# About Supercurrent Direction in London Theory of Supercondutivity

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.

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.