= Magnetic field of a rotating charged annulus
{title2=$B(0)=\mu_0Q\Omega/[2\pi(a+b)]$}
An <annulus> with uniform <surface charge density> $\sigma$, radii $a,b$ and rigid <angular velocity> $\Omega$ carries ring <electric current> $dI=\sigma\Omega r\,dr$. The centre <magnetic field> of each ring is $\mu_0dI/(2r)$ 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.
Back to article page
