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 $d r$

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