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I have difficulty understanding the following vector calculus example. Text can be found here. It is the 5th Q&A -- starting with equation (31.1035).It concerns finding the vector potential of a current loop.

I get the argument up to $$\vec A=\frac{\mu I}{4\pi r^3}\oint \vec r'\cdot \vec r \,\,\,d\vec l'$$. But I don't understand why $$\oint \vec r'\cdot \vec r \,\,\,d\vec l'=\left(\int d\vec a\right)\times \vec r$$

[ADDED]: Definitions of the terms:

This describes a current loop, with a current $I$ flowing in it.

$\vec r$ is the position vector of an arbitrary point in space. $\vec r'$ is a position vector of a point on the loop. $d\vec l'$ is an infinitesimal line element along the loop.

$\vec A(\vec r)$ is the vector potential, given by $$\frac{\mu I}{4\pi}\oint \frac{d\vec l'}{|\vec r'-\vec r|}$$ The above expression is obtained by Taylor expanding this definition.

$d\vec a$ is an infinitesimal area element of the enclosed area in the loop with direction pointig out of the surface enclosed by the loop. So that $$\int d\vec a$$ is the area enclosed by the loop times the unit vector $\hat n$ pointing out of the plane (supposing the loop is planar...)


The first thing I tried is the vector triple product identity $$(\vec r'\cdot \vec r )\,\,\,d\vec l'=\vec r\times(d\vec l'\times \vec r')+(\vec r\cdot d\vec l')\vec r'$$ The first term looks like it's doing the right sort of thing, but I don't see why the second term vanishes (or fits in) ...Maybe I have missed out something quite obvious?

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Welcome to Math.SE. It might help if you more fully defined your terms so that those who may help you don't have to ask all sorts of questions of you. –  Ron Gordon May 6 '13 at 13:41
    
@RonGordon: yup, added! –  val May 6 '13 at 13:50
    
and what is $d \vec{a}$? –  Ron Gordon May 6 '13 at 13:52
    
@RonGordon, oops, added. –  val May 6 '13 at 13:57

1 Answer 1

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This expression relates a loop integral with an area integral, so it has to be related to Green's theorem.

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