A constant horizontal uniform magnetic field $B=5\times {10}^5$ T exist in a uniform vertical gravitational field $(g=10m/s^2)$. Both fields are perpendicular to each other. A particle of mass  $m=50$gm and charge $q=2\mu C$ is released in the space. Find the displacement (in cm) of particle along y-axis when velocity of particle is along x-axis first time.

 $$\begin{gathered} mg-qB{v}_x=m\frac{d{v}_y}{dt} \\ qB{v}_y=m\frac{d{v}_x}{dt} \end{gathered}$$
 Differentiate equation with respect to time
 $0-qB\frac{d{v}_x}{dt}=m\frac{d^2{v}_y}{d{t}^2}$

$\frac{d^2{v}_y}{d{t}^2}=-\frac{q^2{B}^2}{m^2}-v_y=-{\omega }^2{v}_y$ 
Solution of equation  $v_y={\left(v_y\right)}_0\sin \omega t$
 $\frac{d{v}_y}{dt}={\left(v_y\right)}_0\omega \cos \omega tatt=0,\frac{d{v}_y}{dt}=g,so{\left(v_y\right)}_0=\frac{g}{\omega }$
Now form equation  $v_x=\frac{mg}{qB}(1-\cos \omega t)$
So,  $v_y=0,at\omega t=\pi ,t=\frac{\pi }{\omega }{v}_x=\frac{mg}{qB}[1-(-1)]=\frac{2mg}{qB}$
From work energy theorem    $mgy=\frac{1}{2}m{v}_x^2$
 $y=\frac{2m^{2}g}{q^2{B}^2}=0.05meter$