Title 20. Moving Charges and Magnetism Easy Ans.pdf ✅

1. (c) Magnetic field at the centre of current carrying coil is given by r Ni B    2 4 0 =   r N B   1 2 2 1 2 1 r r N N B B =  . The following figure shows that single turn coil changes to double turn coil. N1 = 1 N2 = 2 r1 = r r2 = r / 2 B1 = B B2 = ?  4 / 2 1 2 1 2 =  = r r B B  B2 = 4B Short trick : For such type of problems remember 1 2 B2 = n B 2. (b) If distance is same field will be same       =  r i B 2 4 0    3. (c) Magnetic field lies inside as well as outside the solid current carrying conductor. 4. (b) Because for inside the pipe i = 0 0 2 0  = = r i B   5. (d) 2 0 sin 4 r idl dB    =   3 0 ( ) 4 r i dl r dB  =    6. (c) The magnetic field at the centre of the circle N A m r nq r nq r i o 10 / - 2 ( ) 2 10 2 4 −7 −7 =  =  =       7. (b) The given shape is equivalent to the following diagram The field at O due to straight part of conductor is r i B o 2 . 4 1   = . The field at O due to circular coil is =  r i B    2 . 4 0 2 . Both fields will act in the opposite direction, hence the total field at O. i.e. ( 1) 2 . 4 2 ( 1) 4 2 1  − = −      = − =       r i r i B B B o o 8. (d) R i R i B        − = − = 2 2 4 (2 ) 4 0 0         R i 8 30 = 9. (b) The respective figure is shown below Magnetic field at P due to inner and outer conductors are equal and opposite. Hence net magnetic field at P will be zero. 10. (d) Magnetic field at a point on the axis of a current carrying wire is always zero. 11. (b) A T q i 19 19 1.6 10 2 2 1.6 10 − − =    = = 19 19 10 2 0.8 1.6 10 2 − − =      = = o o o r i B    12. (a) 7 3 4 10 5 1000 2 10 − − B = oni =     =   Tesla 13. (a) r i B 2 4 0 =     1 2 2 1 r r B B =  4 10 12 2 8 = − B  B Tesla 9 2 3.33 10 − =  14. (c) B B r r B B r r B B r B 2 1 / 2 2 1 2 2 2 1   =  =  = 15. (c) Field at the centre of a circular coil of radius r is r I B o 2  = 16. (a) 2 7 0 2 5 10 4 10 100 0.1 2 − −      = =   r Ni B 5 4 10 − =   Tesla 17. (b) Magnetic field inside the solenoid B ni in = 0 18. (a) In the following figure, magnetic fields at O due to sections 1, 2, 3 and 4 are considered as 1 2 3 B , B , B and B4 respectively. 0 B1 = B3 = =  1 0 2 . 4 R i B    2 0 4 . 4 R i B    =  As | | | | B2  B4 So         = −  = − 1 2 0 2 4 1 1 4 R R i Bnet B B Bnet  r1 r2 i i 2 r O 1 i i P r P i Q M 2 a x = − Y 2 a x = x =a R1 O R2 2 1 3 4
19. (b) B ni = o 20. (d) The magnetic induction at O due to the current in portion AB will be zero because O lies on AB when extended. 21. (c) The induction due to AB and CD will be zero. Hence the whole induction will be due to the semicircular part BC . r i B o 4  = 22. (c) The magnetic induction due to both semicircular parts will be in the same direction perpendicular to the paper inwards.         +  = + = + = 1 2 0 1 2 2 0 1 0 1 2 4 4 4 r r i r r r i r i B B B    23. (a) Field at a point x from the centre of a current carrying loop on the axis is 10 3 7 25 3 0 (10 ) 2 10 2 2.1 10 . 4 B − − −    = = x M   32 30 2 2 4.2 10 10 4.2 10 W / m − − =   =  24. (d) At these points, the resultant field = 0 25. (b) e t q i = = 100  r e r i B o o 2 100 . 4 2 . 