Question

A non-conducting sphere of radius $R=50\text{mm}$ charged uniformly with surface density $\sigma =10.0\mu {\text{C/m}}^2$ rotates with an angular velocity $\omega =70rad/s$ about the axis passing through its centre. The magnetic induction $B$ at the centre of the sphere in $\text{pT}$ is $(20+x)$. Find the value of $x$.

Answer 1

As only the outer surface of the sphere is charged, consider the element as a ring, as shown in figure below.
The equivalent current due to the ring element,
$di=\frac{\omega }{2\pi }(2\pi r\sin \theta .r.d\theta )\sigma$ (1)
and magnetic induction due to this loop element at the centre of the sphere, $O$,
$dB=d\frac{{\mu }_0}{4\pi }di\frac{2\pi r\sin \theta .r\sin \theta }{r^3}=\frac{{\mu }_0}{4\pi }di\frac{{\sin }^2\theta }{r}$
Hence, the total magnetic induction due to the sphere at the centre, $O$,
$B=\int dB={\int }_0^{\frac{\pi }{2}}\frac{{\mu }_0}{4\pi }\frac{\omega }{2\pi }\frac{2\pi r^2\sin \theta .d\theta {\sin }^2\theta .\sigma }{r}$ [using (1)]
Hence, $B={\int }_0^{\frac{\pi }{2}}\frac{{\mu }_0\sigma \omega r}{4\pi }{\sin }^3\theta .d\theta =\frac{2}{3}{\mu }_0\sigma \omega r=29pT$