JEE Physics Electromagnetic Induction MCQs Set 14

Answer: (b) 0.1 Wb

Explanation:
1. \( \phi = BA\cos60^\circ = 1\times0.2\times0.5 \).
2. \( \phi = 0.1 \) Wb.

Teacher's Note:
1. The axis is the normal to the disc, so the angle with B is used directly.

Answer: (d) 2 Wb

Explanation:
1. \( \Delta\phi = R\int i\,dt = R\times\text{area under the }i\text{–}t\text{ graph} \).
2. Area \( = \tfrac12\times2\times0.1 = 0.1 \) C.
3. \( \Delta\phi = 20\times0.1 = 2 \) Wb.

Teacher's Note:
1. Charge flowed equals the area under the current graph.
2. Flux change equals R times that charge.

Answer: (a) \( ML^2T^{-2}A^{-2} \)

Explanation:
1. \( U = \tfrac12MI^2 \), so \( [M] = [ML^2T^{-2}]/[A^2] \).
2. \( [M] = ML^2T^{-2}A^{-2} \).

Teacher's Note:
1. Inductance and energy are linked by U = ½LI².

Answer: (c) continuously decrease and finally become zero

Explanation:
1. Eddy currents in the tube oppose the motion (Lenz's law).
2. The opposing force grows with speed until it equals the weight, so the acceleration falls to zero and the magnet reaches terminal velocity.

Teacher's Note:
1. Eddy current braking increases with speed.

Answer: (a) proportional to the rate of change of magnetic flux linked with it

Explanation:
1. Faraday's law states \( \varepsilon = -N\,d\phi/dt \).
2. So the emf depends on the rate of change of flux and on N.

Teacher's Note:
1. Fast change of flux gives a larger emf.

Answer: (c) less than g

Explanation:
1. Induced current in the ring opposes the approach of the magnet (Lenz's law).
2. The upward magnetic force on the magnet reduces its acceleration to below g.

Teacher's Note:
1. Lenz's law always opposes the cause.

Answer: (c) If I increases A will repel the ring

Explanation:
1. When I increases, the flux through the ring increases.
2. By Lenz's law the induced current opposes this, so the ring is repelled.

Teacher's Note:
1. Increase means repel and decrease means attract.

Answer: (a) clockwise

Explanation:
1. The field of the wire to its right points into the paper.
2. Moving away reduces the flux into the paper.
3. The induced current must restore it, so it produces a field into the paper, which is clockwise.

Teacher's Note:
1. Use the right hand grip rule for the wire, then Lenz's law.

Answer: (d) Q increases

Explanation:
1. \( E\propto d\phi/dt \) doubles, so E and I increase.
2. \( Q = \Delta\phi/R \) depends only on the total change of flux, so Q does not change.
3. Therefore the statement 'Q increases' is wrong.

Teacher's Note:
1. Charge does not depend on the rate of change of flux.

Answer: (b) G shows deflection to the left and right but the amplitude steadily decreases

Explanation:
1. The changing flux induces an alternating emf, so G deflects both ways.
2. Induced currents dissipate energy as heat, so the oscillation damps and the deflection steadily decreases.

Teacher's Note:
1. Induced currents take energy from the oscillation.

Answer: (c) 1.6 mV

Explanation:
1. \( B = 0.4\times10^{-4} \) T. The flux changes from \( +BA \) to \( -BA \), so \( \Delta\phi = 2NBA \).
2. \( \Delta\phi = 2(1000)(0.4\times10^{-4})(10^{-2}) = 8\times10^{-4} \) Wb.
3. \( \varepsilon = 8\times10^{-4}/0.5 = 1.6 \) mV.

Teacher's Note:
1. Turning through 180 degrees reverses the flux.
2. 1 Gauss is 10⁻⁴ T.

Answer: (d) Power

Explanation:
1. The iron core raises the permeability, so B, flux and self inductance all increase.
2. The power \( i^2R \) depends only on the current and the resistance, which are unchanged.

Teacher's Note:
1. Iron increases μ but not the resistance.

Answer: (b) the rectangular and square loops

Explanation:
1. The emf is \( B\ell v \), where ℓ is the length of the loop edge in the field perpendicular to v.
2. For rectangular and square loops ℓ is constant while leaving.
3. For circular and elliptical loops ℓ keeps changing.

