Tiêu đề 20. Moving Charges and Magnetism Hard Ans.pdf ✅

1. (c) 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 Bnet B   (2 2 ) 4 0 = 1 = 2. (b) Circular coil Square coil Length L = 2 r Length L = 4a Magnetic field r i B    2 . 4 0 = r i 2 0 4 . 4    = a i B 2 2 . 4 0   = a i B 8 2 . 4 0   = Hence 8 2 2  = square circular B B 3. (a) Case 1 :    = r i . 4 B 0 A r i BB . 4 0   =  r i BC . 4 0   =  So net magnetic field at the centre of case 1 ( ) B1 = BB − BA + BC  r i B    . 4 0 1 =  ..... (i) Case 2 : As we discussed before magnetic field at the centre O in this case =  r i B    . 4 0 2 .....(ii) Case 3 : = 0 BA  − = r BB (2 / 2) . 4 0     =  r i 2 3 . 4 0    r i BC . 4 0   =  So net magnetic field at the centre of case 3        = − 1 2 3 . 4 0 3    r i B ....(iii) From equation (i), (ii) and (iii) B1 : B2 : B3 =   :   :        − 1 2 3       = − − 2 1 4 3 : 2 : 2    4. (d) Magnetic field at P Due to wire 1, d B 2(8) . 4 0 1   = and due to wire 2, d B 2(16) . 4 0 2   =  2 0 2 2 0 2 2 1 12 . 4 16 . 4         +         = + = d d Bnet B B     d d    0 5 0 10 2 4 =   = 5. (b) As shown in the following figure magnetic field at P due to side 1 and side 2 is zero. P B1 B2 Y X (0, 0, d) 6 A 8 A   (A) r i (B) O (C) O 90o r (B) i (C) (A ) r O i (C) (B) (A) 45o 45o O i a/2 a i i i r 45o 45o O i a/2 a i
Magnetic field at P is only due to side 3, Which is 2 3 2 sin 30 . 4 0 1 a i B o   =  a i 3 2 . 4 0   =  = a i   2 3 0  6. (b) According to the question, at centre of coil H BH r i B = B  =    2 . 4 0  5 2 7 7 10 (5 1 ) 2 10 − − − =     i  i = 5.6 amp. 7. (b) Magnetic field at the centre of orbit due to revolution of charge. ; 2 ( ) . 4 0 r q B     = where  = frequency of revolution of charge So, 0.8 2 (100 1) 4 0   =  e B     0 17 10  − B = . 8. (a) By using 1 ; 3 / 2 2 2         = + r x B B axis centre where x = 3R and r = R  (10) 10 10 3 / 2 = = axis centre B B . 9. (b) By using 3 / 2 2 2 1         = + r x B B axis centre , given r = R and Baxis Bcentre 8 1 =  3 / 2 2 2 8 1         = + R x  3 1 / 2 2 2 2 (2) 1                   = + R x  1 / 2 2 2 2 1         = + R x  2 2 4 1 R x = +  x = 3R 10. (c) Magnetic field at any point lying on the current carrying conductor is zero. Here H1 = magnetic field at M due to current in PQ H2 = magnetic field at M due to R + due to QS + due to PQ = 1 1 1 2 3 2 0 H H H + + =  3 2 2 1 = H H 11. (b) According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is half that of ABC i.e. 2 1 1 2 = i i . Also i + i = i 1 2  3 2 1 i i = and 3 2 i i = Magnetic field at centre O due to wire AB and BC (part 1 and 2) =  / 2 2 sin 45 . 4 0 1 1 a i B o   =  a i 0 1 2 2 . 4  and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) a i B B 0 2 3 4 2 2 4  = =  Also i1 = 2i2. So (B1 = B2) > (B3 = B4) Hence net magnetic field at centre O ( ) ( ) Bnet = B1 + B2 − B3 + B4 a i a i 2 3 2 2 . 