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?
$$[vec H]=A/m$$$$[vec J_{SC}]=A/m^2$$

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In the interface between two materials the boundary condition for the $vec H$ field is:
$$vec a_ntimes[vec H_1-vec H_2]=vec J_s$$
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 $A/m$.