question_answer

A particle of mass m and charge q enters a region of magnetic field (as shown) with speed v. There is a region in which the magnetic field is absent, as shown. The particle after entering the region collides elastically with a rigid wall. Time after which the velocity of particle becomes Antiparallel to its initial velocity is

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

B) \[\frac{m}{qB}\left( \pi +2 \right)\]

C) \[\frac{m}{4qB}\left( \pi +2 \right)\]                    

D) \[\frac{m}{4qB}\left( 2\pi +3 \right)\]

Correct Answer: A

Solution :

\[r=\frac{mv}{qB}\]

\[\sin \theta =\frac{x}{r}\]

\[\sin \theta =\frac{\frac{mv}{\sqrt{2}qB}}{\frac{mv}{qB}}=\frac{1}{\sqrt{2}}\]

So, \[\theta =\frac{\pi }{4}\]

Time to complete the circle\[\left( 2\pi  \right)\],\[T=\frac{2\pi m}{qB}\]

\[\therefore \]Time taken to traverses\[\frac{\pi }{4}\], \[t=\frac{2\pi m}{4qB}\]

Time taken to travel horizontal distance\[{{t}_{1}}=\frac{\frac{mv}{\sqrt{2qB}}}{\frac{v}{\sqrt{2}}}=\frac{m}{qB}\]  total time taken\[=2t+2{{t}_{1}}\]\[=\frac{m}{2qB}\left( \pi +4 \right)\]