JEE Physics Electromagnetic Induction MCQs Set 04

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Chapter-wise Objective Questions: Electromagnetic Induction

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Multiple Choice Questions

Question 1: According to Faraday's law of electromagnetic induction,
(a) the direction of the induced current is such that it opposes itself
(b) the induced emf in a coil is proportional to the rate of change of magnetic flux linked with it
(c) the direction of the induced emf is such that it favours the cause
(d) none of the above
Show Answer & Explanation

Answer: (b) the induced emf in a coil is proportional to the rate of change of magnetic flux linked with it

Explanation:
1. Faraday's law: \( |\varepsilon|=\left|\frac{d\phi}{dt}\right| \) (for N turns, multiplied by N).
2. The opposing direction is Lenz's law, not part of the magnitude statement.

Teacher's Note:
1. Faraday gives magnitude, Lenz gives direction.
2. Option (a) is a muddled form of Lenz's law.

Question 2: A square loop of side 22 cm is changed to a circle (same wire length) in 0.4 s in a field of 0.2 T perpendicular to the loop. The magnitude of the induced emf is
(a) 3.3 mV
(b) 6.6 mV
(c) 13.2 mV
(d) 26.4 mV
Show Answer & Explanation

Answer: (b) 6.6 mV

Explanation:
1. Perimeter \( =4\times0.22=0.88 \) m, so circle radius \( r=\frac{0.88}{2\pi}=0.14 \) m.
2. Areas: square \( 0.0484\,m^2 \); circle \( \pi r^2=0.0616\,m^2 \). \( \Delta A=0.01316\,m^2 \).
3. \( \varepsilon=\frac{B\Delta A}{\Delta t}=\frac{0.2\times0.01316}{0.4}\approx6.6\times10^{-3} \) V.

Teacher's Note:
1. A circle encloses the maximum area for a given perimeter.
2. The flux increases, so the induced current opposes this (Lenz).

Question 3: The magnetic flux through a circuit of resistance R changes by Δφ in time Δt. The total charge that passes through the circuit in this time is
(a) \( \frac{\Delta\phi}{R} \)
(b) \( \frac{\Delta\phi}{\Delta t} \)
(c) \( R\frac{\Delta\phi}{\Delta t} \)
(d) \( \frac1R\frac{\Delta\phi}{\Delta t} \)
Show Answer & Explanation

Answer: (a) \( \frac{\Delta\phi}{R} \)

Explanation:
1. \( I=\frac{\varepsilon}{R}=\frac{1}{R}\frac{\Delta\phi}{\Delta t} \).
2. \( Q=I\Delta t=\frac{\Delta\phi}{R} \).

Teacher's Note:
1. The time cancels out.
2. Option (d) is the current, not the charge.

Question 4: When the current through a solenoid increases at a constant rate, the induced emf
(a) is constant and in the direction of the inducing current
(b) is constant and opposite to the direction of the inducing current
(c) increases with time and is in the direction of the inducing current
(d) increases with time and is opposite to the inducing current
Show Answer & Explanation

Answer: (b) is constant and opposite to the direction of the inducing current

Explanation:
1. \( \varepsilon=-L\frac{dI}{dt} \), constant when dI/dt is constant.
2. By Lenz's law it opposes the increase of current.

Teacher's Note:
1. Self-induced emf opposes the change.
2. A constant rate gives a constant emf.

Question 5: For a given solenoid, which statements are correct? (a) An inductor stores energy in the form of a magnetic field. (b) A capacitor stores energy in the form of an electric field. (c) An inductor stores energy in both electric and magnetic fields. (d) A capacitor stores energy in both electric and magnetic fields.
(a) a and b
(b) a and c
(c) b and d
(d) b and c
Show Answer & Explanation

Answer: (a) a and b

Explanation:
1. An inductor stores \( \frac12LI^2 \) in its magnetic field; a capacitor stores \( \frac12CV^2 \) in its electric field.
2. Statements (c) and (d) are false.

Teacher's Note:
1. Each device stores energy in only one field.
2. Remember U = ½LI² for inductors.

