• # question_answer The electric field of a plane electromagnetic wave is given by $\vec{E}={{E}_{0}}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\,\cos \,(kz+\omega t)$ At t = 0, a positively charged particle is at the point $(x,y,z)\,=\,\left( 0,0,\,\frac{\pi }{k} \right)$. If its instantaneous velocity at (t = 0) is${{v}_{0}}\,\hat{k},$, the force acting on it due to the wave is: [JEE MAIN Held on 07-01-2020 Evening] A) Antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$ B) Zero C) Parallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$ D) Parallel to $\hat{k}$

 [a]  $\vec{E}$ at t = 0 at $z=\pi k$ is given by $\vec{E}=\frac{{{E}_{0}}}{\sqrt{2}}\,(\hat{i}+\hat{j})\,cos\,[\pi ]\,=-\,\frac{{{E}_{0}}}{\sqrt{2}}\,(\hat{i}+\hat{j})$ ${{\vec{F}}_{E}}=q\vec{E}$ $\Rightarrow$ Force due to electric field, ${{\vec{F}}_{E}}$ is antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$. ${{\vec{F}}_{mag}}=q(\vec{v}\times \vec{B})$ $\vec{B}\,\,(at\,t=0,\,z\,=\pi k)$is $\frac{{{B}_{0}}}{\sqrt{2}}\,(-\hat{i}+\hat{j})$ $\Rightarrow \,\,\,{{\vec{F}}_{mag}}\,=q\,{{v}_{0}}\hat{k}\times \frac{{{B}_{0}}}{\sqrt{2}}\,(-\hat{i}+\hat{j})$which is antiparallel to $\frac{(\hat{i}+\hat{j})}{\sqrt{2}}$ $\Rightarrow \,\,{{\vec{F}}_{net}}={{\vec{F}}_{E}}+{{\vec{F}}_{B}}$ is Antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$