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Im trying to show that the integral over a closed loop of a crossproduct stays the same if I choose a different origin with $\overrightarrow{r}=\overrightarrow{r}\prime+\overrightarrow{r_0}$ and $\oint{d\overrightarrow{F}}=0$

$$ \overrightarrow{N} = \oint{\overrightarrow{r} \times d\overrightarrow{F}} = \oint{(\overrightarrow{r}\prime+\overrightarrow{r_0} )\times d\overrightarrow{F}} = \oint{\overrightarrow{r}\prime \times d\overrightarrow{F}} + \oint{\overrightarrow{r_0} \times d\overrightarrow{F}}$$

How do I show $\oint{\overrightarrow{r_0} \times d\overrightarrow{F}}= 0 $, or why can I say $\oint{\overrightarrow{r_0} \times d\overrightarrow{F}} = \overrightarrow{r_0} \times\oint{ d\overrightarrow{F}}$?

To clarify, $\overrightarrow{dF}$ is the force on the current $I \overrightarrow{dl}$ in a loop in a uniform and constant magnetic field. So $\overrightarrow{dF} = I \overrightarrow{dl} \times \overrightarrow{B}$.

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You are on the right way with your line of thought. The only part of the puzzle you are missing is, as you have correctly identified, the explanation of $$\oint{\overrightarrow{r_0} \times d\overrightarrow{F}} = \overrightarrow{r_0} \times\oint{ d\overrightarrow{F}}$$ It is actually very simple: $\overrightarrow{r_0}$ is a constant vector (it doesn't depend on coordinates) and as such you can factor it out of the integral. Rewriting the vectors in their component elements does nothing for you.

I am sure you know of the fact that you can factor constants out of integrals, but just for completeness, here is a link to the wikipedia entry for this rule called the Constant factor rule in integration.

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You can extract the constant term out of integral directly, however keep the operation unchanged. See:

$$\oint \vec{r}_0\times \mathrm{d}\vec{F}=\oint\epsilon_{ijk}r_{0i}\mathrm{d}F_j\hat{e}_k=\epsilon_{ijk}r_{0j}\oint\mathrm{d} F_j \hat{e}_k=\vec{r}_0\times\oint\mathrm{d}F$$

Similarly:

$$\int \vec{B_0}\cdot \mathrm{d}\vec{S}= \vec{B_0}\cdot \int\mathrm{d}\vec{S}$$

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  • $\begingroup$ Ah so to see why I can take out r0, I should split it up in its scalars and then its easier to see. Thanks! I dont understand your notation, what is epsilon_ijk? And why is dFj in the direction of e-hat_k? $\endgroup$ – Leo Apr 10 '14 at 14:27
  • $\begingroup$ @Leo This is Levi-Civita symbol, usually use to write vector product in a more compact way. $\endgroup$ – an offer can't refuse Apr 10 '14 at 14:41
  • $\begingroup$ I accepted the other answer because yours went a bit too quickly for me. I didnt understand what you were saying at first. But thanks for also trying and making it more general by showing the same principle with a dot product. $\endgroup$ – Leo Apr 11 '14 at 6:54

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