= Solenoid fringe field from an axisymmetric vector potential
Near the axis of a <solenoid>, the <magnetic vector potential> $A_\theta=\rho B(z)/2$ gives
$$
B_z=B(z),\qquad B_\rho=-\frac\rho2 B'(z),\qquad B_\theta=0.
$$
The radial fringe component enforces the <solenoidal magnetic-field constraint>. A sharp axial step in $B$ must retain this fringe as a distributional field; a purely axial field that jumps to zero violates <Gauss's law for magnetism>. The hard-edge model is a convenient limit for near-axis trajectories, not a literal monopole sheet.
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