The electric field of a plane electromagnetic wave is given by
$\vec{E}=E_{0} \frac{\hat{i}+\hat{j}}{\sqrt{2}} \cos (k z+\omega t)$
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:
Correct Option: , 3
(3) At $t=0, z=\frac{\pi}{k}$
$\therefore \quad \vec{E}=\frac{E_{0}}{\sqrt{2}}(\hat{i}+\hat{j}) \cos [\pi]=-\frac{E_{0}}{\sqrt{2}}(\hat{i}+\hat{j})$
$\vec{F}_{E}=q \vec{E}$
Force due to electric field will be in the direction
$\frac{-(\hat{i}+\hat{j})}{\sqrt{2}}$
Force due to magnetic field is in direction
$q(\vec{v} \times \vec{B})$ and $\vec{v} \| \vec{k}$. Therefore, it is parallel to $\vec{E}$.
$\Rightarrow \quad \vec{F}_{\text {net }}=\vec{F}_{E}+\vec{F}_{B}$ is antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
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