Q 12-05-012NEETJEE MainEasy
A bar magnet of moment $2\ \text{A m}^2$ is placed at $30°$ to a uniform field of $0.3$ T. The torque on it is
Answer: (C) $0.3$ N m
$\tau = MB\sin\theta = 2 \times 0.3 \times 0.5 = 0.3$ N m.
Solution by Sreeraj P, M.Sc Physics