A particle of charge $-16\times 10^{-18}$ coulomb moving with velocity 10 $\mathrm{m}{\mathrm{s}}^{-1}$ along the x-axis enters a region where a magnetic field of induction $\mathrm{B}$ is along the $\mathrm{y}$ -axis, and an electric field of induction $\mathrm{B}$ is along the $\mathrm{y}$-axis, and an electric field of magnitude $10^4\mathrm{V}/\mathrm{m}$ is along the negative $\mathrm{z}$-axis. lf the charged particle continues moving along the $\mathrm{x}$-axis, the magnitude of $\mathrm{B}$ is

Particle travels along $x$ axis 

Hence $V_y=V_z=0$

Electric feild is along the negative $z$ axis

$E_x=E_y=0$

Therefore net force on the particle $\vec{F}=q\left(\vec{E}+V\times \vec{B}\right)$

Resolve the motion along three coordinate axis

$$\begin{gathered} a_x=\frac{F_x}{m}=\frac{q}{m}\left(E_x+V_y{B}_z-V_x{B}_y\right) \\ {a}_y=\frac{F_x}{m}=\frac{q}{m}\left(E_y+V_z{B}_x-V_{xz}\right) \\ {a}_z=\frac{F_x}{m}=\frac{q}{m}\left(E_z+V_x{B}_y-V_y{B}_x\right) \end{gathered}$$

Since, 

$$\begin{gathered} V_y=V_z=0 \\ {E}_x=E_y=0 \\ {B}_x=B_z=0 \end{gathered}$$

$$\begin{gathered} a_x=a_y=0 \\ {a}_z=\frac{q}{m}\left(-E_z+V_x{B}_y\right) \end{gathered}$$

Again $a_z=0$ , as the particle transverse through the region undeflected 

$E_z=V_x{B}_y$

$B_y=\frac{E_z}{V_x}=\frac{{10}^4}{10}={10}^{3}wb/m^2$