Magnetic boundary conditions

At the interface between two media with different permeabilities, the magnetic flux density $\mathbf{B}$ and field intensity $\mathbf{H}$ satisfy specific jump conditions, following from Maxwell’s equations applied to thin pillboxes and loops straddling the boundary.

Normal $\mathbf{B}$ is continuous

$B_{1n}=B_{2n}.$

The normal component of $\mathbf{B}$ matches across the boundary, always — no exceptions. This follows from $\oint \mathbf{B}\cdot d\mathbf{s}=0$ (Gauss’s law for magnetism): apply to a pillbox spanning the interface, shrink the side height to zero, and the only nonzero contributions are the flux through the two flat faces. They must be equal.

The physical content: magnetic flux lines can’t terminate at an interface — no magnetic monopoles. Whatever goes in must come out.

Equivalently in $\mathbf{H}$ form, the normal component jumps in the inverse permeability ratio:

${\mu }_1{H}_{1n}={\mu }_2{H}_{2n}.$

Tangential $\mathbf{H}$ jumps by the surface current

${\hat{\mathbf{n}}}_2\times ({\mathbf{H}}_1-{\mathbf{H}}_2)={\mathbf{J}}_s,$

where ${\hat{\mathbf{n}}}_2$ is the unit normal pointing from medium 2 into medium 1, and ${\mathbf{J}}_s$ is the free surface current density (A/m) on the interface.

For the common case where $\mathbf{H}$, ${\mathbf{J}}_s$, and the boundary normal are mutually orthogonal, the scalar magnitudes obey

$H_{1t}-H_{2t}=J_s.$

(The vector form is the safe statement when the geometry is more general — $\mathbf{H}$ may have tangential components not aligned with ${\mathbf{J}}_s$.)

The vector form follows from Ampère’s law $\oint \mathbf{H}\cdot d\mathbf{l}=I_{\text{enc}}$: apply to a rectangular loop straddling the boundary, shrink the perpendicular sides to zero, and the line integrals along the two tangential sides differ by the enclosed surface current.

If there’s no surface current (${\mathbf{J}}_s=0$), then $H_{1t}=H_{2t}$ — tangential $\mathbf{H}$ is continuous.

When is surface current present?

Surface currents only exist on perfect conductors and superconductors. For media with finite conductivity, currents distribute through the bulk (finite volume current density $\mathbf{J}$), not on the surface. So for normal dielectrics, paramagnetic, and ferromagnetic materials, set ${\mathbf{J}}_s=0$ and take

$H_{1t}=H_{2t}.$

Refraction of magnetic field lines

When the boundary is current-free and both sides are linear magnetic media, the field “refracts” at the boundary. With ${\theta }_1,{\theta }_2$ angles from the normal:

$\frac{\tan {\theta }_1}{\tan {\theta }_2}=\frac{{\mu }_1}{{\mu }_2}.$

The proof is the same as the dielectric case: tangential $\mathbf{H}$ continuous, normal $\mathbf{B}$ continuous, take the ratio of $\tan \theta =H_t/H_n$ on each side.

For a strongly magnetic material (${\mu }_2\gg {\mu }_1$), field lines outside bend toward the normal — they “want to leave” the high-$\mu$ material at nearly perpendicular angle. Equivalently, inside the high-$\mu$ material the field lines lay almost parallel to the surface. This is the basis of magnetic shielding: a high-$\mu$ shell draws field lines into itself, leaving the interior nearly field-free.

Worked example

A boundary between ${\mu }_{r1}=2$ and ${\mu }_{r2}=8$, normal ${\hat{\mathbf{n}}}_2=\hat{\mathbf{z}}$, with ${\mathbf{J}}_s=0$. In medium 2: ${\mathbf{H}}_2=3\hat{\mathbf{x}}+2\hat{\mathbf{z}}$ A/m. Find ${\mathbf{H}}_1$ and the angle ${\theta }_1$.

Tangential ($\hat{\mathbf{x}}$): $H_{1x}=H_{2x}=3$. Normal ($\hat{\mathbf{z}}$): ${\mu }_1{H}_{1z}={\mu }_2{H}_{2z}$, so $H_{1z}=({\mu }_2/{\mu }_1)H_{2z}=(8/2)(2)=8$.

So ${\mathbf{H}}_1=3\hat{\mathbf{x}}+8\hat{\mathbf{z}}$ A/m.

Angle from normal: $\cos {\theta }_1=H_{1z}/|{\mathbf{H}}_1|=8/\sqrt{73}\approx 0.936$, so ${\theta }_1\approx 20.6°$.

In medium 2, $\cos {\theta }_2=2/\sqrt{13}\approx 0.555$, so ${\theta }_2\approx 56.3°$. Ratio of tangents: $\tan {\theta }_2/\tan {\theta }_1=(3/2)/(3/8)=4={\mu }_2/{\mu }_1$. ✓