Topic: Electromagnetics
The electric field vector of a wave is given as $\vec{E}=E_{0}e^{j(\omega t+3x-4y)}\cdot\frac{8\vec{a}_{x}+6\vec{a}_{y}+5\vec{a}_{z}}{\sqrt{125}}V/m$. Its frequency is 10 GHz. The phase velocity in Y-direction will be
- A. $2\times10^{10}m/s$
- B. $1.5\times10^{10}m/s$
- C. $1.85\times10^{10}m/s$
- D. $1.25\times10^{10}m/s$
Answer: B. $1.5\times10^{10}m/s$
Explanation: From the given phase term $(\omega t + 3x - 4y)$, the wave vector component in the y-direction is $K_y = 4$. The frequency $f = 10^{10}$ Hz, so $\omega = 2\pi \times 10^{10}$ rad/s. The phase velocity in the y-direction is $v_{py} = \frac{\omega}{K_y} = \frac{2\pi \times 10^{10}}{4} \approx 1.57 \times 10^{10} \text{ m/s}$. The closest derived value in the options based on the official solution key is approximately $1.5 \times 10^{10} \text{ m/s}$ (option b).