Question

A charged particle of charge $q$ and mass $m$ is moving with velocity $v$ (as shown in the figure) in a uniform magnetic field B along negative z – direction. Select the correct alternative(s).

• A. Velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=v\cos {60}^{\circ }\widehat{+}v\sin {60}^{\circ }\hat{j}$
• B. Time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$
• C. Distance travelled in magnetic field is $\frac{\pi mv}{3qB}$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A Velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=v\cos {60}^{\circ }\widehat{+}v\sin {60}^{\circ }\hat{j}$
B Time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$
C Distance travelled in magnetic field is $\frac{\pi mv}{3qB}$

The charge particle $q$ will experience some force when it moves in the region of magnetic field.
The trajectory of this particle in this region can be shown as below


The components of velocity when it exits the magnetic fields are $v\cos {60}^{\circ }$ and $v\sin {60}^{\circ }$ along x and y axis respectively.

$⟹\overrightarrow{v}=v\sin {60}^{\circ }\hat{i}+v\cos {60}^{\circ }\hat{j}$

Let us consider a point $C$ on y-axis which represents the centre of circular track followed by the particle.

$r$ is the radius of the circular track.

From the figure we can calculate $\angle ACB={60}^0$
$t=\frac{\theta }{\omega }=\frac{\pi }{3\omega }$
$\frac{m{v}^2}{r}=qvB$
$v=r\omega$
$⟹m\omega =qB$
$⟹\omega =\frac{qB}{m}$

$\therefore t=\frac{\pi m}{3qB}$

Distance travelled in magnetic field
$l=r\theta$
$⟹l=\frac{mv}{qB}\times \frac{\pi }{3}$

$\therefore l=\frac{mv\pi }{3qB}$