JEE Physics Electromagnetic Induction MCQs Set 23

Multiple Choice Questions (MCQs) for JEE Physics: Electromagnetic Induction

Access targeted multiple-choice questions for Electromagnetic Induction designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Electromagnetic Induction MCQs for JEE Physics

Access the complete set of multiple-choice questions for Electromagnetic Induction below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: When a circular coil of radius 1 m and 100 turns is rotated in a horizontal uniform magnetic field, the peak value of emf induced is 100 V. The coil is unwound and then rewound into a circular coil of radius 2 m. If it is rotated now, with the same speed, under similar conditions, the new peak value of emf developed is
(a) 50 V
(b) 100 V
(c) 150 V
(d) 200 V
Show Answer & Explanation

Answer: (d) 200 V

Explanation:
1. Wire length \( =100\times2\pi(1) \). With radius 2 m each turn is twice as long, so \( N=50 \).
2. Peak emf \( =NBA\omega\propto Nr^2 \): \( \dfrac{50\times4}{100\times1}=2 \), so the new peak is 200 V.

Teacher's Note:
1. The area per turn becomes four times while the turns halve.

Question 2: A metal disc of radius 100 cm is rotated at a constant angular speed of 60 rad s\(^{-1}\) in a plane at right angles to an external field of magnetic induction 0.05 Wb m\(^{-2}\). The emf induced between the centre and a point on the rim will be
(a) 3 V
(b) 1.5 V
(c) 6 V
(d) 9 V
Show Answer & Explanation

Answer: (b) 1.5 V

Explanation:
1. \( \varepsilon=\tfrac12B\omega r^2=\tfrac12\times0.05\times60\times1^2=1.5 \) V.

Teacher's Note:
1. Use the rotating-rod formula \( \tfrac12B\omega r^2 \).

Question 3: An uniformly wound solenoid coil of self inductance \( 1.8\times10^{-4} \) H and resistance \( 6\ \Omega \) is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant for the current in the circuit is (neglect mutual induction)
(a) \( 0.1\times10^{-4} \) s
(b) \( 0.2\times10^{-4} \) s
(c) \( 0.3\times10^{-4} \) s
(d) \( 0.4\times10^{-4} \) s
Show Answer & Explanation

Answer: (c) \( 0.3\times10^{-4} \) s

Explanation:
1. Each half has \( L'=0.9\times10^{-4} \) H and \( R'=3\ \Omega \).
2. In parallel \( L_{eq}=0.45\times10^{-4} \) H, \( R_{eq}=1.5\ \Omega \).
3. \( \tau=\dfrac{L_{eq}}{R_{eq}}=0.3\times10^{-4} \) s.

Teacher's Note:
1. Halving the coil halves both L and R.

Question 4: A conducting circular loop is placed in a uniform magnetic field of induction B tesla with its plane normal to the field. Now, the radius of the loop starts shrinking at the rate \( \left(\dfrac{dr}{dt}\right) \). Then the induced emf at the instant when the radius is r, is
(a) \( \pi rB\left(\dfrac{dr}{dt}\right) \)
(b) \( 2\pi rB\left(\dfrac{dr}{dt}\right) \)
(c) \( \pi r^2\left(\dfrac{dB}{dt}\right) \)
(d) \( \left(\dfrac{\pi r^2}{2}\right)B\left(\dfrac{dr}{dt}\right) \)
Show Answer & Explanation

Answer: (b) \( 2\pi rB\left(\dfrac{dr}{dt}\right) \)

Explanation:
1. \( \phi=B\pi r^2 \Rightarrow |\varepsilon|=B\dfrac{d(\pi r^2)}{dt}=2\pi rB\dfrac{dr}{dt} \).

Teacher's Note:
1. B is constant, so only the area changes.

Question 5: The flux linked with a circuit is given by \( \phi=t^3+3t-7 \). The graph between time (x-axis) and induced emf (y-axis) will be a
(a) straight line through the origin
(b) straight line with positive intercept
(c) parabola through the origin
(d) parabola not through the origin
Show Answer & Explanation

Answer: (d) parabola not through the origin

Explanation:
1. \( \varepsilon=-\dfrac{d\phi}{dt}=-(3t^2+3) \), a parabola in t that does not pass through the origin (it is \( -3 \) at \( t=0 \)).

