An electron moves through a uniform magnetic field given by $\overline{B}=B_x\widehat{i}+3B_x\widehat{j}$ At a particular instant, the electron has the velocity $\overline{\text{V}}=\left(2\widehat{i}+4\widehat{j}\right)\text{m/s}$and the magnetic force acting on it is $\left(6.4\times {10}^{-19}\text{N}\right)\widehat{K}$Then $B_x$is

$$\begin{gathered} \text{Given\hspace{0.17em}}\overline{B}=B_x\widehat{i}+3B_x\widehat{j} \\ \overline{v}=\left(2\widehat{i}+4\widehat{j}\right)\text{m/s} \\ \text{q}=-\text{1.6}\times {\text{10}}^{-19}\text{c} \\ \overline{F}=\left(6.4\times {10}^{-19}\text{N}\right)\widehat{k} \\ \left(6.4\times {10}^{-19}\right)\widehat{k}=-1.6\times {10}^{-19}\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 2& 4& 0\\ B_x& 3B_x& 0\end{array}\right| \\ \Rightarrow 6.4\times {10}^{-19}\widehat{k}=-1.6\times {10}^{-19}\left[\widehat{i}\left(0\right)-\widehat{j}\left(0\right)+\widehat{k}\left(6B_x-4B_x\right)\right] \\ \Rightarrow 6.4\times {10}^{-19}\widehat{k}=-1.6\times {10}^{-19}\left[2{\text{B}}_x\widehat{k}\right] \\ \Rightarrow {\text{B}}_x=\frac{6.4\times {10}^{-19}}{-1.6\times {10}^{-19}\times 2}=\frac{-6.4}{3.2}=-2.0\text{T} \end{gathered}$$