Question

If magnetic field is witched off after $4s$, then find the angle rotated by the ring before coming to stop after switching off the magnetic field.

Answer 1

Magnitude of induced elecgtric field due to change in magnetic flux is given by
$\oint \overrightarrow{E}.\overrightarrow{dl}=\frac{d\varphi }{dt}=S\cdot \frac{dB}{dt}$
or $E.l=\pi R^2\left(2B_{0}t\right)$ $(\frac{dB}{dt}=2B_{0}t$)
Here $E=$ induced electric field due to change in magnetic flux
or $E\left(2\pi R\right)=2\pi R^2{B}_{0}t$
or $E=B_{0}RT$
Hence $F=QE=B_{0}QRt$
This force is tangential to ring. Ring starts rotating when torque of this force is greater than the torque due to maximum friction ($f_{max}-\mu mg$) or when
${\tau }_F\ge {\tau }_{f_{max}}$
Taking the limiting case
${\tau }_F={\tau }_{f_{max}}$ or $F.R=\left(\mu mg\right)R$
or $F=\mu mg$ or $B_{0}QRt=\mu mg$
It is given that ring starts rotating after $2$. So putting $t=2$, we get
$\mu =\frac{2B_{0}QR}{mg}$
After 2
${\tau }_F>{\tau }_{f_{max}}$
Therefore, net torque is
$\tau ={\tau }_F-{tan}_{f_{max}}=B_{0}Q{R}^{2}t-\mu mgR$
substituting $\mu =\frac{2B_{0}QR}{mg}$ we get,
$\tau =B_{0}Q{R}^2(t-2)$
or $I\left(\frac{d\omega }{dt}\right)=B_{0}Q{R}^2(t-2)$
or $m{R}^2\left(\frac{d\omega }{dt}\right)=B_{0}Q{R}^2(t-2)$
or ${\int }_0^{\omega }d\omega =\frac{B_{0}Q}{m}{\int }_2^4(t-2)dt$
or $\omega =\frac{2B_{0}Q}{m}.....\left(i\right)$
Now magnetic field is switched off and only retarding torque is present due to friction. So, angular retardation will be
$\alpha =\frac{{\tau }_{f_{max}}}{I}=\frac{\mu mgR}{m{R}^2}=\frac{\mu g}{R}$
Therefore applying
${\omega }^2={\omega }_0^2-2\alpha \theta$
or $0={\left(\frac{2B_{0}Q}{m}\right)}^2-2\left(\frac{\mu g}{R}\right)\theta$
or $\theta =\frac{2B_0^2{Q}^{2}R}{\mu m^{2}g}$
Sunstituting $\mu =\frac{2B_{0}QR}{mg}$
we get
$\theta =\frac{B_{0}Q}{m}$