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A particle of mass $m$ and charge $q$, accelerated by a potential difference $V$ enters a region of a uniform transverse magnetic field $B$. If $d$ is the thickness of the region of $B$, the angle $\theta$ through which the particle deviates from the initial direction on leaving the region is given by
$\sin \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$
$\cos \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$
$\tan \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$
$\cot \theta = Bd{\left( {\frac{q}{{2mV}}} \right)^{\frac{1}{2}}}$
Solution
Refer to Fig. Let $v$ be the velocity of the particle.Its kinetic energy is
$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{qV}$ or $\mathrm{v}=\left(\frac{2 \mathrm{q} \mathrm{V}}{\mathrm{m}}\right)^{1 / 2}………(1)$
The particle follows a circular path from $A$ to $\mathrm{B}$ of radius $\mathrm{r}$ which is given by
$\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{q} \mathrm{VB}$ or $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}………(2)$
Using $( 1)$ and $(2),$ we have
$\mathrm{r}=\frac{\mathrm{m}}{\mathrm{qB}}\left(\frac{2 \mathrm{qV}}{\mathrm{m}}\right)^{1 / 2}=\frac{1}{\mathrm{B}}\left(\frac{2 \mathrm{mV}}{\mathrm{q}}\right)^{1 / 2}$
In triangle $\mathrm{BCD}, \sin \theta=\frac{\mathrm{BD}}{\mathrm{BC}}=\frac{\mathrm{d}}{\mathrm{r}} \cdot$ Therefore
$\sin \theta=\operatorname{Bd}\left(\frac{\mathrm{q}}{2 \mathrm{mV}}\right)^{1 / 2},$ which is choice $(a)$.
