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A ring of mass m, radius R having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is

$\begin {array} {1 1} (a)\;\large\frac{qB}{m} & \quad (b)\;\large\frac{2qB}{m} \\ (c)\;\large\frac{qB}{2m} & \quad (d)\;Zero \end {array}$


1 Answer

$ \int E\: d l=-\large\frac{d\phi}{dt}$$ = \pi r^2 \large\frac{dB}{dt}$
Let $\lambda$= q/2$\pi$ R be the charge per unit length of the ring
the torque is given by
$ N = \int \lambda\: E \:d l$
Angular momentum=$I\omega = \int N\: dt$
On solving the above set of sequential equations
$R\lambda \pi ^2 B= I\omega$,
$R\pi R^2 B q/2\pi R =mR^2 \omega$
 one gets the value of $ \omega$ as $ \large\frac{qB}{2m}$
Ans : (c)


answered Feb 22, 2014 by thanvigandhi_1
edited Sep 19, 2014 by thagee.vedartham

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