A circular coil of radius 2\ cm and 125 turns carries a current of 1\ A . The coil is placed in a uniform magnetic field of magnitude 0.4\ T . The axis of the coil makes an angle of 30 ° with the direction of the magnetic field. The torque acting on the coil is α × 10 -4 \ N× m . The value of α is ( π = 3.14 ).

Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 4 April, Evening Shift — Question 23

A circular coil of radius $2\ \mathrm{cm}$ and $125$ turns carries a current of $1\ \mathrm{A}$ . The coil is placed in a uniform magnetic field of magnitude $0.4\ \mathrm{T}$ . The axis of the coil makes an angle of $30^\circ$ with the direction of the magnetic field. The torque acting on the coil is $\alpha \times 10^{-4}\ \mathrm{N\cdot m}$ . The value of $\alpha$ is ( $\pi = 3.14$ ).

Answer: 314

Numerical answer — enter this value.

Step-by-step solution

Torque $\tau = NIAB\sin\theta$ . $N=125$ , $I=1\ \mathrm{A}$ , $A=\pi r^2 = \pi (0.02)^2 = 0.0004\pi\ \mathrm{m}^2$ , $B=0.4\ \mathrm{T}$ , $\theta=30^\circ$ . $\tau = 125 \times 1 \times 0.0004\pi \times 0.4 \times \sin30^\circ = 125 \times 0.0004\pi \times 0.4 \times 0.5 = 125 \times 0.00008\pi = 0.01\pi = 0.0314\ \mathrm{N\cdot m} = 314\times10^{-4}\ \mathrm{N\cdot m}$ . Hence $\alpha = 100\pi = 314$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Force and Torque on Wires and Loops, Magnetic Dipole Moment