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Let us suppose to have two coaxial cylindrical tubes, namely two cylindrical surfaces $\Sigma_1, \Sigma_2$, with a radius of $a$ and $b$ respectively. A current $I$ flows across the surfaces but in the opposite direction. Let us suppose that the current flows in the positive $\hat{z}$ direction of a cylindrical coordinate system for the inner tube and therefore in the negative vertical direction for the outer tube. Now, if the space between the tubes is filled with linear insulating material(with a certain $\chi_m$) I can find $H$ by choosing Amperian loop of radius $s$ between the tubes since I expect $\vec{H}$ to be circumferential. The way in which I go around the loop is given me by the fingers of my right hand after I make the thumb align with the current, in this case the azimuthal direction of our coordinate system, so there’s no ambiguity whatsoever. So, without going further into details, I get that $\vec{H}$ is circumferential and so is the magnetic field $\mathbf{B}$ and the magnetization $\mathbf{M}=\chi_m\mathbf{H}$. If I now want to compute the surface bound current however, I find myself in a quite a troublesome situation, since the formula is simply: $$ \mathbf{K}_b=\mathbf{M} \times \mathbf{\hat{n}}$$ But how am I supposed to know the right expression for the normal unit vector to the surface. I thought that in the case of a cylinder, the normal unit vector for its lateral surface coincides with the radial unit vector of the cylindrical coordinate system, which points radially outwards obviously. But I’ve seen somebody choose the $-\mathbf{\hat{r}}$ for the inner tube, whereas $+\mathbf{\hat{r}}$ for the outer tube. The problem is that this sign ambiguity is actually really problematic, for it causes one to pick the wrong direction for the surface bound current. In our case, for instance, by choosing the outwards radial unit vector for the inner tube, I would get $\mathbf{K}_b=-K_b\mathbf{\hat{z}}$. If someone could help clarify this situation, it would be much appreciated

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When you have a magnetic medium that carries a magnetization, a bound current may reside on its surface, which can be calculated by using the formula $K = M \times \hat{n}$ where $\hat{n}$ points out of the material, and $M$ is the magnetization inside the material on the opposite side of $\hat{n}$. In your scenario, the magnetized material occupies the space between $r = a$ and $r = b$, so at the inner surface ($r = a$) you should take $\hat{n}$ to point inward, and at the outer surface you should take $\hat{n}$ to point outward.

If two magnetic media meet, then each one will contribute to the surface bound current separately, and the total surface bound current will be the sum $M_1 \times \hat{n}_1 + M_2 \times \hat{n}_2$, where $\hat{n}_1$ points away from the material that has magnetization $M_1$, and $\hat{n}_2$ points away from the material that has magnetization $M_2$.

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