Question 166 of 276

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 the initial direction of motion (see figure), the value of $\sin ⁡\alpha$ will be:

A$\dfrac{B}{2}\sqrt{\dfrac{qd}{mV}}$B$Bd\sqrt{\dfrac{q}{2mV}}$C$\dfrac{B}{d}\sqrt{\dfrac{q}{2mV}}$D$qV Bd2m$

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Answer: (B) $Bd\sqrt{\dfrac{q}{2mV}}$

Due to potential difference $V$ speed acquired by proton is ${v}_{0}$,

$\Rightarrow qV=\dfrac{1}{2}m{v}_{0}^{2}=K$

$∵\text{Radius,}R=\dfrac{m{v}_{0}}{qB}=\dfrac{\sqrt{2mK}}{qB}$
$∵K=qV$
$R=\dfrac{\sqrt{2meV}}{qB}$
$\sin \alpha =\dfrac{d}{R}=\dfrac{dqB}{\sqrt{2meV}}$$=Bd\sqrt{\dfrac{q}{2mV}}$