Tiêu đề 21.Magnetism and Matter Easy Ans.pdf ✅

1. (b) On bending a rod it’s pole strength remains unchanged where as its magnetic moment changes. New magnetic moment   L M M m R m 2 2 (2 )  =       = = 2. (d) 3. (c) 3 0 3 0 2 2 4 d M d M Ba     = = 4. (d) 5. (b) If cut along the axis of magnet of length l, then new pole strength 2 m m  = and new length l = l  New magnetic moment 2 2 2 ml M l m M =  = = If cut perpendicular to the axis of magnet, then new pole strength m  = m and new length, l = l / 2  New magnetic moment 2 2 2 l ml M M = m  = = 6. (d) For a magnet 3 0 2 . 4 x M B   = (Nearly) 8 1 2 3 3 2 1 2 1  =      =          = x x x x B B (Approx.) 7. (b) For each part 2 m m  = 8. (c) 2 2 2 2 2 2 1 2 2 2 2 1 2 1 100 400 20 10 1 12.5         − −  =         − − = l l d l d l d d B B  l = 5 cm Hence length of magnet = 2l = 10 cm 9. (a) 10. (c) 2 1 sin 48 25 10 0.15 2 = =     −  MB  = 0.9 N m 11. (d) 1 3 2 3 2 y M and B x M B = = As B1 = B2 Hence 1 / 3 3 3 3 3 or 2 or 2 2 = = = y x y x y M x M 12. (c) Work done = (1 − cos ) W MBH ( ) 3(2 3 ) 2 3 20 0.3 1 cos 30 6 1 = −         =  −  = − 13. (b) Magnetic intensity on end side-on position is twice than broad side on position. 14. (a) Along the axis of magnet guass X M Ba 200 2 3 = = guass X M Ba 100 3  = = 15. (a) 16. (b) 17. (c) 18. (d) Provided length of magnet is <26. (a) Pole strength doesn’t depend upon the length. 27. (a) Torque  = MBH sin 2 1 0.1 10 4 10 sin 30 10 4 3 3 7 =     =   − − −   o =  N m −7 2 10 28. (b) Number of lines of force passing through per unit area normally is intensity of magnetic field, hence option (c) is incorrect. The correct option is (b). 29. (a) Flux 2 ; Weber / m A Flux = B A B = = 30. (a) 31. (b) 2 d m B = in C.G.S. system. 32. (a) (cos cos ) (cos 0 cos 60 ) 1 2 o o W = MB  −  = MB − 2 2 1 1 MB MB  =      = − and 2 3 MB sin MB sin 60 MB o  =  = = W MB 3 3 2   =       =  33. (c) Pole strength of each part =m Magnetic moment of each part 2 M = M = m L = mL = 34. (c) M 2M 2ml. net = = 35. (b) 36. (b) 2 1 2 r m m F  37. (c) ( ) ( ) N r m F 7 2 7 2 2 2 7 10 1 10 1 10 − − − =  = = 38. (b) 2 sin sin 30 MH MH MH o  =  = = 39. (c) 40. (c)        = 4 0 6 4 d MM F   in end-on position between two small magnets. ( ) F 0.6N 0.1 6 10 10 10 4 7 =            = − 41. (b) 42. (a)  = MBH sin or    cos MBH d d = This will be maximum. when . 0 o  = 43. (d) ( ) o o W = MB cos1 − cos 2 ; 1 = 0 and  2 = 360  W = 0 44. (d) 45. (b) W MB MB MB o o 1 = (cos 0 − cos 90 ) = (1 − 0) = 2 2 1 2 (cos 0 cos 60 ) 1 MB W MB MB o o  =      = − = − W1 = 2W2 n = 2 46. (d) In magnetic dipole, force 4 1 r  Hence new force 0.3 N 16 4.8 2 4.8 4 = = = 47. (a) Magnetic moment of bar M 10 J / T 4 = B T 5 4 10 − =  Hence work done W = M.B J o 10 4 10 cos60 0.2 4 5 =    = − 48. (a) 49. (d) 50. (c) ( ) N A m d M B =  −  = =  − − 2 10 / 0.5 2 1.25 10 2 4 6 3 7 3 0   51. (b) S N IV. S S N  N 2L L L S N S N l l M → M →