Teacher's Note:
1. The emf is quadratic in t with a constant term.

Question 6: Lenz's law of electromagnetic induction corresponds to the
(a) law of conservation of charge
(b) law of conservation of energy
(c) law of conservation of momentum
(d) law of conservation of angular momentum
Show Answer & Explanation

Answer: (b) law of conservation of energy

Explanation:
1. The induced current opposes the change that produces it; otherwise energy would be created from nothing, so Lenz's law expresses conservation of energy.

Teacher's Note:
1. The opposing effect ensures work must be done.

Question 7: A conducting rod of length l is moving in a transverse magnetic field of strength B with velocity v. The resistance of the rod is R. The current in the rod is
(a) \( \dfrac{Blv}{R} \)
(b) \( Blv \)
(c) zero
(d) \( \dfrac{B^2v^2l^2}{R} \)
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. An emf \( Blv \) appears across the rod but the rod is not part of a closed circuit, so no current flows.

Teacher's Note:
1. A closed path is needed for a current.

Question 8: Whenever a magnet is moved either towards or away from a conducting coil, an emf is induced, the magnitude of which is independent of
(a) the strength of the magnet
(b) the speed with which the magnet is moved
(c) the number of turns of the coil
(d) the resistance of the coil
Show Answer & Explanation

Answer: (d) the resistance of the coil

Explanation:
1. The induced emf is \( N\dfrac{d\phi}{dt} \), which depends on the magnet's strength, its speed and the number of turns, but not on the coil's resistance.

Teacher's Note:
1. Resistance changes the current, not the emf.

Question 9: According to Lenz's law of electromagnetic induction,
(a) the induced emf is not in the direction opposing the change in magnetic flux
(b) the relative motion between the coil and magnet produces change in magnetic flux
(c) only the magnet should be moved towards the coil
(d) only the coil should be moved towards the magnet
Show Answer & Explanation

Answer: (b) the relative motion between the coil and magnet produces change in magnetic flux

Explanation:
1. Either the coil or the magnet may move; only relative motion is needed to change the flux, and the induced emf opposes that change.

Teacher's Note:
1. Option (a) states the opposite of Lenz's law.

Question 10: A square coil of side 25 cm having 1000 turns is rotated with an uniform speed in a magnetic field about an axis perpendicular to the direction of the field. At an instant t, the emf induced in the coil is \( e=200\sin100\pi t \). The magnetic induction is
(a) 0.02 T
(b) \( 10^{-3} \) T
(c) 0.1 T
(d) 0.01 T
Show Answer & Explanation

Answer: (d) 0.01 T

Explanation:
1. \( e_0=NBA\omega \Rightarrow B=\dfrac{200}{1000\times0.0625\times100\pi}=0.0102\approx0.01 \) T.

Teacher's Note:
1. \( A=0.25^2=0.0625 \) m\(^2\), \( \omega=100\pi \).

Question 11: If a coil of 40 turns and area 4.0 cm\(^2\) is suddenly removed from a magnetic field, it is observed that a charge of \( 2.0\times10^{-4} \) C flows into the coil. If the resistance of coil is \( 80\ \Omega \), the magnetic flux density in Wb m\(^{-2}\) is
(a) 0.5
(b) 1.0
(c) 1.5
(d) 2.0
Show Answer & Explanation

Answer: (b) 1.0

Explanation:
1. \( q=\dfrac{N\Delta\phi}{R}=\dfrac{NBA}{R} \Rightarrow B=\dfrac{qR}{NA}=\dfrac{2\times10^{-4}\times80}{40\times4\times10^{-4}}=1.0 \) T.

Teacher's Note:
1. Charge \( =\Delta\phi\,N/R \).

Question 12: The net magnetic flux through any closed surface, kept in a magnetic field is
(a) zero
(b) \( \dfrac{\mu_0}{4\pi} \)
(c) \( 4\pi\mu_0 \)
(d) \( \dfrac{4\mu_0}{\pi} \)
Show Answer & Explanation

Answer: (a) zero

Explanation:
1. Magnetic field lines are closed loops (no monopoles), so \( \oint\vec B\cdot d\vec S=0 \).

