Question

The electric field of a plane electromagnetic wave is given by
$\overrightarrow{y}=E_0\hat{i}cos\left(kz\right)\left(\omega t\right)$
The corresponding magnetic field $\overrightarrow{B}$ is then given by:

• A. $\overrightarrow{B}=\frac{E_0}{c}\hat{i}sin\left(kz\right)cos\left(\omega t\right)$
• B. $\overrightarrow{B}=\frac{E_0}{c}\hat{j}sin\left(kz\right)cos\left(\omega t\right)$
• C. $\overrightarrow{B}=\frac{E_0}{c}\hat{k}sin\left(kz\right)cos\left(\omega t\right)$
• D. $\overrightarrow{B}=\frac{E_0}{c}\hat{j}cos\left(kz\right)sin\left(\omega t\right)$

Answer 1

The correct option is B $\overrightarrow{B}=\frac{E_0}{c}\hat{j}sin\left(kz\right)cos\left(\omega t\right)$

$\because \overrightarrow{E}\times \overrightarrow{B}||\overrightarrow{V}$
Given that wave is propagating along positive z-axis and $\overrightarrow{E}$ along positive x-axis. Hence $\overrightarrow{B}$ along y-axis.
From Maxwel equation
$\overrightarrow{V}\times \overrightarrow{E}=-\frac{\partial B}{\partial t}$
i.e. $\frac{\partial E}{\partial Z}=-\frac{\partial B}{\partial t}and{B}_0=\frac{E_0}{C}$

so, $\overrightarrow{B}=\frac{\overrightarrow{E_0}}{c}\hat{j}sin\left(kz\right)cos\left(\omega t\right)$