Question

A particle of charge q and mass m starts moving from the origin under the action of an electric field $\vec{E}=E_{0} \hat{i}$and $\vec{B}=B_{0} \hat{i}$ with a velocity $\vec{v}=v_{0} \hat{j}$ . The speed of the particle will become $\frac{\sqrt{5}}{2}v_{0}$ after a time

• A. $\frac{mv_{0}}{qE_{0}}$
• B. $\frac{mv_{0}}{2qE_{0}}$
• C. $\frac{\sqrt{3}mv_{0}}{2qE_{0}}$
• D. $\frac{\sqrt{5}mv_{0}}{2qE_{0}}$

Answer 1

Answer: (B)

$\frac{mv_{0}}{2qE_{0}}$

Answer 2

Here, $\vec{E}$ and $\vec{B}$ are acting along x-axis and $\vec{v}$ is along y-axis i.e. perpendicular to both $\vec{E}$ and $\vec{B}$. Therefore, the path of charged particle is a helix with increasing speed. Speed of particle at time t is $v=\sqrt{v_{x}^{2}+v_{y}^{2}} \quad\quad\quad\quad\quad \quad\quad \quad\quad ...\left(i\right)$ Here, $v_{y}=v_{0} ; v_{x}=\frac{qE_{0}}{m}t$ and $v=\frac{\sqrt{5}}{2}v_{0}$ Putting values in $\left(i\right),$ we get $t=\frac{mv_{0}}{2qE_{0}}$