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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 $A/m$.

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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): $$\vec B_{SC}=(0,0,B_0 e^{-y/\lambda_L})$$ $$\vec J_{SC}=(-\frac{B_0}{\lambda_L \mu_0} e^{-y/\lambda_L},0,0)$$

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