• # question_answer A small compass needle of magnetic moment M and moment of inertia I is free to oscillate in a magnetic field B. It is slightly disturbed from its equilibrium position and then released. Show that it executes simple harmonic motion. Hence, write the expression for its time-period.

As, the needle is displaced from its equilibrium position the torque will try to bring it back in equilibrium position. Hence, acceleration will be related with negative of angular displacement. When compass needle of magnetic moment M and moment of inertia I is slightly disturbed by an angle $\theta$ from the mean position of equilibrium. Then, restoring torque begins to act on the needle which try to bring the needle back to its mean position which is given by $\tau =-MB\sin \theta .$ Since, $\theta$ is small. So, $\sin \theta \approx \theta$ $\therefore$      $\tau =-MB\,\theta$ but $\tau =I\alpha$ where,   $\alpha$= angular acceleration M = magnetic moment of dipole $\Rightarrow$   $I\alpha =-MB\,\theta$ $\Rightarrow \alpha =-\left( \frac{MB}{I} \right)\theta$ $\therefore$      $\alpha \propto -\theta$ $\Rightarrow$   Angular acceleration $\propto -$ (angular displacement) $\Rightarrow$   Therefore, needle executed SHM. Hence, the time-period $T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{MB}{I}}}$ or $T=2\pi \sqrt{\frac{I}{MB}}$ This is the required expression.