A particle of specific charge $\alpha$ enters region with a uniform magnetic field $\overrightarrow{B}=-B_0\hat{k}$ with velocity $\overrightarrow{V}=V_0\hat{i}$ from origin. If the value of $\alpha$ is $\frac{\pi }{3},B_0=1\text{T}$ and $V_0=2\text{m/s}$, then the position vector of particle w.r.t origin at $t=1\text{sec}$ will be:

The correct option is A $(\sqrt{3}\hat{i}+\hat{j})\text{m}$
The scenario decribed in the question can be represented in following diagram.


The position of particle is shown at a given time t, the speed of particle will not change because $W_{magneticforce}=0$

The magnetic force $F_m$ is directed towards centre $\text{O}$.

At that position the velocity of particle is given by:

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

$\Rightarrow \overrightarrow{v}=v_0\cos \theta \hat{i}+v_0\sin \theta \hat{j}.....\left(1\right)$

Here, $\theta$ is the angle subtended by arc at centre.

$\theta =\omega t$

or, $\theta =\left(\frac{q{B}_0}{m}\right)t$

$(\because \omega =\frac{2\pi }{T}),(\alpha =\frac{q}{m})$

or, $\theta =B_0\alpha t.....\left(2\right)$

Now the position vector of particle will be given by,

$\overrightarrow{r}=x\hat{i}+y\hat{j}$

$\overrightarrow{r}=r\sin \theta \hat{i}+(r-r\cos \theta )\hat{j}$

$\overrightarrow{r}=r[\sin \theta \hat{i}+(1-\cos \left)\hat{j}\right]$

$\overrightarrow{r}=\frac{m{v}_0}{q{B}_0}\left[\sin \right({\beta }_0\alpha t)\hat{i}+[1-\cos \left(B_0\alpha \right])\hat{j}$

$\overrightarrow{r}=\frac{v_0}{B_0\alpha }\left[\sin \right(B_0\alpha t)\hat{i}+[1-\cos \left(B_0\alpha t\right)\left]\hat{j}\right]$

Putting $(B_0=1,V_0=2,t=1,\alpha =\frac{\pi }{3}$ we get;
$\overrightarrow{r}=2[\sin \left(\frac{\pi }{3}\right)\hat{i}+(1-\cos \left(\frac{\pi }{3}\right))\hat{j}]$

$\overrightarrow{r}=2[\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}]$

$\therefore \overrightarrow{r}=(\sqrt{3}\hat{i}+1\hat{j})\text{m}$