Maxwell's equations in metals, skin effect

FondamentalDisplacement current in a conductor

Considering Ohm's local law, Maxwell-Ampere's equation is :

\(\overrightarrow {rot} \;\vec B = \mu _0 \left( {\sigma \vec E + \varepsilon _0 \frac{{\partial \vec E}}{{\partial t}}} \right)\)

Let \(T\) be the characteristic time of evolution of the distribution (D) (evolution period).

We can compare conduction current and displacement current :

\(\frac{{\left| {\sigma \vec E} \right|}}{{\left| {\varepsilon _0 \frac{{\partial \vec E}}{{\partial t}}} \right|}} \approx \frac{{\sigma E}}{{\varepsilon _0 \frac{E}{T}}} = \frac{{\sigma T}}{{\varepsilon _0 }}\)

Copper has a conductivity :

\(\sigma = 6.10^7 \;\Omega ^{ - 1} .m^{ - 1}\)

This ratio has an order of magnitude of \(10^{18}\;T\), where \(T\) is in seconds.

Thus, even if \(T\) is of the order of \(10^{-10}\;s\) (for a frequency of \(10\;GHz\)) :

\(\frac{{\left| {\sigma \vec E} \right|}}{{\left| {\varepsilon _0 \frac{{\partial \vec E}}{{\partial t}}} \right|}} \approx 10^8\)

Hence, non steady-states which legitimate the use of Ohm's law, the displacement current is negligible compared to the conduction current inside the conductor.

Maxwell-Ampere's equation becomes :

\(\overrightarrow {rot} \;\vec B = \mu _0 \vec j = \mu _0 \sigma \vec E\)

RappelElectric neutrality

Let us suppose at the time \(t=t_0\), a point M inside the conductor has a density of charges \(\rho (M,t_0)\).

How does this density of charges evolve over time ?

Gauss' law for electricity, Ohm's local law and the conservation of electric charge give :

\(div\;\vec E = \frac{\rho }{{\varepsilon _0 }}\;\;\;\;\;;\;\;\;\;\;\vec j = \sigma \vec E\;\;\;\;\;;\;\;\;\;\;div\;\vec j + \frac{{\partial \rho }}{{\partial t}} = 0\)

Which leads to :

\(div\;\frac{1}{\sigma }\vec j = - \frac{1}{\sigma }\frac{{\partial \rho }}{{\partial t}} = \frac{\rho }{{\varepsilon _0 }}\;\;\;\;\;\;\;so\;\;\;\;\;\;\;\frac{{\partial \rho }}{{\partial t}} + \frac{\sigma }{{\varepsilon _0 }}\;\rho = 0\)

By integration :

\(\rho (M,t) = \rho (M,t_0 )\;\exp \left( { - \frac{{t - t_0 }}{{\tau _d }}} \right)\;\;\;\;\;\;\;with\;\;\;\;\;\;\;\tau _d = \frac{{\varepsilon _0 }}{\sigma }\)

For copper, \(\tau_d \approx 4.10^{-14}\;s\) : very rapidly, the conductor becomes neutral :

\(\rho (M,t)=0\)

Hence, as in steady-state, charges accumulate at the direct proximity of the surface of the conductor.

Thus it is preferable to use the notion of surface charge \(\sigma\).

AttentionMaxwell's equations inside a conductor

Finally, in the steady-state approximation, the electromagnetic field verifies Maxwell's "simplified" equations :

\(\begin{array}{l}div\;\vec B = 0 \\div\;\vec E = 0 \\\overrightarrow {rot} \;\vec E = - \frac{{\partial \vec B}}{{\partial t}} \\\;\overrightarrow {rot} \;\vec B = \mu _0 \vec j = \mu _0 \sigma \;\vec E \\\end{array}\)

Hence, in a conductor, steady-state approximation only differs from real steady-state by the consideration of induction phenomena (Maxwell - Faraday equation).

ComplémentKirchhoff's law under steady-state approximation

Since \(\rho = 0\), the electric charge conservation equation inside the conductor leads to :

\(div\vec j = 0\)

The flux of the density of current vector is conserved, leading to the validation of Kirchhoff's law under the steady-state hypothesis.

Remark :

Do not confuse \(\rho\) and \(\rho_{mobile}\) : the conductor stays globally neutral so inside the conductor, \(\rho=0\).

However, the charge carriers have a charge repartition \(\rho_{mobile}\). They contribute to the current density vector according to the equation :

\(\vec j = \rho_{mobile}\vec v\)

FondamentalSkin effect

The electric field propagation equation can be written using this equation :

\(\overrightarrow {rot} (\overrightarrow {rot} \vec E) = \overrightarrow {grad} (div\vec E) - \Delta \vec E\)

So, with the electric fields :

\(\overrightarrow {rot} (\overrightarrow {rot} \vec E) = - \mu _0 \sigma \frac{{\partial \vec E}}{{\partial t}} = - \Delta \vec E\;\;\;\;\;\;\;so\;\;\;\;\;\;\;\Delta \vec E\; - \mu _0 \sigma \frac{{\partial \vec E}}{{\partial t}} = \vec 0\)

It is an equation of the “diffusion” kind.

It is usually obtained in thermal conductive transfers.

Complex solutions of the kind can :

\(\vec E = \vec E_0 f(z)e^{j\omega t}\)

Thus :

\(f''(z) - j\mu _0 \sigma \omega f(z) = 0\)

We write :

\(k = \pm (1 + j)\sqrt {\frac{{\mu _0 \sigma \omega }}{2}}\)

Hence, by disqualifying the divergent solution in the metal (we suppose the \(Oz\) axis is oriented towards the inside of the metal from \(0\) towards the infinity), and by writing \(\delta\) the skin depth :

\(\delta = \sqrt {\frac{2}{{\mu _0 \sigma \omega }}}\)

We obtain :

\(\vec E = \vec E_0 e^{ - z/\delta } e^{j(\omega t - z/\delta )}\)

The field propagates in the metal but it is attenuated of a factor \(\delta\), called skin depth :

\(\delta\) corresponds to the order of magnitude of the length of penetration of the wave inside the metal.

This depth will be as small as the conductivity of the metal and the frequency of the wave are high.

Here, the wave is absorbed because of the Joule effect inside the conductor, on a depth of the order of a few \(\delta\).

If we consider an electromagnetic wave of an order of frequency of GHz, \(\delta\) is of the order of a few \(\mu m\).

This skin effect provokes the decrease of the density of current as one moves away from the periphery of the conductor.

It leads to an increase in the resistance of the conductor.