Question 6: A constant current i is maintained in a solenoid. Which quantities increase if an iron rod is inserted along its axis? (a) magnetic field at the centre, (b) magnetic flux linked with the solenoid, (c) self-inductance of the solenoid, (d) rate of Joule heating.
(a) a, b, c
(b) c, d
(c) a, b
(d) only b
Show Answer & Explanation

Answer: (a) a, b, c

Explanation:
1. The iron core raises the permeability, so B, flux and \( L=\mu N^2A/l \) all increase.
2. Joule heating rate \( I^2R \) is unchanged at constant current.

Teacher's Note:
1. Only a change in current or resistance changes heating.
2. Iron multiplies L by μ_r.

Question 7: During 0.1 s the current in a coil increases from 1 A to 1.5 A. If the inductance of the coil is 60 µH, the induced current in an external resistance of 600 µΩ is
(a) 1 A
(b) 4/3 A
(c) 2/3 A
(d) 1/2 A
Show Answer & Explanation

Answer: (d) 1/2 A

Explanation:
1. \( \varepsilon=L\frac{\Delta I}{\Delta t}=60\times10^{-6}\times\frac{0.5}{0.1}=3\times10^{-4} \) V.
2. \( I=\frac{\varepsilon}{R}=\frac{3\times10^{-4}}{6\times10^{-4}}=0.5 \) A.

Teacher's Note:
1. Convert µH and µΩ consistently.
2. The induced emf drives current in the external resistance.

Question 8: An inductor L and resistor R are joined to a battery. After some time the battery is disconnected but L and R remain in the closed circuit. The current falls to 37% of its initial value in
(a) RL s
(b) \( \frac RL \) s
(c) \( \frac LR \) s
(d) \( \frac1{LR} \) s
Show Answer & Explanation

Answer: (c) \( \frac LR \) s

Explanation:
1. Decay: \( I=I_0e^{-t/\tau} \) with \( \tau=\frac LR \).
2. At \( t=\tau \): \( I=I_0/e\approx0.37\,I_0 \).

Teacher's Note:
1. 37% is 1/e.
2. The time constant is L/R.

Question 9: An emf of 12 V is induced in a coil when the current in it changes at the rate of 48 A/min. The inductance of the coil is
(a) 0.5 H
(b) 15 H
(c) 1.5 H
(d) 9.6 H
Show Answer & Explanation

Answer: (b) 15 H

Explanation:
1. Rate \( =\frac{48}{60}=0.8 \) A/s.
2. \( L=\frac{12}{0.8}=15 \) H.

Teacher's Note:
1. Convert minutes into seconds.
2. Check units: V s/A = H.

Question 10: A coil of 40 H inductance is connected in series with a resistance of 8 Ω and the combination is joined to the terminals of a 2 V battery. The time constant of the circuit is
(a) 1/5 s
(b) 40 s
(c) 20 s
(d) 5 s
Show Answer & Explanation

Answer: (d) 5 s

Explanation:
1. \( \tau=\frac LR=\frac{40}{8}=5 \) s.

Teacher's Note:
1. The emf does not affect τ.
2. Units: H/Ω = s.

Question 11: An ideal coil of 10 H is connected in series with a 5 Ω resistance and a 5 V battery. Two seconds after the connection is made, the current in the circuit (in amperes) is
(a) e
(b) e − 1
(c) \( 1-e^{-1} \)
(d) 1 − e
Show Answer & Explanation

Answer: (c) \( 1-e^{-1} \)

Explanation:
1. \( I_{max}=1 \) A and \( \tau=\frac{10}{5}=2 \) s.
2. \( I=1-e^{-t/\tau}=1-e^{-1} \) A.

Teacher's Note:
1. At t = τ the current is 63% of maximum.
2. Current cannot exceed 1 A.

Question 12: Electric power is transmitted to a far distance at high potential because
(a) it stops wire theft
(b) it minimises power loss
(c) a generator gives only high potential
(d) electric power is transferred early due to high potential
Show Answer & Explanation

Answer: (b) it minimises power loss

Explanation:
1. For power P = VI, a higher V means a smaller current I.
2. Line loss \( =I^2R=\frac{P^2R}{V^2} \), which falls as \( 1/V^2 \).

