Question

The electric field of a plane electromagnetic wave is given by
$\overrightarrow{E}=E_0\frac{\hat{i}+\hat{j}}{\sqrt{2}}\cos (kz+\omega t)$
At $t=0$, a positively charged particle is at the point $(x,y,z)=(0,0,\frac{\pi }{k})$. 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

The correct option is C antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

At $t=0,z=\frac{\pi }{k}$

$\Rightarrow \overrightarrow{E}=\frac{E_0}{\sqrt{2}}(\hat{i}+\hat{j})\cos \left[\pi \right]$

$\Rightarrow \overrightarrow{E}=-\frac{E_0}{\sqrt{2}}(\hat{i}+\hat{j})$

The force on the charged particle $q$ due to Electric field is given by,

$\overrightarrow{F_E}=q\overrightarrow{E}$

Therefore, force due to electric field is in direction
$\frac{-(\hat{i}+\hat{j})}{\sqrt{2}}$.

In an electromagnetic wave, the magnetic field exists in the direction perpendicular to both the electric field and the direction of propagation.

Therefore, force due to magnetic field is in direction
$\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})$ and here $\overrightarrow{v}||\overrightarrow{k}$

Therefore, $\overrightarrow{F_B}$ is parallel to $\overrightarrow{E}$.

$\Rightarrow {\overrightarrow{F}}_{net}=\overrightarrow{F_E}+\overrightarrow{F_B}$

As both $\overrightarrow{F_E}\text{and}\overrightarrow{F_B}$ are parallel to $\overrightarrow{E}$,

$\Rightarrow {\overrightarrow{F}}_{net}$ is antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Hence, option $\left(C\right)$ is correct.