About Supercurrent Direction in London Theory of Supercondutivity

Question

I'm studying superconductivity starting from London's approach. After you obtain the two London's eqs you get for the static case: $\quad \nabla^2 \textbf{B}=\frac{1}{\lambda_L^2}\textbf{B}$

If there is a superconductor from x = 0 to $+\infty$, and a magnetic field $\textbf{B}_{app}$ in the y-dir,you find that the magnetic field inside sc is $\textbf{B}=B_{app}e^{-x/\lambda_L}\textbf{y}$.

Everything is well up to here. $\textbf{B}$ decays exponentially up to the characteristic length $\lambda_L$. Now we recall the 4th Maxwell eq. (and assume that $\frac{d\textbf{E}}{dt}=0$) $\nabla \times \textbf{B} = \frac{4\pi}{c}\textbf{j }$, and solve for $\textbf{j}$.

It gives $\textbf{j}(x)=-\frac{c}{4\pi \lambda_L}B e^{-x/\lambda_L}\textbf{z}$.

Ok, the current also decays as the magnetic field, and it is in the z-dir. My question is: If the superconductor is a surface in the xy-plane, how can the current flow to the z-dir? In which media does it flow?

I must be missing some idea. I will be grateful for any answer. Thank you.

Answer 1

You defined the superconductor to be in the half-space $x=0\dots\infty$, so the surface of your superconductor is parallel to $yz-plane$, so no problems there - current flows parallel to surface.

You can treat more complicated geometries of course (with superconductor surfaces in both xy and yz planes), but then you need to change your initial conditions, which will also change the magnetic field