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
  • A$qV\,\sqrt {\frac{{Bd}}{{2m}}} $
  • B$\frac{B}{2}\sqrt {\frac{{qd}}{{mV}}} $
  • C$\frac{B}{d}\sqrt {\frac{{q}}{{2mV}}} $
  • D$Bd\sqrt {\frac{q}{{2mV}}} $
JEE MAIN 2015, Diffcult
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