# Perfect conductor in a time-dependent magnetic field

I have read about how a perfect conductor responses to a time-dependent magnetic field. It was stated that the "equation of motion" of such a perfect conductor is given by $$\frac{dj(r, t)}{dt}=ne^2E(r,t),$$ as well as this equation follows from $$j(r,t)=nev(r,t)$$, where $$v$$ is the drift velocity. I don't understand why the first equation follows from the second one. Can somebody explain this to me?

• It doesn't follow from the second one. The first is derived from the drude model of conductivity. It is the steady state solution Commented Apr 26, 2022 at 10:22
• That's kind of confusing me. I thought, the Drude model results in a linear relationship between current density and the electric field? And here, we have a linear relationship between the change in time of the current density and the electric field? Does this mean E has an exponentiell form here? Commented Apr 26, 2022 at 18:22
• Sorry, misread the equation. I have made an answer that derives this expression Commented Apr 26, 2022 at 18:41

This is one of the London equations.

This comes from modeling the electrons as free electrons.

As such the motion of a single electron is

$$\vec{F} = e\vec{E}$$

$$m\frac{d \vec{v}}{dt} = e\vec{E}$$

$$m\frac{d ne\vec{v}}{dt} = ne^2\vec{E}$$

$$m\frac{d \vec{J}}{dt} = ne^2\vec{E}$$

$$\frac{d\vec{J}}{dt} = \frac{ne^2}{m}\vec{E}$$

Behaviour of magnetic fields in perfect conductors

$$\frac{d \nabla × \vec{J}}{dt} = \frac{ne^2}{m}\nabla × \vec{E}$$

$$\frac{d \nabla × \vec{J}}{dt} = - \frac{ne^2}{m} \frac{\partial \vec{B}}{\partial t}$$

$$\nabla × \vec{J} = -\frac{ne^2}{m}\vec{B}$$

Finding $$\nabla × \vec{J}$$:

$$\nabla × (\nabla × \vec{B} = \mu_0 \vec{J})$$

$$\nabla (\nabla \cdot \vec{B}) - \nabla^2 \vec{B} = \mu_0 \nabla × \vec{J}$$

$$- \nabla^2 \vec{B} = \mu_0 \nabla × \vec{J}$$

Sub in:

$$\nabla^2 \vec{B} = \frac{\mu_0 n e^2}{m} \vec{B}$$

Whose solution is an exponential decay of the magnetic field. Which is why "there is no magnetic field inside a superconductor". This makes sense, as any changing external magnetic field induces an electric field, which generates a current to oppose the flux

There are a few assumptions in this derivation however.

• Thank you very much! I think, I get it now :3 Commented Apr 28, 2022 at 15:45
• Instead of modeling $\vec{J}$ as some FIXED value like most situations. Using the solution to F=ma to determine the velocity of a charge density, models the current density in relation to the electric field. If n Is fixed, at acts as though at EVERY point in space, the velocity of charge follows f=ma. , substituting this into the definition of current density, turns it in into a velocity FIELD, instead of a velocity of a single particle at a particular moment in time Commented Apr 28, 2022 at 15:50
• As $\vec{J}(x,y,z,t) = \rho \vec{V}(x,y,z,t)$ . The velocity we're substituting in is $\vec{V}(t)$ so when substituting, it acts as if that velocity is at all points in space ( a velocity FIELD independant in space) Commented Apr 28, 2022 at 15:51