Q 12-05-025JEE MainMedium
A short bar magnet placed with its axis at $30°$ to a uniform field of $0.25$ T experiences a torque of $4.5 \times 10^{-2}$ N m. Find its magnetic moment in $\text{A m}^2$.
Numerical value type. Enter your answer.
Answer: 0.36
Concept. Torque on a magnetic dipole: $\tau = MB\sin\theta$.
Step 1. Solve for $M$.
$$M = \frac{\tau}{B\sin\theta} = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = 0.36\ \text{A m}^2$$
Answer. $0.36$.
Solution by Sreeraj P, M.Sc Physics