Teacher's Note:
1. This is Gauss' law for magnetism.

Question 13: A circular coil of diameter 21 cm is placed in a magnetic field of induction \( 10^{-4} \) T. The magnitude of flux linked with the coil, when the plane of the coil makes an angle \( 30^\circ \) with the field is
(a) \( 1.44\times10^{-6} \) Wb
(b) \( 1.732\times10^{-6} \) Wb
(c) \( 3.1\times10^{-6} \) Wb
(d) \( 4.2\times10^{-6} \) Wb
Show Answer & Explanation

Answer: (b) \( 1.732\times10^{-6} \) Wb

Explanation:
1. \( A=\pi(0.105)^2=0.03464 \) m\(^2\). The plane makes \( 30^\circ \) with B, so \( \phi=BA\sin30^\circ=10^{-4}\times0.03464\times0.5=1.732\times10^{-6} \) Wb.

Teacher's Note:
1. The normal makes \( 60^\circ \) with the field.

Question 14: When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. This is due to
(a) diffraction of the signal received from the antenna
(b) interference of the direct signal received from the antenna with the weak signal reflected by the passing aircraft
(c) change of magnetic flux occurring due to the passage of aircraft
(d) vibration created by the passage of aircraft
Show Answer & Explanation

Answer: (b) interference of the direct signal received from the antenna with the weak signal reflected by the passing aircraft

Explanation:
1. The aircraft reflects a weak signal that interferes with the direct signal, and the changing path difference makes the picture flicker.

Teacher's Note:
1. It is an interference effect.

Question 15: A copper disc of radius 0.1 m is rotated about its centre with 20 rev s\(^{-1}\) in a uniform magnetic field of 0.1 T with its plane perpendicular to the field. The emf induced across the radius of the disc is
(a) \( \dfrac{\pi}{20} \) V
(b) \( \dfrac{\pi}{10} \) V
(c) \( 20\pi \) mV
(d) \( 10\pi \) mV
Show Answer & Explanation

Answer: (c) \( 20\pi \) mV

Explanation:
1. \( \varepsilon=\tfrac12B\omega r^2=\tfrac12\times0.1\times(2\pi\times20)\times0.01=0.02\pi \) V \( =20\pi \) mV.

Teacher's Note:
1. \( \omega=2\pi f \).

Question 16: In 0.1 s, the current in a coil increases from 1 A to 1.5 A. If inductance of coil is 60 mH, then induced current in external resistance of \( 3\ \Omega \) will be
(a) 1 A
(b) 0.5 A
(c) 0.2 A
(d) 0.1 A
Show Answer & Explanation

Answer: (d) 0.1 A

Explanation:
1. \( \varepsilon=L\dfrac{\Delta I}{\Delta t}=0.06\times\dfrac{0.5}{0.1}=0.3 \) V, so \( I=\dfrac{0.3}{3}=0.1 \) A.

Teacher's Note:
1. Find the induced emf first.

Question 17: Induced emf in the coil depends upon
(a) conductivity of coil
(b) amount of flux
(c) rate of change of linked flux
(d) resistance of coil
Show Answer & Explanation

Answer: (c) rate of change of linked flux

Explanation:
1. Faraday's law: \( \varepsilon=-N\dfrac{d\phi}{dt} \).

Teacher's Note:
1. The emf depends on how fast the flux changes.

Question 18: A magnet N-S is suspended from a spring and when it oscillates, the magnet moves in and out of the coil C. The coil is connected to a galvanometer G. Then, as the magnet oscillates,
(a) G shows no deflection
(b) G shows deflection to the left and right but the amplitude steadily decreases
(c) G shows deflection to the left and right with constant amplitude
(d) G shows deflection on one side
Show Answer & Explanation

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

Explanation:
1. The alternating induced current drives the galvanometer both ways. The induced current opposes the motion and removes energy, so the oscillation, and hence the deflection, steadily decays.

