JEE (Advanced) · Physics, Class 12

Electromagnetic Induction and Alternating Currents: JEE Advanced PYQ, 2007–2026

Electromagnetic Induction and Alternating Currents (Physics, Class 12) has been asked 44 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.1 questions a paper, 6% of all Physics questions. It is the 7th most-asked of 19 Physics chapters. The two 2026 papers had 2 from it. Last checked 28 October 2026.

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Questions per year, both papers

Year by year.

Electromagnetic Induction and Alternating Currents: JEE Advanced questions in each year
Year	Questions
2007	2in 2 papers
2008	0in 2 papers
2009	2in 2 papers
2010	2in 2 papers
2011	4in 2 papers
2012	3in 2 papers
2013	2in 2 papers
2014	1in 2 papers
2015	0in 2 papers
2016	3in 2 papers
2017	4in 2 papers
2018	1in 2 papers
2019	2in 2 papers
2020	2in 2 papers
2021	4in 2 papers
2022	2in 2 papers
2023	3in 2 papers
2024	2in 2 papers
2025	3in 2 papers
2026	2in 2 papers

Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.

Inside the chapter

Topics asked most.

Faraday's law18
AC circuits10
Self and mutual inductance9
Lenz's law and eddy currents4
LCR circuits and resonance2
Transformers1

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10 Electromagnetic Induction and Alternating Currents questions to practise.

Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.

