 Magnetic field of a rotating charged annulus

An annulus with uniform surface charge density $\sigma$, radii $a,b$ and rigid angular velocity $\Omega$ carries ring electric current $dI=\sigma \Omega rdr$. The centre magnetic field of each ring is ${\mu }_{0}dI/(2r)$ by the Biot-Savart law. Integrating gives $\mathbf{B}(0)={\mu }_0\sigma \Omega (b-a)\hat{\mathbf{n}}/2$, equivalent to the title expression with $Q=\pi \sigma (b^2-a^2)$. The oriented normal follows the positive rotation convention, with the sign of $Q$ retained.