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Why are electromagnetic fields zero inside perfect conductors? I do not understand why we can make these assumptions in electrodynamics.


SOLUTION:

Firstly, we are going to analize the electric field.

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In the first conductor we can appreciate that the total charge distribution is electrically compensated ($\rho=0$). In consequence, the electric field will be completely null ($∇⋅\vec{D}=\rho=0$).

Later on we apply an external electric field that will distribute the charges inside the conductor almost instantaneously ($\sigma=\infty$). This redistribution will cause another electric field that will cancel the previously applied field.

We can conclude that the elctric field is zero unless we vary the magnetic field ($∇×\vec{E}=-\frac{∂\vec{B}}{∂t}$). The magnetic fields are defined by the following equations:

$∇⋅\vec{B}=0$

$∇×\vec{H}=\vec{J}+\frac{∂\vec{D}}{∂t}=\vec{J}$ext$+\sigma \vec{E}+\frac{∂\vec{D}}{∂t}$

We know that $\vec{D}=0$, so the induced current densities ($\vec{J}$ind$=\sigma\vec{E}$) will be zero. In addition, we will assume that the external current densities ($\vec{J}$ext) applied in the conductor are also zero. Finally we know that $\frac{∂\vec{D}}{∂t}=0$. So we can rewrite the Maxwell equations into the following ones:

$∇⋅\vec{B}=0$

$∇×\vec{H}=0$

Thanks to these equations we can conclude that the magnetic field is constant inside a perfect conductor. So the electric field will be allways zero ($∇×\vec{E}=-\frac{∂\vec{B}}{∂t}=0$).

As we have demonstrated before, we know that there can be a constant magnetic field within a perfect conductor. This field will be non-zero as long as it has been present since before the material transitioned to an infinite conductivity state. However adding some magnetic field would be an unnecessary complication. Therefore, the $\vec{B}=0$ condition is reasonable.

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  • $\begingroup$ What is the context of this statement? $\endgroup$ – J. Murray Sep 26 '19 at 13:49
  • $\begingroup$ You should add that context to the question body - it's extremely important. $\endgroup$ – J. Murray Sep 26 '19 at 14:20
  • $\begingroup$ The depth of the skin effect depends on sigma and it tends to zero when sigma tends to infinity (a perfect descructive interference of the incident and the excited in metal waves). $\endgroup$ – Vladimir Kalitvianski Sep 26 '19 at 14:39
  • $\begingroup$ Thanks Vladimir, that's the answer I've been looking for. $\endgroup$ – javierhersan Sep 26 '19 at 15:23
  • $\begingroup$ If the conductor is statically charged, the E-field inside is zero. That static situation is crucial to the conclusion. $\endgroup$ – Bill N Oct 7 '19 at 16:09
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In principle, a perfect conductor with conductivity $\sigma\rightarrow \infty$ can have a magnetic field in the interior, as long as the magnetic field has been present since before the material transitioned to an infinite conductivity state. Infinite conductivity implies perfect screening of any changes to the interior magnetic field (so the field is "frozen-in"), but does not forbid a steady field.

However, this question refers to the incidence of electromagnetic waves on a perfect conductor. This is a simplified model to describe light reflecting from the surface of a metal, and adding in some frozen-in magnetic field would be an unnecessary complication. Therefore, the $\vec B=0$ boundary condition is reasonable.

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    $\begingroup$ Also note that the expulsion of magnetic fields from superconductors upon reaching the super conductive state is one of the specific hallmarks of superconductivity. So what you describe here is exactly the difference between a perfect conductor and a superconductor. $\endgroup$ – Dale Sep 26 '19 at 16:55
  • $\begingroup$ Sorry, I do not undestand your answer, What did you mean when you said 'the field is frozen'? $\endgroup$ – javierhersan Sep 26 '19 at 20:34
  • $\begingroup$ @JavierHernándezSánchez if $\sigma \rightarrow \infty$ then $\frac{\partial }{\partial t}\mathbf B=0$ in the interior of the conductor. This does not mean that $\mathbf B=0$, only that it is fixed. $\endgroup$ – J. Murray Sep 26 '19 at 20:37
  • $\begingroup$ @J.Murray Thank you, your answer was really useful. Please, can you check the solution I have written in? $\endgroup$ – javierhersan Sep 29 '19 at 0:23

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