A disc (of radius $r$) carrying positive charge $q$is rotating with angular speed $\omega$ in a uniform magnetic field $B$about a fixed axis (as shown in figure below), such that angle made by axis of disc with magnetic field is $\theta$. Torque applied by axis on the disc is

$$\begin{gathered} The current, di =\frac{ dq}{dt}=\frac{\left(2\mathrm{\pi rdr}\right)\sigma }{\left({\displaystyle \frac{2\mathrm{\pi }}{\omega }}\right)}=\sigma \omega rdr \\ dm=\left(di\right)\times {\mathrm{\pi r}}^2={\mathrm{\pi \sigma \omega r}}^3\mathrm{dr} \\ \mathrm{\tau }={\int }_0^{\mathrm{r}}\left(\mathrm{dM}\right)\mathrm{Bsin\theta } \\ =\mathrm{B\pi \sigma \omega } \mathrm{sin\theta }{\int }_0^{\mathrm{r}}{\mathrm{r}}^3\mathrm{dr} \\ =\mathrm{B\pi \sigma \omega } \mathrm{sin\theta } \frac{{\mathrm{r}}^4}{4} \\ =\left({\mathrm{\sigma \pi r}}^2\right)\frac{{\mathrm{\omega r}}^2\mathrm{Bsin\theta }}{4} \\ =\frac{{\mathrm{q\omega r}}^2\mathrm{Bsin\theta }}{4} \end{gathered}$$