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A charged particle of charge $q$ and mass $m$, 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?
$\frac{q B}{m} a \cot \theta$
$\frac{q B}{m} a \tan \theta$
$\frac{q B}{m} a \cot ^2 \theta$
$\frac{q B}{m} \alpha \tan ^2 \theta$
Solution
(a)
In given condition, if $r$ is radius of curved path travelled by changed particle.
Then,
$\sin \theta \approx \tan \theta=\frac{a}{r}$
or $\quad r=a \operatorname{cosec} \theta \approx a \cot \theta \quad \dots(i)$
Also, $\quad r=\frac{m v}{B q} \quad \dots(ii)$
Combining Eqs. $(i)$ and $(ii)$, we get
$v=\frac{q B}{m} a \cot \theta$
