A particle of mass m and charge q, accelerate | vedclass

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Question

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

Vedclass · Accepted Answer

$\sin 	heta  = Bd{\left( {\frac{q}{{2mV}}} ight)^{\frac{1}{2}}}$

Vedclass · Answer

Refer to Fig. Let $v$ be the velocity of the particle.Its kinetic energy is

$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{qV}$ or $\mathrm{v}=\left(\frac{2 \mathrm{q} \mathrm{V}}{\mathrm{m}}ight)^{1 / 2}.........(1)$

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

$\mathrm{r}=\frac{\mathrm{m}}{\mathrm{qB}}\left(\frac{2 \mathrm{qV}}{\mathrm{m}}ight)^{1 / 2}=\frac{1}{\mathrm{B}}\left(\frac{2 \mathrm{mV}}{\mathrm{q}}ight)^{1 / 2}$

In triangle $\mathrm{BCD}, \sin 	heta=\frac{\mathrm{BD}}{\mathrm{BC}}=\frac{\mathrm{d}}{\mathrm{r}} \cdot$ Therefore

$\sin 	heta=\operatorname{Bd}\left(\frac{\mathrm{q}}{2 \mathrm{mV}}ight)^{1 / 2},$ which is choice $(a)$.

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