Question

A uniform disc of radius $R$ having charge $Q$ distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field $B=Kx{t}^2$, where $K=$ constant, $x$ is the distance (in metre) from the centre of the disc and $t$ is the time (in second), is switched on which is perpendicular to the plane of the disc. The torque (in $\text{Nm}$) acting on the disc after $15\text{sec}$ is
(Take $2KQ=1$ S.I. unit and $R=1$ metre)

Answer 1

Consider a ring of thickness $dx$
Torque on this ring =$QE\times x$
Applying Faraday's law of induction,
$\oint E.dl=\frac{d\varphi }{dt}$
$E\times 2\pi x=\pi x^2\times \frac{dB}{dt}$
$E=\frac{x}{2}\times 2Kxt=K{x}^{2}t$
Charge on ring $=\frac{Q}{\pi R^2}\times 2\pi xdx=\frac{2Q}{R^2}\times dx$
Torque on ring $=\frac{2Q}{R^2}x\times K{x}^{2}t\times xdx$
$=\frac{2KQ}{R^2}{x}^{4}tdx$


Total torque $=\underset{0}{\overset{R}{\int }}\frac{2KQ}{R^2}{x}^{4}tdx={\left[\frac{2KQt{x}^5}{R^2\times 5}\right]}_0^R$
$=\frac{2KQ{R}^{3}t}{5}$
at $t=15\text{s}T=3\text{N- m}$