The electric field vector of a wave is given as \overrightarrow{ E }= E _{0} e ^{ j (\omega t+3 x-4 y)} \frac{8 \vec{a}_{x}+6 \vec{a}_{y}+5 \vec{a}_{z}}{\sqrt{125}} V / m
Its frequency is 10 GHz
(i) Investigate if this wave is a plane wave,
(ii) Determine its propagation constant, and
(iii) Calculate the phase velocity in y-direction
(a) P_{a \nu g_{1}}=\frac{\left|E_{1}\right|^{2}}{2 \eta_{0}}=0.997 w / m ^{2}
(b) \varepsilon=\varepsilon^{\prime}-j \varepsilon^{\prime \prime} \varepsilon^{\prime}=9 \varepsilon_{0}
\begin{aligned}&\varepsilon^{\prime \prime}=0.09 \varepsilon_{0}=10^{-3} \varepsilon^{\prime} . \\&\alpha=\frac{\omega}{2} E^{\prime \prime} \sqrt{\frac{40}{E^{\prime}}}=6.29 \times 10^{-3} n / m\end{aligned}
Skin depth = α = 1/α = 159 m
(c) E_{2}=E_{1} e^{-\alpha x}
\begin{aligned}x &=5 f=5 / \alpha \\E_{2} &=E_{1} e^{-\alpha 5 / \alpha}=E_{1} e^{-5} \\\eta_{2} &=\sqrt{\frac{\mu_{0}}{E^{1}}}=\sqrt{\frac{\mu_{0}}{9 E_{0}}}=\frac{1}{3} \times 120 \pi=40 \pi \\P_{2} &=\frac{E_{2}^{2}}{2 \eta_{2}}=\frac{339.24}{80 \pi} e^{-10} \\P_{2} &=3 e^{-10} w / m ^{2}\end{aligned}
(i) \beta_{x}=-3, \beta_{y}=+4, \beta_{z}=0
\hat{\eta}=\frac{8 \hat{a}_{x}}{\sqrt{125}}+\frac{6 \hat{a}_{y}}{\sqrt{125}}+\frac{5 \hat{a}_{z}}{\sqrt{125}} , which conclude that the given field vector represents a plane wave in direction of x^{n}.
(ii) \beta_{x}^{2}+\beta_{y}^{2}+\beta_{z}^{2}=\beta^{2}\left(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C\right)
\begin{gathered}=\beta^{2}\left(\frac{64}{125}+\frac{36}{125}\right)=\frac{4}{5} \beta^{2} \\\frac{4}{5} \beta^{2}=(-3)^{2}+(4)^{2}=25^{2}, \Rightarrow \beta=5.59 \\\gamma=\alpha+_{j} \beta=0+j 5.59 \\=5.39 j\end{gathered}
(iii) Phase velocity in the y-direction.
V_{y}=\frac{\omega}{\beta_{y}}=\frac{\omega}{\beta \cos \beta}=\frac{2 \pi \times 10^{10}}{4}=1.57 \times 10^{10} m / sec