Question

A bar magnet of magnetic moment At and moment $M$ of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle $ \theta , $ angular acceleration is (magnetic induction of earth's horizontal field $=B_{H}$)

• A. $ \frac{MB_{H}\theta }{I} $
• B. $ \frac{IB_{H}\theta }{M} $
• C. $ \frac{M\theta }{IB_{H}} $
• D. $ \frac{I\theta }{MB_{H}} $

Answer 1

Answer: (A)

Restoring torque on the magnet,
$\tau=-2 m B_{H} l \sin \theta=-M B_{H} \sin \theta$
if $\theta$ is small $\sin \theta \approx \theta$
$I \alpha= -M B_{H} \theta ; \alpha=\frac{M B_{H} \theta}{I}$