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Ampere's force law between current elements is:

$d^2\vec{F}=k\dfrac{1}{r^2}[\vec{ds}\times(\vec{ds'}\times\hat{r})]ii'$

Is this force field conservative?

$$OR$$

Does the work done to move $\vec{ds}$ in the presence of $\vec{ds'}$ is path independent if we doesn't change the orientation of $\vec{ds}$ and $\vec{ds'}$?

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I believe it to be non-conservative. Consider $\vec{ds}$ (parallel to $\vec{ds'}$) taken around the rectangular path ABCD. A is on a perpendicular bisector of $\vec{ds'}$; B is on the same perpendicular bisector, but further away from $\vec{ds'}$. C is along the line through B parallel to $\vec{ds'}$, and D is the 'fourth' corner of rectangle ABCD. No work is done moving $\vec{ds}$ along BC and DA, and more work is done against the attractive force on AB than is gained from it on CD.

Having made this pronouncement, I'm not at all confident that it means very much, as there can't be isolated current elements, can there?

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  • $\begingroup$ Please can you tell why no work is done moving $\vec{ds}$ along BC and DA? $\endgroup$ – Joe Aug 2 '17 at 18:58
  • $\begingroup$ Because $\vec{ds}$ itself is parallel to the lines BC and DA, so the force on it is perpendicular to these lines, so the dot product $\vec{F}.\vec{dx}$ is zero all the way along these lines. Have you drawn a diagram of the set-up I'm using? $\endgroup$ – Philip Wood Aug 2 '17 at 21:29

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