# The electric field of a plane electromagnetic wave is given by

Question:

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:

1. (1) parallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

2. (2) zero

3. (3) antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

4. (4) parallel to $\hat{k}$

Correct Option: , 3

Solution:

(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 $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$