The electric field of a plane electromagnetic wave is given by
$$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$

At t = 0, a positively charged particle is at the point (x, y, z) = $\left( {0,0,{\pi \over k}} \right)$.
If its instantaneous velocity at (t = 0) is ${v_0}\widehat k$ , the force acting on it due to the wave is :

$$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$ <br><br>$\overrightarrow E$ at t = 0 at z = ${\pi \over k}$ is given by <br><br>$\overrightarrow E =$$${E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\cos \left( {k{\pi \over k} + \omega \left( 0 \right)} \right)$$ <br><br>= $- {E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$ <br><br>As $\overrightarrow E \times \overrightarrow B = \overrightarrow c$ <br><br>Force due to magnetic field is in direction $q\left( {\overrightarrow v \times \overrightarrow B } \right)$ and given ${\overrightarrow v }$ parallel to $\widehat k$. <br><br>$\overrightarrow F =$ $$q\left( {\overrightarrow E + \overrightarrow v \times \overrightarrow B } \right)$$ <br><br>$\therefore$ $\overrightarrow F$ is in direction of ${\overrightarrow E }$ <br><br>which is antiparallel to ${{\widehat i + \widehat j} \over {\sqrt 2 }}$