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

Answer 1

Answer: (A)

$\vec{E}$ is electric field vector, $\vec{B}$ is magnetic field vector perpendicular to $\vec{E}$. The direction of propagation is $(\vec{E}\times \vec{B})$.
The direction of propagation of wave is along $+y$ axis, then $\vec{E}$ is along $+z$ axis and $\vec{B}$ is along $+x$ axis.

$(E\hat{k}\times B\hat{t})=EB(\hat{k}\times \hat{i})=EB\hat{j}$
As wave is travelling along $+y$ -axis with time, we will use $(y-ct)$ in wave equation. Also intensity is given by
$I=\frac{1}{2}c{\epsilon }_0{E}_0^2$
And $E=cB$
So, $\left|E_0\right|=\sqrt{\frac{2I}{c{\epsilon }_0}}$
Therefore, $\vec{E}=\sqrt{\frac{2I}{c{\epsilon }_0}}\cos \left[\frac{2\pi }{\lambda }(y-ct)\right]\hat{k}$
And $\vec{B}=\frac{E}{c}\hat{i}\Rightarrow \frac{1}{c}E\hat{i}$