The electric fields of two plane electromagnetic plane waves in vacuum are given by

Question:

The electric fields of two plane electromagnetic plane waves in vacuum are given by

$\overrightarrow{\mathrm{E}}_{1}=\mathrm{E}_{0} \hat{j} \cos (\omega t-k x)$ and $\overrightarrow{\mathrm{E}}_{2}=\mathrm{E}_{0} \hat{k} \cos (\omega t-k y)$

At $t=0$, a particle of charge $q$ is at origin with a velocity $\vec{v}=0.8 c \hat{j}$ ( $c$ is the speed of light in vacuum). The instantaneous force experienced by the particle is:

1. (1) $\mathrm{E}_{0} q(0.8 \hat{i}-\hat{j}+0.4 \hat{k})$

2. (2) $\mathrm{E}_{0} q(0.4 \hat{i}-3 \hat{j}+0.8 \hat{k})$

3. (3) $\mathrm{E}_{0} q(-0.8 \hat{i}+\hat{j}+\hat{k})$

4. (4) $\mathrm{E}_{0} q(0.8 \hat{i}+\hat{j}+0.2 \hat{k})$

Correct Option: , 4

Solution:

(4) Given: $\vec{E}_{1}=E_{0} \hat{j} \cos (\omega t-k x)$

i.e., Travelling in $+v e x$-direction $\vec{E} \times \vec{B}$ should be in $x$-direction

$\therefore \quad \vec{B}$ is in $\hat{K}$

$\therefore \vec{B}_{1}=\frac{E_{0}}{C} \cos (\omega t-k x) \hat{k} \quad\left(\because B_{0}=\frac{E_{0}}{C}\right)$

$\vec{E}_{2}=E_{0} \hat{k} \cos (\omega t-k y)$

$\vec{B}_{2}=\frac{E_{0}}{C} \hat{i} \cos (\omega t-k y)$

$\therefore \quad$ Travelling in +ve $y$-axis

$\vec{E} \times \vec{B}$ should be in $y$-axis

$\therefore \quad$ Net force $\vec{F}=q \vec{E}+q(\vec{v} \times \vec{B})$

$q\left(\vec{E}_{1}+\vec{E}_{2}\right)+q\left(0.8 c \hat{j} \times\left(\vec{B}_{1}+\vec{B}_{2}\right)\right.$

If $\mathrm{t}=0$ and $\mathrm{x}=\mathrm{y}=0$

If $\mathrm{t}=0$ and $\mathrm{x}=\mathrm{y}=0$

$\vec{E}_{1}=E_{0} \hat{j} \quad \vec{E}_{2}=E_{0} \hat{k}$

$\vec{B}_{1}=\frac{E_{0}}{c} \hat{k} \quad \vec{B}_{2}=\frac{E_{0}}{c} \hat{i}$

$\therefore \vec{F}_{\text {net }}=q E_{0}(\hat{j}+\hat{k})+q \times 0.8 c \times \frac{E_{0}}{C} \hat{j} \times(\hat{k}+\hat{i})$

$=q E_{0}(\hat{j}+\hat{k})+0.8 q E_{0}(\hat{i}-\hat{k})$

$=q E_{0}(0.8 \hat{i}+\hat{j}+0.2 \hat{k})$