The electric field of a plane electromagnetic wave is given by \(\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})\) At \(\mathrm{t}=0,\) a positively charged particle is at the point \((\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .\) If its instantaneous velocity at \((t=0)\) is \(v_{0} \hat{\mathrm{k}},\) the force acting on it due to the wave is

\(\overrightarrow{\mathrm{E}}=\mathrm{E}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right) \cos \pi\)

\(=-\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)

as \(\quad \overrightarrow{E} \times \overrightarrow{B}=\overrightarrow{\mathrm{c}}\)

\(+\mathrm{E}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right) \times \overrightarrow{\mathrm{B}}=\mathrm{c} \hat{\mathrm{k}}\)

\(\Rightarrow \overrightarrow{\mathrm{B}}=-\left(\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}\right) \frac{\mathrm{E}_{0}}{\mathrm{c}}\)

\(\overline{\mathrm{F}}=\mathrm{q}\left(-\mathrm{E}_{0} \frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{\sqrt{2}}-\frac{\mathrm{v}_{0} \hat{\mathrm{k}}}{\mathrm{c}} \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \mathrm{E}_{\mathrm{o}}\right)\)

since \(\frac{v_{0}}{c}<<1\)

\(\Rightarrow \mathrm{F}\) is antiparallel to \(\frac{\mathrm{i}+\mathrm{j}}{\sqrt{2}}\)