I think first without formulation and equation it is hard to give a correct

answer.

I find on the web:

The same technique does not work in magnetostatics since, in general, curl H ̸= 0. However, since

div B = 0

we conclude the existence of a vector potential A : R3 → R3 with B = −curlA in D. Substituting this into Ampere’s Law yields (for homogeneous media Ω) after multiplication with μ0

−μ0J = curlcurlA = ∇div A − ∆A.

Since curl ∇ = 0 we can add gradients ∇u to A without changing B. We will see later that we can choose u such that the resulting potential A satisfies div A = 0. This choice of normalization is called Coulomb gauge.

With this normalization we also get in the magnetostatic case the Poisson equation ∆A = −μ0J .

We note that in this case the Laplacian has to be taken component wise.

In your case J is normal to the 2d plan so A will be a scalar field in the plan.

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=> 2 case scalaire avec A_\theta et le vectoriel A_z et A_r