If we have two current-free spaces and separated by a surface current, we can solve the magnetic problem by solving two magnetic scalar potentials and then using matching conditions. My question is, is the general scalar magnetic potential continuous? Why?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ As long as the magnetic medium can be modeled as linear, then the magnetic permeability is piecewise constant and satisfies $\nabla\cdot(\mu \Phi_{m})=0$, i.e., you can match boundary conditions. $\endgroup$– Cinaed SimsonCommented Jun 14, 2019 at 20:58
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
The potential for a static magnetic problem is defined through $$ {\bf B} = - \nabla \phi $$ (or you can define another potential for $\bf H$). Then since $\nabla \cdot {\bf B} = 0$ we have Laplace's equation for $\phi$ and that is why it is useful: we have lots of good methods for solving Laplace's equation. (Of course it won't work if $\nabla \times {\bf B} \ne 0$; in that case one should adopt another approach.)
The above equation implies that the answer to your question is that $\phi$ is continuous if and only if $\bf B$ is finite. At a boundary such as a surface you expect to find finite $\bf B$. At a current-carrying wire of arbitarily small radius, on the other hand, $\bf B$ can diverge.
answered Jun 14, 2019 at 17:14
$\begingroup$
$\endgroup$
1
A potential is essentially an integral of work to carry a particle (magnetic or electric one) from an infinite distance up to the point you need to know the potential value. If the force making the work is not a Dirac's Delta, then the integral (i.e. the potential) must be a continuous function.
-
$\begingroup$ Careful: this isn't quite right for magnetic fields because it is not a case of work being done. The potential is here simply a mathematical device useful for finding $B$ (and $\bf H$) field in static problems. $\endgroup$ Commented Jun 14, 2019 at 17:08