Question

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

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

Answer 1

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

Given:
$\overrightarrow{E}=E_0\hat{i}\cos \left(kz\right)\cos \left(\omega t\right)$
As we know that both magnetic field vector and electric field vectors are perpendicular to each other for a electromagnetic wave.
Here, as the electric field is along $x-$Axis so the direction of the magnetic field will be along $y-$axis.
To find the magnitude of the magnetic field,
Using Maxwell's equations

$\frac{\delta E}{\delta z}=-\frac{\delta B}{\delta t}$

$\therefore \frac{\delta E}{\delta z}=E_0(-\sin (kz\left)\right).k\cos \left(\omega t\right)$

$\Rightarrow \frac{\delta E}{\delta z}=-E_{0}k\left(\sin \right(kz\left)\right)\cos \left(\omega t\right)=-\frac{\delta B}{\delta t}$

$\Rightarrow \frac{\delta B}{\delta t}=E_{0}k\left(\sin \right(kz\left)\right)\cos \left(\omega t\right)$

Integrating with respect to $t$,

$B=E_{0}k\left(\sin \right(kz\left)\right)\sin \left(\omega t\right)\times \frac{1}{\omega }$

$B=\frac{E_0}{C}\sin \left(kz\right)\sin \left(\omega t\right)(\because \frac{\omega }{k}=C)$

$\overrightarrow{B}=\frac{E_0}{C}\hat{j}\sin \left(kz\right)\sin \left(\omega t\right)$



Alternate solution:
$\frac{E_0}{B_0}=C$
$\Rightarrow B_0=\frac{E_0}{C}$
Given that $\overrightarrow{E}=E_0\cos \left(kz\right)\cos \left(\omega t\right)\hat{i}$
$\overrightarrow{E}=\frac{E_0}{2}\left[\cos \right(kz-\omega t)\hat{i}-\cos (kz+\omega t\left)\hat{i}\right]$
Correspondingly
$\overrightarrow{B}=\frac{B_0}{2}\left[\cos \right(kz-\omega t)\hat{j}-\cos (kz+\omega t\left)\hat{j}\right]$
$\overrightarrow{B}=\frac{B_0}{2}\times 2\sin kz\sin \omega t$
$\overrightarrow{B}=\left(\frac{E_0}{C}\sin kz\sin \omega t\right)\hat{j}$

Hence option $\left(A\right)$ is correct.