question_answer

A particle having specific charge \[\sigma \] is projected in \[xy\]plane with a speed V. There exists a uniform magnetic field in z-direction having a fixed magnitude\[{{B}_{0}}\]. The field \[2n\] is made to reverse its direction after every interval of \[\frac{2\sigma }{\sigma B}\]. The maximum separation between two positions of the particle during its course of motion will be

A) \[\frac{2\sqrt{2}v}{\sigma {{B}_{0}}}\]             

B)        \[\frac{4v}{\sigma {{B}_{0}}}\]    

C) \[\frac{2v}{\sigma {{B}_{0}}}\]              

D)        \[\frac{v}{\sigma {{B}_{0}}}\]

Correct Answer: B

Solution :

[b] Path is as shown. Distance AB is the required maximum separation. \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,AB=4R=4\frac{mv}{q{{B}_{0}}}=\frac{4v}{\sigma {{B}_{0}}}\]