# Tangential boundary condition for magnetic field in London equation

Im solving the London equations for an infinte superconductor in the y>0 region while applying a magnetic field $$\vec B$$ in the z axis. The solution is well known, the field in the superconductor decays as an exponential characterized by the London penetration lenght $$\lambda_L$$, however Im trying to do it properly by solving the London and Maxwell equations. I found that the magnetic fields and the superconducting current are: $$\vec B_{ext}=(0,0,B_0)$$ $$\vec B_{SC}=(\alpha e^{-y/\lambda_L},0,\beta e^{-y/\lambda_L})$$ $$\vec J_{SC}=(-\frac{\beta}{\lambda_L \mu_0} e^{-y/\lambda_L},0,\frac{\alpha}{\lambda_L \mu_0} e^{-y/\lambda_L})$$ However im stuck because I need to apply the $$\vec H$$ boundary condition to the magnetic field in the SC but I found that the units of the $$J_{SC}$$ are $$A/m^2$$, while $$\vec H$$ are A/m. How do I apply the boundary conditions here? Is something wrong? $$[\vec H]=A/m$$$$[\vec J_{SC}]=A/m^2$$

In the interface between two materials the boundary condition for the $$\vec H$$ field is: $$\vec a_n\times[\vec H_1-\vec H_2]=\vec J_s$$ However doing a unit analysis I found that $$[\vec H]=A/m$$, while for $$[\vec J]=A/m^2$$, so I suposse that $$J_s$$ must have units of $$A/m$$.

Solved, the $$J_sc$$ does not create any problem at the boundary because is not a surface current (there it is were I was wrong) so making:
$$\vec a_n\times[\vec H_1-\vec H_2]=0$$
we obtain $$\alpha=0$$ and $$\beta=B_0$$ and we arrive at the solution (not explained in the books): $$\vec B_{SC}=(0,0,B_0 e^{-y/\lambda_L})$$ $$\vec J_{SC}=(-\frac{B_0}{\lambda_L \mu_0} e^{-y/\lambda_L},0,0)$$