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