Question

Find the magnetic induction at the centre of the disc.

• A. $B=\frac{{\mu }_0}{2}\sigma \omega 2R$
• B. $B=\frac{{\mu }_0}{2}\sigma \omega 4R$
• C. $B=\frac{{\mu }_0}{2}\sigma \omega R$
• D. $B=\frac{{\mu }_0}{2}\sigma \omega 3R$

Answer 1

The correct option is C $B=\frac{{\mu }_0}{2}\sigma \omega R$

Here, 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 $\omega$.
Let, $q$ be the uniform charge over one side of disc.
The current induced on the small area of disc having charge $dq$ in time $dt$ is
$dI=\frac{dq}{dt}$ .......................................................(1)
$\sigma =\frac{dq}{ds}$ $\Rightarrow dq=\sigma ds$
Also, $dt=\frac{1}{n}$ and $\omega =2\pi n\Rightarrow n=\frac{\omega }{2\pi }$
Using above equations we can rewrite equation (1) as
$dI=\frac{\sigma ds\omega }{2\pi }$
Integrating both sides we get,
$I=\sigma \omega \frac{2\pi R}{2\pi }$
$I=\sigma \omega R$.................................................. (2)
According to Biot-Savart's law, the magnetic induction at the center of the disc due to small length $dl$ is
$B=\int dB=\int \frac{{\mu }_{0}I}{4\pi R}dl$
$B=\frac{{\mu }_0}{4\pi }\int \left(\sigma \omega R\right)\frac{dl\sin \theta }{R^2}$....................................................................from (2)
Integrating both sides we get
$B=\frac{\sigma \omega {\mu }_0}{4\pi }\int \frac{\left(R\right)\left(2\pi R\right)\sin {90}^{\circ }}{R^2}dR$..............$($since $dl$ is almost straight, $\theta ={90}^{\circ })$
$B=\frac{\sigma \omega {\mu }_0}{2}\int dR$
$B=\frac{{\mu }_0\left(\sigma \omega R\right)}{2}$