Given a wave for which \(\mathbf{E}_{s}=15 e^{-j \beta z} \mathbf{a}_{x}+18 e^{-j \beta z} e^{j \phi} \mathbf{a}_{y} \mathrm{~V} / \mathrm{m}\) in a medium characterized by complex intrinsic impedance, \(\eta(a)\) find \(\mathbf{H}_{s} ;(b)\) determine the average power density in \(\mathrm{W} / \mathrm{m}^{2}\).

Short Answer

Expert verified
Using the given electric field \(\mathbf{E}_{s}\) and the steps provided, determine the magnetic field \(\mathbf{H}_{s}\) and the average power density \(S_{av}\). Provide the results in their respective units.

Step by step solution

01

Calculating the magnetic field \(\mathbf{H}_{s}\)

To find the magnetic field, we need to divide the electric field by the intrinsic impedance \(\eta\). We are given the electric field as: \(\mathbf{E}_{s}=15 e^{-j \beta z} \mathbf{a}_{x}+18 e^{-j \beta z} e^{j \phi} \mathbf{a}_{y} \mathrm{~V} / \mathrm{m}\). So, the magnetic field \(\mathbf{H}_{s}\) can be calculated using the formula: \(\mathbf{H}_{s} = \frac{\mathbf{E}_{s}}{\eta}\). Using the given \(\mathbf{E}_{s}\) and intrinsic impedance \(\eta\), we can calculate \(\mathbf{H}_{s}\).
02

Determining the average power density

To find the average power density, we will use the Poynting vector \(S_{av}\). The Poynting vector is the cross product of electric and magnetic fields: \(\vec{S}=\mathbf{E} \times \mathbf{H}^* \). The average power density can be found by taking the real part of the Poynting vector component in the direction normal to the surface. To find the Poynting vector, we need to take the cross product of the electric and magnetic fields, \(\vec{S}=\mathbf{E} \times \mathbf{H}^* \), and then take the real part to find the average power density.
03

Calculate the cross product \(\mathbf{E} \times \mathbf{H}^*\)

Now, we need to find the cross product of the electric field \(\mathbf{E}\) and the complex conjugate of the magnetic field \(\mathbf{H}\): \(\vec{S}=\mathbf{E} \times \mathbf{H}^* \). For that, we can use the formula: \(\vec{S}=\left(\mathbf{a}_{x} \cdot \mathbf{E}_{x}+\mathbf{a}_{y} \cdot \mathbf{E}_{y}\right) \times\left(\mathbf{a}_{x} \cdot \mathbf{H}^*_{x}+\mathbf{a}_{y} \cdot \mathbf{H}^*_{y}\right)\). Using the given \(\mathbf{E}_{s}\) and \(\mathbf{H}_{s}\) calculated before, we can find \(\vec{S}\).
04

Find the real part of \(\vec{S}\) to get the average power density

After finding the Poynting vector \(\vec{S}\), we need to take the real part of its z-component to find the average power density in \(\mathrm{W} / \mathrm{m}^{2}\). The z-component of \(\vec{S}\) is \(S_z\), and we can find the average power density by taking the real part: \(S_{av} =\operatorname{Re}(S_{z})\). Using the \(\vec{S}\) found in the previous step, we can calculate the average power density \(S_{av}\). Now, we have calculated both the magnetic field \(\mathbf{H}_{s}\) and the average power density \(S_{av}\) as required.

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