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 t1 = 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