A particle of specific charge $\alpha$ enters a uniform magnetic field $B=-B_0\hat{k}$ with velocity $v=v_0\hat{i}$ from the origin. Find the time dependence of velocity and position of the particle.

How to proceed: In such type of questions first of all see the angle between $v$ and $B$ because only this angle decides the path of the particle. Here, the angle is $90°$. Therefore, the path is a circle. If it is a circle, see the plane of the circle (perpendicualr to the magnetic field). Here the plane is $xy$. Then see the sense of the rotation. Here, it will be anti-clockwise as shown int he figure, because at origin the magnetic force is along positive $y$-direction (which can be seen from Fleming's left hand rule). Find the deviation and radius of the particle.

Now,according to the figure find $v\left(t\right)$ and $r\left(t\right)$. 

Velocity of the particle at any time $t$ is, $v\left(t\right)=v_x\hat{i}+v_y\hat{j}=v_0 \cos \theta \hat{i}+v_0\sin \theta \hat{j}$

or $v\left(t\right)=v_0\cos \left(B_0\alpha t\right)\hat{i}+v_0\sin \left(B_0\alpha t\right)\hat{j}$ o

Position of the particle at any time $t$ is, $r\left(t\right)=x\hat{i}+y\hat{j}= r \sin \theta \hat{i}+\left(r-r \cos \right) \hat{j}$

Substituting the values of $r$ and $\theta$, we have, $r\left(t\right)=\frac{v_0}{B_{0}x}\left[\sin \left(B_0\alpha t\right)\hat{i}+\left\{1{-}_0\cos \left(B_0\alpha t\right)\hat{j}\right\}\right]$