JEE Physics Electromagnetic Induction MCQs Set 06

Answer: (b) anticlockwise while entering

Explanation:
1. While entering, the flux into the page increases, so by Lenz's law the induced current produces a field out of the page: anticlockwise.
2. When the loop is wholly inside the field, there is no emf. While leaving, the flux decreases and the current is clockwise.
3. Partially outside means the flux is changing, so the current is not zero.

Teacher's Note:
1. Lenz's law: the induced current opposes the change in flux.
2. Right-hand rule: anticlockwise current gives a field out of the page.

Answer: (a) \( r^2 \)

Explanation:
1. Charge \( q=\frac{\Delta\Phi}{R}=\frac{B\pi r^2}{R} \).
2. This is directly proportional to \( r^2 \) (and to B, inversely to R).

Teacher's Note:
1. Total charge flown = change of flux / resistance.
2. It is independent of the time taken.

Answer: (d) \( 1.6\times10^{-10}\ \text{V} \)

Explanation:
1. Initial field at the ring \( B=\frac{\mu_0I}{2\pi d}=\frac{2\times10^{-7}\times4}{0.1}=8\times10^{-6}\ \text{T} \).
2. Flux \( \Phi=BA=8\times10^{-6}\times2\times10^{-6}=1.6\times10^{-11}\ \text{Wb} \).
3. \( \varepsilon=\frac{\Delta\Phi}{\Delta t}=\frac{1.6\times10^{-11}}{0.1}=1.6\times10^{-10}\ \text{V} \).

Teacher's Note:
1. The ring is small, so the field is uniform over it.
2. Convert mm² to m².

Answer: (c) \( \frac{Ba^2}{R} \)

Explanation:
1. Initial flux \( \Phi_i=Ba^2 \), final flux \( \Phi_f=0 \).
2. \( q=\frac{\Delta\Phi}{R}=\frac{Ba^2}{R} \).

Teacher's Note:
1. Charge = change in flux / resistance.
2. When the loop straightens its area is zero.

Answer: (b) \( 2B\omega a^2 \)

Explanation:
1. The emf between the ends depends on the straight line joining them (the diameter) of length \( 2a \), rotating about its end.
2. Emf of a rod rotating about one end \( =\tfrac12B\omega\ell^2=\tfrac12B\omega(2a)^2=2B\omega a^2 \).

Teacher's Note:
1. The emf between two points depends only on the positions of the ends in a uniform field.
2. The chord replaces the curved wire.

Answer: (a) \( \mu_0nN\pi a^2 \)

Explanation:
1. Field of the solenoid \( B=\mu_0nI \). Flux linked with the small coil \( N\Phi=N(\mu_0nI)\pi a^2 \).
2. \( M=\frac{N\Phi}{I}=\mu_0nN\pi a^2 \).

Teacher's Note:
1. Mutual inductance = total flux linkage divided by the current.
2. The coil has N turns, but the solenoid's n enters once.

Answer: (d) \( \frac{f_1}{f_2} \)

Explanation:
1. \( e_{max}=L\omega I_0=2\pi fLI_0 \), so \( e\propto f \) for the same current amplitude.
2. \( \frac{e_1}{e_2}=\frac{f_1}{f_2} \).

Teacher's Note:
1. Induced emf = L dI/dt, whose maximum is LωI0.
2. Same coil and same current amplitude.

Answer: (c) 0.2 H

Explanation:
1. \( \varepsilon=L\frac{\Delta I}{\Delta t}\Rightarrow L=\frac{8}{(4-2)/0.05}=\frac{8}{40}=0.2\ \text{H} \).

Teacher's Note:
1. Rate of change of current = ΔI/Δt.
2. ε = L (dI/dt).

Answer: (b) remain the same

Explanation:
1. The emf between the ends depends only on the effective length perpendicular to the velocity, which is the distance between the end points (the rails' separation).
2. The effective length and the circuit resistance are unchanged, so the induced current remains the same.

Teacher's Note:
1. Motional emf = B v × (effective length between the ends).
2. The shape of the wire between the ends does not matter.

Answer: (a) 1 : 2

Explanation:
1. Growth of current \( I=I_0\left(1-e^{-t/\tau}\right) \).
2. 90%: \( e^{-t_1/\tau}=0.1\Rightarrow t_1=\tau\ln10 \). 99%: \( e^{-t_2/\tau}=0.01\Rightarrow t_2=2\tau\ln10 \).
3. \( \frac{t_1}{t_2}=\frac12 \).

Teacher's Note:
1. ln 100 = 2 ln 10.
2. The ratio does not depend on τ.

Answer: (d) Current will decrease in each loop.

Explanation:
1. Bringing the loops closer increases the flux of each loop through the other (the fields add, since the currents are in the same direction).
2. By Lenz's law, the induced emf opposes this increase, so the current in each loop decreases.

Teacher's Note:
1. Lenz's law opposes the change in flux.
2. Mutual flux increases as the loops approach.

Answer: (c) 3.5 ms

Explanation:
1. Time constant \( \tau=\frac LR=\frac{50\times10^{-3}}{10}=5\ \text{ms} \).
2. Half of the steady value: \( 1-e^{-t/\tau}=\frac12\Rightarrow t=\tau\ln2=5\times0.693=3.5\ \text{ms} \).

Teacher's Note:
1. Convert mH to H before using τ = L/R.
2. t = τ ln 2 for half of the final current.

Answer: (b) negatively charged

Explanation:
1. Force on a positive charge: \( \vec F=q\vec v\times\vec B \) with \( \vec v=v\hat x,\ \vec B=B\hat z \): \( \hat x\times\hat z=-\hat y \).
2. Positive charges move towards −y, that is, towards B, so B becomes positive and A (the top end) becomes negatively charged.

Teacher's Note:
1. Use v × B for the direction of the force on positive charges.
2. The end where positive charges collect is at the higher potential.

Answer: (a) 5 V, 0.5 A

Explanation:
1. \( \varepsilon=N\frac{\Delta\Phi}{\Delta t}=1000\times\frac{5.5\times10^{-4}-5\times10^{-5}}{0.1}=1000\times\frac{5\times10^{-4}}{0.1}=5\ \text{V} \).
2. \( I=\frac\varepsilon R=\frac{5}{10}=0.5\ \text{A} \).

Teacher's Note:
1. Multiply by the number of turns.
2. Change in flux = 5.5 × 10⁻⁴ − 0.5 × 10⁻⁴ = 5 × 10⁻⁴ Wb.