Maxwell's fourth equation states that div B = 0. Use this law to show that the
component of B perpendicular to the surface is continuous (i.e., the same just below
and just above the surface), regardless of the surface current Js(r, t).
The volume integral of div · B over a box equals, by Gauss' theorem the integral of B dot dO over the surface of the box. Here the surface element dO points in the direction of the outward normal. So, we have:
Maxwell's fourth equation thus implies:
Integral of B dot dO = 0
Take a box consisting of two sides of area A ...
A detailed solution is given.