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 $dr$ 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
$di=\frac{dq}{T}=\frac{dq\omega }{2\pi }$
$\left(\because T=\frac{2\pi }{\omega }\right)\dots (ii)$

From Eqs. (i) and (ii), we get
$\Rightarrow di=\sigma \omega rdr$
As, the magnetic field $dB$ at centre due to the element,
$dB=\frac{{\mu }_{0}di}{2r}-\frac{{\mu }_0\sigma \omega rdr}{2r}\to dB=\frac{{\mu }_0\sigma \omega }{2}dr$
$\Rightarrow B_{\text{net }}=\frac{{\mu }_0\sigma \omega }{2}\underset{0}{\overset{R}{\int }}dr$
$\Rightarrow B_{\text{net }}=\frac{{\mu }_0\sigma \omega R}{2}$