Question

The electric fields of two plane electromagnetic plane waves in vacuum are given by
${\vec{E}}_1=E_0\hat{j}cos\left(\omega t—kx\right)$ and
$\overrightarrow{E_2}=E_0\hat{k}cos\left(\omega t-ky\right)$
At $t=0$, a partide of charge q is at origin with a velocity $\vec{v}=0.8c\hat{j}$ (c is the speedof light in vaccum). The instantaneousforce experienced by the partide is :

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

Answer 1

Answer: (D)

${\vec{E}}_1=E_0\hat{j}cos\left(\omega t-kx\right)$
Its corresponding magnetic field will be
$\overrightarrow{B_1}=\frac{E_0}{c}\hat{k}cos\left(\omega t-kx\right)$
${\vec{E}}_2=E_0\hat{k}cos\left(\omega t-ky\right)$
$\overrightarrow{B_2}=\frac{E_0}{c}\hat{i}cos\left(\omega t-ky\right)$
Net force on charge particle
$=q\overrightarrow{E_1}+q\overrightarrow{E_2}+q\vec{v}\times B_1+q\vec{v}\times \overrightarrow{B_2}$
$=q{E}_0\hat{j}+q{E}_0\hat{k}+q\left(0.8c\hat{j}\right)\times \left(\frac{E_0}{c}\hat{k}\right)+q\left(0.8c\hat{j}\right)\times \left(\frac{E_0}{c}\hat{i}\right)$
$=q{E}_0\hat{j}+q{E}_0\hat{k}+0.8q{E}_0\hat{i}-0.8q{E}_0\hat{k}$
$\vec{F}=q{E}_0\left[0.8\hat{i}+1\hat{j}+0.2\hat{k}\right]$