4.2.3 Magnetostatic boundary conditions

Real devices are made of several media: the iron core of a transformer and the air around it, the pole piece of a loudspeaker and its gap. At the surface where two media meet, $\mu$ jumps, and so may the fields. As in Section 3.4 the differential equations cannot be applied across such a surface, but the integral forms can, and from them we now derive how $H$ and $B$ on the two sides of an interface are related. The plan is the same as for the electric field: a small loop straddling the interface tells us about the tangential components, and a small pillbox about the normal components.

Let the surface S separate medium 1, of permeability ${\mu }_1$ (region I), from medium 2, of permeability ${\mu }_2$ (region II). At a point of S let $a_n$ be the unit normal pointing from medium 1 into medium 2 and $a_{\tau }$ a unit vector tangent to S (Figure 4.2.7). Near S we decompose the fields into tangential and normal components,

and label the values just below S (in medium 1) with the subscript 1 and those just above S (in medium 2) with the subscript 2. The interface may carry a free surface current $J_S$ [A/m], flowing in a thin sheet along S, as the winding of a solenoid or the current on the surface of a good conductor does.

The tangential component of H

Apply the first fundamental equation in integral form, ${\oint }_{C}H\cdot dl={\int }_{S}J\cdot dS$, to the small rectangular loop C of Figure 4.2.7a. The surface $S$ bounded by $C$ is the rectangle $\Delta l\times \Delta h$, and its normal, by the right-hand rule for the sense $C_1\to C_2\to C_3\to C_4$, is $a_{\tau }\times a_n$, the direction perpendicular to the loop. Now let the loop shrink onto the interface, $\Delta h\to 0$, keeping $\Delta l$ fixed but small enough for the fields to be constant along each side.

• On the right-hand side, a finite volume current density gives $J\Delta l\Delta h\to 0$. A surface current on S, however, is caught by the loop however thin the loop is. Its component along the normal of the loop, which we call $J_{nS}$, contributes $J_{nS}\Delta l$:

  $$\begin{array}{rl}\underset{\Delta h\to 0}{\lim }{\oint }_{C}H\cdot dl& =\underset{\Delta h\to 0}{\lim }{\int }_{S}J\cdot dS\\ & =J_{nS}\Delta l.\end{array}$$

• On $C_1$, in medium 1 and traversed along $+a_{\tau }$, $dl=a_{\tau }dl$ and

  $$\underset{\Delta h\to 0}{\lim }{\int }_{C_1}H\cdot dl=H_{1\tau }\Delta l\text{on S}.$$

• On $C_2$ and $C_4$ the field is finite while the path length $\Delta h$ tends to zero:

  $$\underset{\Delta h\to 0}{\lim }{\int }_{C_2,C_4}H\cdot dl=0.$$

• On $C_3$, in medium 2 and traversed along $-a_{\tau }$,

  $$\underset{\Delta h\to 0}{\lim }{\int }_{C_3}H\cdot dl=-H_{2\tau }\Delta l\text{on S}.$$

Adding the four pieces and equating to the right-hand side,

and since $\Delta l\ne 0$,

The sign convention is worth a second look. Take $a_{\tau }=a_x$ and $a_n=a_y$, so that the loop lies in the $xy$-plane with its normal $a_x\times a_y=a_z$. Then Eq. (4.2.11) says $H_{1x}-H_{2x}=J_{Sz}$: a sheet current flowing along $+z$ makes the $x$-component of $H$ larger below the sheet than above it. This is just the field of a current sheet, $\pm \frac{1}{2}{J}_S$ on the two sides, directed by the right-hand rule, and the infinitely long solenoid of Problem 4.2.3 is the standard example: $H=NI$ inside, $0$ outside, and the jump $NI$ is the sheet current of the winding.

The normal component of B

Apply the second fundamental equation, ${\oint }_{S}B\cdot dS=0$, to the small pillbox of Figure 4.2.7b, with faces $\Delta S$ parallel to S and side $S_3$ of height $\Delta h$. Again let $\Delta h\to 0$.

• On the face $S_1$, in medium 1, the outward normal is $-a_n$, so

  $$\begin{array}{rl}\underset{\Delta h\to 0}{\lim }{\int }_{S_1}B\cdot dS& =B_1\cdot \left(-a_n\right)\Delta S\\ & =-B_{1n}\Delta S\text{on S}.\end{array}$$

• On the face $S_2$, in medium 2, the outward normal is $+a_n$:

  $$\underset{\Delta h\to 0}{\lim }{\int }_{S_2}B\cdot dS=B_{2n}\Delta S\text{on S}.$$

• The area of the side $S_3$ is proportional to $\Delta h$, so its flux tends to zero.

Hence

and therefore

Boundary conditions at work: the refraction of B

Let a line of $B$ cross an interface without surface current, making the angle ${\theta }_1$ with the normal in medium 1 and ${\theta }_2$ in medium 2 (Figure 4.2.8a). The components are $B_n=B\cos \theta$ and $B_{\tau }=B\sin \theta$, and the two boundary conditions give

Dividing the second by the first,

Drag ${\theta }_1$ inside the iron towards 89° and watch the flux still leave almost along the normal, then try air to steel.