Electro Magnetic Induction And Alternating Currents Question 515
Question: A nonconducting ring of mass m and radius R has a charge Q uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field $ B=B_{0}t^{2} $ tesla is switched on. After 2 second from switching on the magnetic field the ring is just about to rotate about vertical axis through its centre. Then
Options:
A) the induced electric field is quadratic in time t
B) the force tangential to the ring is $ 9B_{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{2B_{0}RQ}{mg} $ .
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Answer:
Correct Answer: D
Solution:
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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}(2B_{0}t) $ $ ( \frac{dB}{dt}=2B_{0}t ) $
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{2B_{0}RQ}{mg} $
