JEE Physics Electromagnetic Induction MCQs Set 10

Answer: (a) 25 V

Explanation:
1. Motional emf: \( \varepsilon=(\vec v\times\vec B)\cdot\vec\ell \) with \( \vec v=\hat i \) and \( \vec\ell=5\hat j \).
2. Compute \( \hat i\times(3\hat i+4\hat j+5\hat k)=4\hat k-5\hat j \).
3. Dot with \( 5\hat j \): \( -25 \), so the magnitude is 25 V.

Teacher's Note:
1. Only the components of B perpendicular to both v and the rod matter: here the z-component, 5 T, gives \( 5\times1\times5=25 \) V.

Answer: (b) \( \frac{2a}{v} \)

Explanation:
1. The loop is wider than the field region, so the flux changes only while the leading edge crosses the region and then while the trailing edge crosses it.
2. Each crossing takes time \( \frac av \).
3. In between, the whole region lies inside the loop and the flux is constant, so the total time is \( \frac{2a}{v} \).

Teacher's Note:
1. Emf exists only while the flux is changing.

Answer: (d) 1 J

Explanation:
1. Speed: \( v=\frac{1}{1}=1 \) m/s. Emf: \( B\ell v=1 \) V; current \( i=\frac{1}{1}=1 \) A.
2. Power dissipated \( i^2R=1 \) W for 1 s.
3. Work = heat = \( 1\times1=1 \) J.

Teacher's Note:
1. Work done equals the heat produced in the loop as it leaves the field.

Answer: (a) 2L

Explanation:
1. For a solenoid \( L=\frac{\mu_0N^2A}{\ell} \). With a fixed wire length the number of turns is \( N\propto\frac1r \).
2. New solenoid: \( N'=2N \), \( A'=\frac A4 \), length \( \frac\ell2 \).
3. So \( L'=\frac{\mu_0(2N)^2(A/4)}{\ell/2}=2L \).

Teacher's Note:
1. Smaller radius means more turns from the same wire: N doubles.

Answer: (c) \( \frac{L_2}{L_1} \)

Explanation:
1. Both inductors have the same voltage: \( L_1\frac{di_1}{dt}=L_2\frac{di_2}{dt} \).
2. Integrating from zero current: \( L_1i_1=L_2i_2 \).
3. So \( \frac{i_1}{i_2}=\frac{L_2}{L_1} \).

Teacher's Note:
1. In parallel inductors the larger inductance carries the smaller current.

Answer: (d) \( \frac{2}{\ln(10/9)} \)

Explanation:
1. Decay: \( i=i_0e^{-t/\tau} \), so \( 18=20e^{-2/\tau} \).
2. Then \( e^{-2/\tau}=\frac{9}{10}\Rightarrow\frac2\tau=\ln\frac{10}{9} \).
3. Hence \( \tau=\frac{2}{\ln(10/9)} \) s, about 19 s.

Teacher's Note:
1. Take the natural logarithm of the ratio of currents.

Answer: (b) 1 ms

Explanation:
1. Maximum induced emf in the inductor (at t = 0) equals the battery emf: \( \varepsilon=6 \) V.
2. Final current \( \frac{\varepsilon}{R}=2 \) A, so \( R=3 \) ohm.
3. Time constant: \( \tau=\frac LR=\frac{3\ \mathrm{mH}}{3}=1 \) ms.

Teacher's Note:
1. At t = 0 the whole battery emf appears across the inductor.

Answer: (c) increases

Explanation:
1. Removing the core reduces the inductance abruptly from L to a smaller value L'.
2. The flux linkage \( Li \) cannot change abruptly for a circuit with only finite emfs, so \( L'i'=Li \).
3. Since L' < L, the current i' is larger: it increases (before settling back to the same final value E/R).

Teacher's Note:
1. Think of flux linkage as the quantity that cannot jump.

Answer: (a) 12 A

Explanation:
1. The resistor branch carries a steady current \( \frac{10}{5}=2 \) A.
2. The ideal-battery voltage is directly across the inductor, so \( i_L=\frac{E}{L}t=\frac{10}{2}\times2=10 \) A at t = 2 s.
3. Battery current: \( 2+10=12 \) A.

Teacher's Note:
1. With a constant voltage across an ideal inductor, the current grows linearly.

