Electromagnetic waves
Reflection of an electomagnetic wave on a non perfect metal
Take 20 minutes to prepare this exercise.
Then, if you lack ideas to begin, look at the given clue and start searching for the solution.
A detailed solution is then proposed to you.
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Ohmic conductor of conductivity occupies the halfspace , the space being vacuum.
An incident wave of the form :
propagates in a vacuum.
It gives birth to a transmitted wave of the form (see the lecture notes on the skin effect) :
With :
And a reflected wave of the form :
Question
Determine the corresponding magnetic fields.
Hint
Solution
The structure of a progressive harmonic plane wave gives to the incident wave and the reflected wave, the following expressions of magnetic fields :
And :
For the transmitted wave, the structure of relationship is written as :
Question
Assume there are no surface currents.
By writing the boundary condition of the electrical and magnetic fields, to establish the expression of as a function of .
Ensure that for (what is assumed hereafter), we have :
limiting the calculations to the order in .
Hint
Solution
The boundary condition of the electrical and magnetic fields in leads to the following two equations :
so
These two equations allow to deduce the complex transmission coefficient :
The first order in :
Hence :
Question
In fact, the conductor has a surface in the plane .

Calculate the time average flux of the Poynting vector in .
What does this magnitude represent ?

Show that the time average of the power dissipated by Joule effect in a volume of conductive element is :
Deduce the temporal average power dissipated by Joule effect throughout the conductor.
Compare with the results of the previous question.
Hint

One can calculate the average value of the Poynting vector with the expressions of the complex fields using the relationship :
Use local Ohm's law ( )
Solution

The electric field is transmitted :
Using complex expressions of and , we get :
Thus, for :
Similarly, the transmitted magnetic field is, at :
The mean value of the transmitted Poynting vector, at , is :
Hence :
The flux of this vector through the conductor surface is the electromagnetic power transmitted to the metal :

The power density dissipated by Joule effect is :
The average value of the power density dissipated by Joule effect within the metal (the depth ) is :
So :
And the time average of the power dissipated by Joule effect in a volume of conductive element is :
The total average power dissipated in the conductor will be :
And, finally :
We find the same expression as before :
In time average, the electromagnetic power transmitted from the wave to the metal is completely dissipated by Joule effect.