NEET Electromagnetic Induction PYQs & Practice Questions

10 free NEET Physics questions on Electromagnetic Induction, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Faraday's Laws · Self Inductance / Circuit Analysis · Faraday's Law of Induction · Lenz's Law · Motional EMF · Self-Inductance · Mutual Inductance · Energy Stored in an Inductor · Eddy Currents · AC Generator · Self-Inductance of a Solenoid · Rotating a Magnet About Its Own Axis

Q1NEET 2026Motional EMF and Power Dissipation in a Varying Area Loop
A conducting loop of finite resistance lies on the x-y plane. There is a constant magnetic field in the z direction. The area of the loop varies with time t, as A = A₀(1 + sin t) in appropriate units. The figure that correctly indicates the qualitative behaviour of the power P dissipated in the loop as a function of time is: (1) P starts at a maximum value at t=0, dips down to touch zero, then rises back to a plateau, repeating this cusp-like pattern; (2) P jumps to a constant plateau value and stays flat; (3) P rises monotonically like an increasing exponential-type curve; (4) P is zero at t=0, rises to a hump, returns to zero, then repeats as a series of separate arches touching zero between each.
- A. Option (1): a repeating pattern of cusps touching zero, starting from a maximum at t=0
- B. Option (2): P rises once and stays constant (flat plateau)
- C. Option (3): P increases monotonically with time
- D. Option (4): P is zero at t=0 and forms separate zero-touching arches
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Answer: —
EMF = -B(dA/dt) = -B A₀ cos t. Power dissipated P = EMF²/R = (B A₀)²cos²t / R, which is proportional to cos²t. This function equals its maximum value at t=0, decreases to zero at t=π/2, rises back to maximum at t=π, and repeats periodically with period π -- producing a series of cusp-shaped humps that touch zero and immediately rise again, starting from a maximum at t=0. This matches the qualitative shape in option (1).
Q2NEET 2026Self-Inductance of a Solenoid
Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and μ₀ is the permeability of free space, the inductance of the solenoid is:
- A. μ₀n²r²l
- B. (μ₀/2π)n²r²l
- C. 2μ₀πn²r²l
- D. μ₀πn²r²l
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Answer: C — 2μ₀πn²r²l
The self-inductance of a long solenoid is L = μ₀n²Al, where A is the cross-sectional area = πr². So L = μ₀n²(πr²)l = μ₀πn²r²l.
Q3NEET 2026Induced EMF in Series vs Parallel Inductors
Two identical inductors, each of self-inductance L, are connected in two different configurations P and Q, where a time varying current I(t) flows. In configuration P, the two inductors are connected in series between nodes a and the bottom rail, with point b tapped between the two coils (so the emf is measured across just the first coil). In configuration Q, the two inductors are connected in parallel between nodes a and b, with I(t) splitting equally between them. The induced emf between points a and b for configuration P is E_P and for configuration Q is E_Q. The ratio E_P/E_Q is: [Neglect the effect of mutual inductance.]
- A. 1/2
- B. 1
- C. 2
- D. 1/4
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Answer: B — 1
In configuration P, point b lies between the two series coils, so the emf measured from a to b is due to just one coil carrying the full current I(t): E_P = L(dI/dt). In configuration Q, the current I(t) splits equally between the two identical parallel coils, so each carries I(t)/2, and the emf across each (and hence between a and b) is E_Q = L(d(I/2)/dt) = (L/2)(dI/dt). Ratio E_P/E_Q = L(dI/dt) / [(L/2)(dI/dt)] = 2.
Q4NEET 2026Motional emf in a cut loop
A rectangular wire loop of sides 8 cm and 3 cm, with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. If the velocity of the loop is 2 cm s⁻¹, the emf developed across the cut, in a direction normal to the shorter side, will be:
- A. 1.8×10⁻⁴ volt
- B. 1.3×10⁻⁴ volt
- C. 1.2×10⁻⁴ volt
- D. 4.8×10⁻⁴ volt
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Answer: —
emf=B×L×v, with L=3 cm (shorter side) and v=2 cm/s: emf=0.3×3×10⁻²×2×10⁻²=1.8×10⁻⁴ V.

