Irodov 3.248: Charge Follows Mass, Ratio Follows – Amit & Deepali Sinha Foundation

Problem Irodov 3.248

A uniform ball of mass $m$ and radius $R$ carries a charge $q$ spread evenly through its volume. The ball spins with angular velocity $\omega$ about an axis through its center. Find its magnetic moment, and the ratio of the magnetic moment to the ball’s angular momentum.

The uniform ball of mass m and radius R with the charge q spread through its volume, spinning with angular velocity ω about the axis through its center O (drawn for q > 0). — The uniform ball of mass $m$ and radius $R$ with the charge $q$ spread through its volume, spinning with angular velocity $\omega$ about the axis through its center $O$ (drawn for $q > 0$ ).

Given

charge $q$ , uniform in the volume
mass $m$ , uniform
radius $R$
angular velocity $\omega$

Find

the magnetic moment

p_m

and the ratio

p_m/M

Solution

Step 1The rotating charge as ring currents

The charge density is uniform:

\rho = \frac{q}{\frac{4}{3}\pi R^3} = \frac{3q}{4\pi R^3}

Divide the ball into thin spherical layers of radius $r$ and thickness $dr$ , and each layer into rings at the polar angle $\theta$ from the axis. The ring between $\theta$ and $\theta + d\theta$ has radius $r\sin\theta$ and cross-section $r\,d\theta\,dr$ , so it holds the charge

dq = \rho\,(2\pi r\sin\theta)(r\,d\theta)(dr) = 2\pi\rho\,r^2\sin\theta\,d\theta\,dr

Turning once every $2\pi/\omega$ , it is a circular current

dI = \frac{\omega\,dq}{2\pi} = \rho\omega\,r^2\sin\theta\,d\theta\,dr \tag{1}

The spherical layer of radius r inside the ball (dashed) and on it the ring at the polar angle θ, a current loop dI circulating with the rotation. — The spherical layer of radius $r$ inside the ball (dashed) and on it the ring at the polar angle $\theta$ , a current loop $dI$ circulating with the rotation.

Step 2The moment of one ring, and of the ball

A loop’s moment is its current times the area it encloses, $\pi (r\sin\theta)^2$ here, and every ring’s moment points along the axis. With (1):

dp_m = dI\,\pi r^2\sin^2\theta = \pi\rho\omega\,r^4\sin^3\theta\,d\theta\,dr

The two integrals separate. The angular one, with $\sin^2\theta = 1 - \cos^2\theta$ , is

\int_0^{\pi}\sin^3\theta\,d\theta = \left[-\cos\theta + \frac{\cos^3\theta}{3}\right]_0^{\pi} = \frac{4}{3}

so that

p_m = \pi\rho\omega\int_0^R r^4\,dr\int_0^{\pi}\sin^3\theta\,d\theta = \pi\rho\omega\cdot\frac{R^5}{5}\cdot\frac{4}{3} = \frac{4\pi\rho\omega R^5}{15}

Substituting the density:

p_m = \frac{4\pi\omega R^5}{15}\cdot\frac{3q}{4\pi R^3}

p_m = \frac{1}{5}\,qR^2\omega

(2)

For $q > 0$ the moment points along $\vect{\omega}$ .

Step 3The ratio to the angular momentum

The ball’s angular momentum about its axis is its moment of inertia times $\omega$ , also along $\vect{\omega}$ :

M = \frac{2}{5}\,mR^2\omega

Dividing (2) by it:

\frac{p_m}{M} = \frac{\frac{1}{5}\,qR^2\omega}{\frac{2}{5}\,mR^2\omega}

\frac{p_m}{M} = \frac{q}{2m}

(3)

Check

Each charge element $dq$ moving on a circle of radius $s$ has $dp_m = dq\,\omega s^2/2$ , while its mass $dm$ has $dM = dm\,\omega s^2$ . When charge and mass are spread in the same proportion, $dq/dm = q/m$ everywhere, so every element has the ratio $q/(2m)$ and so does their sum, whatever the shape: the ratio in (3) needs neither $R$ nor $\omega$ , as it should. Dimensionally, $qR^2\omega$ is a current times an area.

Remark

The ratio $q/(2m)$ is the classical gyromagnetic ratio. The electron’s spin breaks it: its magnetic moment is about twice what $e/(2m_e)$ times its spin angular momentum would give (the $g$ factor of very nearly 2), which is one of the signs that spin is not a little charged ball turning.