question_answer

A particle of mass m and charge q enters a region of magnetic field (as shown) with speed v at t = 0. There is a region in which the magnetic field is absent as shown. The particle after entering the region collide elastically with a rigid wall. Time t after which the velocity of particle become antiparallel to its initial velocity is -

A) \[\frac{m(\pi +4)}{2qB}\]          

B)        \[\frac{m}{4qB}(\pi +2)\]

C) \[\frac{m}{qB}(\pi +2)\]

D)        \[\frac{m}{4qB}(2\pi +3)\]

Correct Answer: A

Solution :

[a] \[R=\frac{mV}{qB}\] \[son\theta =\frac{R}{\sqrt{2}\times R}=\frac{1}{\sqrt{2}},\theta =45{}^\circ \] After coming out of B q will collide with the wall Time to exit B \[{{t}_{1}}=\frac{\pi m}{4qB}\] Time to travel in region where B is absent is \[{{t}_{2}}\] \[\frac{R}{\sqrt{2}s}=\cos 45\] S=R \[{{t}_{2}}=\frac{s}{v}=\frac{R}{V}=\frac{mv}{vqB}=\frac{m}{qB}\] total time \[2{{t}_{2}}=\frac{2m}{qB}\] time for returning journey in B \[{{t}_{3}}=\frac{\pi m}{4qB}\] total time \[=\frac{\pi m}{4qB}+\frac{\pi m}{4qB}+2q\frac{m}{qB}\] \[\frac{m}{qB}\left[ \frac{\pi }{2}+2 \right]=\frac{m}{2qB}[\pi +4]\]