A particle of charge -16 x10^-18C moving with velocity 10 m/s along the X-axis enters a region where a magnetic field of induction B is along Y-axis and electric field of magnitude 104 V/m is a. long the negative Z-axis. If the charged particle continues moving along the X-axis, the magnitude of B is

The situation is shown in fig .We know that,
F= q (E+ V x B)
or F = qE + q (v x B)
Now Fe = gE
$\begin{array}{r}=-16\times {10}^{-18}\times {10}^4(-\mathbf{k})\\ =16\times {10}^{-14}\mathbf{k}\end{array}$
And ${\mathbf{F}}_{\mathrm{m}}=-16\times {10}^{-18}(10\mathbf{i}\times \mathrm{B}\mathbf{j})$
$=-16\times {10}^{17}\mathrm{B}\left(\mathbf{k}\right)$
$\therefore \mathrm{F}={\mathrm{F}}_{\mathrm{e}}+{\mathrm{F}}_{\mathrm{m}}=16\times {10}^{-14}\mathrm{k}-16\times {10}^{-17}\mathrm{Bk}$
Since, the particle will continue to move along +X-axis, so resultant force is zero. Therefore, F_e + F_m = 0
or $16\times {10}^{-14}\mathrm{k}=16\times {10}^{-17}\mathrm{Bk}$
$\therefore \mathrm{B}={10}^3\mathrm{Wb}/{\mathrm{m}}^2$