= Axisymmetric magnetic field near a regular axis
{title2=$B_r(r,z)=-\frac r2B_0'(z)+O(r^3)$}
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 $\partial_r(rB_r)=-r\partial_zB_z$. 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 $B_z(r,z)=B_0(z)-r^2B_0''(z)/4+O(r^4)$, which follows by equating $\partial_zB_r=\partial_rB_z$.
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