Geometric perspective of the vector potential

[PerpStudent]

I'm struggling with trying to visualize the vector potential as in the identity:

B = ∇⨯A

For starters, how does A relate to, say, a uniform magnetic field, which is quite easy to visualize. Then, how about the magnetic field around a bar magnet -- where is A?
Any help would be appreciated.

 

[Muphrid]

The formula should look similar to something you're already familiar with:

\mu_0 j = \nabla \times B

If you can imagine the magnetic field that goes with a specific current density, then the same picture applies to the vector potential that goes with the magnetic field.

 

[vanhees71]

It's not easy to visualize \vec{A} or give it a physical meaning since it is a gauge-dependent quantity. What's physical is the magnetic field, \vec{B} which is given by

\vec{B}=\vec{\nabla} \times \vec{A}.

For a constant field, it's easy to get the vector potential. So let's consider

\vec{B}=B_0 \vec{e}_z.

You have quite some freedom to choose the vector potential. You can take always one constraint since it is only defined from \vec{B} up to the gradient of a scalar field. Here, I'd choose the spatial axial gauge

A_z=0.

Then you have

\vec{B}=\begin{pmatrix}<br /> 0 \\ 0 \\ B_0<br /> \end{pmatrix} = \vec{\nabla} \times \vec{A}=\begin{pmatrix}<br /> -\partial_z A_y \\ \partial_z A_x \\ \partial_x A_y-\partial_y A_x<br /> \end{pmatrix}.

Obviously our constraint doesn't fix the solutions completely, and you have some more freedom. You can, e.g., set A_x=0 and A_y=B_0 x. Then you have

\vec{A} = B_0 x \vec{e}_y.

Around a bar magnet you have to solve the magnetostatic Maxwell equations for a given magnetization of your bar. You find some calculations about this in Sommerfeld's Lectures on Theoretical Physics, vol. III.