Question

A non-conducting disc of radius $R$ has surface charge density which varies with distance from the centre as $\sigma (r)={\sigma }_0\left[1+\sqrt{\frac{r}{R}}\right]$, where ${\sigma }_0$ is a constant. The disc rotates about its axis with angular velocity $\omega .$ If $B$ is the magnitude of magnetic induction at the centre, then $\frac{B}{{\mu }_0{\sigma }_0\omega R}$ will be

• A. $\frac{3}{4}$
• B. $\frac{4}{5}$
• C. $\frac{5}{6}$
• D. $\frac{6}{7}$

Answer 1

Answer: (C)

Consider an elemental ring of radius $r$ of thickness $dr$

Charge on ring $dq=$ Area of ring $\times$ Charge density
$\Rightarrow dq=(2\pi r\cdot dr)\sigma$
$\Rightarrow$ Current through ring, $di=\frac{dq}{T}$, as angular
velocity of disc is $\omega$.
So, $\frac{2\pi }{\omega }=T$
$\Rightarrow di=\frac{\omega dq}{2\pi }=\frac{\omega (2\pi r\cdot dr)\sigma }{2\pi }=\omega r\sigma dr$
Now, magnetic induction due to elemental ring at the centre,
$dB=\frac{{\mu }_{0}di}{2r}=\frac{{\mu }_0\omega r\sigma }{2r}\cdot dr=\frac{{\mu }_0\omega \sigma }{2}\cdot dr$
As $\sigma ={\sigma }_0\left(1+\sqrt{\frac{r}{R}}\right)$
For magnetic induction due to entire disc, we integrate from $r=0$ to $r=R$
$\Rightarrow B=\frac{{\mu }_0\omega {\sigma }_0}{2}{\int }_0^R\left(1+\sqrt{\frac{r}{R}}\right)\cdot dr$
IInd Part
$\Rightarrow B=\frac{{\mu }_0\omega {\sigma }_0}{2}{\left[r+\frac{2r^{3/2}}{3\sqrt{R}}\right]}_0^R$
$\Rightarrow B=\frac{{\mu }_0\omega {\sigma }_0}{2}\left[R+\frac{2}{3}R\right]$
$\Rightarrow B=\frac{5}{6}{\mu }_0\omega {\sigma }_{0}R$
$\Rightarrow \frac{B}{{\mu }_0\omega {\sigma }_{0}R}=\frac{5}{6}$