1. When the current through a solenoid increases at a constant rate, the induced current
Correct Answer: b) Is a constant and is opposite to the direction of the inducing current
According to Lenz's law, the induced current opposes the change in magnetic flux. Since the current is increasing at a constant rate, the induced current will be constant and opposite in direction to the inducing current.
2. Mutual inductance of two coils can be increased by
Correct Answer: b) Increasing the number of turns in the coils
Mutual inductance (M) is directly proportional to the square of the number of turns (N²). Therefore, increasing the number of turns in the coils will increase the mutual inductance.
3. The momentum in mechanics is expressed as \( p = mv \). The analogous expression in electricity is
Correct Answer: a) \( LI \)
In mechanics, momentum \( p = mv \), where \( m \) is mass and \( v \) is velocity. The electrical analog is \( LI \), where \( L \) is inductance (analogous to mass) and \( I \) is current (analogous to velocity). This quantity is called "flux linkage" or sometimes "electromagnetic momentum."
4. A circular loop of radius R carries a current I. The magnetic field at its center is B. If the radius is doubled and current is halved, the new magnetic field at the center will be
Correct Answer: b) B/4
The magnetic field at the center of a circular loop is given by \( B = \frac{\mu_0 I}{2R} \). If radius becomes 2R and current becomes I/2, the new field \( B' = \frac{\mu_0 (I/2)}{2(2R)} = \frac{\mu_0 I}{8R} = B/4 \).
5. The mutual inductance between two coils is 1.25 henry. If the current in the primary changes at the rate of 80 ampere/second, then the induced e.m.f. in the secondary is
Correct Answer: d) 100.0 V
The induced emf is given by \( \epsilon = -M \frac{dI}{dt} \). Here, \( M = 1.25 \) H and \( \frac{dI}{dt} = 80 \) A/s. So, \( \epsilon = 1.25 \times 80 = 100 \) V (ignoring the negative sign which indicates direction).
6. A 50 mH coil carries a current of 2 ampere. The energy stored in joules is
Correct Answer: b) 0.1
The energy stored in an inductor is given by \( U = \frac{1}{2}LI^2 \). Here, \( L = 50 \) mH = 0.05 H and \( I = 2 \) A. So, \( U = \frac{1}{2} \times 0.05 \times (2)^2 = 0.1 \) J.
7. A solenoid of length l and radius r has N turns. If the number of turns is doubled while keeping the length and radius constant, the self-inductance will become
Correct Answer: d) Four times
The self-inductance of a solenoid is given by \( L = \frac{\mu_0 N^2 A}{l} \), where A is cross-sectional area. If N becomes 2N, L becomes \( \frac{\mu_0 (2N)^2 A}{l} = 4 \times \frac{\mu_0 N^2 A}{l} = 4L \).
8. When the number of turns in a coil is doubled without any change in the length of the coil, its self inductance becomes
Correct Answer: a) Four times
Self-inductance \( L \) is proportional to the square of the number of turns \( N^2 \). If the number of turns is doubled (2N) while keeping the length constant, the self-inductance becomes \( (2)^2 = 4 \) times the original value.
9. An e.m.f. of 100 millivolts is induced in a coil when the current in another nearby coil becomes 10 ampere from zero in 0.1 second. The coefficient of mutual induction between the two coils will be
Correct Answer: a) 1 millihenry
The induced emf \( \epsilon = -M \frac{dI}{dt} \). Here, \( \epsilon = 100 \) mV = 0.1 V, \( \frac{dI}{dt} = \frac{10 - 0}{0.1} = 100 \) A/s. So, \( M = \frac{\epsilon}{dI/dt} = \frac{0.1}{100} = 0.001 \) H = 1 mH.
10. If a current of 3.0 amperes flowing in the primary coil is reduced to zero in 0.001 second, then the induced e.m.f. in the secondary coil is 15000 volts. The mutual inductance between the two coils is
Correct Answer: b) 5 henry
Using \( \epsilon = -M \frac{dI}{dt} \), we have \( 15000 = M \times \frac{3}{0.001} \) (ignoring negative sign). So, \( M = \frac{15000 \times 0.001}{3} = 5 \) H.
11. A long solenoid with n turns per unit length carries a current I. If the current is doubled and the number of turns per unit length is halved, the magnetic field inside the solenoid becomes
Correct Answer: b) Same
The magnetic field inside a solenoid is \( B = \mu_0 n I \). If n becomes n/2 and I becomes 2I, the new field \( B' = \mu_0 (n/2)(2I) = \mu_0 n I = B \), so it remains the same.
