The flat insulating disc of radius $r$ carries an excess charge on its surface is of $\sigma C/m^2$. Consider disc to rotate around the axis passing through its center and perpendicular to its plane with angular speed $\omega rad/s$. If a magnetic field $\overrightarrow{B}$is directed perpendicular to the rotation axis, then find the torque acts on the disc.

Suppose the disc is placed in $xy$-plane and is rotated about the $z$-axis. Consider an annular ring of radius $r$and of thickness $dr$, the charge on this ring

$dq = \sigma (2\pi r dr)$

As the ring rotates with angular velocity $Z,$ so the current

$i=\frac{dq}{dt}=\frac{\sigma \left(2\pi rdr\right)}{\frac{2\pi }{\omega }}=\sigma \omega rdr$

The torque on the current loop $\tau =i\overrightarrow{ A}\times \overrightarrow{B}$

Hence the torque on this annular ring 

$$\begin{gathered} d\tau = i(d\overrightarrow{A}\times \overrightarrow{B}) \\ =\sigma \omega r dr(\pi r^{2}B \sin {90}^{°}) \\ =\pi \sigma \omega r^{3}Bdr \\ \tau =\frac{\pi \sigma \omega B{a}^4}{4} \end{gathered}$$