Question

A non conducting ring of radius $r$ has a charge $Q$. A magnetic field perpendicular to the plane of the ring changes at the rate $\frac{dB}{dt}$. The torque experienced by the ring is:

Answer 1

The varying magnetic field $\left(B\right)$ creates an electric field field $\left(E\right)$ which is related to it as:

$\oint \overrightarrow{E}.\overrightarrow{d}l=-\frac{d\varphi }{dt}$ $(\varphi =$ Magnetic Flux through the ring$)$

$\Rightarrow \oint Edlcos0=-\frac{d}{dt}\left(BAcos0\right)$ $(\overrightarrow{E},\overrightarrow{d}l$ and $\overrightarrow{B},\overrightarrow{A}$ are in same direction$)$

$\Rightarrow E\oint dl=-\pi r^2\frac{dB}{dt}$ $(A=\pi r^2)$

$\Rightarrow E\left(2\pi r\right)=-\pi r^2\frac{dB}{dt}$ $(\oint dl=2\pi r)$

$\Rightarrow E=-\frac{r}{2}\frac{dB}{dt}$

So Torque acting on the ring $\left(\tau \right)=r\times F$ (where $F=$ net force acting on ring)

Now $F=EQ$

$\Rightarrow F=-\frac{r}{2}\frac{dB}{dt}Q$

Hence, $\tau =-\frac{Q{r}^2}{2}\frac{dB}{dt}$