 Solution

For steady electromagnetic fields, Maxwell's equations give $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ and $\nabla \cdot \mathbf{B}=0$. Introduce a magnetic vector potential with $\mathbf{B}=\nabla \times \mathbf{A}$ and the permitted Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the double curl then gives $-\Delta \mathbf{A}={\mu }_0\mathbf{j}$. For localized currents and the decaying solution,Here $-\Delta (1/(4\pi |\mathbf{r}|))=\delta$ follows from harmonicity away from zero and the inward flux of the radial derivative through a small sphere, whose magnitude is one. It verifies the vector Poisson solution componentwise. Moreover steady electric charge conservation gives ${\nabla }^{\prime }\cdot \mathbf{j}=0$; integration by parts confirms the stated potential is divergence-free. Taking its curl, using $\nabla |\mathbf{r}-{\mathbf{r}}^{\prime }{|}^{-1}=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$, proves the Biot-Savart law:The zero-field condition at infinity selects the magnetic field produced by these currents, excluding an externally imposed homogeneous magnetic field.

At the origin write the source point as $r{\mathbf{e}}_r+z{\mathbf{e}}_z$. Its current is $kr{\mathbf{e}}_{\varphi }$, so the numerator of the Biot-Savart law is $k{r}^2{\mathbf{e}}_z-krz{\mathbf{e}}_r$. The radial term cancels after angular integration. Using $d{V}^{\prime }=rdrd\varphi dz$ givesThe inner integral is $2h/[r^2\sqrt{h^2+r^2}]$, as is verified by differentiating $z/[r^2\sqrt{r^2+z^2}]$. HenceFor $b,h>0$, its magnitude is ${\mu }_0|k|h(\sqrt{h^2+b^2}-h)$; it points along $+z$ for positive azimuthal current $k>0$, and along $-z$ for $k<0$. The limit $h\to \infty$ is ${\mu }_{0}k{b}^2/2$, agreeing with direct integration of the infinitely long cylindrical current distribution.