Magnetic vector potential

Remarks

If $U$ is star-shaped, a vector potential exists because $\nabla \cdot B = 0$ by Gauss's law for magnetism; this is the Poincaré lemma for $2$-forms. If $A$ is a vector potential, so is $A + \nabla \chi$ for every twice continuously differentiable function $\chi$, with gradient $\nabla \chi$. On a simply connected $U$ these are all of them. This freedom is the gauge freedom, and a condition such as $\nabla \cdot A = 0$ selects one potential. Example: the uniform field $B = (0, 0, B_0)$ has the vector potential $A(x) = \tfrac{B_0}{2} (-x_2, x_1, 0)$; the check is $\partial_1 A_2 - \partial_2 A_1 = B_0$ and the other components of the curl vanish. Another vector potential of the same field is $(0, B_0 x_1, 0)$. Non-example: $A(x) = (x_1, 0, 0)$ is not a vector potential of any nonzero field, since its curl is $0$. For a steady current density $J$ as in the Biot–Savart law, $A(x) = \frac{\mu_0}{4 \pi} \int J(y) / \abs{x - y} \, dy$ is a vector potential of its field with $\nabla \cdot A = 0$.