Obtain an expression for the period of a bar magnet vibrating in a uniform magnetic field, performing S.H.M.

Answer: \(T = 2\pi\sqrt{\dfrac{I}{mB}}\)

Consider a bar magnet of magnetic dipole moment \(m\) and moment of inertia \(I\) (about the axis of suspension), freely suspended in a uniform magnetic field \(B\). In equilibrium its axis is along \(B\). If it is turned through a small angle \(\theta\) and released, a restoring torque acts on it:

For small \(\theta\), \(\sin\theta \approx \theta\), so \(\tau = -mB\theta\).

Also \(\tau = I\alpha\), where \(\alpha\) is the angular acceleration. Hence

Since \(\alpha \propto -\theta\), the magnet performs angular S.H.M. Comparing with \(\alpha = -\omega^2\theta\):

The period of oscillation is