Electromagnetic Induction-08-Answersheet

1. (b) 2. (c) 3. (a) 4. (c) 5. (d) 6. (d) 7. (a) 8. (c) 9. (b) 10. (a) 11. (d) 12. (c) 13. (d) 14. (d) 15. (c) 1. (d) 2. (a) 3. (b) 4. (c) 5. (c) 6. (d) 7. (d) 8. (a) 9. (b) 10. (c) 11. (a) 12. (c) 13. (d) 14. (c) 15. (d) SUBJECTIVE LEVEL - I (C.B.S.E.) 1. (a) along abcd (b) along acba (c) along abcda 2. (a) along pq (b) along qp, along xy (c) along xyz (d) along zyx (e) along xy (f) no induced current. 3. (a) along adcb (b) along adcb. 4. 1.6 103 V 5. 7.5106 V 6. 100V 7. max  0.603V, avg  0 Power Loss = 0.036 W, Source is external rotor. 8. Induced emf  6.28 105 V 9. L12  10. 1  a2 M12  0 2 b 11. 2.88106 A 12. B = 0.75 T 13. B = 6.5 V 0 a  a  14. M  2 ln 1  x    1.7 105 V .  B0 a2 ˆ 15. mR k SUBJECTIVE LEVEL - II 1. I   B 3 2. V  0I ln  d  2r cos   2  d  2r cos   ‘O’ is at higher potential than B. 3.   4. i min  mg 2Ba 5. VAB  VBC  3 a2k 8 3 a2k 8 VCA  0 VLOOP  3 a2k 4 6. (a) 3.2  H (b) 6.4  V (c) 1.28  C B2𝑙2     B2𝑙2  7.   0  Rm x , 0 exp  mR .t  8.  P  1 (a 2B)2 / R 2 9. 3.9 105 A 10. (a) 2.5103 V s (b) 17 V s (c) 0 V s SUBJECTIVE LEVEL - III 1. I  1 B (a  b) /  4 0 = 0.5 A. 2. I  1 (b2  a 2 )h /  4 3.   1 (1)n Ba2t where n = 1, 2, 3 ..... is the number of half-revolution that the loop i 2 performs at the the given moment t. i O t tn  2 n /  4. I   [1 ( 1)etR / L ] R 8 𝑙2B 5. 3 R 6. (a) RF  d2B2 mR . (b) RF d2 B2 .   b    I ln  b    0 I ln  a   0  a  F      7. (a) i    (b) R  2  . 2R     8. (a) 100 rad/s (b) 0.05 watt 10. i  mg tan  . 2aB PROBLEMS ANSWERS 1. (a) Current should be from P to Q (b) Required force = mg 10 6 (c) I  mg 6b B0 2. (a) d  L di  iR dt dt (b) q  1  0 I0l ln 2  Li  R  2 1    (c) L  T . R ln 4 3. (a) (b) | → | I L2B   3 I0 B (t)2 4 M 4. Power 2.5W T = 35.63ºC 0.8s Time 5. (a) VA  VB  5V; 24.5J (b) (i) 0.6 A (ii) 1.386 103s, 4.5104 J