The Magnetic Dipole Moment — Question 3

Question 3

A released dipole converts magnetic potential energy into rotational kinetic energy

A rigid dipole of moment .0200 ampere square metre pivots in a plane containing a uniform .500 T field. Its rotational inertia is $1.00\times10^{-5}$ kilogram square metre. It is released from rest at theta0=60.0 degrees. Ignore damping and keep the moment magnitude fixed.

1. Derive angular speed as a function of theta while it moves toward alignment.

2. Find the maximum angular speed and identify where it occurs.

Question 3 - Solution

A released dipole converts magnetic potential energy into rotational kinetic energy

(a) Method and calculation

$$\tfrac12J\dot\theta^2-\mu B\cos\theta=-\mu B\cos\theta_0.$$$$|\dot\theta|=\sqrt{\frac{2\mu B}{J}(\cos\theta-\cos\theta_0)}.$$

(b) Interpretation and further calculation

$$\cos\theta\le1\Rightarrow\omega_{\max}=\sqrt{\frac{2\mu B}{J}(1-\cos\theta_0)}.$$$$\theta=0:\quad\omega_{\max}=\sqrt{1000}=31.6228\,\mathrm{rad/s}.$$

Physical check

The dipole is fastest at minimum potential energy; without damping it overshoots rather than remaining aligned.