Question

The electric fields of two plane electromagnetic plane waves, in vacuum, are given by,
$\overrightarrow{E_1}=E_0\hat{j}\cos (\omega t-kx)$ and, $\overrightarrow{E_2}=E_0\hat{k}\cos (\omega t-ky)$.
At $t=0$, a particle of charge $q$ is at origin with a velocity $\overrightarrow{v}=\left(0.8c\right)\hat{j}$ (Here, $c$ is the speed of light in vacuum). The instantaneous force experienced by the particle is:

• A. $E_{0}q(0.8\hat{i}-\hat{j}+0.4\hat{k})$
• B. $E_{0}q(0.4\hat{i}-3\hat{j}+0.8\hat{k})$
• C. $E_{0}q(-0.8\hat{i}+\hat{j}+\hat{k})$
• D. $E_{0}q(0.8\hat{i}+\hat{j}+0.2\hat{k})$

Answer 1

The correct option is D $E_{0}q(0.8\hat{i}+\hat{j}+0.2\hat{k})$

Given:
$\overrightarrow{E_1}=E_0\hat{j}\cos (\omega t-kx)$

This wave is travelling along the +ve x-direction.

$\Rightarrow \overrightarrow{E}\times \overrightarrow{B}$ should be directed along positive x-direction.

$\therefore {\overrightarrow{B}}_1\text{is directed along}\hat{k}$

$\therefore \overrightarrow{B_1}=\frac{E_0}{c}\hat{k}\cos (\omega t-kx)(\because B_0=\frac{E_0}{c})$


Similarly, $\overrightarrow{E_2}=E_0\hat{k}\cos (\omega t-ky)$

$\overrightarrow{B_2}=\frac{E_0}{c}\hat{i}\cos (\omega t-ky)$

$\therefore$ The second wave is travelling along +ve y-axis.

$\Rightarrow \overrightarrow{E}\times \overrightarrow{B}$ should be along +ve y-axis.

$\therefore$ Net force, ${\overrightarrow{F}}_{\text{net}}=q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})$

$=q(\overrightarrow{E_1}+\overrightarrow{E_2})+q(0.8c\hat{j}\times (\overrightarrow{B_1}+\overrightarrow{B_2}\left)\right)$

At $t=0,$ $x=y=0$
$\begin{array}{cc}\overrightarrow{E_1}=E_0\hat{j}& \overrightarrow{E_2}=E_0\hat{k}\\ \overrightarrow{B_1}=\frac{E_0}{c}\hat{k}& \overrightarrow{B_2}=\frac{E_0}{c}\hat{i}\end{array}$
$\therefore {\overrightarrow{F}}_{\text{net}}=q{E}_0(\hat{j}+\hat{k})+q\times 0.8c\times \frac{E_0}{c}\hat{j}\times (\hat{k}+\hat{i})$

$=q{E}_0(\hat{j}+\hat{k})+0.8q{E}_0(\hat{i}-\hat{k})$

$=q{E}_0(0.8\hat{i}+\hat{j}+0.2\hat{k})$

Hence, $\left(D\right)$ is the correct answer.