JEE Physics Electromagnetic Induction MCQs Set 09

Answer: (a) 1.0 V, anticlockwise

Explanation:
1. Average emf: \( |\varepsilon|=\frac{\Delta\Phi}{\Delta t}=\frac{0.9-0.4}{0.5}=1.0 \) V.
2. The field of the north pole points away from the magnet, into the coil, and is increasing.
3. By Lenz's law the coil opposes the increase, so it produces a field pointing back towards the magnet. That needs an anticlockwise current when seen from the magnet side.

Teacher's Note:
1. Lenz's law decides the sense, Faraday's law the magnitude.
2. A north-pole face of a loop looks anticlockwise when seen from that face.

Answer: (d) 2.5 mV

Explanation:
1. Initial flux: \( BA=0.2\times25\times10^{-4}=5\times10^{-4} \) Wb; final flux is zero.
2. Average emf: \( \frac{5\times10^{-4}}{0.2}=2.5\times10^{-3} \) V = 2.5 mV.

Teacher's Note:
1. Convert cm squared to m squared using a factor of \( 10^{-4} \).

Answer: (c) 0.078 C

Explanation:
1. On turning through 180 degrees the flux changes from \( +NBA \) to \( -NBA \), a change of \( 2NBA \).
2. Area: \( A=\pi(0.25)^2=0.196 \) \( m^2 \).
3. Charge: \( q=\frac{2NBA}{R}=\frac{2\times200\times10^{-2}\times0.196}{10}=0.0785 \) C.

Teacher's Note:
1. Remember that the flux reverses, so the change is twice the original flux, not equal to it.

Answer: (b) \( 1.26\times10^{-7} \) V/m

Explanation:
1. Inside the solenoid, \( E\cdot2\pi r=\pi r^2\frac{dB}{dt} \), so \( E=\frac r2\frac{dB}{dt} \).
2. With \( \frac{dB}{dt}=2.51\times10^{-5} \) T/s and \( r=0.01 \) m: \( E=0.005\times2.51\times10^{-5}=1.26\times10^{-7} \) V/m.

Teacher's Note:
1. Inside the field region E increases linearly with r; outside it decreases as 1/r.

Answer: (d) \( 2.5\times10^{-14} \) J

Explanation:
1. Field: \( B=\frac{\mu_0 I}{2\pi r}=\frac{2\times10^{-7}\times4}{0.1}=8\times10^{-6} \) T.
2. Energy density: \( u=\frac{B^2}{2\mu_0}=\frac{(8\times10^{-6})^2}{2\times4\pi\times10^{-7}}=2.55\times10^{-5} \) J/m\( ^3 \).
3. Energy: \( u\times V=2.55\times10^{-5}\times10^{-9}=2.5\times10^{-14} \) J.

Teacher's Note:
1. 1 \( mm^3 \) is \( 10^{-9}\ m^3 \).

Answer: (a) \( (42+20t) \) V

Explanation:
1. Kirchhoff's loop rule: \( \varepsilon=iR+L\frac{di}{dt} \).
2. Here \( iR=4(3+5t)=12+20t \) and \( L\frac{di}{dt}=6\times5=30 \).
3. So \( \varepsilon=42+20t \) V.

Teacher's Note:
1. Add the resistive drop and the inductive drop; the inductive drop is constant because di/dt is constant.

Answer: (c) 1.11 s and 640 J

Explanation:
1. For growth, \( i=i_0(1-e^{-t/\tau}) \) with \( \tau=\frac{80}{50}=1.6 \) s.
2. Half value: \( e^{-t/\tau}=0.5\Rightarrow t=\tau\ln2=1.6\times0.693=1.11 \) s.
3. Final current \( \frac{200}{50}=4 \) A, so maximum energy \( \frac12\times80\times16=640 \) J.

Teacher's Note:
1. Time to reach half of the final current is always \( 0.693\,\tau \).

Answer: (d) 3.47 s

Explanation:
1. Energy is proportional to \( i^2 \). One quarter of the energy means \( i=i_0/2 \).
2. So \( 1-e^{-t/\tau}=\frac12\Rightarrow t=\tau\ln2 \).
3. With \( \tau=\frac{10}{2}=5 \) s: \( t=5\times0.693=3.47 \) s.

Teacher's Note:
1. Quarter energy corresponds to half current, not half time.

Answer: (b) 0.66 V

Explanation:
1. Time constant: \( \tau=\frac{L}{R}=\frac{5}{100}=0.05 \) s, so \( t/\tau=0.4 \).
2. Current: \( i=\frac{2}{100}(1-e^{-0.4})=0.02\times0.33=6.6\times10^{-3} \) A.
3. Voltage across R: \( iR=6.6\times10^{-3}\times100=0.66 \) V.

