A nonconducting plate of mass m and radius r has a charge q...

For a uniform distribution of charge and mass, the ratio of magnetic moment $\mu$ to angular momentum $l$ can be found by considering an elemental ring of radius $x$ and thickness $dx$ on the plate.

The charge on this ring is $dq = \dfrac{q}{\pi r^2} (2\pi x dx) = \dfrac{2q}{r^2} x dx$

The equivalent current due to the rotation of this ring is $di = \dfrac{dq}{T} = \dfrac{dq \omega}{2\pi} = \dfrac{q \omega}{\pi r^2} x dx$

The magnetic moment of this elemental ring is $d\mu = di (\pi x^2) = \dfrac{q \omega}{r^2} x^3 dx$

Integrating from $x = 0$ to $x = r$, the total magnetic moment is:

$\mu = \int_{0}^{r} \dfrac{q \omega}{r^2} x^3 dx = \dfrac{q \omega r^2}{4}$

The angular momentum of the circular plate (disk) about its axis is:

$l = I\omega = \dfrac{mr^2}{2} \omega$

Taking the ratio of magnetic moment to angular momentum:

$\dfrac{\mu}{l} = \dfrac{\dfrac{q \omega r^2}{4}}{\dfrac{m r^2 \omega}{2}} = \dfrac{q}{2m}$

$\mu = \dfrac{q}{2m} l$