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) \[\left( \frac{{{v}_{0}}\pi }{{{B}_{0}}\alpha },\,\,0,\,\,\frac{-\,2{{v}_{0}}}{{{B}_{0}}\alpha } \right)\]
Correct Answer: D
Solution :
Radius of projection of helix will be \[r=\frac{{{v}_{0}}}{\alpha {{B}_{0}}}\] and time period of projection will be \[T=\frac{2\pi }{\alpha {{B}_{0}}},\] projected circle will be formed on y-z plane. It will make half circle in time\[t=\frac{\pi }{{{B}_{0}}\alpha }.\] x-coordinate \[={{\upsilon }_{0}}t={{\upsilon }_{0}}\cdot \frac{\pi }{{{B}_{0}}\alpha }.\]You need to login to perform this action.
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