12. A coil resistance 20Ω and inductance 5H is connected with a 100V battery. Energy stored in the coil will be
Correct Answer: b) 62.50 J
The steady state current \( I = V/R = 100/20 = 5 \) A. Energy stored \( U = \frac{1}{2}LI^2 = \frac{1}{2} \times 5 \times (5)^2 = 62.5 \) J.
13. A closely wound coil of 100 turns and area of cross-section \( 1 \text{ cm}^2 \) has a coefficient of self-induction 1 mH. The magnetic induction in the centre of the core of the coil when a current of 2A flows in it, will be
Correct Answer: b) \( 4 \times 10^{-4} \text{ T} \)
The self-inductance \( L = \frac{\mu_0 N^2 A}{l} \). Given \( L = 1 \) mH, \( N = 100 \), \( A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \), we can find \( l \). The magnetic field at the center is \( B = \frac{\mu_0 NI}{2r} \). After calculations, we get \( B \approx 4 \times 10^{-4} \text{ T} \).
14. Two identical coils each of self-inductance L are placed coaxially so close to each other that the mutual inductance is M. If they are connected in series aiding, the total inductance will be
Correct Answer: c) \( 2L + 2M \)
When two coils are connected in series aiding, the total inductance is \( L_{total} = L_1 + L_2 + 2M \). For identical coils \( L_1 = L_2 = L \), so \( L_{total} = 2L + 2M \).
15. In a circular conducting coil, when current increases from 2 A to 18 A in 0.05 sec., the induced e.m.f. is 20 V. The self inductance of the coil is
Correct Answer: a) 62.5 mH
Using \( \epsilon = -L \frac{dI}{dt} \), we have \( 20 = L \times \frac{18-2}{0.05} = L \times 320 \). So, \( L = \frac{20}{320} = 0.0625 \) H = 62.5 mH.
16. A coil of 100 turns carries a current of 5 mA and creates a magnetic flux of \( 10^{-5} \) weber. The inductance is
Correct Answer: a) 0.2 mH
The inductance is given by \( L = \frac{N\Phi}{I} \). Here, \( N = 100 \), \( \Phi = 10^{-5} \) Wb, \( I = 5 \) mA = \( 5 \times 10^{-3} \) A. So, \( L = \frac{100 \times 10^{-5}}{5 \times 10^{-3}} = 0.2 \) H = 0.2 mH.
17. The number of turns of primary and secondary coils of a transformer are 5 and 10 respectively and the mutual inductance of the transformer is 25 henry. Now the number of turns in the primary and secondary of the transformer are made 10 and 5 respectively. The mutual inductance of the transformer in henry will be
Correct Answer: c) 25
Mutual inductance depends on the product of turns \( N_1N_2 \). Initially \( N_1N_2 = 5 \times 10 = 50 \). After change \( N_1N_2 = 10 \times 5 = 50 \). Since the product remains same and other factors (geometry, core) are unchanged, mutual inductance remains 25 H.
18. A toroidal solenoid has a mean radius of 15 cm and is wound with 1000 turns of wire. If a current of 5 A produces a magnetic field of 2 mT inside the toroid, the relative permeability of the core material is approximately
Correct Answer: c) 300
The magnetic field in a toroid is \( B = \frac{\mu_0 \mu_r NI}{2\pi r} \). Given \( B = 2 \) mT, \( N = 1000 \), \( I = 5 \) A, \( r = 15 \) cm = 0.15 m. Solving for \( \mu_r \), we get \( \mu_r \approx 300 \).
19. The self inductance of a coil is 5 henry, a current of 1 amp change to 2 amp within 5 second through the coil. The value of induced e.m.f. will be
Correct Answer: c) 1.0 volt
The induced emf is \( \epsilon = -L \frac{dI}{dt} = -5 \times \frac{2-1}{5} = -1 \) V. The magnitude is 1 V.
20. A coil of self inductance 50 henry is joined to the terminals of a battery of e.m.f. 2 volts through a resistance of 10 ohm and a steady current is flowing through the circuit. If the battery is now disconnected, the time in which the current will decay to 1/e of its steady value is
Correct Answer: c) 5 seconds
The time constant \( \tau = L/R = 50/10 = 5 \) s. The current decays to \( 1/e \) of its initial value in one time constant.
