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}{\epsilon_{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}{\epsilon_{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}{\epsilon_{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}{\epsilon_{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.

$(E \hat{k} \times B \hat{t})=E B(\hat{k} \times \hat{i})=E B \hat{j}$
As wave is travelling along $+y$ -axis with time, we will use $(y-c t)$ in wave equation. Also intensity is given by
$I=\frac{1}{2} c \varepsilon_{0} E_{0}^{2}$
And $E=c B$
So, $\left|E_{0}\right|=\sqrt{\frac{2 I}{c \varepsilon_{0}}}$
Therefore, $\vec{E}=\sqrt{\frac{2 I}{c \varepsilon_{0}}} \cos \left[\frac{2 \pi}{\lambda}(y-c t)\right] \hat{k}$
And $\vec{B}=\frac{E}{c} \hat{i} \Rightarrow \frac{1}{c} E \hat{i}$