33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Electromagnetic Induction
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59
31. Two coils have a mutual inductance 0.005 H. The
current changes in the first coil according to equation
I = I 0 sinwt, where I 0 = 10 A and w = 100p rad/sec. The
maximum value of e.m.f. in the second coil is
(a) p
(b) 5p
(c) 2p
(d) 4p(1998)
32. If N is the number of turns in a coil, the value of self
inductance varies as
(a) N 0
(b) N
(c) N 2 (d) N –2 (1993)
33. What is the self-inductance of a coil which produces
5 V when the current changes from 3 ampere to
2 ampere in one millisecond?
(a) 5000 henry (b) 5 milli-henry
(c) 50 henry (d) 5 henry (1993)
34. If the number of turns per unit length of a coil of
solenoid is doubled, the self-inductance of the
solenoid will
(a) remain unchanged (b) be halved
(c) be doubled (d) become four times
(1991)
35. A 100 millihenry coil carries a current of 1A. Energy
stored in its magnetic field is
(a) 0.5 J
(b) 1 J
(c) 0.05 J (d) 0.1 J (1991)
36. The current in self inductance L = 40 mH is to
be increased uniformly from 1 amp to 11 amp in
4 milliseconds. The e.m.f. induced in inductor
during process is
(a) 100 volt
(b) 0.4 volt
(c) 4.0 volt (d) 440 volt (1990)
37. An inductor may store energy in
(a) its electric field (b) its coils
(c) its magnetic field
(d) both in electric and magnetic fields (1990)
6.10 AC Generator
38. A wire loop is rotated in a magnetic field. The
frequency of change of direction of the induced
e.m.f. is
(a) four times per revolution
(b) six times per revolution
(c) once per revolution
(d) twice per revolution (NEET 2013)
39. In a region of magnetic induction B = 10 –2 tesla, a
circular coil of radius 30 cm and resistance p 2 ohm
is rotated about an axis which is perpendicular to
the direction of B and which forms a diameter of the
coil. If the coil rotates at 200 rpm the amplitude of
the alternating current induced in the coil is
(a) 4p 2 mA
(b) 30 mA
(c) 6 mA (d) 200 mA (1988)
ANSWER KEY
1. (c) 2. (a) 3. (d) 4. (a) 5. (b) 6. (a) 7. (d) 8. (a) 9. (c) 10. (b)
11. (a) 12. (c) 13. (a) 14. (b) 15. (b) 16. (d) 17. (d) 18. (b) 19. (b) 20. (b)
21. (c) 22. (a) 23. (a) 24. (d) 25. (b) 26. (d) 27. (a) 28. (a) 29. (d) 30. (a)
31. (b) 32. (c) 33. (b) 34. (d) 35. (c) 36. (a) 37. (c) 38. (d) 39. (c)
Hints & Explanations
1
1. (c) : B = ⎛ ⎝ ⎜ Wb ⎞
π 2 ⎠
⎟
m
Area of the disc normal to B is pR 2 cos60°.
Flux = B × Area normal
1
∴ Flux = × 004 . = 0.
02 Wb
2
2. (a) : Here N = 800, A = 0.05 m 2 , Dt = 0.1 s
B = 5 × 10 –5 T
Induced emf, e = – ∆ φ ( φf
− φi)
=−
∆t
∆t
f i = N ( B. A ) = 800 × 5 × 10 –5 × 0.05 × cos 0°
= 2 × 10 –3 T m 2
f f = 0
\ ε= − ( − × − 3
0 2 10 )
= 2 × 10 –2 V = 0.02 V
01 .
3. (d) : Emf generated in loop 1,
dφ
d
ε1 =− =− ⋅ =− =− ×
dt dt B A
d
dB
( ) ( BA)
A
dt dt
2
ε1
=− ⎛ π
⎝ ⎜ ⎞
r dB ⎠
⎟
dt
(Q A = pr 2 because dB is restricted upto radius r.)
dt
Emf generated in loop 2,
d
ε 2 =− =− 0× = 0
dt BA d
( ) ( A)
dt
4. (a) : Here, f = 50t 2 + 4 Wb, R = 400 W