$A$ particle of mass $m$ and charge $q$,accelerated by a potential difference $V$,enters a region of a uniform transverse magnetic field $B$. If $d$ is the thickness of the region of $B$,the angle $\theta$ through which the particle deviates from the initial direction on leaving the region is given by

• A
  $\sin \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$
• B
  $\cos \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$
• C
  $\tan \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$
• D
  $\cot \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$