Magnets — Question 4

Question 4

A magnet pendulum measures a field through its small-angle period

A magnet with dipole magnitude m and rotational inertia J pivots without friction in a uniform horizontal field B. It is displaced through a small horizontal angle theta from stable alignment.

1. Derive its angular frequency and period.

2. The period becomes T/2 after a parallel field is added. Find the added field and the small-angle oscillation energy at amplitude theta0.

Question 4 - Solution

A magnet pendulum measures a field through its small-angle period

(a) Method and calculation

$$J\ddot\theta=-mB\sin\theta\simeq-mB\theta.$$$$\boxed{\omega_0=\sqrt{mB/J},\qquad T=2\pi\sqrt{J/(mB)}}.$$

(b) Interpretation and further calculation

$$T_{\mathrm{new}}=T/2\Rightarrow B_{\mathrm{new}}=4B\Rightarrow\boxed{B_{\mathrm{added}}=3B}.$$$$\boxed{E_{\mathrm{osc}}\simeq\tfrac12mB\theta_0^2}\quad\text{in the original field}.$$

Physical check

The period formula requires a small angle in radians and a fixed dipole magnitude.