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Let S be a piecewise smooth oriented surface in $\mathbb{R}^3$ with positive oriented piecewise smooth boundary curve $\Gamma:=\partial S$ and $\Gamma : X=\gamma(t), t\in [a,b]$ a rectifiable parametrization of $\Gamma$. Imagine $\Gamma$ is a wire in which a current I flows through. Then

$$m:=\frac{I}{2}\int_a^b\gamma(t)\times \dot{\gamma}(t)dt$$

is the magnetic moment of the current.

Show that for an arbitrary $u\in \mathbb{R}^3$

$$m\cdot u=I\int_Su\cdot d\Sigma$$ is true.

I tried doing this with Stokes but I can't seem to get to the desired equation. The teacher gave us a hint: $k_u(x):=\frac{1}{2}u\times x$ is a vector field and $\operatorname{curl}k_u = u$.

Any tips or hints? I would appreciate it.

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$$\vec m=\frac{I}{2}\int_a^b \vec \gamma \times \frac{d\vec \gamma}{dt}dt$$

Then,

$$\begin{align} \vec u \cdot \vec m&=\frac{I}{2}\vec u \cdot \left(\int_a^b \vec \gamma \times \frac{d\vec \gamma}{dt}dt\right)\\\\ &=\frac{I}{2}\int_a^b \vec u \cdot \left(\vec \gamma \times \frac{d\vec \gamma}{dt}\right)dt\\\\ &=\frac{I}{2}\int_a^b \vec (\vec u \times \gamma) \cdot \frac{d\vec \gamma}{dt}dt\\\\ &=\frac{I}{2}\int_S \nabla \times (\vec u \times \gamma) \cdot d\vec \Sigma\\\\ &=\frac{I}{2}\int_S 2 \vec u \cdot d\vec \Sigma\\\\ &=I\int_S \vec u \cdot d\vec \Sigma \end{align}$$

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Such $m$ is the magnetic moment of the current loop. The formula you want to prove is a consequence of Stokes theorem:

$$ \int_S u \cdot d\Sigma =\int_{S}\nabla \times k_u (x) \cdot d\Sigma = \int_{\partial S} k_u(\gamma)\times d\gamma = \int_{\Gamma}\frac{1}{2} (u \times \gamma)\cdot d\gamma $$ Recall the cyclic property of the triple scalar product, using it we can rewrite the last integral as:

$$ \frac{1}{2} \int_{\Gamma}\gamma \times d\gamma \cdot u = \frac{1}{2}\int_{a}^{b} \gamma(t) \times \dot{\gamma} (t) ~dt \cdot u $$

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