A charged particle of specific charge $\alpha$ is released from origin at time $t=0$ with velocity $\overrightarrow{v}=v_0(\hat{i}+\hat{j})$ in magnetic field $\overrightarrow{B}=B_0\hat{i}$. The coordinate of the particle at time $t=\frac{\pi }{B_0\alpha }$ will be

The correct option is A $(\frac{v_0\pi }{B_0\alpha },0,\frac{-2v_0}{B_0\alpha })$
Given:
$\overrightarrow{B}=B_0\hat{i}$

The velocity of the charged particle $(q,m)$ is,

$\overrightarrow{v}=v_0\hat{i}+v_0\hat{j}$

The magnetic force on the particle is given by

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

$\overrightarrow{F}=q\left[\right(v_0\hat{i}+v_0\hat{j})\times (B_0\hat{i}\left)\right]$

$\overrightarrow{F}=q\left[B_0{v}_0\right(-\hat{k})+0]=-q{B}_0{v}_0\hat{k}$

It is clear that magnetic force is acting along $-vez-$direction. Thus, the component of velocity along $y-$ direction is perpendicular to ${\overrightarrow{F}}_{magnetic}$ which gives circular motion and presence of velocity $v_0\hat{i}$ will produce a helical path for the particle.


And the time period of circular motion is given by

$T=\frac{2\pi m}{qB}=\frac{2\pi }{\alpha B_0}$

The particle is revolving in $y-z$ plane & propagating along $+vex-$axis.

At time, $t=\frac{\pi }{B_0\alpha }=\frac{T}{2}$

So, at time $t=T/2$, it's half round, hence $y-$coordinate $=0$.

And for $x-$ coordinate, applying equation of motion,

$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$x=\left(v_0\right)(\frac{\pi }{B_0\alpha })+0(\because a_x=0)$

$\therefore x=\frac{\pi v_0}{B_0\alpha }$

Since, magnetic force is acting in $-vez-$ direction, so the displacement along $z-$direction for $t=T/2$ will be

$z=-2R=-2\times \frac{m{v}_{\perp }}{qB}=\frac{-2m{v}_0}{q{B}_0}$

$\therefore z=\frac{-2v_0}{B_0\alpha }$

Thus, coordinate of particle is $(\frac{\pi v_0}{B_0\alpha },0,\frac{-2v_0}{B_0\alpha })$