A magnetic field of $\left(4.0\times {10}^{-3}\hat{k}\right)T$ exerts a force $\left(4.0\hat{i}+3.0\hat{j}\right)\times {10}^{-10}N$ on a particle having a charge ${10}^{-9}C$ and moving in the x-y plane. Find the velocity of the particle. 

Given $B=\left(4\times {10}^{-3}\hat{k}\right)T, q={10}^{-9}C$

and magnetic force is $F_m=\left(4.0\hat{i}+3.0\hat{j}\right)\times {10}^{-10}N$

Let velocity of the particle in x-y plane be $v=v_x\hat{i}+v_y\hat{j}$.

Then from the relation,

$F_m=q\left(v\times B\right)$

We have, $\left(4.0\hat{i}+3.0\hat{j}\right)\times {10}^{-10}={10}^{-9}\left[\left(v_x\hat{i}+v_y\hat{j}\right)\times \left(4\times {10}^{-3}\hat{k}\right)\right]$

                                                   $=\left(4v_y\times {10}^{-12}\hat{i}-4v_x\times {10}^{-12}\hat{j}\right)$

Comparing the coefficients of $\hat{i}$ and $\hat{j}$, we have 

$$\begin{gathered} 4\times {10}^{-10}=4v_y\times {10}^{-12} \\ \therefore v_y={10}^{2}m/s=100m/s \end{gathered}$$

and $3.0\times {10}^{-10}=-4v_x\times {10}^{-12}$

$\therefore v_x=-75m/s$

$\therefore v=\left(-75\hat{i}+100\hat{j}\right)m/s$