= Solution
For a steady filamentary <electric current> $I$, the <Biot-Savart law> is
$$
\boxed{\mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int
\frac{I\,d\boldsymbol\ell'\times(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}.}
$$
Here $\mathbf r$ is the observation position, $\mathbf r'$ the source position, $d\boldsymbol\ell'$ points in the direction of positive <electric current>, and $\mu_0$ is the <vacuum permeability>. A circular loop has centre <magnetic field> $\mu_0I/(2r)$: the cross product has magnitude $r\,d\ell'$ and a constant axial direction, and $\int d\ell'=2\pi r$.
Introduce the constant signed <angular velocity> $\Omega$ about the positive axis. A ring's <electric charge> $dq=2\pi\sigma r\,dr$ crosses any radial cut once per rotation, so its <electric current> is $dI=\Omega\,dq/(2\pi)=\sigma\Omega r\,dr$. Adding loop contributions yields the <magnetic field of a rotating charged annulus>:
$$
\mathbf B(0)=\hat{\mathbf x}\int_a^b\frac{\mu_0}{2r}\sigma\Omega r\,dr,
\qquad
\boxed{\mathbf B(0)=\frac{\mu_0\sigma\Omega(b-a)}2\hat{\mathbf x}
=\frac{\mu_0Q\Omega}{2\pi(a+b)}\hat{\mathbf x}.}
$$
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 $\Omega$.
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