I've managed to solve 5.34 (a) and (b) of Jackson's electrodynamics. Right this minute I believe I know how to solve (c) but I am not certain yet and need some help on this.
It is not a trivial problem. The geometry is as such,
Two identical circular loops of radius $a$ are initially located a distance $R$ apart on a common axis perpendicular to their planes.
Part (c) asks to use the techniques of section 3.3 for solutions of the Laplace equation to show that the mutual inductance for two coplanar identical circular loops of radius $a$ whose centers are separated by a distance $R> 2a$ is
$$M_{12} = - \frac{\mu \pi a}{4}[(\frac{a}{R})^3 + \frac{9}{4} (\frac{a}{R})^5 + \frac{375}{64} (\frac{a}{R})^7 + . . . .] $$
Now you might ask what is the secret sauce of section 3.3?
i) $$\phi(z = r) = \sum_{l =0}^\infty [ A_l r^l + B_l r^{-(l+1)}]$$
ii) $$\phi(z = r) = V [1 - \frac{r^2 - a^2}{r \sqrt{r^2 + a^2}}]$$
Expanding in powers of $\frac{a^2}{r^2}$
iii) $$\phi(z = r) = \frac{V}{\sqrt{\pi}} \sum_{j=1}^\infty (-1)^{j-1} \frac{(2j - \frac{1}{2}) \Gamma(j - \frac{1}{2})}{j!} (\frac{a}{r})^{2j}$$
iv) $$\phi(r , \theta) = \frac{V}{\sqrt{\pi}} \sum_{j=1}^\infty (-1)^{j-1} \frac{(2j - \frac{1}{2}) \Gamma(j - \frac{1}{2})}{j!} (\frac{a}{r})^{2j} P_{2j -1} (\cos \theta)$$
There are other equations in 3.3 but I suspect the listed should suffice (I could be wrong, please refer to Jackson if you can).
How can I follow through using the follow plan, the analogous thing for magnetism,
- Use the potential to solve for $B$
- compute flux through the second loop
- Read off $M_{12}$ from $\Phi_2 = M_{12} I_1$
