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

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

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

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

Correct Answer: A

Solution :

[a] \[r=\frac{mv}{qB}\] \[\sin \,\theta =\frac{x}{r}=\frac{\frac{mv}{\sqrt{2}qB}}{\frac{mv}{qB}}=\frac{1}{\sqrt{2}}\] or \[\theta =\frac{\pi }{4}\] Time to complete the circle\[(2\pi )\], \[T=\frac{2\pi m}{qB}\]         \[\therefore \]  time taken to traverses \[\frac{\pi }{4}\], \[t=\frac{\pi m}{4qB}\] Time taken to travel horizontal distance \[{{t}_{1}}=\frac{\frac{mv}{\sqrt{2}qB}}{\frac{v}{\sqrt{2}}}=\frac{m}{qB}\] Total time taken\[=2t+2{{t}_{1}}=\frac{m}{2qB}(\pi +4)\]