Q148 / 160 · TS EAMCET 2007 · Physics

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 $=B_H$ )

  • Correct answer
    A
    $\frac{M B_H \theta}{I}$
  • B
    $\frac{I B_H \theta}{M}$
  • C
    $\frac{M \theta}{I B_H}$
  • D
    $\frac{I \theta}{M B_H}$

Solution

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 $\begin{aligned} \alpha & =\text { angular acceleration } \\ & =\frac{M B_H \theta}{I}\end{aligned}$

Hence, the correct answer is option: 'A'.

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