Question

A non-conducting thin disc of radius $R$ charged uniformly over one side with surface density $\sigma$ rotates about its axis with an angular velocity $\omega$. Find:
(a) the magnetic induction at the centre of the disc;
(b) the magnetic moment of the disc.

Answer 1

(a) Let us a ring element of radius r and thickness dr,

then charge on the ring element.,

$d\rho =$ $\sigma$ $2\pi$ $r$ $dr$

and current, due to this element,

di = $\frac{\left(\sigma 2\pi rdr\right)\omega }{2\pi }$ $=\pi$ $\omega$ r dr

So, magnetic induction at the center, due to this element :

dB = $\frac{{\mu }_0}{2}$ $\frac{di}{r}$

B = $\int dB$

${\int }_0^R\frac{{\mu }_0\sigma \omega rdr}{r}$ $

$\frac{{\mu }_0}{2}$ $\sigma \omega R$

and hence, from symmetry :

(b) Magnetic moment of the element, considered,

$d{p}_m=\left(di\right)\pi r^2=\sigma \omega dr\pi r^2$

= $\sigma \pi \omega r^3$ dr

Hence, the sought magnetic moment,

$P_m=\int d{p}_m$

= ${\int }_0^R\sigma \pi \omega r^{3}dr$

= $\sigma \omega \pi \frac{R^4}{4}$