The electric field part of an electromagnetic wave in a medium is represented by : $\begin{aligned} &…
The electric field part of an electromagnetic wave in a medium is represented by :
$\begin{aligned}
& \mathrm{E}_{\mathrm{x}}=0 \\
& \mathrm{E}_{\mathrm{y}}=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) \mathrm{t}-\left(\pi \times 10^{-2} \frac{\mathrm{rad}}{\mathrm{s}}\right) \mathrm{x}\right]
\end{aligned}$
$E_z=0$. The wave is :
Moving along - $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
Moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $100 \mathrm{~m}$
Moving along $x$ direction with frequency $10^6$ $\mathrm{Hz}$ and wavelength $200 \mathrm{~m}$
Solution
As the coefficient of $x$ is negative, it is moving along +ve $x$-axis and equating the equation
$\begin{aligned}
\mathrm{E}_{\mathrm{y}} & =2.5 \cos \left[\left(2 \pi \times 10^6\right) \mathrm{t}-\left(\pi \times 10^{-2}\right) \mathrm{x}\right] \\
\text { with } \mathrm{y} & =\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx}) \\
\omega & =2 \pi \times 106 \\
\Rightarrow \mathrm{f} & =\frac{\omega}{2 \pi}=10^6 \mathrm{~Hz} \\
\mathrm{k} & =\pi \times 10^{-2} \\
\Rightarrow \quad \lambda & =\frac{2 \pi}{\mathrm{k}} \\
& =\frac{2 \pi}{\pi \times 10^{-2}}=200 \mathrm{~m}
\end{aligned}$