Answer: (b) \( \frac{2U}{P} \)

Explanation:
1. Energy: \( U=\frac12Li^2 \). Heating rate: \( P=i^2R \).
2. Ratio: \( \frac UP=\frac{L}{2R} \).
3. So \( \tau=\frac LR=\frac{2U}{P} \).

Teacher's Note:
1. Divide the two given expressions to eliminate the current.

Answer: (d) 200 mH

Explanation:
1. Mutual inductance satisfies \( M=k\sqrt{L_1L_2} \) with \( k\le1 \).
2. Maximum (perfect coupling): \( \sqrt{100\times400}=200 \) mH.

Teacher's Note:
1. The coupling coefficient k cannot exceed 1.

Answer: (a) zero

Explanation:
1. The wire's field is circular around it and lies in the plane of the ring.
2. No field line passes through the ring's surface, so the flux is zero for any current.
3. Hence the mutual inductance is zero.

Teacher's Note:
1. Check whether the field is parallel to the surface of the loop.

Answer: (c) anticlockwise and clockwise

Explanation:
1. When the switch closes, the flux through Q due to P's clockwise current grows. Q opposes the growth with a current opposite to P's: \( I_{Q1} \) is anticlockwise.
2. While the current is steady nothing is induced.
3. When the switch opens, P's flux decays, and Q tries to maintain it with a current in the same sense as P's: \( I_{Q2} \) is clockwise.

Teacher's Note:
1. Growth: induced current opposes the original; decay: induced current has the same sense as the original.

Answer: (d) the same

Explanation:
1. Emf is proportional to the number of turns: \( \varepsilon\to4\varepsilon \).
2. Resistance \( R=\rho\frac{L}{a} \): the length of wire is 4 times and the cross-section area is one quarter, so \( R\to16R \).
3. Power \( \frac{\varepsilon^2}{R}\to\frac{16\varepsilon^2}{16R} \) is unchanged.

Teacher's Note:
1. Track both emf (squared) and resistance when the wire is changed.

Answer: (b) \( \frac{8qR}{mg} \)

Explanation:
1. Induced electric field at the ring: \( E\cdot2\pi R=\pi R^2\frac{dB}{dt}=\pi R^2(8t) \), so \( E=4Rt \).
2. Torque of the electric force on the charge: \( \tau=qER=4qR^2t \), equal to \( 8qR^2 \) at t = 2 s.
3. Limiting friction torque on the ring: \( \mu mgR \). Equating, \( \mu=\frac{8qR}{mg} \).

Teacher's Note:
1. The ring starts to rotate when the electric torque equals the maximum friction torque.

Answer: (c) \( B_0\ell\sqrt{R^2-\ell^2} \)

Explanation:
1. Join the ends of the chord to the centre to form a triangle. The emf around the triangle equals \( \frac{dB}{dt}\times\text{area} \) and the radial sides contribute nothing, because the induced field is circular.
2. Distance of the chord from the centre: \( \sqrt{R^2-\ell^2} \), so the triangle area is \( \frac12(2\ell)\sqrt{R^2-\ell^2}=\ell\sqrt{R^2-\ell^2} \).
3. Hence \( V=B_0\ell\sqrt{R^2-\ell^2} \).

Teacher's Note:
1. Radial elements carry no emf because the induced electric field is tangential.

Answer: (a) 4 H

Explanation:
1. Increasing: \( iR+L(10)=140 \). Decreasing: \( iR-L(10)=60 \).
2. Subtract: \( 20L=80 \), so \( L=4 \) H.
3. (Also \( iR=100 \) V, so \( R=20 \) ohm.)

Teacher's Note:
1. The inductive drop changes sign with di/dt; the resistive drop does not.

Answer: (b) zero at t = 0, rises to a maximum and then decays towards zero

Explanation:
1. Inductor energy \( U=\frac12Li^2 \) so \( \frac{dU}{dt}=Li\frac{di}{dt} \).
2. With \( i=\frac ER(1-e^{-t/\tau}) \) and \( \frac{di}{dt}=\frac{E}{L}e^{-t/\tau} \): \( \frac{dU}{dt}=\frac{E^2}{R}e^{-t/\tau}(1-e^{-t/\tau}) \).
3. This is zero at t = 0 (no current) and at t \( \to\infty \) (no di/dt), with a maximum in between at \( t=\tau\ln2 \).

