Context: It is commonly stated (such as in Tinkham's "Introduction to Superconductivity") that the current in a superconducting wire is concentrated in a layer around the surface of thickness of the order of the penetration depth $\lambda$. I'm trying to prove this statement.

I started with the assumption that all the current is flowing parallel to the wire and the system has translation and rotational symmetry along $\hat{\mathbf{z}}$, \begin{align} \mathbf{j} = j(r) \hat{\mathbf{z}}, \end{align} so that the magnetic field generated by the current is along $\hat{\mathbf{\phi}}$ and depends only on $r$: \begin{align} \mathbf{B} = B^{\phi}(r)\hat{\mathbf{\phi}}. \end{align}

I then tried to find the field generated by this current by solving the Helmholtz equation \begin{align} \mathbf{\nabla}^2 \mathbf{B} = \frac{\mathbf{B}}{\lambda^2}, \end{align} which in this case reduces to solving the ODE \begin{align} r^2 \frac{\partial^2 B^{\phi}}{\partial r^2} + r \frac{\partial B^{\phi}}{\partial r} + \left(\frac{r^2}{\lambda^2} - 1\right)B^{\phi} = 0. \end{align}

This equation, in turn, has as solution the Bessel functions $J_1$ and $Y_1$ with imaginary arguments: \begin{align} B^{\phi}(r) = c_1 J_1\left(\frac{ir}{\lambda}\right) + c_2 Y_1\left(- \frac{ir}{\lambda}\right). \end{align}

One of the two variables can be eliminated by using the boundary conditions on the surface, $B^{\phi}(R) = B_{surface}$. This leaves still a constant to be determined. This is where I'm puzzled.


As far as I'm aware, the idea here is to say the field must be zero at the center of the superconductor, which then sets $c_2 = 0$. The problem with this approach, from my point of view, is that we don't know that the field is zero at the center yet, that is part of what we are trying to prove! On the other hand, if we consider any other value for $c_2$ we find that $B^{\phi}$ diverges at the origin.

  • One could argue that this is unphysical and so we should discard. But the electric field diverges at point charges, for example, so I don't think this is a very good argument. Moreover, if we integrate the flux around the origin up to a finite radius we find that the total flux is finite.
  • One could also argue that it would lead to a infinite current density at the center of the wire. But also in this case, the total current inside any finite region will still be finite: $I_{tot}(r) = 2\pi r B^{\phi}(r)$ is finite for $r>0$.


  1. Is there a more solid argument of why we should throw away the second part of the solution?

  2. Also, are there experiments that can be made and maybe have been done to determine the field and current densitiy as a function of the position inside superconducting wires?


1 Answer 1


"the idea here is to say the field must be zero at the center of the superconductor"

This could be true and thought of like this:

At the center, the field would have no reason to point in any particular direction, so by symmetry, should be zero.

For a cylindrical line of charge at the center, the field is $$E = \frac{kQ}{r}$$ where $k$ is a constant - but the charge within that radius is proportional to ${r^2}$, so the overall effect is that the field from within the cylinder is proportional to ${r}$ and approaches zero at the center.

The above is for a uniform charge density. This charge density may be caused like this:

The other moving charges cause a magnetic field, that in turn causes a force on a moving charge in the wire, directed towards the center.

The moving charges tend to be pushed towards the center, but this force is balanced by electrostatic repulsion.

  • $\begingroup$ You are completely right that by symmetry the field cannot point anywhere in the center, but I would add that this is only valid for the radial and angular components, as the current could break parity on the $z$ axis (but this doesn't matter unless we consider currents in the angular direction). Moreover, the unit vector $\hat{\phi}$ is not even defined for $r=0$, so it's a really bad oversight of mine. $\endgroup$ Commented Jul 5, 2021 at 21:42
  • $\begingroup$ As for the second part of your answer, I'm not sure I follow. Why is the charge proportional to $r^2$? What is your $r$ - the radius of the cylinder or the position coordinate in the radial direction? $\endgroup$ Commented Jul 5, 2021 at 21:43
  • $\begingroup$ If you can clarify this I will accept your answer. $\endgroup$ Commented Jul 5, 2021 at 21:43
  • $\begingroup$ It's just an idea, it was modelled as uniform charge density, $r$ is the position coordinate in the radial direction. Another idea is that other moving charges create a magnetic field (right hand grip rule), and then this field moves the moving charges sideways (seems to be towards the center of wire) until electrostatic repulsion balances this magnetic field... $\endgroup$ Commented Jul 6, 2021 at 7:57
  • $\begingroup$ Thanks for the clarification, I see what you mean now. I'm accepting your answer, but could you include in your answer this idea of the balancing of forces? I think it provides a clearer view of why the fields must be zero at the center and could be useful for future readers. $\endgroup$ Commented Jul 7, 2021 at 2:18

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