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A charged particle of charge $\mathrm{q}$ and mass $\mathrm{m}$, gets deflected through an angle $\theta$ upon passing through a square region of side 'a' which contains a uniform magnetic field B normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?
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Verified Answer
The correct answer is:
$\frac{\mathrm{qB}}{\mathrm{m}} \mathrm{acot} \theta$
Now $\sin \theta=\frac{\mathrm{a}}{\mathrm{R}}, \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$
$$
\therefore \mathrm{v}=\frac{\mathrm{q} \mathrm{Bacot} \theta}{\mathrm{m}}
$$
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