Question

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

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

Answer 1

Answer: (A)

$\because \overrightarrow{E}\times \overrightarrow{B}||\overrightarrow{v}$
Given that wave is propagating along positive $z$ -axis and $\overrightarrow{E}$ along positive $x$ -axis. Hence $\vec{B}$ along $y$ -axis. From Maxwel equation $\overrightarrow{V}\times \overrightarrow{E}=-\frac{\partial B}{\partial t}$
i.e. $\frac{\partial E}{\partial Z}=-\frac{\partial B}{\partial t}{\mathrm{andB}}_0=\frac{E_0}{C}$
so, $\overrightarrow{B}=\frac{\overrightarrow{E_0}}{c}\hat{j}\sin (kz)\cos (\omega t)$