Teacher's Note:
1. The inductor takes the rest of the battery voltage: 1.34 V at this instant.

Answer: (c) 0.74 V

Explanation:
1. Time constant \( \tau=\frac{L}{R}=0.05 \) s, so t = 50 ms is one time constant.
2. The inductor's voltage decays as \( L\frac{di}{dt}=\varepsilon e^{-t/\tau}=2e^{-1} \).
3. This is \( 2\times0.368=0.74 \) V (and \( di/dt=0.74 \) A/s).

Teacher's Note:
1. Inductor voltage plus resistor voltage always equals the battery emf.

Answer: (a) \( \frac{2\pi B R^2}{L} \)

Explanation:
1. A superconducting loop keeps its total flux constant, since the emf must be zero.
2. Initial flux: \( \pi R^2B \). After rotation the external flux is \( -\pi R^2B \) and the loop's own flux is \( Li \).
3. Conservation: \( Li-\pi R^2B=\pi R^2B \), so \( i=\frac{2\pi BR^2}{L} \).

Teacher's Note:
1. Zero resistance means zero emf, so the flux linked can never change.

Answer: (b) 0.01 H

Explanation:
1. Emf magnitude: \( e=M\frac{\Delta i}{\Delta t} \).
2. So \( M=\frac{0.1}{(2-1)/0.1}=\frac{0.1}{10}=0.01 \) H.

Teacher's Note:
1. Mutual inductance is emf divided by the rate of current change in the other coil.

Answer: (d) 1 J and 159 Hz

Explanation:
1. Energy: \( U=\frac{q^2}{2C}=\frac{(10^{-2})^2}{2\times50\times10^{-6}}=1 \) J.
2. Angular frequency: \( \omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.02\times5\times10^{-5}}}=1000 \) rad/s.
3. So \( f=\frac{\omega}{2\pi}=159 \) Hz.

Teacher's Note:
1. The energy stays constant in an ideal LC circuit; it only moves between the capacitor and the inductor.

Answer: (a) 0.785 ms

Explanation:
1. With \( q=q_0\cos\omega t \), the capacitor energy is half the total when \( \cos^2\omega t=\frac12 \).
2. So \( \omega t=\frac\pi4 \) and \( t=\frac{\pi}{4\times1000}=0.785\times10^{-3} \) s.
3. (All the energy is magnetic first at \( T/4=1.57 \) ms.)

Teacher's Note:
1. Equal sharing occurs at T/8, complete transfer at T/4.

Answer: (c) 4 m/s and 3 V

Explanation:
1. At constant speed the applied force equals the magnetic force: \( F=\frac{B^2\ell^2v}{R} \), so \( v=\frac{FR}{B^2\ell^2}=\frac{1\times4}{1}=4 \) m/s.
2. Emf of the side cutting the field: \( B\ell v=4 \) V; current \( i=\frac{4}{4}=1 \) A.
3. Each side has resistance 1 ohm, so the potential difference across the side in the field (a source of emf 4 V with internal resistance 1 ohm) is \( 4-1\times1=3 \) V.

Teacher's Note:
1. The side cutting the field acts as a battery; the other three sides are its external circuit.

Answer: (d) \( \frac{\varepsilon}{B\ell} \)

Explanation:
1. Net emf in the circuit: \( \varepsilon-B\ell v \), so the current is \( i=\frac{\varepsilon-B\ell v}{r} \).
2. The wire reaches constant velocity when the force \( iB\ell \) vanishes, that is, when \( i=0 \).
3. Then \( \varepsilon=B\ell v \), giving \( v=\frac{\varepsilon}{B\ell} \).

Teacher's Note:
1. Here the rod behaves as a motor: back emf grows until it cancels the battery.

Answer: (b) \( \frac{B_0v_0L}{2} \)

Explanation:
1. The emf of a small element dx is \( (\vec v\times\vec B)\cdot d\vec x=v_0\frac{B_0x}{L}dx \).
2. Integrating from 0 to L: \( \varepsilon=\frac{v_0B_0}{L}\cdot\frac{L^2}{2}=\frac{B_0v_0L}{2} \).

Teacher's Note:
1. When B varies along the rod, integrate \( vB\,dx \) instead of multiplying.

