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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 reversed, the angular velocity acquired by the ring is $n (\frac{q B}{m}),$ then $n$ is______

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Detailed Solution

$\int E d l = \frac{- d ϕ}{d t} = - π r^{2} \frac{d B}{d t};$

$\Rightarrow E = \frac{r}{2} \frac{d B}{d t}$

$d q = \frac{q}{π r^{2}} \times 2 π r d r = \frac{2 q}{r} d r$

$τ = (d q E) (r)$

$I ω = \int τ d t$

$\Rightarrow m R^{2} ω = q B R^{2}$

$\Rightarrow ω = \frac{q B}{m}$

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