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A bar magnet of magnetic moment $M$ and moment 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 $(\theta)$, the angular acceleration is (Magnetic induction of earth's horizontal field $=B_H$ )
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The correct answer is:
$\frac{M B_H \theta}{I}$
When magnet is displaced by a very small angle $\theta$, then restoring couple acting on the magnet is
$\tau=-M B_H \sin \theta$
Negative sign shows the restoring nature of torque. Now since $\tau=I \alpha$ and $\sin \theta \approx \theta$ for small angular displacement
$\therefore \quad I \alpha=M B_H \theta$
or
$\begin{aligned} \alpha & =\text { angular acceleration } \\ & =\frac{M B_H \theta}{I}\end{aligned}$
$\tau=-M B_H \sin \theta$
Negative sign shows the restoring nature of torque. Now since $\tau=I \alpha$ and $\sin \theta \approx \theta$ for small angular displacement
$\therefore \quad I \alpha=M B_H \theta$
or
$\begin{aligned} \alpha & =\text { angular acceleration } \\ & =\frac{M B_H \theta}{I}\end{aligned}$
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