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
JEE MAIN 2015, Diffcult
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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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