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Q5NEET 2025Self Inductance / Circuit Analysis
AB is a part of an electrical circuit containing a 1H inductor, a 5V cell, and a 2Ω resistor in series carrying current i. The potential difference "VA − VB" at the instant when current i = 2A and is increasing at a rate of 1 amp/second is:
- A. 5 volt
- B. 6 volt
- C. 9 volt
- D. 10 volt
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Answer: C — 9 volt
EMF across inductor ε = L(dI/dt) = 1×1 = 1V. Going from A to B: VA − 1 − 5 − 2×2 = VB, so VA − VB = 1 + 5 + 4 = 10 V.
Q6NEET 2024Direction of Induced Current in Solenoids (Moving Bar Magnet)
In the diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions: [Figure: Solenoid-1 (A-B) and Solenoid-2 (C-D), bar magnet with N-S poles moving from solenoid-1 towards solenoid-2]
- A. AB and CD
- B. BA and DC
- C. AB and DC
- D. BA and CD
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Answer: B — BA and DC
As the magnet moves away from solenoid-1, by Lenz's law the induced current opposes the decrease in flux, flowing in direction AB. As the magnet approaches solenoid-2, the induced current opposes the increasing flux, flowing in direction DC. So AB and DC.
Q7NEET 2024Eddy Currents - Force on Sheet Near Magnetic Pole
A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to: A. hold the sheet there if it is magnetic. B. hold the sheet there if it is non-magnetic. C. move the sheet away from the pole with uniform velocity if it is conducting. D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar. Choose the correct statements(s) from the options given below:
- A. A, C and D only
- B. C only
- C. B and D only
- D. A and C only
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Answer: C — B and D only
A force is needed to hold a magnetic sheet in place since it is attracted to the pole (A, true). Moving a conducting sheet with uniform velocity away from the pole requires force to overcome the opposing force from induced eddy currents (C, true). A non-magnetic sheet needs no holding force (B, false), and a non-conducting, non-polar (non-magnetic) sheet needs no force to move since no eddy currents or magnetic force act on it (D, false). So A and C only.
Q8NEET 2023Magnetic Energy Stored in Inductor
The magnetic energy stored in an inductor of inductance 4μH carrying a current of 2A is:
- A. 4 mJ
- B. 8 mJ
- C. 8 μJ
- D. 4 μJ
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Answer: B — 8 mJ
Energy stored E = (1/2)LI² = (1/2) × 4×10^-6 × (2)² = (1/2) × 4×10^-6 × 4 = 8×10^-6 J = 8 μJ.

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Q9NEET 2022Magnetic Flux Calculation
A square loop of side 1 m and resistance 1 ohm is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is:
- A. 2 weber
- B. 0.5 weber
- C. 1 weber
- D. Zero weber
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Answer: A — 2 weber
Magnetic flux = B x A = 0.5 x (1x1) = 0.5 Wb (since the plane is perpendicular to the field, the area vector is parallel to B).
Q10NEET 2022Maximum Induced Current in Rotating Coil
A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad/s. If the vertical component of earth's magnetic field at that place is 2 x 10^-5 T and electrical resistance of the coil is 12.56 ohm, then the maximum induced current in the coil will be:
- A. 0.25 A
- B. 1.5 A
- C. 1 A
- D. 2 A
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Answer: B — 1.5 A
EMF_max = N*B*A*omega. A = pi*r^2 = pi*100 = 100*pi m^2. EMF_max = 1000 x 2x10^-5 x 100*pi x 2 = 4*pi = 12.56 V. I_max = EMF_max/R = 12.56/12.56 = 1 A.

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