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 a will be : <br>

Question

https://clay6.com/mpaimg/1082_a.jpg

( A ) $qV \sqrt{\frac{Bd}{2m}}$
( B ) $Bd \sqrt{\frac{q}{2mV}}$
( C ) $\frac{B}{d} \sqrt{\frac{q}{2mV}}$
( D ) $\frac{B}{2} \sqrt{\frac{qd}{mV}}$