Question

A uniform plane wave of 10 kHz travelling in free space strikes a large block of a
material having ε = 9ε0, μ = 4μ0 and σ = 0 normal to the surface. If the incident
magnetic field vector is given by
B =10^− 3sin (ωt −βy) kˆ
tesla
Write the complete expressions for the incident, reflected, and transmitted field vectors.

Answer 1

1) The incident wave is written in the form of

\textbf{B}_i =10^{− 3}\text{sin} (ωt −βy) \hat\textbf{k}\space\text{ T}.

It means that the electric field has magnitude

$$E=\frac{B}{\sqrt{\mu_0\epsilon_0}}=300\cdot10^3\space\frac{\text{V}}{\text{m}},\\ \space\\ H=\frac{B}{\mu_0}=\frac{2500}{\pi}\space\frac{\text{A}}{\text{m}}.$$ $$\beta=2\pi f\sqrt{\mu_0\epsilon_0}=\frac{2\pi f}{c}=\frac{\pi}{15000}\text{ m}^{-1},\\ \space\\ \omega=2\pi f=2\pi\cdot10^4\text{ s}^{-1}.$$

Hence the electric field vector is

\textbf{E}_i =300\cdot10^3\cdot\text{sin}\Big(20\pi\cdot10^3 t-\frac{\pi}{15000}y\Big)(-\hat\textbf{i})\space\frac{\text{V}}{\text{m}},\\ \space\\ \textbf{H}_i=\frac{2500}{\pi}\cdot\text{sin}\Big(20\pi\cdot10^3 t-\frac{\pi}{15000}y\Big)\hat\textbf{k}\space\frac{\text{V}}{\text{m}}.

2) Prior to calculate the reflection coefficient for the reflected wave, calculate the wave impedances:

$$\eta_0=\sqrt{\frac{\mu_0}{\epsilon_0}}=376.73\space\Omega,\\ \eta_2=\sqrt{\frac{4\mu_0}{9\epsilon_0}}=251.15\space\Omega,\\ \space\\ r=\frac{\eta_2-\eta_1}{\eta_2+\eta_1}=-0.2.$$

The magnitudes of E- and H-components will be

$$E_r=rE,\\ H_r=rH.$$

Keep in mind that since the wave is reflected, the wave number beta here has the same magnitude but opposite sign. This gives us:

\textbf{E}_r=-60\cdot10^3\cdot\text{sin}\Big(20\pi\cdot10^3 t+\frac{\pi}{15000}y\Big)\hat\textbf{i}\space\frac{\text{V}}{\text{m}},\\ \space\\ \textbf{H}_r=\frac{500}{\pi}\cdot\text{sin}\Big(20\pi\cdot10^3 t+\frac{\pi}{15000}y\Big)\hat\textbf{k}\space\frac{\text{A}}{\text{m}}.

3) To obtain equations for the transmitted wave, calculate the transmission coefficient:

$$t=\frac{2\eta_2}{\eta_1+\eta_2}=1+r=0.8.$$

The magnitudes are:

$$E_t=tE,\\ H_t=tH.$$

The equations, thus, are:

\textbf{E}_t=240\cdot10^3\cdot\text{sin}\Big(20\pi\cdot10^3 t-\frac{\pi}{2500}y\Big)(-\hat\textbf{i})\space\frac{\text{V}}{\text{m}},\\ \space\\ \textbf{H}_t=\frac{2000}{\pi}\cdot\text{sin}\Big(20\pi\cdot10^3 t-\frac{\pi}{2500}y\Big)\hat\textbf{k}\space\frac{\text{A}}{\text{m}}.