A particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly in a region of uniform magnetic field $B$ and executes circular arc of radius $R$, then $\frac{q}{m}$ equals
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(a)
$r=\frac{\sqrt{2 m k}}{q B}=\frac{\sqrt{2 m q V}}{q B}$
$\Rightarrow r=\frac{\sqrt{2 V}}{B} \sqrt{\frac{m}{q}} \Rightarrow \frac{q}{m}=\frac{2 V}{R^2 B^2}$
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