A) the induced electric field is quadratic in time t
B) the force tangential to the ring is\[9{{B}_{0}}QRt\]
C) until 2 seconds, the friction force does not come into play
D) the friction coefficient between the ring and the surface is \[\frac{2{{B}_{0}}RQ}{mg}\].
Correct Answer: D
Solution :
[d] Magnitude of induced electric field due to change in magnetic flux is given by \[\oint{\overset{\to }{\mathop{E}}\,}.d\overset{\to }{\mathop{\ell }}\,=\frac{d\phi }{dt}=A\frac{dB}{dt}\] \[(\because \,\,N=1\,\,\,and\,\,\,\cos \,\theta =1)\] or \[E.\ell =\pi {{R}^{2}}(2{{B}_{0}}t)\] \[\left( \frac{dB}{dt}=2{{B}_{0}}t \right)\] Here, E = induced electric field due to change in magnetic flux \[E(2\pi R)=2\pi {{R}^{2}}{{B}_{0}}t\] 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 \[F.R=(\mu mg)R\] It is given that ring starts rotating after 2 seconds So, putting t=2 seconds, we get \[\mu =\frac{2{{B}_{0}}RQ}{mg}\]
You need to login to perform this action.
You will be redirected in
3 sec