4 centre  = =       o o   17 19 10 4 0.8 200 1.6 10 − − =     = 26. (d) 3 3 2 2 1 . 4 r B r NiR B o =      27. (c) r q B 2 ( ) 4 0     =  10 19 15 7 0.53 10 2 3.14 (1.6 10 6.6 10 ) 10 − − −       =  2 = 12.5 Wb/m 28. (a) 29. (b) 30. (d) Two coils carrying current in opposite direction, hence net magnetic field at centre will be difference of the two fields. i.e.       =  − 2 2 1 0 1 2 4 r i r Ni Bnet N   0 0 4 5 0.4 0.3 0.2 0.2 2 10   =       = − 31. (b) Because B ni = 0  B  ni . 32. (a) 33. (a) n B B ni i 0 0  =   = 4 10 20 100 20 10 7 3     = − −  =7.9 amp = 8amp 34. (d) Directions of currents in two parts are different, so directions of magnetic fields due to these currents are opposite. Also applying Ohm’s law across AB 1 1 2 2 1 2 2 2 i R = i R  i l = i l       = A l  R  Also 2 1 1 1 4 r i l B o =    and 2 2 2 2 4 r i l B o =    ( l = r ) 1 2 2 1 1 1 2  = = i l i l B B Hence, two field induction’s are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of  . 35. (d) The magnetic field at any point on the axis of wire be zero. 36. (d) Magnetic field inside the hollow conductor (tube) is zero. 37. (b) If a wire of length l is bent in the form of a circle of radius r then 2r = l  2 l r = 2 2 2 2     = = = l r Magnetic field due to straight wire 2 0 0 1 1 10 2 2 4 2 4 −   =  =      r i B also magnetic field due to circular loop / 2 2 2 4 2 4 0 0 2         =  =  r i B  50 1 1 2 = B B 38. (c) See the following figure 39. (a) r i B 2 4 0 =     (10 10 ) 2 10 10 2 5 7 − − −  =  i  i = 5A 40. (c) r Ni B    2 4 0 =   (10 10 ) 10 2 3.14 10 3.14 10 2 7 3 − − −       = N  N = 50  r r i2 A l2 B l1 i1 l =  2m i = 2A r Magnetic lines of forces i Plane perpendicular to conductor
41. (c) The magnetic field in the solenoid along its axis (i) At an internal point ni = o 7 3 2 4 10 5000 4 25.1 10 Wb / m − − =     =  (Here n = 50 turns/ cm = 5000 turns/ m) (ii) At one end 2 25.1 10 2 2 1 3 0 −  = = = ni Bend Bin  3 2 12.6 10 Wb /m − =  42. (b) Magnetic field at the centre of solenoid B ni 0 ( ) =  Where n = Number of turns /meter 7 2 2 B 4 10 4250 5 2.7 10 Wb / m − −  =     =  43. (d) Use Right hand palm rule, or Maxwell’s Cork screw rule or any other. 44. (b) B n B B (4) B B 16 B 2 2  =   = =  = 45. (d) 1 1 0 7 5.2 10 2 12.56 10 2 . 4 − −   =  =  i r i B     i A 3 1.04 10 −  =  46. (b) R i R i r i B 4 4 2 8 0 0  0      = =   = 47. (a) B T r B r B B r B 0.01 10 1 0.04 40 2 1 2 2 2 1   =  =  = 48. (a) 49. (b) r i B   2 0 = or r B 1  50. (d) i L N B ni = 0 = 0 51. (c) Here B ni =  0 where n is number of turns per unit length l N = 52. (b) 5 7 0 7 10 0.05 2 (10 ) 2 3.142 4 − − =      =  i H r i   5.6 2 3.142 35 2 3.142 10 7 0.05 10 7 5 =  =      = − − i amp 53. (c) T r Ni B 4 7 0 6.28 10 2 0.1 4 10 1000 0.1 2 − − =      = =   54. (b) T r Ni B 4 7 0 1.25 10 2 0.5 4 10 50 2 2 − − =      = =   55. (d) r i B  56. (a) . . 1 . . 2 0 i e r i e B r i B =    when r is doubled, B is halved. 57. (b) Applying Ampere’s law  B dl = i 0 .  to any closed path inside the pipe we find no current is enclosed. Hence B = 0 . 