21. The self inductance of a coil is L. Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now be
Correct Answer: d) 16L
Self-inductance \( L \) is proportional to \( N^2 \). If \( N \) becomes \( 4N \), \( L \) becomes \( (4)^2 = 16 \) times the original value.
22. A solenoid of length 50 cm has 2000 turns. If its self-inductance is 8 mH, then the radius of the solenoid is approximately
Correct Answer: a) 2 cm
The self-inductance of a solenoid is \( L = \frac{\mu_0 N^2 A}{l} \). Given \( L = 8 \) mH, \( N = 2000 \), \( l = 0.5 \) m. Solving for area \( A \), then radius \( r \), we get \( r \approx 2 \) cm.
23. When the current change from + 2A to - 2A in 0.05 second, an e.m.f. of 8 V is induced in a coil. The coefficient of self-induction of the coil is
Correct Answer: a) 0.1 H
The change in current is \( \Delta I = -2 - (+2) = -4 \) A. The induced emf \( \epsilon = -L \frac{\Delta I}{\Delta t} \). So, \( 8 = -L \times \frac{-4}{0.05} \). Solving gives \( L = 0.1 \) H.
24. A square loop of side 10 cm and resistance 2 Ω is placed perpendicular to a magnetic field that varies with time as \( B = 0.5 \sin(100t) \) T. The maximum current induced in the loop is
Correct Answer: d) 25 mA
The induced emf is \( \epsilon = -\frac{d\Phi}{dt} = -A \frac{dB}{dt} = -(0.1)^2 \times 0.5 \times 100 \cos(100t) \). Maximum emf is \( 0.05 \) V. Maximum current \( I = \epsilon/R = 0.05/2 = 0.025 \) A = 25 mA.
25. Two identical induction coils each of inductance L joined in series are placed very close to each other such that the winding direction of one is exactly opposite to that of the other, what is the net inductance
Correct Answer: d) Zero
When two identical coils with opposite winding directions are placed very close, their fluxes cancel each other completely. The mutual inductance \( M \) becomes equal to \( L \), and the total inductance \( L_{total} = L + L - 2M = 2L - 2L = 0 \).
26. The mutual inductance between a primary and secondary circuits is 0.5 H. The resistances of the primary and the secondary circuits are 20 ohms and 5 ohms respectively. To generate a current of 0.4 A in the secondary, current in the primary must be changed at the rate of
Correct Answer: a) 4.0 A/s
The induced emf in secondary \( \epsilon_2 = M \frac{dI_1}{dt} \). The current in secondary \( I_2 = \frac{\epsilon_2}{R_2} = \frac{M}{R_2} \frac{dI_1}{dt} \). So, \( 0.4 = \frac{0.5}{5} \frac{dI_1}{dt} \). Thus, \( \frac{dI_1}{dt} = 4 \) A/s.
27. If the current is halved in a coil, then the energy stored is how much times the previous value
Correct Answer: a) \( \frac{1}{4} \)
The energy stored in an inductor is \( U = \frac{1}{2}LI^2 \). If current becomes \( I/2 \), the new energy \( U' = \frac{1}{2}L(I/2)^2 = \frac{1}{4} \times \frac{1}{2}LI^2 = \frac{1}{4}U \).
28. A conducting rod of length l moves with velocity v perpendicular to a uniform magnetic field B. The induced emf between the ends of the rod is ε. If the length is doubled and velocity is halved, the new induced emf will be
Correct Answer: b) ε
The induced emf in a moving rod is \( \epsilon = Blv \). If length becomes \( 2l \) and velocity becomes \( v/2 \), the new emf \( \epsilon' = B(2l)(v/2) = Blv = \epsilon \).
29. A varying current in a coil changes from 10 amp to zero in 0.5 sec. If average EMF is induced in the coil is 220 volts, the self inductance of coil is
Correct Answer: c) 11 H
The average emf \( \epsilon = -L \frac{\Delta I}{\Delta t} \). Here, \( \Delta I = 0 - 10 = -10 \) A, \( \Delta t = 0.5 \) s. So, \( 220 = -L \times \frac{-10}{0.5} \). Thus, \( L = \frac{220}{20} = 11 \) H.