Teacher's Note:
1. Loss goes as the square of the current.
2. Step-up transformers raise V at the plant.

Question 13: The mutual inductance of two coils when the magnetic flux changes by 2 × 10⁻² Wb and the current changes by 0.01 A is
(a) 2 H
(b) 3 H
(c) 4 H
(d) 8 H
Show Answer & Explanation

Answer: (a) 2 H

Explanation:
1. \( M=\frac{\Delta\phi}{\Delta I}=\frac{2\times10^{-2}}{0.01}=2 \) H.

Teacher's Note:
1. M = flux linked per unit current in the other coil.
2. Check units: Wb/A = H.

Question 14: The input voltage of a transformer is 2500 V and the output current is 80 A. The ratio of turns in the primary to the secondary is 20 : 1. If the efficiency of the transformer is 100%, the voltage in the secondary is
(a) \( \frac{2500}{20} \) V
(b) \( 2500\times20 \) V
(c) \( \frac{2500}{80\times20} \) V
(d) \( \frac{2500\times20}{80} \) V
Show Answer & Explanation

Answer: (a) \( \frac{2500}{20} \) V

Explanation:
1. \( \frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac1{20} \).
2. \( V_s=\frac{2500}{20}=125 \) V.

Teacher's Note:
1. Voltage follows the turns ratio.
2. The current value is not needed.

Question 15: Two coaxial solenoids are made by winding thin copper wire over a pipe of cross-sectional area 10 cm² and length 20 cm. One has 300 turns and the other 400 turns. Their mutual inductance is
(a) \( 2.4\pi\times10^{-5} \) H
(b) \( 4.8\pi\times10^{-4} \) H
(c) \( 4.8\pi\times10^{-5} \) H
(d) \( 2.4\pi\times10^{-4} \) H
Show Answer & Explanation

Answer: (d) \( 2.4\pi\times10^{-4} \) H

Explanation:
1. \( M=\frac{\mu_0N_1N_2A}{l}=\frac{4\pi\times10^{-7}\times300\times400\times10^{-3}}{0.2} \).
2. \( M=2.4\pi\times10^{-4} \) H.

Teacher's Note:
1. Convert cm² to m² (10 cm² = 10⁻³ m²).
2. Wound one over the other so flux is shared fully.

Question 16: A conducting rod rotates with constant angular velocity ω about an axis through point O, perpendicular to its length. The rod extends 2l on one side of O and 4l on the other, and a uniform field B is parallel to the axis. The potential difference between the two ends is
(a) \( 6B\omega l^2 \)
(b) \( B\omega l^2 \)
(c) \( 10B\omega l^2 \)
(d) zero
Show Answer & Explanation

Answer: (a) \( 6B\omega l^2 \)

Explanation:
1. Emf from O to the 4l end \( =\tfrac12B\omega(4l)^2=8B\omega l^2 \); to the 2l end \( =\tfrac12B\omega(2l)^2=2B\omega l^2 \).
2. Both ends are at higher potential than O (same sense), so the difference is \( 8-2=6B\omega l^2 \).

Teacher's Note:
1. Both ends are pushed the same way relative to the pivot.
2. Subtract, do not add.

Question 17: Two coils of self-inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between them is
(a) 10 mH
(b) 6 mH
(c) 4 mH
(d) 16 mH
Show Answer & Explanation

Answer: (c) 4 mH

Explanation:
1. For perfect coupling \( M=\sqrt{L_1L_2}=\sqrt{2\times8}=4 \) mH.

Teacher's Note:
1. Perfect coupling means k = 1.
2. M never exceeds √(L1L2).

Question 18: A long solenoid has 500 turns. When a current of 2 A is passed through it, the magnetic flux linked with each turn is 4 × 10⁻³ Wb. The self-inductance of the solenoid is
(a) 1.0 H
(b) 4.0 H
(c) 2.5 H
(d) 2.0 H
Show Answer & Explanation

Answer: (a) 1.0 H

Explanation:
1. \( L=\frac{N\phi}{I}=\frac{500\times4\times10^{-3}}{2}=1 \) H.

