Question

A thin non-conducting disc of radius $R$ and mass $M$ is held horizontally and is capable of rotation about an axis passing through its centre and perpendicular to its plane. A charge $Q$ is distributed uniformly over the surface of the disc. A time varying magnetic field $B=Kt$ (where $K$ is a constant and $t$ is the time) directed perpendicular to the plane of the disc is applied to it. If the disc is stationary initially, find the torque acting on the disc.

• A. $KQ{R}^2$
• B. $\frac{KQ{R}^2}{4}$
• C. $\frac{KQ{R}^2}{2}$
• D. Zero

Answer 1

The correct option is B $\frac{KQ{R}^2}{4}$


We know that,

$E\cdot dl=A\frac{dB}{dt}$

For an arbitrary radius $x$,

$E\cdot \left(2\pi x\right)=\left(\pi x^2\right)\frac{d\left(Kt\right)}{dt}=\pi x^{2}K$

$\Rightarrow E=\frac{Kx}{2}$

Force acting on the element is

$dF=dq\times E=\frac{Q}{\pi R^2}\times 2\pi xdx\times \frac{Kx}{2}$

$dF=\frac{KQ}{R^2}{x}^{2}dx$

Torque acting on the disc is

$\tau ={\int }_0^{R}xdF=\frac{KQ}{R^2}{\int }_0^R{x}^{3}dx$

$\therefore \tau =\frac{KQ{R}^2}{4}\text{Nm}$

Hence, option $\left(\text{B}\right)$ is correct.