Question

A dielectric circular disc of radius $R$ carries a uniform surface charge density $\sigma$. If it rotates about its axis with angular velocity $\omega$, the magnetic field at the cente of disc is :

• A. $\frac{\mu_{0} \sigma \omega R^{2}}{2 \pi}$
• B. $\frac{\mu_{0} \sigma \omega R}{2}$
• C. $\frac{\mu_{0} \sigma \omega R^{2}}{4}$
• D. $\frac{\mu_{0} \sigma \omega R^{2}}{2 \sqrt{2}}$

Answer 1

Answer: (B)

Let us assumed a ring like element of the disc of radius $r$ and thickness $d r$ If $\sigma$ is the surface charge density, then the charge on the element.
$dq=\sigma(2 \pi r) dr \,\,\,\,\,\,\dots(i)$
Current di associated with rotating charge $dq$ is
$d i=\frac{d q}{T}=\frac{d q \omega}{2 \pi}$
$\left(\because T=\frac{2 \pi}{\omega}\right) \ldots( ii )$

From Eqs. (i) and (ii), we get
$\Rightarrow \, d i=\sigma \omega r \,dr$
As, the magnetic field $dB$ at centre due to the element,
$dB = \frac{\mu_{0} d i}{2 r}-\frac{\mu_{0} \sigma \omega r d r}{2 r} \rightarrow d B=\frac{\mu_{0} \sigma \omega}{2} d r $
$\Rightarrow \,B_{\text {net }}=\frac{\mu_{0} \sigma \omega}{2} \int\limits_{0}^{R} d r $
$\Rightarrow B_{\text {net }}=\frac{\mu_{0} \sigma \omega R}{2}$