Teacher's Note:
1. Constant width of the cutting edge gives constant emf.

Answer: (b) \( 3.2\pi\ \mu V \)

Explanation:
1. \( \phi = B\pi r^2 \), so \( \varepsilon = 2\pi Br\,dr/dt \).
2. \( \varepsilon = 2\pi(0.04)(0.02)(0.002) = 3.2\pi\times10^{-6} \) V.

Teacher's Note:
1. Differentiate the area with respect to time.

Answer: (c) 0.5 A

Explanation:
1. \( \varepsilon = d\phi/dt = 200t = 200 \) V at t = 1 s.
2. \( i = 200/400 = 0.5 \) A.

Teacher's Note:
1. Differentiate the flux, then use Ohm's law.

Answer: (b) 5 µC

Explanation:
1. \( q = \Delta\phi/R \).
2. \( q = (0.5\times10^{-4})/10 = 5\times10^{-6} \) C = 5 µC.

Teacher's Note:
1. Charge is independent of time taken.

Answer: (d) 0.566 T

Explanation:
1. \( q = NBA/R_{total} \), so \( B = qR/(NA) \).
2. \( A = \pi(6\times10^{-3})^2 = 1.131\times10^{-4} \) m².
3. \( B = (32\times10^{-6})(200)/(100\times1.131\times10^{-4}) = 0.566 \) T.

Teacher's Note:
1. Total resistance is coil plus galvanometer.

Answer: (b) a square wave, alternating between a positive and a negative constant value

Explanation:
1. \( \varepsilon = -M\,di/dt \).
2. The slope of the triangular wave is constant but alternates in sign, so the emf is an alternating constant: a square wave.

Teacher's Note:
1. The emf follows the slope of the current graph.

Answer: (d) π V

Explanation:
1. \( \varepsilon = M\,dI/dt = MI_0\omega\cos\omega t \).
2. Maximum \( = 0.005\times2\times100\pi = \pi \) V.

Teacher's Note:
1. The maximum of cosine is 1.

Answer: (a) A.C.W. concentric circular field lines in the plane of the paper

Explanation:
1. Induced electric field lines form closed circles in the plane perpendicular to B.
2. The induced current must produce a field out of the paper to oppose the increase, which requires anticlockwise circulation.

Teacher's Note:
1. Apply Lenz's law to find the circulation.

Answer: (b) 32 µA

Explanation:
1. \( B = \mu_0nI \), so \( \Delta B = 4\pi\times10^{-7}\times2\times10^4\times4 = 32\pi\times10^{-3} \) T.
2. Flux change \( = N\,\Delta B\,\pi r^2 = 200\times32\pi\times10^{-3}\times\pi\times10^{-4} = 64\pi^2\times10^{-6} \) Wb.
3. \( \varepsilon = 64\pi^2\times10^{-6} \) V and \( i = \varepsilon/R = 32\times10^{-6} \) A = 32 µA.

Teacher's Note:
1. The solenoid diameter is not needed.
2. Use the small coil area.

Answer: (d) \( Bv(R) \)

Explanation:
1. The emf of an arc equals that of the straight line joining its ends.
2. Effective length perpendicular to v is \( 2R\sin30^\circ = R \).
3. \( \varepsilon = BvR \).

Teacher's Note:
1. Use the chord, not the arc length.

Answer: (d) \( \dfrac1{x(x+a)} \)

Explanation:
1. Only the two sides parallel to the wire have emf.
2. \( \varepsilon = \dfrac{\mu_0I}{2\pi}av\left[\dfrac1x-\dfrac1{x+a}\right] = \dfrac{\mu_0Iva^2}{2\pi x(x+a)} \).

Teacher's Note:
1. Field of a long wire falls as 1/r, so the two sides give different emfs.

Answer: (c) \( \pi\ \mu V \)

Explanation:
1. \( \varepsilon = \tfrac12B\omega r^2 \).
2. \( = \tfrac12(0.2\times10^{-4})(2\pi\times20)(0.05)^2 = \pi\times10^{-6} \) V.

Teacher's Note:
1. ω = 2π times the frequency.