question_answer

A charged particle of specific charge (charge/ mass) \[\alpha \] is released from origin at time t = 0 with velocity \[\overset{\to }{\mathop{v}}\,={{v}_{0}}(\hat{i}+\hat{j})\] in uniform magnetic field \[\overset{\to }{\mathop{B}}\,={{B}_{0}}\hat{i}\]. Coordinates of the particle at time  \[t=\pi /({{B}_{0}}\alpha )\]

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 \[\left( \frac{{{v}_{0}}\pi }{{{B}_{0}}\alpha },0,\frac{-2{{v}_{0}}}{{{B}_{0}}\alpha } \right)\]