# Magnetic field in perfect conductor : how to decide the “initial value” as $B=cste$

In a superconductor, we have no magnetic field because of Meissner effect.

In a perfect conductor, as $\vec{E}=\vec{0}$, we have $\frac{\partial \vec{B}}{\partial t}=\vec{0}$.

Thus the magnetic field is constant.

An experiment used to distinguish a superconductor with a perfect conductor is the following :

We start at $T>T_c$ ($T_c$ is the critical temperature for the superconductor), we apply a non $0$ magnetic field $B_0$. This field goes inside the material.

We lower the temperature to go below $T_c$.

The superconductor will expell the magnetic field, but the perfect conductor will still have the magnetic field $B_0$ inside as $\frac{\partial \vec{B}}{\partial t}=\vec{0}$.

What I don't understand : How to decide which magnetic field is the "initial one" in the perfect conductor. I mean, we could say "well, we started the magnetic field at a given time $t_0$, so before the perfect conductor had'nt any magnetic field inside". Thus the magnetic field inside it will always be $0$.

I don't understand how we decide what is the initial magnetic field inside the perfect conductor.

For simplicity, let us assume a conductor with uniform conductivity and light propagation speed. From charge conservation, we get $$\frac{\partial\rho }{\partial t} = - \nabla\cdot\mathbf{J}.$$ Since $\mathbf{J}=\sigma\mathbf{E}$, and $\nabla\cdot\mathbf E=4\pi\rho$, this leads to $$\frac{\partial \rho}{\partial t}=-4\pi\sigma\rho,$$ so that the charge density relaxes at timescales inversely proportional to the conductivity (fast). As I've stated, but not yet proved, the magnetic field relaxes at much slower timescales, so it is safe to assume that the charge density is always equal to zero.
From Faraday's law $$\nabla \times \mathbf E =- \frac{\partial B}{\partial t}$$ We see that the slowly varying magnetic field induces a slowly varying solenoidal electric field, so that $$\left|\frac{\partial \mathbf E}{\partial t}\right|\ll \sigma \mathbf E$$. Therefore, Ampere's law gives us $$\nabla \times \mathbf B = \frac{1}{c}\left[4\pi\mathbf J+\frac{\partial \mathbf E}{\partial t}\right] = \frac{1}{c}\left[4\pi \sigma \mathbf E+\frac{\partial \mathbf E}{\partial t}\right] \approx \frac{4\pi \sigma }{c}\mathbf E,$$
or $$\nabla\times\nabla\times\mathbf{B}= -\frac{4\pi \sigma }{c} \frac{\partial \mathbf B}{\partial t}.$$ From the vector identity $$\nabla\times\nabla\times\mathbf{B} = \nabla(\nabla \cdot \mathbf{B}) - \nabla^2 \mathbf B$$ and Gauss's law for the magnetic field $\nabla \cdot \mathbf B = 0$, we get $$\nabla^2 \mathbf B = \frac{4\pi \sigma }{c} \frac{\partial \mathbf B}{\partial t},$$ which is a diffusion equation for the magnetic field. In an initial configuration with spatial variation scale $L$ the characteristic relaxation time is $$\tau_D \sim \frac{4\pi \sigma }{c L^2}.$$ In the limit of a perfect conductor, $\tau_D$ would go to infinity. In realistic cases, it varies vastly. I quote from Jackson's "Classical Electrodynamics":
For a copper sphere of radius 1cm, the decay time of some initial $\mathbf B$ field inside is of the order of 5-10 miliseconds; for the molten iron core of the earth it is of the order of $10^5$ years.