Question

A uniformly charged disc whose total charge has magnitude q and whose radius is r rotates with constant angular velocity of magnitude $\omega$. What is the magnetic dipole moment?

Answer 1

The surface charge density is $q/\pi r^2$. Hence the charge within a ring radius R and width dR is
$dq=\frac{q}{\pi r^2}\left(2\pi RdR\right)=\frac{2q}{r^2}\left(RdR\right)$
The current carried by this ring is its charge divided by the rotation period,
$di=\frac{dq}{2\pi /\omega }=\frac{q\omega }{\pi r^2}[R.dR]$
The magnetic moment contributed by this ring has the magnitude
$dM=a|dl|$, where a is the area of the ring.
$dM=\pi R^2|di|=\frac{q\omega }{r^2}.R^{3}dR$.
$M=\int dM={\int }_{R=0}^{r}q.\frac{\omega }{r^2}\left(R^{3}dR\right)=q\omega \frac{r^2}{4}$