Q127 / 180 · AIIMS 2018 (27 May) · 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 $\theta$, angular acceleration is (magnetic induction of earth's horizontal field is $B_H$ )

  • 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

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

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

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