Irodov 3.246: Spinning Charge Becomes Ring Currents – Amit & Deepali Sinha Foundation

Problem Irodov 3.246

A thin insulating disc of radius $R$ carries a uniform surface charge of density $\sigma$ on one face and spins about its own axis with angular velocity $\omega$ . Find:

the magnetic induction at the center of the disc
the magnetic moment of the disc

The insulating disc of radius R, charged with surface density σ, spinning with angular velocity ω about its axis (drawn for σ > 0). — The insulating disc of radius $R$ , charged with surface density $\sigma$ , spinning with angular velocity $\omega$ about its axis (drawn for $\sigma > 0$ ).

Given

radius $R$
surface density $\sigma$
angular velocity $\omega$

Find

(a)

B

at the center · (b) the magnetic moment

p_m

Solution

Step 1The rotating charge as ring currents

Cut the disc into thin rings. The ring between radii $r$ and $r + dr$ carries the charge on its area $2\pi r\,dr$ .

dq = \sigma\,2\pi r\,dr

This charge passes any fixed radius once per revolution, every $T = 2\pi/\omega$ , so the ring is a circular current

dI = \frac{dq}{T} = \frac{\sigma\,2\pi r\,dr\,\omega}{2\pi} = \sigma\omega r\,dr \tag{1}

The ring element of the disc between r and r + dr is a current loop dI; its field dB at the center lies along the axis, in the sense of ω for σ > 0. — The ring element of the disc between $r$ and $r + dr$ is a current loop $dI$ ; its field $d\vect{B}$ at the center lies along the axis, in the sense of $\omega$ for $\sigma > 0$ .

Step 2(a) The field at the center

A circular current $dI$ of radius $r$ gives at its center the field $\mu_0\,dI/(2r)$ along its axis, and every ring gives it in the same direction, so the contributions add. With (1):

dB = \frac{\mu_0\,dI}{2r} = \frac{\mu_0\sigma\omega r\,dr}{2r} = \frac{\mu_0\sigma\omega}{2}\,dr

Integrating from the center to the rim:

B = \frac{\mu_0\sigma\omega}{2}\int_0^R dr

B = \frac{1}{2}\,\mu_0\sigma\omega R

(2)

For $\sigma > 0$ the field points along $\vect{\omega}$ : the charge circulates in the sense of the rotation, and the right-hand rule then gives $\vect{B}$ along the rotation axis in the sense of $\vect{\omega}$ .

Step 3(b) The magnetic moment

A current loop has the moment current times enclosed area, so the ring of (1) contributes

dp_m = dI\,\pi r^2 = \pi\sigma\omega r^3\,dr

all of them along the axis. Adding the rings:

p_m = \pi\sigma\omega\int_0^R r^3\,dr

p_m = \frac{1}{4}\,\pi\sigma\omega R^4

(3)

Like the field, the moment points along $\vect{\omega}$ for $\sigma > 0$ .

Check

With the total charge $q = \sigma\pi R^2$ the moment is $p_m = q\omega R^2/4$ . The mechanical angular momentum of a uniform disc of mass $m$ is $M = m\omega R^2/2$ , so $p_m/M = q/(2m)$ , the gyromagnetic ratio every body with charge distributed like its mass must have. Dimensionally, $\mu_0\sigma\omega R$ is $\mu_0$ times a current per unit length, $\sigma\omega R$ in A/m, which is a field in tesla, and $\sigma\omega R^4$ is a current times an area.

Remark

The field at the center is linear in $R$ while the moment goes as $R^4$ : the field at the center favors no ring over another, but the moment weights the outer rings by their area as well as their current. That is why the outer rim dominates how the disc looks from far away, where only the moment matters.