# [Physics] Force between parallel current carrying loops

homework-and-exercisesmagnetic fields

How to derive the force between two parallel current carrying loops?(current flows in the same sense,say clockwise,in both the loops)
Radius of both the loops is R and each has a current I flowing through it.Both the loops are at a distance of 'd' from each other.

(a) Not a very profound contribution, but we'd expect (wouldn't we?) that when the separation, $$d$$, between the loops is much less than the radius, $$r$$, the force is approximately that between long parallel straight wires a distance $$d$$ apart. So... $$F=\mu_0 I^2 \frac rd\ \ \ \ \ \ \text {if}\ \ \ \ \ \ d << r.$$
(b) We can derive an exact equation for the force between two 'co-axial' square loops using pretty simple mathematics, starting with $$\vec F_{1,2}=\frac{\mu_o I_1I_2}{4\pi r^3}\left[(\vec{\delta l_1}.\vec r_{2,1}) \vec{\delta l_2}- (\vec{\delta l_1}.\vec{\delta l_2})\vec r_{2,1}\right]$$ I find the attractive force to be of magnitude $$F=\frac{2\mu_o I_1I_2}{\pi}\left[-\frac{t \sqrt{2+t^2}}{1+t^2} + \frac{1+2t^2}{t\ \sqrt{1+t^2}}-1 \right]$$ in which $$t=\frac{\text{separation of loops}}{\text{side length of loop}}$$ This predicts limiting expressions for large and small $$t$$ quite nicely. For example, for small $$t$$ (loops much closer than side length), the second term in the square brackets dominates, and the expression simplifies to $$F=\frac{2\mu_o I_1I_2}{\pi t}$$ that is $$F=\frac{4a\mu_o I_1I_2}{2\pi b}$$ in which $$a$$ is the side length of the loop and $$b$$ is the loop separation. So we have the same force as between parallel straight wires when the loop separation is much less than the side length.
For large separations, $$b\gg a$$, that is $$t\gg 1$$, we can show by expanding the terms in the square bracket binomially in powers of $$1/t$$, that $$F$$ varies as $$b^{-4}$$.