Question

A particle of charge q and mass m starts moving from the origin under the action of an electric field $\overrightarrow{E}=E_o\hat{i}and\overrightarrow{B}=B_o\hat{i}$ with the velocity $\overrightarrow{v}=v_o\hat{j}$. The speed of the particle will become $2v_o$ after a time

• A. $t=\frac{2m{v}_o}{qE}$
• B. $t=\frac{2Bq}{m{v}_o}$
• C. $t=\frac{\sqrt{3}Bq}{m{v}_o}$
• D. $t=\frac{\sqrt{3}m{v}_0}{qE}$

Answer 1

The correct option is D $t=\frac{\sqrt{3}m{v}_0}{qE}$

Given :A particle of charge q and mass m starts moving from the origin under the action of an electric field .

$\overrightarrow{E}=E_0\hat{i}$ and $\overrightarrow{B}=B_0\hat{i}$ with velocity $\overrightarrow{v}=v_0\hat{j}$. Final speed of particle becomes $2v_0$

Solution :

Here, $\overrightarrow{E}=E_0\hat{i}$ and $\overrightarrow{B}=B_0\hat{i}$ are acting along x-axis and $\overrightarrow{v}=v_0\hat{j}$ is along y-axis i.e. perpendicular to both $\overrightarrow{E}=E_0\hat{i}$ and $\overrightarrow{B}=B_0\hat{i}$ . Therefore, the path of charged particle is a helix with increasing speed. Speed of particle at time t is

$\overrightarrow{v_0}=\sqrt{v_x^2+v_y^2}$ eq-1

Here $v_y=v_0$ , $v_x=\frac{q{E}_{0}t}{m}$ v=2$v_0$

Putting values in (i),(i), we get

$t=\frac{\sqrt{3}{v}_{o}m}{q{E}_o}$

The Correct Opt: D