MCQ
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}}} $
- ✓$Bd\sqrt {\frac{q}{{2mV}}} $

