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Let's say we have a current loop of radius $a$ in $x-y$ plane. Let's consider the flux ($\phi=BA$) through the loop, and the flux through the rest of the plane, outside the loop. (to clarify, if the loop covers the area A, here I'm referring to the rest of the $x-y$ plane, so the entirety of $x-y$ plane from $-\infty$ to $+\infty$ - A).

I think the two fluxes should have the same magnitude (different sign):

  1. All B field have to form a loop because there is no magnetic monopole, hence all the B field going through the loop has to come back down , through somewhere outside the loop.

  2. Also using $\nabla \cdot \textbf{B}=0$ and Gauss's. Consider a cylinder with infinite dimensions, with its upper surface in the $x-y$ plane containing the loop, and all the other surfaces will be at infinity. Hence, the only contribution to the surface integral would be from the surface containing the loop, which has to be zero.

However, I believe my conclusion has to be wrong; using this conclusion one can calculate the self inductance of a single wire loop independent of its thickness/height, by calculating the flux through the rest of the plane, which we can do if we know the current and the radius of the loop. But the self inductance and the B field through the loop has to depend on the thickness of the loop(by Ampere's and as other sources suggest).

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  • $\begingroup$ You are right up to your last paragraph. But regarding this last paragraph, have you actually tried to calculate the inductance of a loop by integrating up the 'external' flux, as you propose? I can't do so (which may not be saying very much). $\endgroup$ Dec 13, 2017 at 22:48
  • $\begingroup$ @PhilipWood yes I did and I got an expression independent of the thickness. I used the scalar magnetic potential and the magnetic moment of the wire loop. $\endgroup$ Dec 14, 2017 at 8:11
  • $\begingroup$ An approximate formula for the flux linked with a circular current-carrying loop of radius $๐‘Ÿ$ made from wire of radius ๐‘Ž is $$\Phi=\mu_0 I๐‘Ÿ \ln(1.39๐‘Ÿ/๐‘Ž).$$ Is this what you arrived at (maybe without the 1.39) by considering the 'returning' flux outside the loop? $\endgroup$ Jul 1, 2019 at 23:01
  • $\begingroup$ @RelativisticDolphin Your last sentence. Do you mean thickness of the loop, or area within the loop? You should provide a formula. $\endgroup$
    – Leviathan
    Jul 15, 2021 at 17:27
  • $\begingroup$ I think this is a bit of a magnetic equivalent of the self-energy of a point charge. Nature simply doesn't allow us to reduce the size of charge/current distributions to zero without including, at least, a logarithmic divergence. Maybe the problem is a little deeper, though, than the usual static analysis suggests. In a relativistic universe no such thing as a "static" field can exist. The formation transient always takes infinite time to spread to all of space. At the "small end", the formation of an infinitesimal charge/current density would cause an infinite spike of em radiation. $\endgroup$ Apr 15, 2023 at 1:22

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Start with the vector potential $\textbf{B}=\textrm{curl}\textbf{A}$. For a magnetic dipole $\textbf{A}=\frac{\mu_0}{4\pi}\frac{\mathbf{\mathfrak{m}}\times\textbf{r}^0 }{|\textbf{r}|^2}$, this is also the asymptotic field at large distances away of any finite sized source. Now from Stokes' theorem the flux through any surface $S$ with boundary $\mathcal{C}=\partial S$ is $$\Phi = \int_{\mathcal{S}}\textbf{B}\cdot d\textbf{S}=\int_\mathcal{S} \textrm{curl} \textbf{A} \cdot d{\textbf{S}} \\ =\oint_{\mathcal{C}} \textbf{A} \cdot d{\textbf{c}}$$ Let $\mathcal{C}$ be a circle of arbitrary large radius, that is we are looking at the flux through an infinite sheet of surface, then the latter integral approaches zero for a dipole field because the perimeter is proportional to the radius while the integrand is proportional to the inverse squared radius. Then it also follows that the flux of an infinite sheet is also zero for any finite sourced magnetic field, in other words the flux threaded by any simple loop is as exactly equal to the (negative) flux threaded by the outside of that loop.

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