Q. A particle of charge $q\ and\ mass\ m\ $starts moving from the origin under the action of an electric field $\overrightarrow{E} = E_{0}\widehat{i}$ and $\overrightarrow{B} = B_{0}\widehat{i}\ \ $with a velocity $\overrightarrow{V} = v_{0}\widehat{j}.$ The speed of the particle will becomes $\frac{\sqrt{5}}{2}v_{0}$ after a time
A) $\frac{mv_{0}}{qE}$
B) $\frac{m\ v_{0}}{2q\ E}$
C) $\frac{\sqrt{3}\ m\ v_{0}}{2q\ E}$
D) $\frac{\sqrt{5}\ m\ v_{0}}{2q\ E}$
Explanation
(b) Here, $\overrightarrow{E}$ and $\overrightarrow{B}$ are acting along $x$-axis and $\overrightarrow{V}$ is acting along y-axis $ie,$ perpendicular to both $\overrightarrow{E}$ and $\overrightarrow{B}$. Therefore, the path of charged particle is a helix with increasing speed. Speed of particle at time $t$ is
$v = \ \sqrt{v_{x}^{2} + v_{y}^{2}}$ (i)
Here, $v_{y} = v_{0};v_{x} = \frac{qE}{m}\ t\ \ \ \ and\ \ v = \frac{\sqrt{5}}{2}\ v_{0}$
Putting values in Eq. (i), we get $t = \frac{m\ v_{0}}{2\ q\ E}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.