Q. A short bar magnet, placed with its axis at $30{^\circ}$ with an external magnetic field of 0.16 T, experiences a torque of magnitude 0.032 J. The magnetic moment of the bar magnet is (in units of J/T)
A) 4
B) 0.2
C) 0.5
D) 0.4
Explanation
(d) The instantaneous moment of the deflecting couple is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \tau = MB\sin\theta$
Where M is magnetic moment of magnet.
Given, $\tau = 0.032\ J,\ \ B = 0.16\ T,\ \ \theta = 30{^\circ}$
$\ \ \ \ \ \ \ \ \ \ \ M = \frac{\tau}{B\sin{30{^\circ}}}$
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$\ \ \ \ \ \ \ \ \ \ \ M = \frac{0.032}{0.16 \times \sin{30{^\circ}}} = \frac{0.032 \times 2}{0.16}$
$\ \ \ \ \ \ \ \ \ \ \ M = 0.4\ JT^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.