Q 12-04-028NEETJEE MainMedium
A rectangular coil of $50$ turns and area $4 \times 10^{-3}\ \text{m}^2$ carries a current of $2$ A in a uniform magnetic field of $0.5$ T. The plane of the coil makes an angle of $30^\circ$ with the field. The torque on the coil is about
Answer: (A) $0.17$ N m
The torque depends on the angle between the field and the coil's normal (its magnetic moment). If the plane is at $30^\circ$ to $B$, the normal is at $60^\circ$.
$\tau = NIAB\sin 60^\circ = 50 \times 2 \times 4 \times 10^{-3} \times 0.5 \times 0.866 \approx 0.17$ N m.
Solution by Sreeraj P, M.Sc Physics