Q: A charged ring of mass m = 50 gm, charge 2 coulomb and radius R = 2 m is placed on a smooth horizontal surface. A magnetic field varying with time at a rate of (0.2 t) Tesla/sec is applied on to the ring in a direction normal to the surface of ring. Find the angular speed attained in a time t_{1} = 10 sec.

Solution :Induced EMF is $\displaystyle \oint \vec{E}.\vec{dl} = \frac{d \phi}{dt}$

$\displaystyle E (2 \pi R) = \pi R^2 \frac{d B}{dt}$

$\displaystyle E = \frac{R}{2} \times 0.2 t $

$\displaystyle E = \frac{2}{2} \times 0.2 t = 0.2 t$

As Force F = q E = 0.4 t

Torque τ = I α = F R

α = F R/I

$\displaystyle \alpha = \frac{F R}{m R^2}$

$\displaystyle \alpha = \frac{F}{m R} = \frac{0.4 t}{50 \times 10^{-3} \times 2}$

α = 4 t

$\displaystyle \frac{d\omega}{dt} = 4 t $

$\displaystyle \omega = \int_{0}^{10}4 t dt $

$\displaystyle \omega = [\frac{4t^2}{2}]_{0}^{10} $

= 200 rad/sec