52. (b) 53. (c) 54. (b) o  = MBH sin  0.032 = M  0.16  sin 30  M = 0.4 J / tesla 55. (b) 3 0 4 r M Bequatorial   = 56. (c) Inside a magnet, magnetic lines of force move form south pole to north pole. 57. (b) Magnetic moment of circular loop carrying current ( ) I M L L IL M IA I R I      4 2 4 2 2 2  =  =      = = = 58. (c) 59. (b) Concept of magnetic screening. 60. (b) Repulsion is the sure test of magnetism. 61. (d) 62. (a) 63. (a) 5 4 2 Cmax = MB  4 10 = M 10  M = 0.4 Am − − 64. (c) Magnetic flux Tesla m Weber A = BA  B = = = 2   65. (b) Mnet M M 2M 2 2  = + = 66. (b) Suppose magnetic field is zero at point P. Which lies at a distance x from 10 unit pole. Hence at P ( ) x cm x x 10 30 40 . 4 10 . 4 2 0 2 0  = − =     So from stronger pole distance is 20 cm. 67. (b)  = MB sin  = (mL )B sin  =       − − (40 10 10 ) 2 10 sin 45 2 4 =  N − m −3 0.565 10 68. (a) Potential energy U = −MB cos max (at 180 ) o  U = MH  = 69. (b)  = MB sin  = 200 0.25 sin 30 = 25 N m. 70. (c) If pole strength, magnetic moment and length of each part are m  , M and L respectively then 2 m m  = m  = m L = L 2 L L = 2 M  M = 2 M  M = 71. (b) ) ˆ 3 ˆ (0.5 ˆ  = M  B  = 50i  i + j ( ) . ˆ 150 ˆ ˆ = 150 i  j = kN m 72. (c)  = MB sin   sin 2 2 1 2 1 sin sin 90 sin / 2 sin         =  = o 30 2 1  sin 2 =  2 =  angle of rotation = 0 Å − 30 = 60 73. (d) 74. (d) B L M F = mB  F =  2 10 0.1 . 3 6 10 4 5 L m L   =    = − − 75. (a)  = MB sin   = (mL)B sin 25 10 ( 5 10 ) 5 10 sin30 6 2 2   =      − − − m m =  A − m −2 2 10 . S N S N P • B2 B1 Bnet 1 2  M → Mnet → M → S N N S 10 unit 40 unit x (30 – x) 30 cm P S N S N Q N P S S N
76. (d) 77. (c) Monopole do not exists. 78. (d) 79. (a) 80. (a) 81. (b) W = MB(1 − cos ); where o  = 180 W MB W J 3 2 2 2 2 5 10 2 10 − −  =  =    =  82. (a) Torque on a bar magnet in earths magnetic field (BH) is  sin. = MBH  will be maximum if sin  = maximum i.e.  = 90o . Hence axis of the magnet is perpendicular to the field of earth. 83. (b) 84. (a) Both points A and B lying on the axis of the magnet and on axial position 3 1 d B   1 8 24 48 3 3  =      =         = A B B A d d B B 85. (b) W MB J o = (1 − cos) = 2  0.1(1 − cos 90 ) = 0.2 86. (a) 2 2 M = mL = 4 10 10 = 0.4 A m − 87. (a) Similar to solution (1) New magnetic moment 2 2 0.1 3.14 2 2 2 0.5 31.4 10 ' amp m M mL M =     = = = −   88. (d) Magnetic potential at a distance d from the bar magnet on it's axial line is given by 2 0 . 4 d M V   =  V  M  2 1 2 1 M M V V =  2 M / 4 M V V =  4 2 V V = 89. (b) ( ) B T d M B 4 3 7 3 0 2.4 10 0.1 2 1.2 10 2 . 4 − − =   =  =    90. (d) From figure 2 2 Bnet = Ba + Be 2 3 0 2 3 0 . 4 2 . 4         +         = d M d M     3 0 . 4 5 . d M   = 3 7 (0.1) 10 = 5  10  − = 3 5 10 −  Tesla. 91. (c)  = MB sin  = m (2l) B sin  =    − 10 0.1 30 sin 30 4 Nm 4 1.5 10 − =  92. (b) At neutral point B = BH ( ) 0.3 20 2 3  = M 1.2 10 . 3  M =  emu 93. (d) No magnetic lines of force passes through the steel box. 94. (b) At magnetic poles, the angle of dip is 90o . Hence the horizontal component B = Bcos = 0. H 95. (a) 96. (c) 97. (c) 98. (d) 3 , BH = BV also o H V B B 30 3 1 tan = =  = 99. (d) At magnetic equator, the angle of dip is 0o . Hence the vertical component V = Isin = 0. 100. (b) N E W S Magnetic Axis Magnetic Meridian S N S N S N P • B1 B2 0.1m 0.1m 1 2 W N E S S N BH N1 N2 B BH B 20cm 20cm