Question

A plane electromagnetic wave of wavelength $\lambda$ has an intensity $I$. It is propagating along the positive $Y-direction$. The allowed expressions for the electric and magnetic fields are given by

• A. $\overrightarrow{E}=\sqrt{\frac{I}{{ϵ}_{0}C}}\cos \left[\frac{2\pi }{\lambda }\right(y-ct\left)\right]\hat{i};\overrightarrow{B}=\frac{1}{c}E\hat{k}$
• B. $\overrightarrow{E}=\sqrt{\frac{I}{{ϵ}_{0}C}}\cos \left[\frac{2\pi }{\lambda }\right(y-ct\left)\right]\hat{k};\overrightarrow{B}=-\frac{1}{c}E\hat{i}$
• C. $\overrightarrow{E}=\sqrt{\frac{2I}{{ϵ}_{0}C}}\cos \left[\frac{2\pi }{\lambda }\right(y-ct\left)\right]\hat{k};\overrightarrow{B}=+\frac{1}{c}E\hat{i}$
• D. $\overrightarrow{E}=\sqrt{\frac{2I}{{ϵ}_{0}C}}\cos \left[\frac{2\pi }{\lambda }\right(y+ct\left)\right]\hat{k};\overrightarrow{B}=\frac{1}{c}E\hat{i}$

Answer 1

Answer: (C)

$\overrightarrow{E}=\sqrt{\frac{2I}{{ϵ}_{0}C}}\cos \left[\frac{2\pi }{\lambda }\right(y-ct\left)\right]\hat{k};\overrightarrow{B}=+\frac{1}{c}E\hat{i}$
is the electric field vector, and is the magnetic field vector of the EM wave. For electromagnetic waves and are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of x .

So, if the wave propagates in the +Y direction then the direction of E and B should be in +X and +Z or vice versa i.e +Z and +Xrespectively.

$Case1.$

Let us suppose $\overrightarrow{E}$ is in $\hat{i}$ and $\overrightarrow{B}$ is in $\hat{k}$

Then $\overrightarrow{E}\times \overrightarrow{B}$ will be in $-\hat{j}$

Not Possible.

$Case2.$

Let us suppose $\overrightarrow{E}$ is in $\hat{k}$ and $\overrightarrow{B}$ is in $\hat{i}$

Then $\overrightarrow{E}\times \overrightarrow{B}$ will be in $\hat{j}$

This is satisfying option (3)

as the electric and magnetic field also propagate in positive y direction with time so $(y-ct)$ should be there in wave equation.

Also $I=\frac{c{ϵ}_o}{2}|E_o{|}^2$

$|E_o|=\sqrt{\frac{2I}{c{ϵ}_o}}$

From these, we can say that option (c) would be the best option.