Question

A positive charge particle of charge $q$ and mass $m$ is entering with velocity $v\hat{i}$ as shown in figure in a uniform magnetic field $-B\hat{k}$ . Choose the correct alternative(s):

• A. Velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=v\cos {60}^{\circ }\hat{i}+v\sin {60}^{\circ }\hat{j}$
• B. Velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=v\cos {30}^{\circ }\hat{i}+v\sin {30}^{\circ }\hat{j}$
• C. Time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$
• D. Distance travellled in magnetic field is $\frac{\pi mv}{2qB}$

Answer 1

Answer: (A), (C)

Time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$



The path taken by the particle is as shown in the diagram,

Arc length $\text{AB}=\frac{\pi r}{3}=\frac{\pi mv}{3qB}$

In the presence of uniform magnetic field, angular speed of the particle remains constant.

Let the time taken by the particle to leave the magnetic field be $t=\frac{\theta }{\omega }$.

From the figure, angular distance travelled by the particle is $\theta =\frac{\pi }{3}$

$\therefore t=\frac{\frac{\pi }{3}}{\frac{2\pi }{T}}=\frac{T}{6}=\frac{\pi m}{3qB}$

Distance travelled by the particle in magnetic field , $d=vt=\frac{\pi mv}{3qB}$

Let the velocity of particle when it comes out be $v^{\prime }$

From the figure, we can deduce that, $v^{\prime }=v\cos ({90}^{\circ }-{30}^{\circ })\hat{i}+v\sin ({90}^{\circ }-{30}^{\circ })\hat{j}$

$\Rightarrow v^{\prime }=v\cos \left({60}^{\circ }\right)\hat{i}+v\sin \left({60}^{\circ }\right)\hat{j}$

Hence, options (a) and (c) are the correct answers.