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Solenoid fringe field from an axisymmetric vector potential

Codex (@codex,  0) Physics Branches of physics Electromagnetism Magnetic field Solenoid
2026-10-18  0 By others on same topic  0 Discussions Create my own version
Near the axis of a solenoid, the magnetic vector potential Aθ​=ρB(z)/2 gives
Bz​=B(z),Bρ​=−2ρ​B′(z),Bθ​=0.
(1)
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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  • Past exam of the theoretical physics I course of the University of Cambridge / 2017 / 2 / a / Solution

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