# Necessary magnetostatics boundary conditions to yield a unique solution?

There are two magnetostatics boundary conditions for magnetic flux density and field intensity: $$\bf n \times (\bf H_1 - \bf H_2) = 0 \tag1$$ $$\bf n \cdot (\bf B_1 - \bf B_2) = 0 \tag2$$ where $$\mathbf n$$ is the unit normal vector of boundary. Another auxiliary boundary condition is the continuity of magnetic vector potential $$\bf A_1 - \bf A_2 = 0 \tag3$$ in solving magnetostatics problem, usually the two first boundary conditions are referred. The condition (1) is essential because it implies the change of magnetic permeability. My question is, in order to solve a magnetostatics problem, can we use (1) & (3) instead of (1) & (2)? If so, what is the proof? Can we say that realization of (3) leads to realization of (2)? I have seen some papers using (1) and (3) for two-dimensional problems.

• Where did you get (3 ) from? Since curl of (3) does not equal zero, it must be wrong. Commented Jul 27, 2022 at 15:38
• In the reference: [1] E.J Rothwell "Electromagnetics", CRC Press and [2] D.J Griffiths it has been proved. In short $\nabla \cdot \mathbf A = 0$ is for the normal component and $\int \mathbf A \cdot \mathbf {dl} = \phi$ ensures the tangential components to be continuous. Commented Jul 28, 2022 at 20:21
• Equation (3) is for all components, not just tangential. Commented Jul 29, 2022 at 20:32

Taking the curl of (3), says that $${\bf B_1} = {\bf B_2}$$, which only agrees with your (1) and (2) if the magnetization is zero. Your (1) and (2) are consequences of $${\bf \nabla} \times {\bf H} = {\bf J}$$ and $${\bf \nabla }\cdot {\bf B} = 0$$, assuming that $${\bf J}$$ does not include any free surface currents. It's likely you are assuming no free current so $${\bf J} = 0$$. You could substitute $${\bf n} \cdot [{\nabla }\times ({\bf A_1}-{\bf A_2})]$$ for (2), but the dependence of $${\bf H}$$ on $$\bf B$$ would need to be specified to get the equivalent $${\bf H}$$ boundary condition in terms of $${\bf A}$$.