# Magnetic flux through a loop the same as flux through the plane outside the loop?

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).

• 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). – Philip Wood Dec 13 '17 at 22:48
• @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. – RelativisticDolphin Dec 14 '17 at 8:11
• An approximate formula for the flux linked with a circular current-carrying loop of radius $𝑟$ made from wire of radius 𝑎 is $$\Phi=\mu_0 I𝑟 \ln(1.39𝑟/𝑎).$$ Is this what you arrived at (maybe without the 1.39) by considering the 'returning' flux outside the loop? – Philip Wood Jul 1 '19 at 23:01

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.