A charged particle is projected in a magnetic field $\overline{B}=\left(3\widehat{i}+4\widehat{j}\right)\times {10}^{-2}\text{T}$. The acceleration of the particle is found to be $\overline{a}=\left(x\widehat{i}+2\widehat{j}\right)m{s}^{-2}$.The value of x is (SI unit)                        

$$\begin{gathered} \overline{B}=\left(3\widehat{i}+4\widehat{j}\right)\times {10}^{-2}\text{T} \\ \overline{a}=\left(x\widehat{i}+2\widehat{j}\right){\text{m/s}}^{\text{2}} \\ \text{Here }\overline{a}\&\overline{B}are perpendicular to each other hence the d at product = 0 \\ \overline{a}\perp \overline{B}\left(\because \overline{F}\perp \overline{B}\right) \\ \therefore \overline{a}.\overline{B}=0 \\ i.e.,\left(x\widehat{i}+2\widehat{j}\right).\left(3\widehat{i}+4\widehat{j}\right)\times {10}^{-2}=0 \\ 3x\left(\widehat{i}.\widehat{i}\right)\times {10}^{-2}+4\times 2\left(\widehat{j}.\widehat{j}\right)\times {10}^{-2}=0 \\ 3x\left({10}^{-2}\right)+8\times {10}^{-2}=0 \\ \Rightarrow 3x+8=0 \\ \Rightarrow x=-\text{8/3\hspace{0.17em}} \end{gathered}$$