Q. A short bar magnet placed with its axis at$30{^\circ}$, with a uniform external magnetic field of 0.25 T experiences a torque of $4.5 \times 10^{- 2}$N-m. Magnetic moment of the magnet is
A) $0.36\ J\ T^{- 1}$
B) 0.72 $J\ T^{- 1}$
C) 0.18 $J\ T^{- 1}$
D) Zero
Explanation
(a) From $\tau = MB\ sin\theta,M = \frac{\tau}{B\ sin\theta}$
$= \frac{4.5 \times 10^{- 2}}{0.25\sin{30{^\circ}}} = 0.36\ JT^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.