A) \[\left( \frac{{{v}_{0}}}{2{{B}_{0}}\alpha },\frac{\sqrt{2}{{v}_{0}}}{\alpha {{B}_{0}}},\frac{-{{v}_{0}}}{{{B}_{0}}\alpha } \right)\]
B) \[\left( \frac{-{{v}_{0}}}{2{{B}_{0}}\alpha },0,0 \right)\]
C) \[\left( 0,\frac{2{{v}_{0}}}{{{B}_{0}}\alpha },\frac{{{v}_{0}}\pi }{2{{B}_{0}}\alpha } \right)\]
D) \[\left( \frac{{{v}_{0}}\pi }{{{B}_{0}}\pi },0\frac{-2{{v}_{0}}}{{{B}_{0}}\alpha } \right)\]
Correct Answer: D
Solution :
[d] \[\alpha =\frac{q}{m}\], path of the particle will be a helix of time period, \[T=\frac{2\pi m}{{{B}_{0}}q}=\frac{2\pi }{{{B}_{0}}\alpha }\] The give time \[t=\frac{\pi }{{{B}_{0}}\alpha }=\frac{T}{2}\] \[\therefore \]Coordinates of particle at time \[t=T/2\] would be \[(vx\,T/2,\,0,-2r)\] Here, \[r=\frac{m{{v}_{0}}}{{{B}_{0}}q}=\frac{{{v}_{0}}}{{{B}_{0}}\alpha }\] \[\therefore \] The coordinate are \[\left( \frac{{{v}_{0}}\pi }{{{B}_{0}}\alpha },0,\frac{-2{{v}_{0}}}{{{B}_{0}}\alpha } \right)\]You need to login to perform this action.
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