Question

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

• A. $\frac{qB}{m}a\cot \theta$
• B. $\frac{qB}{m}a\tan \theta$
• C. $\frac{qB}{m}a{\cot }^2\theta$
• D. $\frac{qB}{m}a{\tan }^2\theta$

Answer 1

Answer: (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 $r=a\text{cosec}\theta \approx a\cot \theta ...(i)$
Also, $r=\frac{mv}{Bq}...(ii)$
Combining Eqs. (i) and (ii), we get
$v=\frac{qB}{m}a\cot \theta$