Q 12-05-028JEE MainJEE Main 2026 (24 Jan, Shift 1)Easy
A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of $0.016$ N.m. The work done in moving it from most stable to most unstable position is $\alpha\times10^{-3}$ J. The value of $\alpha$ is ______ .
Numerical value type. Enter your answer.
Answer: 64
$B = 800$ G $= 0.08$ T. From $\tau = mB\sin\theta$:
$$m = \frac{0.016}{0.08\times0.5} = 0.4\ \text{A·m}^2$$
From $\theta = 0^\circ$ (most stable) to $180^\circ$ (most unstable):
$$W = mB(\cos0^\circ - \cos180^\circ) = 2mB = 2\times0.4\times0.08 = 0.064\ \text{J} = 64\times10^{-3}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics