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 $\overrightarrow{B}=B_0\hat{i}$ with a velocity $\overrightarrow{v}=v_0\hat{j}$. The speed of the particle will become $2v_0$ after a time

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

Answer 1

Answer: (C)

$\overrightarrow{E}=E_0\hat{i}\parallel \overrightarrow{B}=B_0\hat{i}$

$\overrightarrow{v}=v_0\hat{j}\perp \overrightarrow{E}$

$\overrightarrow{v}\perp \overrightarrow{B}$

Hence, the path of the particle is a helix with speed remaining constant for the circular path in the $yz$ plane.

Magnitude of velocity of particle at any time $t:$ $v=\sqrt{v_x^2+v_y^2+v_z^2}=\sqrt{(a_{x}t{)}^2+v_0^2}$ where $v_y^2+v_z^2=v_0^2$

Given $v=2v_0$

$\Rightarrow (2v_0{)}^2=v_x^2+v_0^2$

$\Rightarrow (v_x{)}^2=3(v_0{)}^2$

$\Rightarrow v_x=\sqrt{3}{v}_0$

$\Rightarrow a_{x}t=\sqrt{3}{v}_0$

$\Rightarrow \left(\frac{qE}{m}\right)t=\sqrt{3}{v}_0$

$\Rightarrow t=\frac{\sqrt{3}{v}_{0}m}{qE}$