Question

A positive charge particle with charge $q$ and mass $m$ is moving with velocity $v=v_0\hat{i}$ in a uniform magnetic field $\overrightarrow{B}=-B_0\hat{k}$. The time spent by the particle in the magnetic field if $d$ is equal to $\frac{mv}{2qB}$ will be.

• A. $\frac{\pi m}{qB}$
• B. $\frac{\pi m}{4qB}$
• C. $\frac{\pi m}{3qB}$
• D. $\frac{\pi m}{6qB}$

Answer 1

The correct option is D $\frac{\pi m}{6qB}$

Radius of circular path $R$ is, $R=\frac{mv}{qB}$

Given $d=\frac{mv}{2qB}=\frac{R}{2}$

$\Rightarrow d<R$ , width of magnetic field region is insufficient to describe a complete circle. Hence the particle will leave the magnetic field region as shown in the figure.

Thus, the angle of deflection is $\theta$.

From the figure,

$\sin \theta =\frac{d}{R}$

$\sin \theta =\frac{\left(\frac{R}{2}\right)}{R}=\frac{1}{2}$

$\therefore \theta =\frac{\pi }{6}$

Distance travelled by particle in the magnetic field region is arc $\text{SQ}$ substending an angle $\theta$ at the center of the circular path.

$\therefore$ Time spent in magnetic field region is;

$t=\left(\frac{2\pi m}{qB}\right)\left(\frac{\theta }{2\pi }\right)$

or, $t=\left(\frac{2\pi m}{qB}\right)\times \left[\frac{\pi }{6\left(2\pi \right)}\right]$

$\therefore t=\frac{\pi m}{6qB}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (d) is the correct answer.