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
Concept. Torque on a coil $\tau=NIAB\sin\theta$, $\theta$ between the axis (moment) and $\vec B$.
Step 1. Value. $125\times1\times\pi(0.02)^2\times0.4\times\tfrac12=0.0314$ N m.
Answer. $314$.
Solution by Sreeraj P, M.Sc Physics
