A uniform plane wave in free space has electric field vector given by \(\mathbf{E}_{s}=\) \(10 e^{-j \beta x} \mathbf{a}_{z}+15 e^{-j \beta x} \mathbf{a}_{y} \mathrm{~V} / \mathrm{m} .(a)\) Describe the wave polarization. (b) Find \(\mathbf{H}_{s} .(c)\) Determine the average power density in the wave in \(\mathrm{W} / \mathrm{m}^{2}\).

First, let's examine the phases of E_y and E_z components of the electric field: \(E_y = 15e^{-j\beta x} \ \mathrm{V/m}\) \(E_z = 10e^{-j\beta x} \ \mathrm{V/m}\) Both components have the same phase, so we can safely assume that the wave polarization is linear. The angle of polarization can be calculated as: \(\theta = \arctan{\left(\frac{E_y}{E_z}\right)} = \arctan{\left(\frac{15}{10}\right)}\) \(\theta \approx 56.31^\circ\) The wave is linearly polarized with an angle of approximately 56.31 degrees.

Use the relation between the electric and magnetic fields in free space: \(\frac{\mathbf{E}_{s}}{\mathbf{H}_{s}} = \eta_0\) Rearrange this equation to solve for H_s: \(\mathbf{H}_{s} = \frac{\mathbf{E}_{s}}{\eta_0} = \frac{1}{120\pi} (10 e^{-j \beta x} \mathbf{a}_{z} + 15 e^{-j \beta x} \mathbf{a}_{y})\) Now, split the components: \(H_y = -\frac{1}{120\pi} (10 e^{-j \beta x})\) \(H_z = \frac{1}{120\pi} (15 e^{-j \beta x})\) Thus, the magnetic field vector is: \(\mathbf{H}_{s} = -\frac{10}{120\pi} e^{-j \beta x} \mathbf{a}_{y} + \frac{15}{120\pi} e^{-j \beta x} \mathbf{a}_{z} \ \mathrm{A/m}\)

To calculate the average power density, we will use the Poynting vector, which is given by: \(\mathbf{S} = \frac{1}{2}Re(\mathbf{E}_{s} \times \mathbf{H}_{s}^*)\) Substitute the values of E_s and H_s: \(\mathbf{S} = \frac{1}{2}Re\left(\left[10 e^{-j \beta x} \mathbf{a}_{z} + 15 e^{-j \beta x} \mathbf{a}_{y}\right] \times \left[-\frac{10}{120\pi} e^{j \beta x} \mathbf{a}_{y} + \frac{15}{120\pi} e^{j \beta x} \mathbf{a}_{z}\right]\right)\) Simplify and perform the cross product: \(\mathbf{S} = \frac{1}{2}\frac{1}{120\pi}Re\left(150 e^{-j \beta x} e^{j \beta x} \mathbf{a}_{x}\right)\) After simplification, we get: \(\mathbf{S} = \frac{1}{2}\frac{1}{120\pi}(150) \mathbf{a}_{x}\) Now, we can calculate the average power density: \(S = \frac{1}{2}\frac{150}{120\pi} \ \mathrm{W/m^2}\) \(S \approx 0.398 \ \mathrm{W/m^2}\) The average power density in the wave is approximately 0.398 W/m².