##### Magnetic Boundary Conditions
ems.ms.vp.mbc

We know that electrostatic fields are discontinuous at location of suface charges.

In a similar vein, magnetostatic fields will be discontinuous at the location of surface currents.

Let us follow a treatment greatly reminiscent of the electrostatic one. Our starting point is equation divB0 rewritten in integral form (using the divergence theorem) as \(\oint d{\bf a} \cdot {\bf B} = 0\) and applied to a wafer-thin pillbox straddling a surface carrying surface current density \({\bf K}\). This means that the normal component is continuous,

\[ B^{\perp}_{above} = B^{\perp}_{below}. \] To get the tangential component, let us consider an amperian loop of side length \(l\) oriented perpendicular to the direction of the surface current: \[ \oint d{\bf l} \cdot {\bf B} = (B^{\parallel}_{above} - B^{\parallel}_{below}) l = \mu_0 I_{enc} = \mu_0 K l, \] and therefore

\[ B^{\parallel}_{above} - B^{\parallel}_{below} = \mu_0 K \] Thus, the component of \({\bf B}\) that is parallel to surface but perpendicular to the current flow is discontinuous, whereas the one parallel to the flow is continuous. In vector notation, this becomes

- Gr (5.76)

\[ {\bf B}_{above} - {\bf B}_{below} = \mu_0 {\bf K} \times \hat{\bf n}, \tag{Bdisc}\label{Bdisc} \]

where \(\hat{\bf n}\) points upwards. For the vector potential, the relations are

- Gr (5.77)

\[ {\bf A}_{above} = {\bf A}_{below} \tag{Adisc}\label{Adisc} \]

This can be seen first from the condition \({\boldsymbol \nabla} \cdot {\bf A} = 0\), which guarantees that the normal component is continuous. Second, \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) leads to \[ \oint {\bf A} \cdot d{\bf l} = \int {\bf B} \cdot d{\bf a} = \Phi, \] where the loop is vanishingly small and straddles the surface. Since the flux then goes to zero, the tangential components of \({\bf A}\) are also continuous.

The derivative of \({\bf A}\) however inherits the discontinuity of \({\bf B}\): explicitly,

\begin{align} {\bf B}_{above} - {\bf B}_{below} &= {\boldsymbol \nabla} \times {\bf A}_{above} - {\boldsymbol \nabla} \times {\bf A}_{below} \nonumber \\ &= \left| \begin{array}{ccc} \hat{\bf x} & \hat{\bf y} & \hat{\bf z} \\ \partial_x & \partial_y & \partial_z \\ A_{x, above} & A_{y, above} & A_{z, above} \end{array} \right| - \left| \begin{array}{ccc} \hat{\bf x} & \hat{\bf y} & \hat{\bf z} \\ \partial_x & \partial_y & \partial_z \\ A_{x, below} & A_{y, below} & A_{z, below} \end{array} \right| \end{align}

We put the normal direction along \(\hat{\bf z}\) and the current along \(\hat{\bf x}\). Since the normal component is continuous, \(\partial_{x,y} A_z\) is the same above and below, and we can drop these terms. The \(\hat{\bf z}\) term vanishes since it just involves the difference of \(B^{\perp}\), which is the same above and below. Similary, the \(\hat{\bf x}\) term is the magnetic field parallel to the surface current, which also isn't discontinuous. What is left is \[ %\hat{\bf x} (-\partial_z A_{y,above} + \partial_z A_{y, below}) + \hat{\bf y} (\partial_z A_{x, above} - \partial_z A_{x, below}) %+ \hat{\bf z} (\partial_x A_{y, above} - \partial_x A_{y below} - \partial_y A_{x, above} + \partial_y A_{x, below}) = \mu_0 K \hat{\bf x} \] Therefore, reidentifying the normal component, we get

- Gr (5.78)

\[ \frac{\partial {\bf A}_{above}}{\partial n} - \frac{\partial {\bf A}_{below}}{\partial n} = -\mu_0 {\bf K} \tag{dAdisc}\label{dAdisc} \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42