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Imagine we have a circuit on the xy-plane, with a random geometry (it just need to be closed). I want to calculate the magnetic moment of this setup:

$$\vec{m} = \frac{1}{2} \int_V d^3 x' \vec{x}' \times \vec{j}(\vec{x}')$$

Assuming that current is constant we get $I d \vec{x}' = \vec{j}(\vec{x}')$ which turns the integral into:

$$\vec{m} = \frac{1}{2} \int_V d^3 x' \vec{x}' \times (I d\vec{x}')$$

Next step in the computation, I'm trying to follow, is immediately writing down: $$\vec{m} = \frac{1}{2} I \int_{\delta A} \vec{x}' \times d\vec{x}'$$

I probably flunked an important lesson and can't understand what is happening here. We don't use Stokes, because then the "$\times$" would go away.

My question is: how does the integral over the volume/surface transform into one over its edge?

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I think you should consider Ostrogradsky's theorem:

$\iiint_V\left[\mathbf{G}\cdot\left(\nabla\times\mathbf{F}\right) - \mathbf{F}\cdot \left( \nabla\times\mathbf{G}\right)\right]\, dV = \underbrace{\oint\oint}_\textrm{Surface} \mathbf F\times\mathbf{G}\cdot d\mathbf{S}.$

which is special case of the Stokes' theorem, but saves cross-product.

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  • $\begingroup$ @user17574 Is this question still actual? Maybe, you found other solution? $\endgroup$
    – sigrlami
    Commented Jul 19, 2013 at 13:26