Townsend - Problem 1.2

Chapter 1 - Stern Gerlach Experiments

Problem 1.2
Show for a solid spherical ball of mass $m$ rotating about an axis through its center with a charge $q$ uniformly distributed on the surface of the ball that the magnetic moment $\mu$ is related to the angular momentum $L$ by the relation
$$\mu = \frac{5q}{6mc}L$$

Solution 
We are going to work in Gaussian units.

Recall that the magnetic moment is given by $$\mu = \frac{IA}{c}$$. Since we are dealing with uniformly charged sphere, we will cut up the sphere into infinitely many rings that are infinitely small in height. Without loss of generality we can cut them horizontally so that they are parallel to the x-axis.

Because the ball is rotating around the y-axis, there is an induced current. The current for each ring is given by $$I = \frac{\sigma dA}{T}$$, where $\sigma = \frac{q}{4\pi R^2}$ is the surface charge density for a sphere of radius $R$ and $T = \frac{2\pi}{\omega}$ is the period of the sphere's rotation.

Assuming the sphere has radius R, each ring has radius $r = Rsin\theta$ and thus $dA = 2\pi r Rd\theta$ is the area of each ring.

Thus $d\mu = \frac{dIA}{c} = \frac{dI}{c}\pi R^2sin^2\theta$.

Then to obtain $\mu$ of the whole sphere we need to integrate over all angles to get $$\mu = \int d\mu = \int \frac{\sigma dA}{Tc}\pi R^2sin^2\theta = \int\limits_0^{\pi} \frac{\sigma}{Tc}2{\pi}^2 R^4sin^3\theta d\theta \\ = \int\limits_0^{\pi} \frac{\pi q}{2Tc} R^2sin^3\theta d\theta \\ = \frac{q}{3c}\frac{2\pi}{T}R^2 \\ = \frac{q}{3c}\omega R^2 $$

Since the angular momentum of sphere is given by $$L = \frac{2}{5}mR^2\omega$$ we finally get $$\mu = \frac{5q}{6mc}L$$ as required.