Chapter 12, Problem 12.1P

To determine

The percentage of the incident power density transmitted into the copper.

Answer to Problem 12.1P

The percentage of the incident power density transmitted into the copper is $=0.01\%$.

Explanation of Solution

Given:

A uniform plane wave in air $E_{x1}^{+}=E_{x10}^{+}\cos ({10}^{10}t-\beta z)\text{V/m}$ incident normally on copper surface $z=0$.

Concept Used:

First, we find out the intrinsic impedance for the given conductor (copper in our case).

For a good conductor like copper, the intrinsic impedance is,

   ${\eta }_c=\sqrt{\frac{j\omega \mu }{\sigma }}$

Then, we calculate the reflection coefficient by using the below formula,

   $\Gamma =\frac{{\eta }_c-{\eta }_0}{{\eta }_c+{\eta }_0}$

Calculation:

First, we calculate the intrinsic impedance for copper.

For a good conductor like copper, the intrinsic impedance is

   ${\eta }_c=\sqrt{\frac{j\omega \mu }{\sigma }}$

   $\sigma$ is electric permeability

   $\mu$ is magnetic permeability

   ${\eta }_c=\sqrt{\frac{j\omega \mu }{\sigma }}$ can also be represented as $(1+j)\sqrt{\frac{\omega \mu }{2\sigma }}$

For copper $\sigma =5.8\times 10^7$ and

And $\mu =(4\pi \times {10}^{-7})$

   $\begin{array}{l}=(1+j)\sqrt{\frac{{10}^{10}(4\pi \times {10}^{-7})}{2(5.8\times 10^7)}}\\ =(1+j)(0.0104)\end{array}$

Using ${\eta }_0=376.7288\Omega$ (free space wave impedance)

The reflection coefficient can be calculated as below:

   $\Gamma =\frac{{\eta }_c-{\eta }_0}{{\eta }_c+{\eta }_0}=\frac{0.0104-376.7288+j0.0104}{0.0104+376.7288+j0.0104}=-0.9999+j0.0001$

   ${\left|\Gamma \right|}^2=0.9999$ and so, the transmitted power fraction is $1-{\left|\Gamma \right|}^2=1-0.9999=0.0001=0.01\%$