A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width ‘d’. If $\alpha$ ’ be the angle of deviation of proton from initial direction of motion (see figure), the value of $\sin \alpha$ will be :

Question

$\begin{array}{1 1} \frac{B}{2} \sqrt {\frac{qd}{mV}} \\ \frac{B}{d} \sqrt {\frac{q}{2mV}}\\ Bd \sqrt {\frac{q}{2mV}} \\ qV \frac{Bd}{2m} \end{array} $