Q.Obtain an expression for period of a bar magnet vibrating in a uniform magnetic field and performing angular S.H.M.

A bar magnet displaced from alignment with a field experiences a restoring torque $-MB\sin\theta$; for small oscillations this gives SHM with $T=2\pi\sqrt{I/MB}$.

Consider a bar magnet of magnetic moment $M$ and moment of inertia $I$ (about the vertical axis through its centre), free to oscillate in a uniform magnetic field $B$. When its axis makes a small angle $\theta$ with the field direction, the field exerts a restoring torque on it:

$\tau = -MB\sin\theta$

(negative sign because the torque tends to reduce $\theta$, i.e. restore alignment). For small angular displacements, $\sin\theta \approx \theta$, so:

$\tau \approx -MB\theta$

By Newton's second law for rotation, $\tau = I\dfrac{d^2\theta}{dt^2}$, so:

$I\frac{d^2\theta}{dt^2} = -MB\theta \quad\Longrightarrow\quad \frac{d^2\theta}{dt^2} = -\left(\frac{MB}{I}\right)\theta$

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