Skip to content
doubtly
Sign up

1 answer

Log in to upvote 0 Log in to downvote

Best answer

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

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

\[\tau = -mB\sin\theta \approx -mB\,\theta \quad(\text{for small }\theta)\]

By Newton’s second law for rotation, \(\tau = I\alpha\):

\[I\alpha = -mB\theta \Rightarrow \alpha = -\frac{mB}{I}\theta\]

Since \(mB/I\) is constant, angular acceleration \(\propto\) (−angular displacement), so the motion is angular S.H.M. Comparing with \(\alpha = -\omega^2\theta\):

\[\omega^2 = \frac{mB}{I} \Rightarrow T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{mB}}\]
Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report
Was this answer helpful?
Sign in to comment

Your answer

Log in or create a free account to answer this question.