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\[=2t+2{{t}_{1}}=\frac{m}{2qB}(\pi +4)\]You need to login to perform this action.
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