Question

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

Hint:

This derivation is very similar to that of a simple pendulum. Just as \(T = 2\pi\sqrt{L/g}\) depends on length and gravity, the magnetic pendulum's period depends on its rotational inertia (\(I\)) and the magnetic restoring force (\(MB\)).

Answer 1

When a bar magnet of magnetic moment \(M\) and moment of inertia \(I\) is suspended in a uniform magnetic field \(B\), and displaced by a small angle \(\theta\), it experiences a restoring torque \(\tau\).
The restoring torque is given by: \[ \tau = -MB\sin\theta \] For small angular displacements (\(\theta\)), \(\sin\theta \approx \theta\). \[ \tau \approx -MB\theta \] This torque is also related to the angular acceleration \(\alpha\) by the equation \(\tau = I\alpha\). \[ I\alpha = -MB\theta \] \[ \alpha = -\left(\frac{MB}{I}\right)\theta \] This equation is of the form \(\alpha = -\omega^2\theta\), which is the condition for angular Simple Harmonic Motion (S.H.M.).
By comparing the two equations, we get the angular frequency \(\omega\): \[ \omega^2 = \frac{MB}{I} \quad \Rightarrow \quad \omega = \sqrt{\frac{MB}{I}} \] The period of oscillation \(T\) is related to the angular frequency by \(T = \frac{2\pi}{\omega}\).
Substituting the expression for \(\omega\): \[ T = 2\pi\sqrt{\frac{I}{MB}} \] This is the expression for the period of a bar magnet vibrating in a uniform magnetic field.