Question

The electric field 0f a plane electromagnetic wave is given by
$\vec{E}=E_{0} \frac{\hat{i}+\hat{j}}{\sqrt{2}}\,cos\,\left(kz+\omega t\right)$
At $t=0$ , a positively charged particle is at the point $\left(x,\,y,\,z\right)=\left(0, 0, \frac{\pi}{k}\right)$. If its instantaneous velocity at $( t= 0)$ is $v_{0}\hat{k}$, the force acting on it due to the wave is :

• A. parallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
• B. zero
• C. antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
• D. parallel to $\hat{k}$

Answer 1

Answer: (C)

$\vec{F} = q\left(\vec{E}+\vec{v} \times \vec{B}\right)$
$\vec{E} = E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) cos \,\pi$
$= -E_{0} \frac{\hat{i}+\hat{j}}{\sqrt{2}}$
$\vec{E} \times \vec{B} = \vec{c}$
$+ E_{0} \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\times \vec{B} = c\hat{k}$
$\Rightarrow \vec{B} = -\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \frac{E_{0}}{c}$
$\vec{F} = q\left(E_{0} \frac{\hat{i}+\hat{j}}{\sqrt{2}}-\frac{v_{0}\hat{k}}{c}\times \left(\hat{i}-\hat{j}\right)E_{0}\right)$
since $\frac{v_{0}}{c} < < 1$
$\Rightarrow $ F is antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$