Answer: (c) \( \frac{BV\sin\alpha}{r(1+\sin\alpha)} \)

Explanation:
1. When the rod is at distance y from the vertex its length is \( 2y\tan\alpha \) and each arm of the loop is \( \frac{y}{\cos\alpha} \).
2. Emf: \( 2By\tan\alpha\,V \). Resistance: \( r\left(\frac{2y}{\cos\alpha}+2y\tan\alpha\right)=\frac{2ry(1+\sin\alpha)}{\cos\alpha} \).
3. Current: \( i=\frac{2BVy\tan\alpha\cos\alpha}{2ry(1+\sin\alpha)}=\frac{BV\sin\alpha}{r(1+\sin\alpha)} \), independent of y.

Teacher's Note:
1. Both emf and resistance grow in proportion to y, so the current stays constant.

Answer: (a) \( 9.4\times10^{-6} \) V

Explanation:
1. The wheel plane faces north-south, so the relevant field (perpendicular to the wheel plane) is the horizontal component.
2. Angular speed: \( \omega=\frac{2\pi\times50}{60}=5.24 \) rad/s.
3. Emf: \( \frac12B\omega\ell^2=\frac12\times4\times10^{-5}\times5.24\times0.09=9.4\times10^{-6} \) V, that is \( 3\pi\times10^{-6} \) V.

Teacher's Note:
1. A bicycle wheel standing east-west lies in the north-south vertical plane, so the horizontal field is perpendicular to it.

Answer: (b) \( B\ell\sqrt{g\ell}\,\sin\frac{\theta}{2} \)

Explanation:
1. The wire pivots about its upper end like a rotating rod, with emf \( \frac12B\omega\ell^2=\frac12B\ell v \) where v is the bob's speed.
2. Maximum speed at the bottom: \( v^2=2g\ell(1-\cos\theta)=4g\ell\sin^2\frac\theta2 \), so \( v=2\sqrt{g\ell}\sin\frac\theta2 \).
3. Therefore \( \varepsilon_{max}=\frac12B\ell\cdot2\sqrt{g\ell}\sin\frac\theta2=B\ell\sqrt{g\ell}\sin\frac\theta2 \).

Teacher's Note:
1. Use energy conservation to find the speed at the lowest point; the emf is largest there.

Answer: (d) \( \pi\ \mathrm{mV} \) and 2 mV

Explanation:
1. Angular speed: \( \omega=2\pi\times5=10\pi \) rad/s. Maximum emf: \( NBA\omega=50\times2\times10^{-4}\times10^{-2}\times10\pi=\pi\times10^{-3} \) V.
2. In half a turn the flux goes from \( NBA \) to \( -NBA \): change \( 2NBA=2\times10^{-4} \) Wb in \( 0.1 \) s.
3. Average emf: \( \frac{2\times10^{-4}}{0.1}=2\times10^{-3} \) V = 2 mV.

Teacher's Note:
1. Average over a full turn is zero; average over a half turn is \( \frac2\pi\varepsilon_{max} \).

Answer: (a) anticlockwise in A and clockwise in B

Explanation:
1. For a current upward, the field is out of the plane on the left of the wire and into the plane on the right.
2. As the current decreases, ring A (field out of the plane) loses flux, so it makes more outward field: anticlockwise current.
3. Ring B (field into the plane) loses flux and makes more inward field: clockwise current.

Teacher's Note:
1. Use the right-hand grip rule for the wire's field, then oppose the change in each ring.

Answer: (c) zero

Explanation:
1. The solenoid field is along its axis.
2. The plane of the loop contains the axis, so the field lies in the plane of the loop and no field line passes through it: flux is always zero.
3. No flux change means no emf and no induced current.

Teacher's Note:
1. Flux depends on the component of B perpendicular to the loop's plane.

Answer: (d) 2 Wb

Explanation:
1. Charge passed equals the area of the current-time graph: \( q=\frac12\times4\times0.1=0.2 \) C.
2. Since \( q=\frac{\Delta\Phi}{R} \), \( \Delta\Phi=qR=0.2\times10=2 \) Wb.

Teacher's Note:
1. Area under an i-t graph is charge; charge times resistance gives the change in flux.

Answer: (b) there is repulsion between A and B

Explanation:
1. Increasing the current in A increases the flux through B.
2. B develops an induced current whose field opposes that increase, which makes B's near face the same pole as A's near face.
3. Like poles repel, so B is repelled. Reversing the current in A reverses both currents and gives the same result.

Teacher's Note:
1. The result does not depend on current direction: the induced current always opposes the change.

Answer: (c) the current in each loop will decrease

Explanation:
1. Each loop's flux through the other adds to its own flux when currents are in the same direction.
2. As they approach, the mutual flux grows, so each loop's induced emf opposes the increase.
3. The opposing emf reduces the current in each loop.

Teacher's Note:
1. Same-direction currents have positive mutual inductance; approaching increases the flux linkage.