 Solution

For a steady filamentary electric current $I$, the Biot-Savart law isHere $\mathbf{r}$ is the observation position, ${\mathbf{r}}^{\prime }$ the source position, $d{\mathbf{\ell }}^{\prime }$ points in the direction of positive electric current, and ${\mu }_0$ is the vacuum permeability. A circular loop has centre magnetic field ${\mu }_{0}I/(2r)$: the cross product has magnitude $rd{\ell }^{\prime }$ and a constant axial direction, and $\int d{\ell }^{\prime }=2\pi r$.

Introduce the constant signed angular velocity $\Omega$ about the positive axis. A ring's electric charge $dq=2\pi \sigma rdr$ crosses any radial cut once per rotation, so its electric current is $dI=\Omega dq/(2\pi )=\sigma \Omega rdr$. Adding loop contributions yields the magnetic field of a rotating charged annulus:The direction follows the right-hand rule for positive electric charge; reversing either $Q$ or $\Omega$ reverses the magnetic field. The source supplies no rotation rate, so the answer necessarily retains $\Omega$.