Teacher's Note:
1. Use flux per turn × turns.
2. Check: Wb-turn/A = H.

Question 19: A coil of resistance 400 Ω is placed in a magnetic field. The magnetic flux φ (Wb) linked with the coil varies with time t (s) as φ = 50t² + 4. The current in the coil at t = 2 s is
(a) 2 A
(b) 1 A
(c) 0.5 A
(d) 0.1 A
Show Answer & Explanation

Answer: (c) 0.5 A

Explanation:
1. \( \varepsilon=\frac{d\phi}{dt}=100t=200 \) V at t = 2 s.
2. \( I=\frac{200}{400}=0.5 \) A.

Teacher's Note:
1. Differentiate first, then substitute.
2. Ohm's law for the coil.

Question 20: A solenoid having 500 turns and length 2 m has a radius of 2 cm. Its self-inductance is
(a) \( 4\times10^{-4} \) H
(b) \( 2\times10^{-4} \) H
(c) \( 8\times10^{-4} \) H
(d) \( 16\times10^{-4} \) H
Show Answer & Explanation

Answer: (b) \( 2\times10^{-4} \) H

Explanation:
1. \( L=\frac{\mu_0N^2A}{l}=\frac{4\pi\times10^{-7}\times500^2\times\pi(0.02)^2}{2} \).
2. \( L\approx1.97\times10^{-4}\approx2\times10^{-4} \) H.

Teacher's Note:
1. Compute area from the radius.
2. Ignore end effects.

Question 21: A copper disc of radius 10 cm rotates in a magnetic field B = 0.4 gauss at 10 rev/s. The potential difference between two points on its periphery is
(a) \( 20\pi\,\mu V \)
(b) \( 10\pi\,\mu V \)
(c) zero
(d) \( 5\pi\,\mu V \)
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. Every radial element contributes the same emf, so all points on the rim are at the same potential.
2. The emf exists between the centre and the rim, not between two rim points.

Teacher's Note:
1. Rim points are equipotential.
2. The centre-to-rim emf here would be ½Bωr².

Question 22: A wire is sliding on two parallel conducting rails 1 m apart, joined by a 50 Ω resistor, in a perpendicular field of 2 T. The force needed to keep the wire moving at a constant 1 cm/s is
(a) \( 0.08\times10^{-3} \) N
(b) \( 8\times10^{-6} \) N
(c) \( 4\times10^{-6} \) N
(d) \( 8\times10^{-4} \) N
Show Answer & Explanation

Answer: (d) \( 8\times10^{-4} \) N

Explanation:
1. \( \varepsilon=Blv=2\times1\times0.01=0.02 \) V, \( I=\frac{0.02}{50}=4\times10^{-4} \) A.
2. \( F=BIl=2\times4\times10^{-4}\times1=8\times10^{-4} \) N.

Teacher's Note:
1. External force equals the magnetic braking force.
2. Equivalent: F = B²l²v/R.

Question 23: A solenoid of radius R and length L carries a current I = I0 cos ωt. The magnitude of the induced electric field at a distance r (greater than R) outside the solenoid is
(a) \( \frac{\mu_0nI_0\omega R^2}{2r}\sin\omega t \)
(b) \( \frac{\mu_0nI_0\omega R^2}{r}\sin\omega t \)
(c) \( \frac{\mu_0nI_0R^2}{2r}\sin\omega t \)
(d) zero
Show Answer & Explanation

Answer: (a) \( \frac{\mu_0nI_0\omega R^2}{2r}\sin\omega t \)

Explanation:
1. Flux \( \phi=\mu_0nI\,\pi R^2 \). \( E(2\pi r)=\left|\frac{d\phi}{dt}\right|=\mu_0n\pi R^2I_0\omega\sin\omega t \).
2. \( E=\frac{\mu_0nI_0\omega R^2}{2r}\sin\omega t \).

Teacher's Note:
1. The field is zero outside, but the changing flux still induces E.
2. E ∝ 1/r outside.

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Multiple Choice Questions (MCQs) for JEE Physics Electromagnetic Induction

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