), , v > m > x, , (3), , m > v > x, , (4), , v > x > m, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 123 :
Solution of Assignment, , Electromagnetic Waves, , 123, , 15. For a medium with permittivity and permeability , the velocity of light is given by, (1), , , , , (2), , , , 1, , , (3), , , , , (4), , Sol. Answer (3), Speed of light in vacuum C , , in any other medium, V , , 1,  0 0, , 1, , , 16. Which of the following electromagnetic radiations have the smallest wavelength?, (1) X-rays, , (2), , -rays, , (3), , UV waves, , (4), , Microwaves, , Sol. Answer (2), -rays  minimum wavelength., 17. If 0 and 0 are the electric permittivity and magnetic permeability in a free space,  and  are the corresponding, quantities in medium, the index of refraction of the medium is, , (1), ,  0 0, , , (2), , ,  0 0, , (3), , 0 ,  0, , (4), , (3), , X-rays, , (4), , , 0, , Sol. Answer (2), Refractive index =,  V , , , , C, V, , 1, , , C, , 1,  0 0, , , , ,  0 0, , 18. What is the cause of “Greenhouse effect”?, (1) Infra-red rays, , (2), , Ultraviolet rays, , Radiowaves, , Sol. Answer (1), Green house effect is mainly due to the infrared radiation, methane gas, SO2 etc., 19. The velocity of electromagnetic wave is parallel to,  , (1) B  E, , (2), ,  , E B, , (3), , , E, , (4), , , B, , Sol. Answer (2), ,  , E  B  Direction of wave propagation, , , perpendicular to the plane of E and B ., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 124 :
124, , Electromagnetic Waves, , Solution of Assignment, , SECTION - D, Assertion-Reason Type Questions, 1., , A : Different electromagnetic waves differ considerably in their mode of interaction with matter., R : Different electromagnetic waves have different wavelength or frequency., , Sol. Answer (2), 2., , A : All electromagnetic waves travel through vacuum with same speed but they have different wavelength or, frequency., R : The wavelength of the electromagnetic waves is often correlated with the characteristic size of the system that, produces and radiates them., , Sol. Answer (1), 3., , A : High frequency electromagnetic waves are detected by some means based on the physical effects they, produce on interacting with matter., R : The oscillating fields of an electromagnetic wave can accelerate charges and can produce oscillating, currents therefore, an apparatus designed to detect EM waves is based on this fact., , Sol. Answer (1), 4., , A : Infrared waves are often called heat waves., R : Infrared waves vibrate not only the electrons, but entire atoms or molecules of a substance which increases, the internal energy and temperature of the substance., , Sol. Answer (1), 5., , A : The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun., R : Humans have evolved with visions most sensitive to the strongest wavelength from the sun., , Sol. Answer (1), 6., , A : Long distance radio broadcasts use short-wave bands., R : Ionosphere reflectes waves in these bands., , Sol. Answer (1), 7., , A : It is necessary to use satellites for long distance TV transmission., R : Television signals are not properly relfected by the ionosphere therefore, relfection is effected by satellites., , Sol. Answer (1), 8., , A : Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from satellites, orbiting the earth., R : Atmosphere absorbs X-rays, while visible and radiowaves can penetrate it., , Sol. Answer (1), 9., , A : If the earth did not have an atmosphere, its average surface temperature would have been lower., R : In the absence of atmosphere, the green house effect will be absent., , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 125 :
Solution of Assignment, , Electromagnetic Waves, , 125, , 10. A : It has been predicted that a global nuclear war on the earth would be followed by a severe 'nuclear winter' with, a devastating effect on life on earth., R : The clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar, light from reaching many parts of the globe causing winter., Sol. Answer (1), 11. A : In an EM wave the magnitude of the electric field vector is more than the magnitude of the magnetic field, vector., R : Energy of the EM wave is shared equally between the electric and magnetic fields., Sol. Answer (2), 12. A : The displacement current goes through the gap between the plates of a capacitor when the charge on the, capacitor does not change., R : Displacement current arises only when the electric field is constant., Sol. Answer (4), 13. A : When the frequency of the AC is increased, the displacement current increases., R : The sum of the conduction current and displacement current is constant., Sol. Answer (3), 14. A : When cooking in microwave ovens, metal containers are used., R : Energy of the microwaves can be easily transferred to the food thorugh metal., Sol. Answer (4), 15. A : Food is cooked faster by microwaves than by conventional gas burner., R : Microwaves have more energy than heat waves., Sol. Answer (3), 16. A : Microwaves are commonly used in radar to locate flying objects., R : Microwaves have smaller wavelength than radiowaves., Sol. Answer (1), 17. A : Environmental damage has depleted the ozone layer in the atmosphere., R : Increase in ozone decreases the amount of UV radiation to earth., Sol. Answer (2), 18. A : The electrical conductivity of the earth's atmosphere does not change with altitude., R : Cosmic rays from outer space entering the earth's atmosphere do not affect it., Sol. Answer (4), 19. A : Static crashes are heard on a radio when a lightening flash occurs., R : Light and radiowaves are EM waves and they interfere., Sol. Answer (1), 20. A : TV signals are affected if a low flying aircraft passes by or a petrol vehicle is started next to it., R : Aircarft signals or vehicle's spark plug generate interfering EM waves., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Page 126 :
126, , Electromagnetic Waves, , Solution of Assignment, , 21. A : Light waves can be polarised., R : All electromagnetic waves move with same speed in vacuum., Sol. Answer (1), 22. A : In an electromagnetic wave the energy density in electric field is equal to energy density in magnetic field., R : Electromagnetic waves are transverse in nature., Sol. Answer (2), ,  ,  E B, represents the instantaneous intensity at a point., 23. A : The Poynting vector given as S , 0,  , R : The velocity of an electromagnetic wave is in the direction of the vector E  B ., , Sol. Answer (2), 24. A : The radiation pressure due to light waves is maximum when the surface is a perfect reflector., R : The momentum transfer by the photons to a perfectly reflecting surface is maximum., Sol. Answer (1), 25. A : In a material medium the speed of a particle can be more than the speed of light in that medium., R : In the phenomenon of green house effect, low wavelength radiation is allowed to pass but high wavelength, radiation is not allowed to pass., Sol. Answer (2), , , , , , , , https://t.me/NEET_StudyMaterial, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456