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A flat disk of radius $R$ with uniform surface charge density $\sigma_q = \frac{Q}{\pi R^2}$ and surface mass density $\sigma_m = \frac{M}{\pi R^2}$ is rotating at angular velocity $\omega$.

Show that its magnetic moment $\vec{m} = \mu_B \vec{L}$ with $\mu_B = \frac{Q}{2 M}$.

Hints: So many ways to do this one. One I showed you this morning - consider the relation between $\sigma_m$, differential angular momentum about the $z$-axis of a tiny chunk of the disk of area $dA$, and stuff like $\omega$, $T$, $v$, $r$. Then consider the relation between $\sigma_q$ and the differential $z$-component of the magnetic moment of the same tiny chunk of the disk. The desired result can be derived either as a differential relation (and nominally integrating) or by doing the simple integrals over the disk.

Or you can just evaluate the two (like you did for a homework problem or two) and divide. This isn't difficult either (and you'll find yourself doing the same actual $r$-integral twice with different constants).