A) 2Bqd/m
B) Bqd/m
C) Bqd/2 m
D) Bqm/d
Correct Answer: B
Solution :
The charged particle bends in a circle of radii r given by: \[Bqv\sin \frac{\pi }{2}=\frac{m{{v}^{2}}}{r}\] \[r=\frac{mv}{Bq}\] The particle dies not hit the plate if \[r\le d\]or \[\frac{mv}{Bq}\le d\] or \[v\le \frac{Bqd}{m}\,\therefore \,{{v}_{\max }}=\,\frac{Bqd}{m}\] Hence, the correction option is (b).You need to login to perform this action.
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