Two Magnetic Loops

Very nice problem! Solving it was fun

An arbitrary point on the smaller loop is:

$\vec{r}_1 = R_1\cos{\theta_1} \hat{i} + R_1\sin{\theta_1} \hat{j} + 0 \hat{k}$

One of the larger loop is:

$\vec{r}_2 = R_2\cos{\theta_2} \hat{i} + R_2\sin{\theta_2} \hat{j} + 1 \hat{k}$

$\vec{r} = \vec{r}_2 -\vec{r}_1$

The magnetic field due to a length element on the larger loop at a point on the smaller loop is:

$d\vec{B} = \frac{\mu_oI}{4\pi}\left(\frac{d\vec{r}_1\times\vec{r}}{\mid \vec{r} \mid^3}\right)$

Having found the magnetic field components, the next step is computing the magnetic force experienced by a length element of the smaller loop. The expression is:

$d\vec{F} = dF_x \hat{i} +dF_y \hat{j} +dF_z \hat{k}$

$d\vec{F} = I (d\vec{r}_2\times d\vec{B})$

Substituting all expressions and simplifying gives:

$dF_x = -\frac{1}{4\pi}\left(\frac{\cos\left(\theta_{2}\right)\,\left(2\,\cos\left(\theta_{1}-\theta_{2}\right)-4\right)}{{\left(6-4\,\cos\left(\theta_{1}-\theta_{2}\right)\right)}^{3/2}}\right)d\theta_1 d\theta_2$

$dF_y = -\frac{1}{4\pi}\left(\frac{\sin\left(\theta_{2}\right)\,\left(2\,\cos\left(\theta_{1}-\theta_{2}\right)-4\right)}{{\left(6-4\,\cos\left(\theta_{1}-\theta_{2}\right)\right)}^{3/2}}\right)d\theta_1 d\theta_2$

$dFz = -\frac{1}{4\pi}\left(\frac{2\,\cos\left(\theta_{1}-\theta_{2}\right)}{{\left(6-4\,\cos\left(\theta_{1}-\theta_{2}\right)\right)}^{3/2}}\right)d\theta_1 d\theta_2$

The forces can then be computed by solving double integrals as both $\theta_1$ and $\theta_2$ vary from $0$ to $2\pi$ . The X and Y components of the force are zero. This is expected by virtue of the configuration of the loops. The Z component is:

$\boxed{ F_z =\int_{0}^{2\pi}\int_{0}^{2\pi}-\frac{1}{4\pi}\left(\frac{2\,\cos\left(\theta_{1}-\theta_{2}\right)}{{\left(6-4\,\cos\left(\theta_{1}-\theta_{2}\right)\right)}^{3/2}}\right)d\theta_1 d\theta_2 \approx -0.404}$