Teacher's Note:
1. Lenz's law gives damping.

Question 19: An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along positive Z-axis. The direction of induced current as seen from the Z-axis will be
(a) clockwise of the positive Z-axis
(b) anti-clockwise of the positive Z-axis
(c) zero
(d) along the magnetic field
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. The flux through the cylinder does not change, so no current is induced.

Teacher's Note:
1. No change in flux means no emf.

Question 20: The flux associated with a coil changes from 1.35 Wb to 0.79 Wb within \( \dfrac1{10} \) s. Then, the charge produced by the earth coil, if resistance of coil is \( 7\ \Omega \) is
(a) 0.08 C
(b) 0.8 C
(c) 0.008 C
(d) 8 C
Show Answer & Explanation

Answer: (a) 0.08 C

Explanation:
1. \( q=\dfrac{\Delta\phi}{R}=\dfrac{1.35-0.79}{7}=0.08 \) C.

Teacher's Note:
1. The time interval is not needed.

Question 21: The magnitude of the earth's magnetic field at a place is \( B_0 \) and the angle of dip is \( \delta \). A horizontal conductor of length l lying magnetic North South moves Eastwards with a velocity v, the emf induced across the conductor is
(a) zero
(b) \( B_0lv\sin\delta \)
(c) \( B_0lv \)
(d) \( B_0lv\cos\delta \)
Show Answer & Explanation

Answer: (b) \( B_0lv\sin\delta \)

Explanation:
1. Only the vertical component \( B_0\sin\delta \) is perpendicular to both the rod and its velocity, so \( \varepsilon=B_0\sin\delta\,lv \).

Teacher's Note:
1. The horizontal component is parallel to the rod.

Question 22: Two parallel rails of a railway track insulated from each other and with the ground are connected to a millivoltmeter. The distance between the rails is 1 m. A train is travelling with a velocity of 72 km h\(^{-1}\) along the track. The reading of the millivoltmeter (in mV) is (vertical component of the earth's magnetic induction is \( 2\times10^{-5} \) T)
(a) 1.44
(b) 0.72
(c) 0.4
(d) 0.2
Show Answer & Explanation

Answer: (c) 0.4

Explanation:
1. \( v=20 \) m s\(^{-1}\), so \( \varepsilon=B_vlv=2\times10^{-5}\times1\times20=4\times10^{-4} \) V \( =0.4 \) mV.

Teacher's Note:
1. Convert 72 km/h to 20 m/s.

Question 23: Two coils have a mutual inductance 5 mH. The current changes in the first coil according to equation \( I=I_0\sin\omega t \), where \( I_0=10 \) A and \( \omega=100\pi \) rad s\(^{-1}\). The maximum value of emf in the second coil is
(a) \( 2\pi \)
(b) \( 5\pi \)
(c) \( \pi \)
(d) \( 4\pi \)
Show Answer & Explanation

Answer: (b) \( 5\pi \)

Explanation:
1. \( \varepsilon_{max}=MI_0\omega=5\times10^{-3}\times10\times100\pi=5\pi \) V.

Teacher's Note:
1. Differentiate \( \sin\omega t \) to get \( \omega\cos\omega t \).

Question 24: A solenoid of inductance 2 H carries a current 1 A. What is the magnetic energy stored in a solenoid?
(a) 2 J
(b) 1 J
(c) 4 J
(d) 5 J
Show Answer & Explanation

Answer: (b) 1 J

Explanation:
1. \( U=\tfrac12LI^2=\tfrac12\times2\times1=1 \) J.

Teacher's Note:
1. Use \( \tfrac12LI^2 \).

Question 25: A coil of wire of radius r has 600 turns and a self inductance of 108 mH. The self inductance of a coil with same radius and 500 turns is
(a) 80 mH
(b) 75 mH
(c) 108 mH
(d) 90 mH
Show Answer & Explanation

Answer: (b) 75 mH

Explanation:
1. \( L\propto N^2 \): \( L'=108\times\left(\dfrac{500}{600}\right)^2=75 \) mH.

Teacher's Note:
1. Same geometry, so L scales as \( N^2 \).

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