58. (a) Magnetic field at the centre of current carrying coil is r ni r ni B 2 2 4 0  0   =  = 59. (d) The magnetic field is given by r i B 2 4 0   = . It is independent of the radius of the wire. 60. (d) Magnetic meridian is a vertical N-S plane, the earth’s magnetic field ( ) BH lies in it. (For more details see magnetism). To obtain neutral point at the centre of coil, magnetic field due to current (B) and BH must cancel each other. Hence plane of the coil and magnetic meridian must be perpendicular to each other as shown 61. (c) 1 Tesla Gauss 4 = 10 62. (c) 63. (d) 64. (b) 5 2 0 7 2 10 10 2 1 10 2 4 − − − =   =  =  r i B   Tesla 65. (a) Magnetic field due to one side of the square at centre O / 2 2 sin 45 . 4 0 1 a i B o   =  a i B 2 2 . 4 0 1   = Hence magnetic field at centre due to all side a i B B   (2 2 ) 4 0 = 1 = Magnetic field due to n turns (2 ) 2 2 2 2 0 0 l ni a ni Bnet nB     = = = l ni  20 = (a = 2l) B BH Magnetic meridian Plane of the coil N S
64. (c) 63. (a) 62. (c) Magnetic field on the axis of circular current 2 2 3 / 2 2 0 ( ) 2 4 x r nir B + =      2 2 3 / 2 2 (x r ) nr B +  61. (a) r1 :r2 = 1 : 2 and B1 : B2 = 1 : 3 We know that 6 1 3 2 2 1 1 . 4 2 2 1 1 2 0 1 =   =  = = B r B r i i r ni B    60. (b) T r i B 7 7 8 8 10 5 2 2 10 2 10 − − − =   = =  59. (c) B T r i B 5 2 0 7 6.28 10 5 10 10 10 4 − − − =    =   =      58. (c) Magnetic field due to solenoid is independent of diameter (Because B ni = 0 ). 57. (b) 0 7 5 2 8 10 / 0.0157 2 2 10 2 4 Wb m r i B − − =   = =      56. (b) 2.5 10 200 4 10 2 7 = 0 =    − − B  ni  2 2 6.28 10 Wb / m − =  55. (d) Magnetic field at centre due to smaller loop 1 0 1 1 2 . 4 r i B    = ..... (i) Due to Bigger loop 2 0 2 2 2 . 4 r i B    = So net magnetic field at centre         = − =  − 2 2 1 0 1 1 2 2 4 r i r i B B B    According to question 1 2 1 B =  B 1 0 1 2 2 1 0 1 2 . 2 4 1 .2 4 r i r i r i       =           − 1 2 2 2 1 2 2 1 1 1 1 2 2 1 1 − =  =  = i i r i r i r i r i r i { 2 } 2 1 r = r 54. (b) 53. (b) 2 2 3 / 2 2 2 3 / 2 2 0 ( ) 1 ( ) 2 4 r x B R x NiR B +   + =     0.0025 0.04 1 8 ( ) ( ) 1 8 2 2 / 3 2 2 3 / 2 1 2 2 3 / 2 2 2 + +  =       + +  = R R R x R x 0.0025 0.04 1 4 2 2 + +  = R R . On solving R = 0.1m 52. (b) / 4 5 2 20 10 7 B B B B r i B =   =  = −  51. (c) T r ni B 3 2 7 7 1.257 10 5 10 2 25 4 10 2 10 − − − − =      = =    50. (b) F = Bil  2 1 2 [ ][ ] [ ] [ ] − − − = = = MT A AL MLT i l F B 49. (d) Magnetic field on the axis of conductor is zero. 48. (c) 2 1 2 1 2 2 1   = = = r r r r B B r B 47. (c) T r i B 7 7 7 2 10 1 2 1 10 2 10 − − − =   =  =  46. (d) At midpoint, magnetic fields due to both the wires are equal and opposite. So BNet= 0. 45. (c) ( ) (sin 45 sin 45 ) 4 / 2 4 0 0 =    +  a i B   2 2 2 4 4 0 =    a i   a i  0 2 2 = 44. (b) r i B 2 10 −7 = ; according to question BH = B i A i 4 5 10 2 3.14 5 10 10 2 5 7  =      =  − − − 43. (d) T r Ni B 5 2 7 0 4 10 5 10 2 10 2 100 0.1 . 4 − − − =      = =      42. (a) Corresponding current i = en So ( ) r ne r en B 2 2 . 4 0  0   = = 41. (a) B at ends of solenoid is ni 2 0 40. (b) Use Right hand palm rule or Maxwell’s Cork screw rule. 39. (c) At P  Bnet = B1 − B2 45o 45o a/2 a P r r i i