Gauss's law for magnetism

Gauss's law for magnetism is $\nabla \cdot \mathbf{B}=0$. Its integral form says that the net magnetic flux through every closed surface vanishes, so the normal magnetic field is continuous across an ordinary interface.

Let a smooth magnetic field with axisymmetry have axial value $B_0(z)$ and no line singularity on its axis. Gauss's law for magnetism gives ${\partial }_r(r{B}_r)=-r{\partial }_z{B}_z$. Since $B_z(r,z)=B_0(z)+O(r^2)$, integration from the regular axis gives the displayed expansion. A falling axial field therefore implies outward field lines above a positive-source loop. In a current-free region, Ampère's law also gives $B_z(r,z)=B_0(z)-r^2{B}_0^{\prime \prime }(z)/4+O(r^4)$, which follows by equating ${\partial }_z{B}_r={\partial }_r{B}_z$.

For a smooth axisymmetric magnetic field, write $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$. The absence of magnetic monopoles gives ${\rho }^{-1}{\partial }_{\rho }(\rho B_{\rho })=-B_0^{\prime }(z)+O({\rho }^2)$. Integrating from the axis and excluding a singular $1/\rho$ term proves the displayed relation. An axial field decreasing along $+z$ has an outward radial component, even when the field points purely axially on the axis itself.