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

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

Then, from the relation $\overrightarrow{F_m}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

We have,

$$\begin{gathered} \left(4.0\hat{i}+3.0\hat{j}\right)\times {10}^{-10}={10}^{-9}\left[\left(v_x\hat{i}+v_y\hat{j}\right)\times \left(4\times {10}^{-3}\hat{k}\right)\right] \\ =\left(4v_y\times {10}^{-12}\hat{i}-4v_x\times {10}^{-12}\hat{j}\right) \end{gathered}$$

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

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

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

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