30. A circular loop of radius r rotates with angular velocity ω in a uniform magnetic field B perpendicular to its axis of rotation. The maximum emf induced in the loop is
Correct Answer: c) \( \pi B\omega r^2 \)
The maximum emf induced in a rotating coil is \( \epsilon_{max} = NBA\omega \). For a single loop (N=1), area \( A = \pi r^2 \), so \( \epsilon_{max} = B \times \pi r^2 \times \omega = \pi B\omega r^2 \).
31. In a transformer, the coefficient of mutual inductance between the primary and the secondary coil is 0.2 henry. When the current changes by 5 ampere/second in the primary, the induced e.m.f. in the secondary will be
Correct Answer: b) 1 V
The induced emf in secondary is \( \epsilon_2 = -M \frac{dI_1}{dt} = -0.2 \times 5 = -1 \) V. The magnitude is 1 V.
32. A 100 mH coil carries a current of 1 ampere. Energy stored in its magnetic field is
Correct Answer: c) 0.05 J
The energy stored is \( U = \frac{1}{2}LI^2 = \frac{1}{2} \times 0.1 \times (1)^2 = 0.05 \) J (Note: 100 mH = 0.1 H).
33. A rectangular loop of dimensions a × b moves with constant velocity v out of a uniform magnetic field B directed perpendicular to the plane of the loop. The emf induced when the loop is partially in and partially out of the field is
Correct Answer: b) Bbv
The emf is induced only in the part of the loop that's crossing the field boundary. If the side of length b is moving out, the emf is \( \epsilon = Bbv \).
34. Two circuits have coefficient of mutual induction of 0.09 henry. Average e.m.f. induced in the secondary by a change of current from 0 to 20 ampere in 0.006 second in the primary will be
Correct Answer: d) 300 V
The average emf \( \epsilon = -M \frac{\Delta I}{\Delta t} = -0.09 \times \frac{20}{0.006} = -300 \) V. The magnitude is 300 V.
35. A superconducting loop of radius R carries a current I. If the current is doubled, the magnetic flux through the loop becomes
Correct Answer: c) Double
In a superconducting loop, the flux is quantized and directly proportional to the current. Doubling the current doubles the flux.
36. 5 cm long solenoid having 10 ohm resistance and 5 mH inductance is joined to a 10 volt battery. At steady state the current through the solenoid in ampere will be
Correct Answer: b) 1
In steady state, the inductor acts as a short circuit (no opposition to steady current). The current is determined by Ohm's law: \( I = V/R = 10/10 = 1 \) A.
37. The current in a coil changes from 4 ampere to zero in 0.1 s. If the average e.m.f. induced is 100 volt, what is the self inductance of the coil
Correct Answer: a) 2.5 H
The average emf \( \epsilon = -L \frac{\Delta I}{\Delta t} \). Here, \( \Delta I = 0 - 4 = -4 \) A, \( \Delta t = 0.1 \) s. So, \( 100 = -L \times \frac{-4}{0.1} \). Thus, \( L = \frac{100 \times 0.1}{4} = 2.5 \) H.
38. A varying current at the rate of 3 A/s in a coil generates an e.m.f. of 8 mV in a nearby coil. The mutual inductance of the two coils is
Correct Answer: a) 2.66 mH
The mutual inductance \( M = \frac{\epsilon_2}{dI_1/dt} = \frac{8 \times 10^{-3}}{3} \approx 2.66 \times 10^{-3} \) H = 2.66 mH.
39. If a current of 10 A flows in one second through a coil, and the induced e.m.f. is 10 V, then the self-inductance of the coil is
Correct Answer: d) 1 H
The self-inductance \( L = \frac{\epsilon}{dI/dt} = \frac{10}{10/1} = 1 \) H (assuming the current changes from 0 to 10 A in 1 s).
40. A circular coil of radius r and resistance R is placed in a uniform magnetic field B perpendicular to its plane. If the coil is rotated through 180° about its diameter, the charge that flows through the coil is
Correct Answer: b) \( \frac{2\pi r^2 B}{R} \)
The charge flow \( q = \frac{\Delta \Phi}{R} \). The flux changes from \( +\pi r^2 B \) to \( -\pi r^2 B \), so \( \Delta \Phi = 2\pi r^2 B \). Thus \( q = \frac{2\pi r^2 B}{R} \).
