I have seen the other threads on an infinitely long wires vector potential.Its obvious that really small wires are just infinitely long cylinders:(adsbygoogle = window.adsbygoogle || []).push({});

∇xA=B

∫∇xA.da=∫B.da

∫A.dl = ∫B.da = φ(flux)

For an infinite cylinder

A.2πr_{i}=B.2πr_{i}h

A=Bh

A=μ_{0}*I*h/(2π*r)

Now for a cylinder of radius lim_{r->0}=> μ_{0}I/2π which im sure is wrong (interesting multiply this by r and u get B field of a wire)

Obviously for infinitely long wire:

A = (μ_{0}/I2π)(log(Λρ+√1+Λ2ρ2))

Any help?

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# B Vector potential of an infinitely long cylinder

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