# Is Ampere's force law conservative?

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 $\vec{ds'}$?

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.
• Please can you tell why no work is done moving $\vec{ds}$ along BC and DA? – Joe Aug 2 '17 at 18:58
• 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? – Philip Wood Aug 2 '17 at 21:29