A positively charged particle of specific charge $\left(\frac{q}{m}\right)$ is projected from the origin of coordinate with initial velocity $(u\hat{i}-v\hat{j})$. Uniform electric and magnetic fields exist in the region along the $+y-$direction, of magnitude $E$ and $B$. The particle will definitely return to the origin once if

The correct option is B $\frac{vB}{\pi E}$ is an integer
Given:
$\overrightarrow{E}=E\hat{j};\overrightarrow{B}=B\hat{j}$

So, the force due to $\overrightarrow{E}\&\overrightarrow{B}$ on the charge $+q$ is given by,

$\overrightarrow{F_e}=qE\hat{j}$

$\overrightarrow{F_m}=q\left[\right(u\hat{i}-v\hat{j})\times B\hat{j}]=quB\hat{k}$

Now, the acceleration of the charge is given by,

$a_y=\frac{qE}{m}\hat{j};a_z=\frac{quB}{m}\hat{k}$

Assuming particle will return to origin in time $t$.

Now applying the equation of motion in $y-$ direction for time $t$

$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow 0=-vt+\frac{1}{2}\left(\frac{qE}{m}\right)t^2$

$\therefore t=\frac{2vm}{qE}$ is the time taken by the particle to come to origin due to electric field.

Now due to magnitude field it will move in circular path, time period of its circular motion is given by

$T=\frac{2\pi m}{qB}$

If the particle completes $N$ number of turns in time of return, then

$N\times \frac{2\pi m}{qB}=\frac{2vm}{Eq}$

$\therefore N=\frac{vB}{\pi E}$

So, here the above value must be an integer so that it complete integral number of rounds.

Hence, option $\left(b\right)$ is the correct answer.