4 3 2 2 2 . 4 2 0 0        −        =      = −  =  a i a i     3 2 (2 1) 3 4 2 . 4 0 0 12. (b) 2 2 2 7 0 2.5 6.28 10 / 10 200 B ni 4 10 Wb m − − − =  =     =  . 13. (a) B ni =  0 ; where cm turn n 10 20 = = m turn 2000 . So, 5 20 10 −  = 4  2000  i  i = 8 A. 14. (a) i L N B =  0 L N  B  . 2 2 1 1 4 2 2 1 2 1  =  =  = L l N N L L N N B B 15. (d) B ni =  0 ; where R N n 2 =  0.5 2 0.1 500 4 10 7   =   −  B  5 10 . 4 T − =  16. (a) .2 (sin sin ) 4 0      B = ni + . From figure  = (90o – 30o ) = 60o and  = (90o – 60o ) = 30o i D B A C (1) (2) (3) (4) i1 i2 O P i i 3 30o 30o 1 2 a 2 3 a
 ( 3 1) 4 (sin 60 sin 30 ) 2 0 0 = + = + ni ni B  o o  . 17. (d) By using         − − = 2 2 2 2 0 2 b a r a r i B   here , 2 3R r = a = R, ab = 2R                −  −             = 2 2 2 0 ( ) 2 3 2 3 2 R R R R R i B   r i o   36 5. = . 18. (c) qB mv r =  11 2 27 17 10 1.5 10 6 10 ( / ). −     = q m B v m 2 2.35 10 − =  = 2.35 cm. 19. (d) 19 5 31 8 1.6 10 10 2 2 10 7.2 10 − − − −        = = q q qB mK r = 0.25 cm = 25 cm . 20. (b) By using qB mk r 2 = ; For both particles q → same, B → same, k → same Hence r  m  p e p e m m r r =  p e p e m  m so r  r Since radius of the path of proton is more, hence it's trajectory is less curved. 21. (b) By using qB mv r = ; v → same, B → same  2 m r   p p p q q m m r r    =  2 2 1 4 =  = p p p p q q m m 22. (c) Time period of proton 5 sec 5 25 Tp = =  By using qB m T 2 =     q q m m T T p p p =  p p p p q q m m 2 4 =   2 10 sec . T = Tp = 23. (d) By using F = q(v  B) ; where v j ˆ = 10 and B i ˆ = 0.5  ) ˆ ˆ ) 5 10 ( ˆ 0.5 ˆ 10 (10 11 8 4 F = j  i =  j  i − − ) ˆ 5 10 ( 4 =  −k − i.e., 4 5 10 −  N along . ˆ − k 24. (a) The given situation can be drawn as follows According to figure, for deflecting electron in x-y plane, force must be acting an it towards y-axis. Hence according to Flemings left hand rule, magnetic field directed along positive y − axis. 25. (b) Particles is moving undeflected in the presence of both electric field as well as magnetic field so it's speed B E v = v E  B = 10 / . 10 10 3 2 4 = = Wb m 26. (b) As shown in the following figure, the z − axis points out of the paper and the magnetic fields is directed into the paper, existing in the region between PQ and RS. The particle moves in a circular path of radius r in the magnetic field. It can just enter the region x  b for r  (b − q) Now (b a) qb mv r =  −  m q b a B v ( − )  m q b a B v ( ) min −  = . 27. (c) By using Flemings left hand rule. 28. (a) qB mK r 2 = and 2 A = Aq  2 2 (2 ) q b mK A  =  A  K. 29. (b) The net force acting on the electron is zero because it moves with constant velocity, due to it's motion on straight line.  = e + m = 0 Fnet F F  | Fe | =| F m |  e E = evB  B B E ve 0   = =       = o E    The time of motion inside the capacitor   lB v l t 0 = = . Q Y S O v x = a x = b R X  B P y x z e – e – x-y plane