Q 12-04-041JEE MainJEE Main 2026 (4 Apr, Shift 2)Easy
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 $30^\circ$ with the direction of the magnetic field. The torque acting on the coil is $\alpha \times 10^{-4}$ N.m. The value of $\alpha$ is ______. ($\pi = 3.14$)
Numerical value type. Enter your answer.
Answer: 314
$\tau = NIAB\sin 30^\circ = 125 \times 1 \times 3.14 \times (0.02)^2 \times 0.4 \times 0.5 = 0.0314$ N m $= 314 \times 10^{-4}$ N m.
Solution by Sreeraj P, M.Sc Physics