 Solution

Because $\nabla \cdot \mathbf{B}=0$, a magnetic vector potential can be introduced with $\mathbf{B}=\nabla \times \mathbf{A}$. It is not unique: $\mathbf{A}+\nabla \chi$ gives the same field. In magnetostatics, the Coulomb gauge $\nabla \cdot \mathbf{A}=0$ gives ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{J}$ in vacuum. For localized electric currents,A magnetic scalar potential is more restricted. In a simply connected current-free region of constant permeability, write $\mathbf{B}=-\nabla \psi$; then ${\nabla }^2\psi =0$. Alternatively define the scalar potential for $\mathbf{H}$, differing by a constant permeability factor. A single-valued scalar potential generally cannot cover a domain encircling a net electric current, even where the local curl vanishes. The vector potential does not require the field itself to be curl-free.