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## Homework Statement

A triangular loop of side length a carries a current I. If this loop is placed a distance d away from a very long straight wire carrying a current I', determine the force on the loop. See attachment for diagram.

## Homework Equations

F=([itex]\mu[/itex]II'/2[itex]\pi[/itex]r)L

F=I'LB

B=[itex]\mu[/itex]I/2[itex]\pi[/itex]r

## The Attempt at a Solution

I broke the calculation into three pieces. Force on the horizontal portion of the triangle, and the force on each of the two diagonal pieces.

The force of the horizontal portion is just F=[itex]\mu[/itex]II'*L/2[itex]\pi[/itex]r

dB=[itex]\mu[/itex]I dr/2[itex]\pi[/itex]r

B=[itex]\mu[/itex]I/2[itex]\pi[/itex] [itex]\int[/itex]dr/r

The distance from the wire carrying I' ranges from d to d+a/[itex]\sqrt{3}[/itex]. Thus, I set these as the limits of integration. Integrating I dr/r I get ln(r). Plugging this in with the appropriate limits, I get:

B=([itex]\mu[/itex]I/2[itex]\pi[/itex])*(ln(1+a/d[itex]\sqrt{3}[/itex])

Because both horizontal elements are equal, I will use the total length = a and combine these two forces:

Diagonal Force=([itex]\mu[/itex]II'/2[itex]\pi[/itex])*(ln(1+a/d[itex]\sqrt{3}[/itex])

Total force on the current element is the difference b/w the two forces:

Total Force=([itex]\mu[/itex]II'/2[itex]\pi[/itex])*(a/d-(ln(1+a/d[itex]\sqrt{3}[/itex]))

??? Not sure what is wrong??? Please help and thank you!