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Eight points, A,B,C,D and A',B',C',D' are arranged in space as the vertices of a cube of side a, with the faces ABCD and A'B'C'D' opposite, and A nearest to A' etc. Currents I and I' flow in current carrying wires around the sides of these two faces. What is the force between the two wires.

I know that I need to consider all the adjacent pairings of wires and the result will be zero unless the edges are parallel. This leaves 8 pairs, 4 with edges adjacent like AB and A'B' and 4 with edges opposite like AB and D'C'. I have the equation for a force between two wires, but I'm not sure if I can reduce my workload more and then use it, but I'm also not sure what I actually need to integrate to find the force...

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1 Answer 1

For a pair of parallel wires, $dF=\frac{\mu_0II'}{2\pi r}\mathrm{d}l$, where $r$ is the distance between the wires and $\mu_0$ is the permittivity of free space. The direction of the force is away from the other wire if the currents are parallel. Let $I$ and $I'$ be both currents flowing clockwise, or replace $I$, say, with $-I$ if the current goes anticlockwise. For each side of $ABCD$, the component of that force in the direction perpendicularly away from $A'B'C'D'$ is $F=\frac{\mu_0II'}{2\pi a}a+\frac{\mu_0II'}{2\pi \sqrt{2}a}a$, giving a total force of $(1+\frac{1}{\sqrt{2}})\frac{\mu_0II'}{2\pi}$

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