Question

A long dielectric cylinder of radius $R$ is statically polarized so that at all its points the polarization is equal to $P=\alpha r$, where $\alpha$ is a positive constant, and $r$ is the distance from the axis. The cylinder is set into rotation about its axis with an angular velocity $\omega$. Find the magnetic induction $B$ at the centre of the cylinder.

Answer 1

Because of polarization a space charge is present within the cylinder. It's density is
${\rho }_p=-div\overrightarrow{P}=-2\alpha$
Since the cylinder as a whole is neutral a surface charge density ${\sigma }_p$ must be present on the surface of the cylinder also. This has the magnitude (algebrically)
${\sigma }_p\times 2\pi R=2\alpha \pi R^2$ or, ${\sigma }_p=\alpha R$
When the cylinder rotates, currents are set up which give rise to magnetic fields. The contribution of ${\rho }_p$ and ${\sigma }_p$ can be calculated separately and then added.
For the surface charge the current is (for a particular element)
$\alpha R\times 2\pi Rdx\times \frac{\omega }{2\pi }=\alpha R^2\omega dx$
Its contribution to the magnetic field at the centre is
$\frac{{\mu }_0{R}^2\left(\alpha R^2\omega dx\right)}{2{(x^2+R^2)}^{\frac{3}{2}}}$
and the total magnetic field is
$B_s={\int }_{-\infty }^{\infty }\frac{{\mu }_0{R}^2\left(\alpha R^2\omega dx\right)}{2{(x^2+R^2)}^{\frac{3}{2}}}=\frac{{\mu }_0\alpha R^4\omega }{2}{\int }_{-\infty }^{\infty }\frac{dx}{{(x^2+R^2)}^{\frac{3}{2}}}=\frac{{\mu }_0\alpha R^4\omega }{2}\times \frac{2}{R^2}={\mu }_0\alpha R^2\omega$
As for the volume charge density consider a circle of radius $r$, radial thickness $dr$ and length $dx$.
The current is $-2\alpha \times 2\pi rdrdx\times \frac{\omega }{2\pi }=-2\alpha rdr\omega dx$
The total magnetic field due to the volume charge distribution is
$B_v=-{\int }_0^{R}dr{\int }_{-\infty }^{\infty }dx2\pi r\omega \frac{{\mu }_0{r}^2}{2{(x^2+r^2)}^{\frac{3}{2}}}=-{\int }_0^R\alpha {\mu }_0\omega r^{3}dr{\int }_{-\infty }^{\infty }dx{(x^2+r^2)}^{\frac{3}{2}}$
$=-{\int }_0^R\alpha {\mu }_0\omega rdr\times 2=-{\mu }_0\alpha \omega R^2$ so, $B=B_s+B_v=0$