Answer to Question #93452 in Electricity and Magnetism for MGM

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

For incident wave:

$$\textbf{E}_i=E_oe^{-j\beta_1 z}\hat\textbf{x},\\ \textbf{H}_i=\frac{E_o}{\eta_1}e^{-j\beta_1 z}\hat\textbf{y}.$$

Magnitude of the electric field:

$$E_0=\frac{B_0}{\sqrt{\mu_0\epsilon_0}}=3\cdot10^5\text{ V/m}.$$

The vector equations:

$$\textbf{E}_i=3\cdot10^5\cdot\text{sin}\Big(2\pi f t-\frac{2\pi f}{c}y\Big)(-\hat\textbf{x})=\\ =3\cdot10^5\cdot\text{sin}\Big(2\pi\cdot10^4 t-\frac{\pi}{15000}y\Big)(-\hat\textbf{x})\text{ V/m}.\\ \textbf{H}_i=\textbf{B}_i/\mu_0=\frac{2500}{\pi}\cdot\text{sin}\Big(2\pi\cdot10^4 t-\frac{\pi}{15000}y\Big)\hat\textbf{z}\text{ A/m}.$$

Below we will use the following parameters:

$$\eta_1=\sqrt{\frac{\mu_0}{\epsilon_0}}=120\pi\space\Omega,\\ \eta_2=\sqrt{\frac{4\mu_0}{9\epsilon_0}}=80\pi\space\Omega;\\ \Gamma=\frac{\eta_2-\eta_1}{\eta_2+\eta_1}=-0.2,\\ \tau=1+\Gamma=0.8.$$

Reflected wave with reflection coefficient $\Gamma$:

$$\textbf{E}_r=\Gamma E_oe^{+j\beta_1 z}\hat\textbf{x},\\ \textbf{H}_r=-\Gamma \frac{E_o}{\eta_1}e^{+j\beta_1 z}\hat\textbf{y}.$$

For reflected wave, since it propagates in the same medium, $\beta_\text{reflected}=-\beta_1.$ :

$$\textbf{E}_r=-6\cdot10^4\cdot\text{sin}\Big(2\pi\cdot10^4 t+\frac{\pi}{15000}y\Big)\hat\textbf{x}\text{ V/m}.$$

$$\textbf{H}_r=\frac{500}{\pi}\cdot\text{sin}\Big(2\pi\cdot10^4 t+\frac{\pi}{15000}y\Big)\hat\textbf{z}\text{ A/m}.$$

For transmitted wave with transmission coefficient $\tau$:

$$\textbf{E}_\tau=\tau E_oe^{-\beta_2 z}\hat\textbf{x},\\ \textbf{H}_\tau=\tau\frac{E_o}{\eta_2}e^{-j\beta_2 z}\hat\textbf{y}.$$ $$\beta_2=2\pi f\sqrt{\mu_2\cdot\epsilon_2}=\frac{\pi}{2500}.$$

$$\textbf{E}_\tau=24\cdot10^4\cdot\text{sin}\Big(2\pi\cdot10^4 t-\frac{\pi}{2500}y\Big)(-\hat\textbf{x})\text{ V/m}.$$

$$\textbf{H}_\tau=\frac{2000}{\pi}\cdot\text{sin}\Big(2\pi\cdot10^4 t-\frac{\pi}{2500}y\Big)\hat\textbf{z}\text{ A/m}.$$