JEE (Main) · Physics, Class 12
Electromagnetic Induction and Alternating Currents: JEE Main PYQ, 2020–2025
Electromagnetic Induction and Alternating Currents (Physics, Class 12) has been asked 189 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.7 questions a paper, 6% of all Physics questions. It is the 5th most-asked of 20 Physics chapters. The 2025 shifts had 13 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 16in 14 papers |
| 2021 | 52in 26 papers |
| 2022 | 47in 24 papers |
| 2023 | 24in 12 papers |
| 2024 | 37in 18 papers |
| 2025 | 13in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- AC circuits50
- LCR circuits and resonance49
- Faraday's law47
- Self and mutual inductance31
- Transformers7
- Lenz's law and eddy currents5
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10 Electromagnetic Induction and Alternating Currents questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Electromagnetic Induction and Alternating Currents
A small bar magnet is released from rest and falls vertically through a long copper tube, with its axis along the tube. It falls more slowly than a magnet falling freely through air. Which statement best explains this?
Answer: (A)
Lenz's law says the induced current opposes the change that produces it. As the magnet falls, the flux through the tube below it changes, so eddy currents circulate in the copper. Their magnetic field opposes the magnet's motion, producing an upward force that grows with speed. This converts gravitational potential energy into Joule heat in the tube, so the magnet falls more slowly than free fall.
Q2 · 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.
Q3 · 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.
Q4 · Physics · Electromagnetic Induction and Alternating Currents
A resistor of 3 Ω is connected in series with a capacitor whose reactance is 4 Ω at the supply frequency. What is the magnitude of the impedance of the series combination?
Answer: (C)
Resistance and capacitive reactance are 90 degrees out of phase, so they combine as perpendicular phasors. The impedance is Z = √(R² + X_C²). With R = 3 Ω and X_C = 4 Ω, the sum of squares is 25, so Z = 5 Ω. The 3-4-5 triangle is the standard check.
Q5 · 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.
Q6 · 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.
Q7 · Physics · Electromagnetic Induction and Alternating Currents
A series LCR circuit is driven by an AC source at its resonant frequency. What is the power factor of the circuit?
Answer: (C)
The power factor is cos φ = R/Z. At resonance the inductive and capacitive reactances cancel exactly, X_L = X_C, so Z = R. Therefore cos φ = 1, meaning the voltage and current are in phase and the average power is V_rms I_rms.
Q8 · 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.
Q9 · 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.
Q10 · Physics · Electromagnetic Induction and Alternating Currents
A capacitor of capacitance 20 μF is connected to a 50 Hz AC supply. What is its capacitive reactance?
Answer: (A)
The capacitive reactance is X_C = 1/(ωC) with ω = 2πf. Substituting f = 50 Hz and C = 20 × 10⁻⁶ F gives X_C = 1/(2π × 50 × 20 × 10⁻⁶) ≈ 159 Ω. Note that capacitive reactance falls as frequency or capacitance rises, the opposite of inductive reactance.
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About Electromagnetic Induction and Alternating Currents in JEE Main.
How many questions come from Electromagnetic Induction and Alternating Currents in JEE Main?
Electromagnetic Induction and Alternating Currents (Physics, Class 12) has been asked 189 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.7 questions a paper, 6% of all Physics questions. It is the 5th most-asked of 20 Physics chapters. The 2025 shifts had 13 from it in all.
Which topics of Electromagnetic Induction and Alternating Currents are asked most in JEE Main?
AC circuits (50), LCR circuits and resonance (49), Faraday's law (47), Self and mutual inductance (31), counted over the papers from 2020–2025.
Is Electromagnetic Induction and Alternating Currents important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 5 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
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