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\[=2t+2{{t}_{1}}\]\[=\frac{m}{2qB}\left( \pi +4 \right)\] |
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