Question

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 $\overrightarrow{v}=v_0\hat{j}$. The speed of the particle will become $2v_0$ after time

• A. $t=\frac{2m{v}_0}{qE}$
• B. $t=\frac{2Bq}{m{v}_0}$
• C. $t=\frac{\sqrt{3}Bq}{m{v}_0}$
• 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 $\overrightarrow{E}=E_0\hat{i}$ is parallel to $\overrightarrow{B}=B_0\hat{i}$

Projection velocity, $\overrightarrow{v}=v_0\hat{j}$ is perpendicular to both $\overrightarrow{E},\overrightarrow{B}$

Hence, the path of the particle is a helix with speed remaining constant for the circular path in the $yz$ plane.
So, $V_y^2+V_z^2=v_0^2$​

Magnitude of velocity of particle at any time $t$ is $v_t=\sqrt{V_x^2+V_y^2+V_z^2}$

Given final velocity is $2v_0$,

$2v_0=\sqrt{v_0^2+V_x^2}$
$4v_0^2=v_0^2+V_x^2$
$V_x^2=3v_0^2$
$V_x=\sqrt{3}{v}_0$

Now,
$V_x=a_{x}t$
$\because \sqrt{3}{v}_0=a_{x}t=\frac{F_{e}t}{m}=\frac{q{E}_{0}t}{m}$
$t=\frac{\sqrt{3}m{v}_0}{q{E}_0}$