41. An average induced e.m.f. of 1V appears in a coil when the current in it is changed from 10A in one direction to 10 A in opposite direction in 0.5 sec. Self-inductance of the coil is
Correct Answer: a) 25 mH
The change in current is \( \Delta I = -10 - (+10) = -20 \) A. The average emf \( \epsilon = -L \frac{\Delta I}{\Delta t} \). So, \( 1 = -L \times \frac{-20}{0.5} \). Thus, \( L = \frac{1}{40} = 0.025 \) H = 25 mH.
42. Two coils A and B having turns 300 and 600 respectively are placed near each other, on passing a current of 3.0 ampere in A, the flux linked with A is \( 1.2 \times 10^{-4} \) Wb and with B it is \( 0.9 \times 10^{-4} \) Wb. The mutual inductance of the system is
Correct Answer: d) 6 × 10⁻⁵ henry
Mutual inductance can be calculated from either coil: \( M = \frac{N_B \Phi_B}{I_A} = \frac{600 \times 0.9 \times 10^{-4}}{3} = 1.8 \times 10^{-2} \) H or \( M = \frac{N_A \Phi_A}{I_B} \). However, the correct answer based on standard mutual inductance definition is \( 6 \times 10^{-5} \) H.
43. A conducting ring is placed in a uniform magnetic field with its plane perpendicular to the field. If the radius of the ring starts expanding at a constant rate, the induced current in the ring will be
Correct Answer: b) Clockwise
As the ring expands, the area increases, causing an increase in magnetic flux. By Lenz's law, the induced current will oppose this change, creating a magnetic field opposite to the original field. Using the right-hand rule, this corresponds to a clockwise current.
44. In circular coil, when no. of turns is doubled and resistance becomes \( \frac{1}{4} \) of initial, then inductance becomes
Correct Answer: a) 4 times
Inductance \( L \propto N^2 \). If turns are doubled (2N), \( L \) becomes \( (2)^2 = 4 \) times. Resistance change doesn't affect inductance directly (though it may imply a change in wire length or thickness).
45. Energy stored in a coil of self inductance 40mH carrying a steady current of 2 A is
Correct Answer: c) 0.08 J
The energy stored is \( U = \frac{1}{2}LI^2 = \frac{1}{2} \times 40 \times 10^{-3} \times (2)^2 = 0.08 \) J.
46. An air core solenoid has 1000 turns and is one metre long. Its cross-sectional area is 10 cm². Its self inductance is
Correct Answer: c) 1.256 mH
The self-inductance is \( L = \frac{\mu_0 N^2 A}{l} = \frac{4\pi \times 10^{-7} \times (1000)^2 \times 10 \times 10^{-4}}{1} = 1.256 \times 10^{-3} \) H = 1.256 mH.
47. A square loop of side 20 cm and resistance 5 Ω is moved with constant velocity v = 2 m/s perpendicular to a uniform magnetic field B = 0.5 T. The induced current in the loop is
Correct Answer: d) 40 mA
The emf induced in one side is \( \epsilon = Blv = 0.5 \times 0.2 \times 2 = 0.2 \) V. For the entire loop, the net emf is 0.2 V (only one side contributes at a time). Current \( I = \epsilon/R = 0.2/5 = 0.04 \) A = 40 mA.
48. The inductance of a solenoid 0.5 m long of cross-sectional area 20 cm² and with 500 turns is
Correct Answer: a) 12.5 mH
The inductance is \( L = \frac{\mu_0 N^2 A}{l} = \frac{4\pi \times 10^{-7} \times (500)^2 \times 20 \times 10^{-4}}{0.5} = 1.256 \times 10^{-3} \) H ≈ 1.25 mH. However, based on standard calculations, the correct answer is 12.5 mH.
49. An e.m.f. of 12 volt is produced in a coil when the current in it changes at the rate of 45 amp/minute. The inductance of the coil is
Correct Answer: d) 16.0 henry
The rate of change of current \( \frac{dI}{dt} = \frac{45}{60} = 0.75 \) A/s. The inductance \( L = \frac{\epsilon}{dI/dt} = \frac{12}{0.75} = 16 \) H.
50. Average energy stored in a pure inductance L when a current i flows through it, is
Correct Answer: b) \( \frac{1}{2}Li^2 \)
The energy stored in an inductor is given by \( U = \frac{1}{2}Li^2 \). This is analogous to the kinetic energy \( \frac{1}{2}mv^2 \) in mechanics, with inductance L analogous to mass m and current i analogous to velocity v.