A particle of charge $+q$ and mass $m$ is released from rest in the crossed electric field $E$ and magnetic field  $B$ as shown in Fig. Find the trajectory of the particle.

The given electric and magnetic fields can be expressed as
      $\overrightarrow{E}=E\hat{\mathrm{j}}, \overrightarrow{B}=\mathrm{B}\hat{k}$
The electric force is always along the electric field, i.e., $y-$axis in this case. The magnetic force will be in $xy-$plane. Thus the particle can never come out of the $xy-$plane. Let the velocity of particle at any time $t$ be
$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$
The lorentz force on the particle is
$\overrightarrow{F}=q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})=q\left[\left(E\hat{j}+\left(v_x\hat{i}+v_y\hat{j}\right)B\hat{k}\right]\right.$

$m\left[\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}\right]=q\left[\left(E-v_{x}B\right)\hat{j}+B{v}_y\hat{i}\right]$
Component wise we can write
$\frac{d{v}_x}{dt}=\frac{q}{m}B{v}_y$              …………..$\left(1\right)$

$\frac{d{v}_y}{dt}=\frac{q}{m}\left(E-v_{z}B\right)$      ………….$\left(2\right)$
This is pair of coupled differential equations. We can decouple the equations by differentiating one of them w.r.t. time and substituting the result into the other equation.
Differentiating eqn. $\left(2\right)$ w.r.t. time,
$\frac{d^2{v}_y}{d{t}^2}=-\frac{q}{m}B\frac{d{v}_x}{dt}$    …………$\left(3\right)$
Now substituting $\frac{d{v}_x}{dt}$ from eqn. $\left(1\right) in \left(3\right)$, we get
$\frac{d^2{v}_y}{d{t}^2}=-{\left(\frac{qB}{m}\right)}^2{v}_y=-{\omega }^2{v}_y$    ……..$\left(4\right)$

Where $\omega =\frac{qB}{m}$
Similarly by differentiating eqn. $\left(1\right)$, we get
$\frac{d^2{v}_x}{d{t}^2}=\frac{q}{m}B\frac{d{v}_y}{dt}$          ……….$\left(5\right)$
On substituting $\frac{d{v}_y}{dt}$ from eqn. $\left(2\right)$, we get
$\frac{d^2{v}_x}{d{t}^2}=\frac{q^2{B}^2}{m^2}\left(E-v_{z}B\right)$        ……..$\left(6\right)$
The eqn. $\left(4\right)$ is similar to equation of particle executing simple harmonic motion. Hence its solution is the form
$v_y-v_0\sin \omega t$          ………..$\left(7\right)$

$\frac{d{v}_y}{dt}=v_0\omega \cos \omega t$     ……….$\left(8\right)$

From equation  $\left(8\right)$, at $t=0, \frac{d{v}_v}{dt}=v_0\omega$
Also ${\mathrm{v}}_{\mathrm{x}}=0$ at $\mathrm{t}=0$.
Hence from eqn. $\left(2\right)$,  $\frac{d{v}_y}{dt}=\frac{qE}{m}$
Thus $v_0\omega =\frac{qE}{m}$ or $v_0=\frac{E}{B}$ 

and $v_y=\frac{E}{B}\sin \omega t=\frac{dy}{dt}$

or ${\int }_0^{y}dy=\frac{E}{B}{\int }_0^t\sin \omega tdt$
      $y=\frac{E}{B\omega }(1-\cos \omega t)$             ………..$\left(9\right)$
On substituting the expression for $\frac{\mathrm{dy}}{\mathrm{dt}}$ in eqn. $\left(1\right)$, we get
$\frac{d{v}_x}{dt}=\left(\frac{q}{m}B\right)\left(\frac{E}{B}\sin \omega t\right)$

${\int }_0^{v_x}d{v}_x=\frac{qE}{m}{\int }_0^t\sin \omega t$

$v_x=\frac{E}{B}(1-\cos \omega t)=\frac{\text{dx}}{\text{dt}}$

${\int }_0^{x}dv=\frac{E}{E}{\int }_0^t(1-\cos \omega t)\text{dt}$

      $x=\frac{E}{B\omega }(\omega t-\sin \omega t)$        ………$\left(10\right)$
The equations $\left(9\right)$ and $\left(10\right)$ are the parametric equations of a cycloid.
$\Rightarrow$ The maximum value of $y$ is $y_{\max }=\frac{2E}{\omega B}.$
$\Rightarrow$ If a circle of radius $R-\frac{E}{\omega B}$ were to roll on the $x-$axis, a fixed point on the circle would generate this cycloid.
$\Rightarrow$ An observer at the Centre of this circle will be moving with velocity $R\omega =E/B$ along $\mathrm{x}-$axis, Tle particle will appear to move along the circumference of a circle with speed $R\omega =E/B$.
$v^2=v_x^2+v_y^2\Rightarrow v=\frac{E}{B}\sqrt{{(1-\cos \omega t)}^2+{\sin }^2\omega t}$

      $=\frac{E}{B}·2\sin \frac{\omega t}{2}=\frac{2E}{B}\sin \frac{\omega t}{2}$
Now $v=0$ when $\frac{\omega t}{2}=0,\pi ,2\pi , t=0.\frac{2\pi }{\omega },\frac{4\pi }{\omega },......$
Thus the two nearest instants at which velocity is zero differ by $2\pi /\omega$.
Thus,$S={\int }_0^{2\pi /\theta }\mathrm{v}dt={\int }_0^{2\pi /\omega }\frac{2E}{B}\sin \frac{\omega t}{2}dt=\frac{\frac{8E}{qB}}{m}B=\frac{8mE}{q{B}^2}$.