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 \[\theta =\frac{\pi }{4}\]You need to login to perform this action.
You will be redirected in
3 sec