A particle of mass $1\times {10}^{-26} \mathrm{kg}$ and charge $+1.6\times {10}^{-19}\mathrm{C}$ traveling with a velocity $1.28\times {10}^6 \mathrm{m}/\mathrm{s}$ in the $+x-$direction enters a region in which uniform electric field $E$ and a uniform magnetic field of induction $B$ are present such that $E_x=E_y=0,E_z=-102.4\mathrm{kV}/\mathrm{m}$, and $B_x=B_z=0,B_y=8\times {10}^{-2}$. The particle enters this region at time $t=0$. Determine the location $(x, y, z coordinates)$ of the particle at $t=5\times {10}^{-6} \mathrm{s}$. If the electric field is switched off at this instant (with the magnetic field present), what will be the position of the particle at $t=7.45\times {10}^{-6} \mathrm{s} ?$

The Lorentz force on the charged particle is
      $F=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$
The electric force on the charged particle, $F_E=q{E}_Z$, which acts towards negative direction $z$ in the direction of electric field.
The magnetic force on the charged particle,
$F_B=q{v}_x{B}_y$
As velocity of charge is in $+\mathcal{x}-$direction and magnetic field is along $+y-$direction. From right hand rule the magnetic force acts along positive z-direction.
The resultant force
       $F=F_E+F_B=q\left(E_2+v_z\times B_y\right)$

         $=q\left[-102.4\times {10}^3+1.28\times {10}^6\times 8\times {10}^{-2}\right]=0$

During time $t-0$ to $t_1=5\times {10}^{-6}$ the resultant force on the particle is zero, it moves with uniform velocity $v_x$. The position of the particle $\left(X_1,Y_1,Z_1\right)$ after time $t_1$ is
$X_1-v_x{t}_1=\left(1.28\times {10}^6\right)\times \left(5\times {10}^{-6}\right]=6.4 \mathrm{m}$
When electric field is switched off, the particle circulates in $xz-$plane under the influence of magnetic field. Radius $R$ of circulation is
$R=\frac{m{v}_x}{q{B}_y}=\frac{{10}^{-26}\times 1.28\times {10}^6}{1.6\times {10}^{-19}\times 8\times {10}^{-2}}=1 \mathrm{m}$
Let the particle rotate by an angle $\theta =\omega \left(t_2-t_1\right)$. The arc length $P_1{P}_2=R\theta =v_x\left(t_2-t_1\right)$, as the particle circulates for
$t_2-t_1=(7.45-5)\times {10}^{-6}=2.45\times {10}^{-6} \text{s}$

$\theta =\frac{v_x\left(t_2-t_1\right)}{R}=\frac{\left(1.28\times {10}^6\right)\times \left(2.45\times {10}^{-6}\right)}{1}=3.136\simeq \pi \text{ radians.}$
The coordinates of the particle are
$X_2=X_1+R\sin \theta =X_1+R\sin \pi =X_1=6.4 \text{m}$

and    $Z_2=R-R\cos \theta =R-R\cos \pi =2R=2 \text{m}$
Note that $\theta =\pi$ implies that $t_2=T/2$ where $T$ is time period of circulation. We could have written the result directly.