Reflection of an electomagnetic wave on a non perfect metal
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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 \(\gamma\) occupies the half-space \(x>0\), the space \(x<0\) being vacuum.
An incident wave of the form :
\(\underline {\vec E}_i=E_0exp(j\omega (t-x/c))\;\vec u_z\)
propagates in a vacuum.
It gives birth to a transmitted wave of the form (see the lecture notes on the skin effect) :
\(\underline {\vec E}_{t}=\underline t\; E_0exp(j(\omega t-\underline k x))\;\vec u_z\)
With :
\(\underline k=\frac{1-j}{\delta}\;\;\;\;\;and\;\;\;\;\;\delta=\sqrt{\frac{2}{\mu_0\gamma\omega}}\)
And a reflected wave of the form :
\(\underline {\vec E}_r=\underline r \;E_0exp(j\omega (t+x/c))\;\vec u_z\)
Question
Determine the corresponding magnetic fields.
Solution
The structure of a progressive harmonic plane wave gives to the incident wave and the reflected wave, the following expressions of magnetic fields :
\(\underline {\vec B}_i=\frac{\vec u_x \wedge \underline {\vec E}_i}{c}=-\frac{E_0}{c}e^{j\omega(t-x/c)}\;\vec u_y\)
And :
\(\underline {\vec B}_r=\frac{-\vec u_x \wedge \underline {\vec E}_r}{c}=\underline r \frac{E_0}{c}e^{j\omega(t+x/c)}\;\vec u_y\)
For the transmitted wave, the structure of relationship is written as :
\(\underline {\vec B}_t=\frac{\underline k \vec u_x \wedge \underline {\vec E}_t}{\omega}=-\underline k \;\underline t \frac{E_0}{ \omega}e^{j(\omega t - \underline k x)}\;\vec u_y\)
Question
Assume there are no surface currents.
By writing the boundary condition of the electrical and magnetic fields, to establish the expression of \(\underline t\) as a function of \(\alpha = \omega\delta/c\).
Ensure that for \(\alpha<<1\) (what is assumed hereafter), we have :
\(\underline t=\alpha(1+j)\)
limiting the calculations to the order \(1\) in \(\alpha\).
Solution
The boundary condition of the electrical and magnetic fields in \(x=0\) leads to the following two equations :
\(1+\underline r = \underline t\)
\(-\frac{1}{c}+\frac{\underline r}{c}=-\frac{\underline k \underline t}{\omega}\;\;\;\;\) so \(\;\;\;\;1-\underline r = \frac{(1-j)\underline t}{\alpha}\)
These two equations allow to deduce the complex transmission coefficient :
\(\underline t = \frac {2\alpha}{1+\alpha -j}=\frac{2\alpha(1+\alpha+j)}{(1+\alpha)^2+1)}\)
The first order in \(\alpha <<1\) :
\(\underline t \approx \frac{2\alpha(1+j)}{2}\)
Hence :
\(\underline t \approx \alpha(1+j)\)
Question
In fact, the conductor has a surface \(S\) in the plane \(x=0\).
Calculate the time average flux of the Poynting vector in \(x=0^+\).
What does this magnitude represent ?
Show that the time average of the power dissipated by Joule effect in a volume \(Sdx\) of conductive element is :
\(\left\langle {dP_J } \right\rangle = \gamma \alpha ^2 E_0^2 \exp ( - 2x/\delta )\;Sdx\)
Deduce the temporal average power dissipated by Joule effect throughout the conductor.
Compare with the results of the previous question.
Indice
One can calculate the average value of the Poynting vector with the expressions of the complex fields using the relationship :
\({\left\langle {\vec \Pi } \right\rangle _t} = \frac{1}{2}{\mathop{\rm Re}\nolimits} \left( {\frac{{{{\underline E }_t} \wedge \underline B _t^{\;*}}}{{{\mu _0}}}} \right)\)
Use local Ohm's law (\(\vec j = \gamma \vec E\))
Solution
The electric field is transmitted :
\(\underline {\vec E}_{t}=\underline t\; E_0exp(j(\omega t-\underline k x))\;\vec u_z\)
Using complex expressions of \(\underline t\) and \(\underline k\), we get :
\(\underline {\vec E}_t=\alpha(1+j)E_0 e^{-x/\delta} e^{j(\omega t -x/\delta)}\;\vec u_z\)
Thus, for \(x=0^+\) :
\(\underline {\vec E}_t(0^+,t)=\alpha(1+j)E_0 e^{j\omega t}\;\vec u_z\)
Similarly, the transmitted magnetic field is, at \(x=0^+\) :
\(\underline {\vec B}_t(0^+,t)=\frac {2\alpha E_0}{\delta \omega} e^{j\omega t}\vec u_y\)
The mean value of the transmitted Poynting vector, at \(x=0^+\), is :
\({\left\langle {\vec \Pi } \right\rangle _t} =\frac {1}{2} Re (\frac{{{{\underline {\vec E} }_t} \wedge {{\underline {\vec B}^* }_t}}}{{{\mu _0}}})\)
Hence :
\({\left\langle {\vec \Pi } \right\rangle _t} = \frac {\alpha E_0^2}{\mu_0 c} \;\vec u_x\)
The flux of this vector through the conductor surface \(S\) is the electromagnetic power \(P\) transmitted to the metal :
\(P={\left\langle { \Pi } \right\rangle _t}S = \frac {\alpha E_0^2}{\mu_0 c}S\)
The power density dissipated by Joule effect is :
\(p_v=\vec j .\vec E=\gamma \vec E\;^2\)
The average value of the power density dissipated by Joule effect within the metal (the depth \(x\)) is :
\(\left\langle {p_v(x)} \right\rangle = \frac{1}{2}\gamma {\left| {\underline E(x,t) } \right|^2}\)
So :
\(\left\langle {p_v(x)} \right\rangle = \gamma \alpha^2 E_0^2 e^{-2x/\delta}\)
And the time average of the power dissipated by Joule effect in a volume \(Sdx\) of conductive element is :
\(\left\langle {dP_J } \right\rangle =\left\langle {p_v(x)} \right\rangle \;Sdx= \gamma \alpha ^2 E_0^2 \exp ( - 2x/\delta )\;Sdx\)
The total average power dissipated in the conductor will be :
\({P_J} = \int_0^\infty {\gamma {\alpha ^2}E_0^2{e^{ - 2x/\delta }}Sdx} = \frac{\gamma \alpha^2 E_0^2 S \delta}{2}=\frac{{{\alpha ^2}E_0^2S}}{{{\mu _0}\omega \delta }}\)
And, finally :
\(P_J= \frac {\alpha E_0^2}{\mu_0 c}S\)
We find the same expression as before :
\(P=P_J\)
In time average, the electromagnetic power transmitted from the wave to the metal is completely dissipated by Joule effect.