A) \[\frac{m{{v}_{0}}}{qE}\]
B) \[\frac{m{{v}_{0}}}{2qE}\]
C) \[\frac{\sqrt{3}m{{v}_{0}}}{2qE}\]
D) \[\frac{\sqrt{5}m{{v}_{0}}}{2qE}\]
Correct Answer: B
Solution :
The E and B are acting along X-axis and v is acting along Y-axis, i.e. perpendicular to both E and B. Therefore, the path of charged particle is a helix with increasing speed. The speed of particle at time t is \[v=\sqrt{v_{x}^{2}+v_{y}^{2}}\] Here,\[{{v}_{y}}={{v}_{0}},{{v}_{x}}=\frac{dE}{m}t\] and\[v\frac{\sqrt{5}}{2}{{v}_{0}}\] Putting these values in Eq. (i), we get time,\[t=\frac{m{{v}_{0}}}{2qE}\]You need to login to perform this action.
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