JEE (Advanced) · Physics, Class 12
Magnetic Effects of Current and Magnetism: JEE Advanced PYQ, 2007–2026
Magnetic Effects of Current and Magnetism (Physics, Class 12) has been asked 43 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.1 questions a paper, 6% of all Physics questions. It is the 8th most-asked of 19 Physics chapters. The two 2026 papers had 3 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 2in 2 papers |
| 2008 | 2in 2 papers |
| 2009 | 1in 2 papers |
| 2010 | 1in 2 papers |
| 2011 | 2in 2 papers |
| 2012 | 4in 2 papers |
| 2013 | 3in 2 papers |
| 2014 | 4in 2 papers |
| 2015 | 3in 2 papers |
| 2016 | 0in 2 papers |
| 2017 | 5in 2 papers |
| 2018 | 3in 2 papers |
| 2019 | 1in 2 papers |
| 2020 | 1in 2 papers |
| 2021 | 3in 2 papers |
| 2022 | 1in 2 papers |
| 2023 | 0in 2 papers |
| 2024 | 3in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 3in 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.
- Force on moving charges and currents18
- Biot-Savart law8
- Magnetic dipole8
- Moving coil galvanometer5
- Ampere's law4
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10 Magnetic Effects of Current and Magnetism 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 · Magnetic Effects of Current and Magnetism
A proton and an alpha particle, with the same kinetic energy, enter a uniform magnetic field at right angles to it. What is the ratio of the radius of the proton's path to that of the alpha particle's?
Answer: (B)
r = √(2mK)/qB. The alpha particle has four times the mass and twice the charge: √4/2 = 1. The two radii are equal.
Q2 · Physics · Magnetic Effects of Current and Magnetism
A bar magnet of magnetic moment 1.6 A m² is placed on a horizontal table with its north pole pointing towards geographic north (along the magnetic meridian). The horizontal component of the Earth's field is 2.0 × 10⁻⁵ T. Neutral points, where the net horizontal field is zero, occur on the equatorial line of the magnet. How far from the centre of the magnet are they? Treat the magnet as a point dipole and take μ₀/4π = 10⁻⁷ T m A⁻¹.
Answer: (A)
With the magnet's north pole pointing north, the field on its equatorial line points from north to south, opposite to M. That is the opposite direction to the Earth's horizontal field H (south to north). They cancel where (μ₀/4π)M/d³ = H. So d³ = (10⁻⁷ × 1.6)/(2.0 × 10⁻⁵) = 8.0 × 10⁻³ m³, and d = 0.20 m. If the magnet's north pole pointed south, its axial field (twice as large at the same distance) would oppose H instead, and the neutral points would lie on the axis.
Q3 · Physics · Magnetic Effects of Current and Magnetism
Three identical pieces of thin wire, each of length L, carry the same steady current I. One is bent into a circular loop, another into an equilateral triangle, and the third into a square. What is the ratio of the magnetic dipole moment of the circular loop to that of the equilateral triangular loop?
Answer: (A)
For a planar loop, μ = I·A, and I is the same for all three, so the ratio equals the ratio of areas. Circle: 2πR = L, so R = L/2π and A = πR² = L²/(4π). Equilateral triangle: side s = L/3, so A = (√3/4)(L/3)² = √3L²/36. The ratio is (L²/4π) ÷ (√3L²/36) = 36/(4π√3) = 9/(π√3) = 3√3/π.
Q4 · Physics · Magnetic Effects of Current and Magnetism
A very long wire carrying a current I = 5 A is bent into an L shape. One arm lies along the positive x-axis from the origin O to infinity. The other arm lies along the positive y-axis from infinity to O. The current enters along the y-arm and leaves along the x-arm. What is the magnitude of the magnetic field at the point P(−5 cm, 0)? (μ₀/4π = 10⁻⁷ T m A⁻¹)
Answer: (A)
The Biot–Savart law gives dB ∝ I dl × r/r³. For the x-arm, every element and every r vector lie along the x-axis, so dl × r = 0 and this arm contributes nothing at P. For the y-arm, P is at perpendicular distance d = 0.05 m from the foot of the arm at O. One end of the arm is at this foot (angle 0) and the other is at infinity (angle 90°), so B = (μ₀I/4πd)(sin 90° − sin 0°) = μ₀I/(4πd). This gives B = 10⁻⁷ × 5/0.05 = 1.0 × 10⁻⁵ T = 10 μT, which is half the field of an infinite wire.
Q5 · Physics · Magnetic Effects of Current and Magnetism
A long solenoid with 500 turns per metre carries a steady current of 0.50 A. A soft ferrite rod of relative permeability 400 and cross-sectional area 2.0 × 10⁻⁴ m² fills the central region of the solenoid, with its axis along the solenoid axis. A search coil of 50 turns is wound tightly around the middle of the rod. The search coil is connected to a circuit of total resistance 5.0 Ω. The solenoid current is then reversed (changed to 0.50 A in the opposite direction). Take μ₀ = 4π × 10⁻⁷ T m A⁻¹ and ignore any flux outside the rod. The total charge that flows through the search-coil circuit is closest to:
Answer: (A)
The magnetising field depends only on the free current: H = nI = 500 × 0.50 = 250 A m⁻¹. In the ferrite, B = μ₀μr H = 4π × 10⁻⁷ × 400 × 250 = 4π × 10⁻² T ≈ 0.126 T. The flux through each turn is Φ = BA = 0.126 × 2.0 × 10⁻⁴ ≈ 2.5 × 10⁻⁵ Wb. Reversing the current reverses B, so the change in flux per turn is 2Φ ≈ 5.0 × 10⁻⁵ Wb. The charge that flows is q = NΔΦ/R = 50 × 5.03 × 10⁻⁵ / 5.0 ≈ 5.0 × 10⁻⁴ C.
