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
A$\frac{2 V}{B^2 R^2}$
B$\frac{V}{2 B R}$
C$\frac{V B}{2 R}$
D$\frac{m V}{B R}$
Medium
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A$\frac{2 V}{B^2 R^2}$
a (a)
$r=\frac{\sqrt{2 m k}}{q B}=\frac{\sqrt{2 m q V}}{q B}$
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