Teacher's Note:
1. The product of a growing current and a decaying di/dt rises and falls.

Answer: (d) \( \frac{LE^2}{2R_2^2} \)

Explanation:
1. In steady state at position 2 the inductor current is \( I_0=\frac{E}{R_2} \) (the inductor has no resistance of its own).
2. On switching to position 1 the inductor discharges through \( R_1 \) only, so all its stored energy becomes heat in \( R_1 \).
3. Heat: \( \frac12LI_0^2=\frac{LE^2}{2R_2^2} \).

Teacher's Note:
1. The heat dissipated in a discharging circuit equals the energy stored initially.

Answer: (a) 4L

Explanation:
1. For perfect coupling \( M=L \).
2. Series aiding: \( L_{total}=L+L+2M=4L \).

Teacher's Note:
1. Aiding series connection: total = \( L_1+L_2+2M \); opposing: \( L_1+L_2-2M \).

Answer: (c) 3

Explanation:
1. Turns are uniformly distributed in radius, so the number of turns between r and r+dr is \( \frac Na\,dr \).
2. Total flux linkage: \( \int_0^a\frac Na\pi r^2B\,dr=\frac{N\pi a^2}{3}B \).
3. Emf amplitude: \( \frac13\pi a^2N\omega B_0 \), so x = 3.

Teacher's Note:
1. Turns of different radii enclose different areas; average the area over the radius.

Answer: (d) 40

Explanation:
1. At terminal velocity the weight's power equals the total electrical power: \( mgv=P_1+P_2 \), so \( 0.2\times9.8\times v=1.96 \) and \( v=x=1 \) m/s.
2. Emf \( =B\ell v=0.6 \) V is across both resistors: \( R_1=\frac{0.36}{0.76}=0.474 \) ohm and \( R_2=\frac{0.36}{1.2}=0.3 \) ohm.
3. So \( x+76y+10z=1+36+3=40 \).

Teacher's Note:
1. Both resistors are in parallel across the same emf.

Answer: (b) 3

Explanation:
1. The force on a small element at distance r is \( dF\propto\frac{dr}{r} \).
2. Point of application (centre of force): \( \bar r=\frac{\int r\,dF}{\int dF}=\frac{\int_a^{3a}dr}{\int_a^{3a}dr/r}=\frac{2a}{\ln3} \).
3. So x = 3.

Teacher's Note:
1. Treat the force distribution like parallel forces and find its centre.

Answer: (c) 12

Explanation:
1. Only the two sides parallel to the wire have emfs: \( \varepsilon=\frac{\mu_0i}{2\pi}\ell v\left(\frac1a-\frac{1}{a+\ell}\right) \).
2. With \( a=2\ell \): \( \frac{\mu_0iv}{2\pi}\left(\frac12-\frac13\right)=\frac{\mu_0iv}{12\pi} \).
3. So x = 12.

Teacher's Note:
1. The near side is in a stronger field, so it dominates and the two emfs oppose.

Answer: (a) 3

Explanation:
1. Flux linkage is conserved at the instant of change: \( L\cdot\frac ER=\frac L4\,i(0^+) \), so \( i(0^+)=\frac{4E}{R} \).
2. Final value \( \frac ER \), so \( i=\frac ER+\left(\frac{4E}{R}-\frac ER\right)e^{-t/\tau} \) with \( \tau=\frac{L/4}{R} \).
3. Thus \( i=\frac ER\left[1+3e^{-4Rt/L}\right] \) and x = 3.

Teacher's Note:
1. The new time constant is \( (L/4)/R \), which gives the exponent \( 4Rt/L \).

Answer: (b) (i), (ii) and (iv) only

Explanation:
1. (i) True: at constant velocity the work done by the agent all appears as heat.
2. (ii) True: a larger force makes the net force positive, so the rod accelerates and kinetic energy rises.
3. (iii) False: Lenz's law always holds. (iv) True: \( P=\frac{B^2\ell^2v_0^2}{R} \) halves when R doubles.

Teacher's Note:
1. Lenz's law is a statement of energy conservation and is never violated.