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 is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 and R = 1m)

Answer 1

$d\varphi =\left(2\pi xdx\right)kx{t}^2$



$\varphi =\frac{2}{3}k{t}^2\pi x^3$

$\frac{d\pi }{dt}=\frac{4k\pi t{x}^3}{3}$

$E\left(2\pi x\right)=\frac{4\pi kt{x}^3}{3};E=\frac{2}{3}kt{x}^2;d\tau =\left(\frac{2}{3}kt{x}^2\right)\frac{Q}{\pi R^2}\left(2\pi xdx\right)x$

$\int d\tau =\frac{4}{3}\frac{ktQ}{R^2}{\int }_0^R{x}^{4}dx\Rightarrow \tau =\frac{4}{3}\frac{ktQ}{R^2}\frac{R^5}{5}$

At t = 15 sec,

$\tau =1N-m$