Does "magnetic charge" fit into vector potential?

[Charles Link]

@Gene Naden The $H$ field from the single pole of the solenoid is spherically symmetric (except at very close range). This can be computed from the "pole" model of magnetism, as opposed to using surface currents and Biot-Savart.$\\$ For the details of how these two methods give the same result, see https://www.overleaf.com/read/kdhnbkpypxfk It's something I wrote up several years ago to submit for publication as a journal article, but the AJP responded that much of this info is already known. $\\$ (Hopefully this is "permissible" under PF rules for me to provide a "link" such as this). $\\$ The other pole from this solenoid is so far away that its observed field is zero.

 

[Demystifier]

But a solenoid does not produce a spherically symmetric field, but rather a cylindrically symmetric field. The vector potential in this thread does produce a spherically symmetric field. So how can it be the limiting case of the field of a solenoid?

What you said above is true inside the solenoid. But you have to concentrate on the field outside of the solenoid, around the place where the solenoid ends. See the attached copy of a page from the Felsager's book. Fig. 9.3 shows a finite solenoid, while Fig. 9.4 shows the limit of an infinitesimally thin solenoid. If you pretend that there is no solenoid (the upper vertical line) in Fig. 9.4, the magnetic field (represented by arrows) looks like that of a magnetic monopole. If the solenoid is not ignored in Fig. 9.4, then the total flux of magnetic field through the circle is zero because the contribution from the solenoid (strong field represented by dense arrows) is canceled by all other contributions, showing that in reality there is no monopole. In my opinion, this is the most physical interpretation of the Dirac singularity.

 

[Gene Naden]

Well, @Demystifier, that diagram is most helpful. Thanks!