Physics - EM Waves Question with Solution | TestHub

Comparing the given equation $$\begin{array}{rl}& {\mathrm{E}}_{\mathrm{y}}=2.5\frac{\mathrm{N}}{\mathrm{C}}\cos [\left(2\pi \times 10^6\frac{\mathrm{rad}}{\mathrm{m}}\right)\mathrm{t}\\ & -\left(\pi \times 10^{-2}\frac{\mathrm{rad}}{\sec }\right)\mathrm{x}]\end{array}$$ With the standard equation $${\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_0\cos (\omega \mathrm{t}-\mathrm{kx})$$ we get $$\begin{array}{ll}& \omega =2\pi \mathrm{f}=2\pi \times 10^6\\ \therefore & \mathrm{f}=10^6\mathrm{Hz}\end{array}$$ Moreover, we know that $$\Rightarrow \begin{array}{rl}\frac{2\pi }{\lambda }& =\mathrm{k}=\pi \times 10^{-2}{\mathrm{m}}^{-1}\\ \Rightarrow \lambda & =200\mathrm{m}\end{array}$$ Hence, the wave is moving along positive $x$-direction with frequency $10^6\mathrm{Hz}$ and wavelength $200\mathrm{m}$.