Q6 · Physics · Magnetic Effects of Current and Magnetism
A moving coil galvanometer has a coil of 50 turns and area 2.0 × 10⁻³ m², placed in a radial magnetic field of magnitude 0.040 T. A current of 5.0 mA produces a steady deflection of 30° (π/6 rad). What is the torsional constant of the suspension spring?
Answer: (D)
In a radial field the magnetic torque is NIAB for every position of the coil. The spring's restoring torque is kφ. At equilibrium NIAB = kφ. NIAB = 50 × 5.0×10⁻³ × 2.0×10⁻³ × 0.040 = 2.0×10⁻⁵ N m. With φ = 30° = π/6 rad ≈ 0.524 rad, k = 2.0×10⁻⁵ / 0.524 ≈ 3.8×10⁻⁵ N m rad⁻¹.
Q7 · Physics · Magnetic Effects of Current and Magnetism
A long thin wire carrying a steady current I is bent into a hairpin. Two parallel semi-infinite straight sections lie a distance 2R apart. Their finite ends are joined by a semicircular arc of radius R, whose diameter is perpendicular to the straight sections. The whole wire lies in one plane, and the current flows in along one straight section, round the arc, and out along the other. Each straight section ends at the arc's end-point, so the straight sections are perpendicular to the diameter joining the arc's ends. The point O is the centre of the semicircle. What is the magnitude of the magnetic field at O?
Answer: (A)
Use the Biot–Savart law, dB = (μ₀/4π) I dl × r̂ / r², for each part of the wire. The semicircular arc has every element perpendicular to the radius, at distance R from O. Its field is half that of a full loop: B_arc = μ₀I/(4R). Each semi-infinite straight section ends at the foot of the perpendicular from O, a distance R away. Such a wire gives half the field of an infinite wire, so B = (1/2)(μ₀I/2πR) = μ₀I/(4πR). Now check directions. Take the current entering along the upper wire toward the arc, going round the arc, and leaving along the lower wire. Applying dl × r̂ gives the same sense (out of the plane) for the upper wire, the arc and the lower wire, so the three fields add. Total B = μ₀I/(4R) + 2·μ₀I/(4πR) = (μ₀I/4R)(1 + 2/π).
Q8 · Physics · Magnetic Effects of Current and Magnetism
A long straight cylindrical conductor of radius R carries a current whose density varies with distance s from the axis as J = J₀(s/R), where J₀ is a constant and the current flows along the axis. What is the magnetic field magnitude at a distance R/2 from the axis?
Answer: (A)
With J varying, the enclosed current must be found by integration, not by J × area. I_enc = ∫₀^r J₀(s/R)·2πs ds = (2πJ₀/R)(r³/3). Ampere's law then gives B(2πr) = μ₀I_enc, so B = μ₀J₀r²/(3R). At r = R/2, B = μ₀J₀R/12.
Q9 · Physics · Magnetic Effects of Current and Magnetism
A moving coil galvanometer of resistance 50 Ω gives full-scale deflection for 5 mA. It is converted into an ammeter of range 0–100 mA by connecting a suitable shunt. This ammeter is then inserted in series with a 47.5 Ω resistor and an ideal 2.5 V cell (negligible internal resistance). By what percentage is the reading of the ammeter lower than the current that flowed in the circuit before the ammeter was inserted?
Answer: (A)
Converted to an ammeter, the galvanometer and shunt form a parallel combination. At full range (100 mA), the voltage across the combination is I_g G = 5 mA × 50 Ω = 0.25 V. So the ammeter resistance is R_A = 0.25 V/100 mA = 2.5 Ω. This is not the shunt value, which is about 2.63 Ω. Without the ammeter, I_true = 2.5 V/47.5 Ω ≈ 52.6 mA. With the ammeter, the total resistance is 47.5 + 2.5 = 50 Ω, so the reading is 2.5/50 = 50 mA. The fractional error relative to the true current is (52.63 − 50)/52.63 = R_A/(R + R_A)... which equals 2.5/50 = 0.05. So the reading is 5.0 % lower than the true current.
Q10 · Physics · Magnetic Effects of Current and Magnetism
Two coplanar concentric circular loops have radii 5 cm and 10 cm. The inner loop carries 3 A in the anticlockwise sense. What current in the outer loop makes the net magnetic field at the common centre zero?
Answer: (D)
The field at the centre of a loop is B = μ₀I/2R. Its direction follows the right-hand rule: anticlockwise gives out of the plane, clockwise gives into it. For the fields to cancel they must be equal in magnitude and opposite in direction. So I₁/R₁ = I₂/R₂, which gives I₂ = 3 × 10/5 = 6 A, and the outer current must be clockwise.
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All 43 real JEE Advanced questions from Magnetic Effects of Current and Magnetism.
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About Magnetic Effects of Current and Magnetism in JEE Advanced.
How many questions come from Magnetic Effects of Current and Magnetism in JEE Advanced?
Magnetic Effects of Current and Magnetism (Physics, Class 12) has been asked 43 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.1 questions a paper, 6% of all Physics questions. It is the 8th most-asked of 19 Physics chapters. The two 2026 papers had 3 from it.
Which topics of Magnetic Effects of Current and Magnetism are asked most in JEE Advanced?
Force on moving charges and currents (18), Biot-Savart law (8), Magnetic dipole (8), Moving coil galvanometer (5), counted over the papers from 2007–2026.
Is Magnetic Effects of Current and Magnetism important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 8 of 19 in Physics; use that to decide how much time it gets, not whether to study it.
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On Nikaalle, free: sign up and practise every past JEE Advanced question from Magnetic Effects of Current and Magnetism with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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