Previous year questions: Electromagnetic Induction

The questions below come from official CBSE board papers and official Sample Question Papers. The year label in each heading is printed on the source, and the marks are the marks printed for that question or part. Try the question before reading the answer.

Q1 (CBSE Board 2025, 1 mark)

The magnetic flux linked with a coil changes with time t as phi = (8t squared + 5t + 7) Wb. The value of the emf induced in the coil at t = 4 s is (A) 32 V (B) 37 V (C) 64 V (D) 69 V

Answer: Option (D), 69 V. By Faraday's law the magnitude of the emf is the rate of change of flux. Differentiating gives dphi/dt = 16t + 5, and at t = 4 s this is 64 + 5 = 69, so the emf has magnitude 69 V. The negative sign from Lenz's law gives the direction, which is not asked here.

Q2 (CBSE Board 2024, 1 mark)

Assertion (A): The mutual inductance between two coils is maximum when the coils are wound on each other. Reason (R): The flux linkage between two coils is maximum when they are wound on each other.

Answer: Both statements are true and the reason correctly explains the assertion, so the answer is option (A). Mutual inductance is the flux linkage in one coil per unit current in the other, so it is largest when the two coils share the same magnetic flux, which happens when they are wound over each other. Any separation or relative tilt reduces the shared flux and hence reduces M.

Q3 (CBSE Board 2024, 1 mark)

A circular conducting loop sits in a magnetic field of 50 mT, with the plane of the loop normal to the field. The radius shrinks steadily at 1 mm per second. When the radius happens to be 4 cm, the emf induced in the loop is

Answer: The flux is B times the area, phi = B pi r squared, and the radius is changing, so the emf is B times 2 pi r times dr/dt. Substituting B = 0.05 T, r = 0.04 m and dr/dt = 1 x 10 to the power -3 m per second gives emf = 0.05 x 2 pi x 0.04 x 0.001 = 1.26 x 10 to the power -5 V, about 12.6 microvolt. The field is perpendicular to the plane, so the full area is used in the flux.

Q4 (CBSE Board 2023, 1 mark)

A circular coil of radius 8.0 cm and 40 turns is rotated about its vertical diameter with an angular speed of 25 rad per second in a uniform horizontal magnetic field of magnitude 3.0 x 10 to the power -2 T. The maximum emf induced in the coil is

Answer: The peak emf of a rotating coil is N B A omega. The area is pi r squared = pi x 0.0064 = 0.0201 square metre. So the peak emf is 40 x 0.03 x 0.0201 x 25 = 0.603 V, about 0.60 V. The emf is maximum when the plane of the coil is parallel to the field, because at that instant the flux through the coil is changing most rapidly.

Q5 (CBSE Board 2023, 3 marks)

A long solenoid of radius r has n turns per unit length. A current I = I0 sin omega t flows in the solenoid. A coil of N turns is wound tightly around it near its centre. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid and the coil.

Answer: The field inside the solenoid is B = mu0 n I, and the flux through one turn of the coil is phi = B times pi r squared = mu0 n I pi r squared. The total flux linkage of the coil is N phi, so the induced emf is -N mu0 n pi r squared times dI/dt. With I = I0 sin omega t, dI/dt = I0 omega cos omega t, so the emf is -N mu0 n pi r squared I0 omega cos omega t. Comparing with emf = -M dI/dt gives the mutual inductance M = mu0 N n pi r squared. The answer does not depend on the current amplitude, which is the point of mutual inductance.

Q6 (CBSE Board 2026, 5 marks)

(a) State Faraday's law of electromagnetic induction. (b) Derive an expression for the self inductance of an air-filled long solenoid of length l and cross-sectional area A having N turns. (c) A conducting rod of length 50 cm, with one end pivoted, is rotated with an angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of the rod. Find the emf induced in the rod.

Answer: (a) The emf induced in a circuit equals the negative rate of change of magnetic flux through it, and for N turns the emf is -N dphi/dt. (b) Inside a long solenoid the field is mu0 n I with n = N/l, so the flux through each turn is mu0 (N/l) I A. The flux linkage is N times this, and dividing by the current gives L = mu0 N squared A / l, or equivalently mu0 n squared A l. (c) For a rod rotating about one end the emf is half B omega l squared. Here l = 0.5 m, B = 4.0 x 10 to the power -3 T, and the angular speed is 60 rpm = 2 pi rad per second. So the emf is 0.5 x 0.004 x 2 pi x 0.25 = 3.14 x 10 to the power -3 V, about 3.1 mV.

Q7 (CBSE SQP 2021-22, 1 mark)

Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon (i) the rate at which current changes in the two coils (ii) the relative position and orientation of the coils (iii) the current in the coils (iv) the voltage across the coils

Answer: Option (ii), the relative position and orientation of the coils. Mutual inductance depends on the geometry of the two coils, their separation, their relative orientation and the magnetic material between them. It does not depend on the current, the rate of change of current or the voltage; those decide the emf produced for a given M, not M itself.

Q8 (CBSE SQP 2021-22, 1 mark)

The magnetic flux linked with a coil in weber is given by phi = 5t squared + 3t + 16. The induced emf in the coil at t = 4 s will be (i) -27 V (ii) -43 V (iii) -108 V (iv) -118 V

Answer: Option (ii), -43 V. Faraday's law gives emf = -dphi/dt. Differentiating the flux gives dphi/dt = 10t + 3, and at t = 4 s this equals 43, so the emf is -43 V. The negative sign shows that the induced emf opposes the increase of flux, which is Lenz's law.

Q9 (CBSE SQP 2022-23, 1 mark)

A rectangular, a square, a circular and an elliptical loop are all in the x-y plane and move out of a uniform magnetic field with constant velocity along the positive x direction. The field is along the negative z direction. The induced emf during the passage of these loops out of the field will not remain constant for (i) any of the four loops (ii) the circular and elliptical loops (iii) the rectangular, circular and elliptical loops (iv) only the elliptical loops

Answer: Option (ii), the circular and elliptical loops. The emf equals B l v, where l is the length of the part of the loop still cutting the field boundary. For a rectangular or square loop that length stays constant while one side is crossing the boundary, so the emf is constant. For circular and elliptical loops the length of the boundary segment changes continuously, so the emf varies during the exit.

Q10 (CBSE SQP 2025-26, 1 mark)

Two small identical magnets are allowed to fall freely, one through a vertical copper solenoid of length 20 m and another in air through the same vertical distance. The time taken by the two magnets to fall will be (A) same in both cases (B) more for the magnet falling in air (C) more for the magnet falling through the solenoid (D) infinite

Answer: Option (C), more for the magnet falling through the solenoid. As the magnet falls through the copper solenoid, the flux through the turns changes and eddy currents are induced in the copper. By Lenz's law those currents oppose the motion, producing an upward magnetic force that reduces the acceleration. The magnet in air has no such opposition, so it lands first; the time in the solenoid is longer, but not infinite.