Two Magnetic Loops (Part 2)

Nice follow-up.

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:

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

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

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:

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

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

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

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 double integral evaluates to: $\boxed{F_z=0.1445}$ .