The electric field part of an electromagnetic | vedclass

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Question

$E_{y}=2.5 \frac{\mathrm{N}}{\mathrm{C}}\left[\left(2 \pi 	imes 10^{6} \frac{\mathrm{rad}}{\mathrm{m}}ight) t-\left(\pi 	imes 10^{-2} \frac{\mathr

Vedclass · Accepted Answer

moving along $x$ direction with  frequency $10^6 Hz$ and wavelength $200\,m$

Vedclass · Answer

$E_{y}=2.5 \frac{\mathrm{N}}{\mathrm{C}}\left[\left(2 \pi 	imes 10^{6} \frac{\mathrm{rad}}{\mathrm{m}}ight) t-\left(\pi 	imes 10^{-2} \frac{\mathrm{rad}}{\mathrm{s}}ight) xight]$

$E_{z}=0$

The wave is moving in the positive direction of $x .$

This is the form $E_{y}=E_{0}(\omega t-k x)$

$\omega= 2 \pi 	imes 10^{6} $

$2 \pi v= 2 \pi 	imes 10^{6} \Rightarrow \mathrm{v}=10^{6}\, \mathrm{Hz} $

$\frac{2 \pi}{\lambda}=k \Rightarrow  \frac{2 \pi}{\lambda}=\pi 	imes 10^{-2} $

$\Rightarrow \lambda=\frac{2 \pi}{\pi 	imes 10^{-2}} =2 	imes 10^{2}=200\, \mathrm{m}$

The electric field part of an electromagnetic wave in a medium is represented by

$E_x=0, E_y=2.5 \frac{N}{C}\, cos\,\left[ {\left( {2\pi \;\times\;{{10}^6}\;\frac{{rad}}{s}\;\;} \right)t - \left( {\pi \;\times\;{{10}^{ - 2}}\;\frac{{rad}}{m}} \right)x} \right]$,and $ E_z=0$ . The wave is

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