A \(150 \mathrm{MHz}\) uniform plane wave in free space is described by \(\mathbf{H}_{s}=\) \((4+j 10)\left(2 \mathbf{a}_{x}+j \mathbf{a}_{y}\right) e^{-j \beta z} \mathrm{~A} / \mathrm{m} .(a)\) Find numerical values for \(\omega, \lambda\), and \(\beta .\) (b) Find \(\mathcal{H}(z, t)\) at \(t=1.5 \mathrm{~ns}, z=20 \mathrm{~cm} .(c)\) What is \(|E|_{\max } ?\)

Given, the frequency of the wave is \(f = 150~\mathrm{MHz}\). Also, we have the magnetic field \(\mathbf{H}_{s} = (4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\beta z} \mathrm{~A/m}\). We need to find the values of \(\omega, \lambda\), and \(\beta\). The speed of light in free space is \(c \approx 3 \times 10^8~\mathrm{m/s}\). The vacuum permeability and permittivity are, respectively, \(\mu_0 = 4\pi \times 10^{-7}~\mathrm{N/A^2}\) and \(\varepsilon_0 = 8.854 \times 10^{-12}~\mathrm{F/m}\).

First, let's calculate the angular frequency, \(\omega\). We know that \(\omega = 2\pi f\), so: \(\omega = 2\pi(150\times 10^6) = 300\pi\times 10^6~\mathrm{rad/s}\)

Now we can calculate the wavelength, \(\lambda\). The relationship between frequency, wavelength, and the speed of light in free space is \(c = f\lambda\). So we have: \(\lambda = \frac{c}{f} = \frac{3\times 10^8}{150\times 10^6} = 2~\mathrm{m}\)

Next, let's calculate the propagation constant, \(\beta\). We know that \(\beta = \frac{2\pi}{\lambda}\), so: \(\beta = \frac{2\pi}{2} = \pi~\mathrm{rad/m}\)

The given magnetic field is in phasor form. We have to convert it to time domain form, which is: \(\mathcal{H}(z, t) = \Re\{\mathbf{H}_{s}e^{j\omega t}\}\) This results in: \(\mathcal{H}(z, t) = \Re\{(4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\pi z}e^{j 300\pi\times10^6t} ~\mathrm{A/m}\}\) Now, we will find \(\mathcal{H}(20~\mathrm{cm}, 1.5~\mathrm{ns})\): \(\mathcal{H}(20~\mathrm{cm}, 1.5~\mathrm{ns}) = \Re\{(4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\pi (0.2)}e^{j 300\pi\times10^6(1.5\times 10^{-9})} ~\mathrm{A/m}\}\) After evaluating real part, we get, \(\mathcal{H}(20~\mathrm{cm}, 1.5~\mathrm{ns}) = -(8+20j)\mathbf{a}_{x} +(20+8j)\mathbf{a}_{y} ~\mathrm{A/m}\)

We will use Ampère's law: \(\nabla\times\mathbf{H} = \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t}\). For a plane wave, the relationship between \(\mathbf{H}\) and \(\mathbf{E}\) is: \(\mathbf{E} = \frac{1}{\eta}\mathbf{H}\times\mathbf{a}_{z}\), where \(\eta\) is the intrinsic impedance, and for free space is given by \(\eta = \sqrt{\frac{\mu_0}{\varepsilon_0}}\). So, first calculation of \(\eta\): \(\eta = \sqrt{\frac{4\pi \times 10^{-7}}{8.854 \times 10^{-12}}} = 376.7~\Omega\) Now, let's calculate the phasor electric field: \(\mathbf{E}_{s} = \frac{1}{\eta}(4+j10)(2\mathbf{a}_{x}+j\mathbf{a}_{y})e^{-j\beta z}\times(-j\mathbf{a}_{z})\) To find the maximum amplitude, we need the magnitude of the electric field phasor: \(|E|_{\max} = \frac{1}{\eta}|\mathbf{H}_{s}| = \frac{1}{376.7}\sqrt{(4)^2+(10)^2} = \frac{10}{376.7}~\mathrm{V/m}\)