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}}} $

Answer

Correct option: D.
$Bd\sqrt {\frac{q}{{2mV}}} $
d
From figure, $sin\,\alpha =dlR$

And we know, $\frac{m v^{2}}{R}=q v B$

$\Rightarrow \quad R=\frac{m v}{q B}$

$\because \sin \alpha=\frac{d q B}{m v}$

$\sin \alpha=B d \sqrt{\frac{q}{2 m V}}\left[\because q V=\frac{1}{2} m v^{2}\right]$

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