Question

A disc of radius $R$ uniformly charged with a charge $Q$ is rotated at angular velocity ${}^{\prime }{\omega }^{\prime }$ about an axis passing through its center and perpendicular to plane disc. Net magnetic field at center of disc is.

• A. $\frac{{\mu }_0}{u}\frac{Q\omega }{R}$
• B. $\frac{{\mu }_{0}Q\omega }{4\pi R}$
• C. $\frac{{\mu }_{0}Q\omega }{2\pi R}$
• D. $\frac{{\mu }_{0}Q\omega }{2R}$

Answer 1

The correct option is C $\frac{{\mu }_{0}Q\omega }{2\pi R}$


Field at 0 due to ring of radius r and charge $dq$ is
$dB=\frac{{\mu }_0}{2}\frac{di}{r}$ here $di=\frac{dq}{t}=\frac{dq}{2\pi }\omega$
$dq=2\pi rdr$

$B=\int dB=\int \frac{{\mu }_0}{2r}\frac{\omega }{2\pi }dq=\frac{{\mu }_0\omega }{4\pi }{\int }_0^R\frac{\sigma 2\pi rdr}{r}$
$B=\frac{{\mu }_0\sigma \omega R}{2}$
As $\sigma =\frac{Q}{\pi R^2}$
$B=\frac{{\mu }_{0}Q\omega }{2\pi R}$