Axially symmetric B field vector potential?

[AxiomOfChoice]

Suppose you have an axially symmetric magnetic field for which the azimuthal component $$B_\phi = 0$$. This is all you know. What are some possible vector potentials $$\vec A$$ (such that $$\vec B = \nabla \times \vec A$$) that would produce this field? (So we're working in cylindrical coordinates.)

The obvious one I've thought of is just $$\vec A = A \hat \phi$$. But I'm not sure what form $$A$$ should take in terms of $$B_r$$ and $$B_z$$, where $$\vec B = B_r \hat r + B_z \hat z$$.

 

[LightHero]

One possible form of A is A = \frac{1}{2} (B_r \hat z - B_z \hat r). This gives B_\phi = 0, and B_r = \frac{\partial A_z}{\partial r} - \frac{\partial A_r}{\partial z}, and B_z = \frac{\partial A_r}{\partial r} - \frac{\partial A_z}{\partial z}.

 

[astrokreng]

One possible vector potential that would produce an axially symmetric magnetic field with no azimuthal component is \vec A = \frac{1}{2}B_r r^2 \hat \phi, where B_r is the radial component of the magnetic field. This potential satisfies the condition of \nabla \times \vec A = \vec B and also takes into account the fact that the magnetic field strength increases with radial distance from the axis of symmetry.

Another potential that could produce this magnetic field is \vec A = \frac{1}{2}B_z z^2 \hat \phi, where B_z is the axial component of the magnetic field. This potential takes into account the fact that the magnetic field strength also increases with axial distance from the axis of symmetry.

In general, there are many possible vector potentials that could produce an axially symmetric magnetic field with no azimuthal component. The choice of which potential to use would depend on the specific characteristics of the magnetic field and the system in question.