Q. A short bar magnet experiences a torque of magnitude 0.64 J. When it is placed in a uniform magnetic field of 0.32 T, taking an angle of $30{^\circ}$ with the direction of the field. The magnetic moment of the magnet is
A) $1\ \ Am^{2}$
B) $4\ Am^{2}$
C) $6\ Am^{2}$
D) None of these
Explanation
(b) Torque, $\tau = 0.64\ J,\ B = 0.32\ T,\ \theta = 30{^\circ}$
Torque, $\tau = MB\ \sin\theta$
$\ \ \ \ \ \ \ \ \ \ \ 0.64 = M \times 0.32\sin{30{^\circ}}$
$\ \ \ \ \ \ \ \ \ \ \ 0.64 = M \times 0.32 \times \frac{1}{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = \frac{2 \times 0.64}{0.32} = 4\ Am^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.