Q1 · Physics · Electromagnetic Induction and Alternating Currents
A lamp rated 60 V, 30 W is to be run from a 100 V (rms) AC mains supply using a pure inductor (choke) in series, so that the lamp operates at its rated values. Treating the lamp as a pure resistor, what must the inductive reactance of the choke be?
Answer: (A)
The lamp current is I = P/V = 30/60 = 0.5 A and its resistance is 120 Ω. In series, V_R is in phase with the current and V_L leads by 90°, so V² = V_R² + V_L². Then V_L = √(100² − 60²) = 80 V and X_L = V_L/I = 160 Ω. A choke is used because it consumes no average power, unlike a resistor.
Q2 · Physics · Electromagnetic Induction and Alternating Currents
In a series LCR circuit with L = 0.2 H, the amplitude of the current is found to be the same when the source frequency is 400 rad/s and when it is 900 rad/s, the source voltage amplitude being constant. What is the value of C?
Answer: (B)
Equal current amplitudes mean equal impedances, so |ω1L − 1/(ω1C)| = |ω2L − 1/(ω2C)|. The circuit is inductive on one side of resonance and capacitive on the other, so ω1L − 1/(ω1C) = −(ω2L − 1/(ω2C)). This gives L(ω1 + ω2) = (1/C)(ω1 + ω2)/(ω1ω2), hence ω1ω2 = 1/(LC), so the resonant frequency is the geometric mean, ω0 = √(ω1ω2) = 600 rad/s. Then C = 1/(ω0²L) = 1/(3.6×10⁵ × 0.2) ≈ 1.39×10⁻⁵ F = 13.9 μF.
Q3 · Physics · Electromagnetic Induction and Alternating Currents
The core of a power transformer is made of many thin sheets of soft iron, each insulated from its neighbours, instead of a single solid block of hard steel. What are the respective purposes of the laminations and of using soft iron?
Answer: (A)
A transformer core loses energy in two main ways. Eddy currents are induced in the core by the changing flux, and cutting the core into thin insulated laminations makes their paths small and resistive. Hysteresis loss is the energy spent re-magnetising the core each cycle, and it is small for soft iron because its hysteresis loop is narrow. So laminations address eddy currents and soft iron addresses hysteresis.
Q4 · Physics · Electromagnetic Induction and Alternating Currents
A coil P of self-inductance 2 H and resistance 5 Ω is connected at t = 0 to a battery of emf 10 V with negligible internal resistance. A second coil S, kept near P with mutual inductance 0.1 H between them, is open-circuited. What is the magnitude of the emf induced in S at t = 0.4 s, which is one time constant of the circuit containing P? (Take e⁻¹ = 0.368.)
Answer: (A)
The emf in S is e_S = M·dI_P/dt. For an LR circuit with a battery, I = (E/R)(1 − e^(−t/τ)) with τ = L/R = 2/5 = 0.4 s. Differentiating gives dI/dt = (E/L)e^(−t/τ) = 5e^(−t/τ) A/s. At t = τ this is 5 × 0.368 = 1.84 A/s. So e_S = 0.1 × 1.84 ≈ 0.18 V.
Q5 · Physics · Electromagnetic Induction and Alternating Currents
A series LCR circuit has R = 20 Ω, L = 0.40 H and C = 10 μF. It is connected to a source e = 200 sin(500t) volts, where t is in seconds. What is the rms current in the circuit?
Answer: (A)
The source has peak voltage 200 V and ω = 500 rad/s. The natural frequency is ω₀ = 1/√(LC) = 1/√(0.40 × 10 × 10⁻⁶) = 1/√(4 × 10⁻⁶) = 500 rad/s. The circuit is therefore at resonance, X_L = X_C and Z = R = 20 Ω. The peak current is 200/20 = 10 A, so I_rms = 10/√2 ≈ 7.1 A.
Q6 · Physics · Electromagnetic Induction and Alternating Currents
The primary of an ideal transformer with turns ratio N_p : N_s = 1 : 5 is connected to an AC source of 20 V (rms) and angular frequency 1000 rad/s. The secondary is connected across a series combination of a 50 Ω resistor, a 0.10 H inductor and a 10 μF capacitor. What rms current does the primary draw?
Answer: (A)
Principle: an ideal transformer fixes V_s/V_p = N_s/N_p and I_p/I_s = N_s/N_p, and a series LCR circuit has impedance Z = √(R² + (X_L − X_C)²). Step 1: V_s = 20 × 5 = 100 V. Step 2: X_L = ωL = 1000 × 0.10 = 100 Ω and X_C = 1/(ωC) = 1/(1000 × 10×10⁻⁶) = 100 Ω. They cancel, so the circuit is at resonance and Z = R = 50 Ω. Step 3: I_s = 100/50 = 2 A. Step 4: for an ideal transformer, V_p I_p = V_s I_s, so I_p = I_s × (N_s/N_p) = 2 × 5 = 10 A. Check: the input power is 20 × 10 = 200 W, equal to I_s²R = 4 × 50 = 200 W.
Q7 · Physics · Electromagnetic Induction and Alternating Currents
A long air-cored solenoid has 1000 turns per metre and a cross-sectional radius of 2.0 cm. A coil of 20 turns and radius 5.0 cm is wound tightly around the middle of the solenoid, with its axis along the solenoid axis. The solenoid current increases uniformly at 5.0 A/s. What emf is induced in the 20-turn coil? (Take μ₀ = 4π × 10⁻⁷ T m/A and neglect the field outside the solenoid.)
Answer: (A)
The field inside the solenoid is B = μ₀nI, so dB/dt = μ₀n(dI/dt) = 4π×10⁻⁷ × 1000 × 5 ≈ 6.28×10⁻³ T/s. The field is zero outside the solenoid, so the flux through each turn of the coil is B times the solenoid's area, π(0.02)² ≈ 1.257×10⁻³ m², not the coil's area. The emf is |E| = N·A·dB/dt = 20 × 1.257×10⁻³ × 6.28×10⁻³ ≈ 1.58×10⁻⁴ V = 158 μV.
Q8 · Physics · Electromagnetic Induction and Alternating Currents
A bar magnet is pushed from far away to the centre of a closed coil of resistance R. In trial 1 this takes time T, and in trial 2 it takes time 2T. In each trial the flux through the coil changes at a uniform rate. Compare the total charge that flows through the coil and the total heat produced in the coil in trial 2 with the values in trial 1.
Answer: (A)
The flux change ΔΦ is the same in both trials. The emf is ε = ΔΦ/t and the current is I = ΔΦ/(Rt). The charge is q = It = ΔΦ/R, so it is independent of t. The heat is H = I²Rt = (ΔΦ)²/(Rt). Doubling t halves H. By Lenz's law the heat comes from the work done against the opposing magnetic force, and this work is smaller when the magnet is pushed more slowly. So the charge is the same and the heat is half as much.
Q9 · Physics · Electromagnetic Induction and Alternating Currents
The secondary winding of an ideal transformer is left open, and the primary is connected to a sinusoidal AC supply. Which statement about the primary circuit is correct?
Answer: (A)
With the secondary open, no current flows in it and there is no reflected load. The ideal primary coil is then a pure inductor connected to an AC source. It carries a magnetising current that sets up the flux, and this current lags the applied voltage by 90°. The average power in a pure inductor is V_rms I_rms cos 90° = 0, so no net energy is drawn from the source.
Q10 · Physics · Electromagnetic Induction and Alternating Currents
A coil of resistance 30 Ω and inductive reactance 40 Ω is connected across an ac source of fixed rms voltage and frequency. When a capacitor is connected in series with the coil, the rms current in the circuit is found to be exactly the same as before the capacitor was added. What is the (non-zero) reactance of the capacitor at this frequency?
Answer: (D)
The current is V/Z, so the current is unchanged only if the impedance is unchanged. Without the capacitor, Z² = R² + X_L². With it, Z² = R² + (X_L − X_C)². Setting them equal gives (X_L − X_C)² = X_L², so X_L − X_C = ±X_L. The root X_C = 0 means no capacitor, so X_C = 2X_L = 80 Ω. The circuit goes from lagging to leading by the same angle, and the value of R does not matter.

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About Electromagnetic Induction and Alternating Currents in JEE Advanced.

How many questions come from Electromagnetic Induction and Alternating Currents in JEE Advanced?

Which topics of Electromagnetic Induction and Alternating Currents are asked most in JEE Advanced?

Faraday's law (18), AC circuits (10), Self and mutual inductance (9), Lenz's law and eddy currents (4), counted over the papers from 2007–2026.

Is Electromagnetic Induction and Alternating Currents important for JEE Advanced 2027?

It is in the syllabus in force, so it can be asked. By past papers it ranks 7 of 19 in Physics; use that to decide how much time it gets, not whether to study it.

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