Neumann formula for mutual inductance

Remarks

A steady current $I_1$ in $C_1$ has the magnetic vector potential $A_1(x) = \frac{\mu_0 I_1}{4 \pi} \oint_{C_1} \frac{d\ell_1}{\abs{x - x_1}}$, whose curl is the field given by the Biot–Savart law. By Stokes' theorem, the flux of $\nabla \times A_1$ through a surface bounded by $C_2$ equals the line integral $\oint_{C_2} A_1 \cdot d\ell_2$. Dividing by $I_1$ gives the formula. The integrand is symmetric under exchange of the two curves, which proves $M_{12} = M_{21}$.

The formula needs the curves to be disjoint. For $C_1 = C_2$ the integrand blows up like $1 / \abs{x_1 - x_2}$ along the diagonal, and the double integral diverges, which is why self-inductance needs a wire of positive thickness.

Example: take two coaxial circles of radii $a$ and $b$ far apart at distance $d$. To leading order in $1/d$ the formula gives $M_{12} \approx \mu_0 \pi a^2 b^2 / (2 d^3)$, the value for two small coaxial magnetic dipoles.

Going deeper

For a student

The Neumann formula is the magnetostatic analogue of writing the interaction energy of two charge distributions as a double integral against the Coulomb kernel $1 / \abs{x - y}$. The magnetic energy of two currents is $\frac{1}{2} L_{11} I_1^2 + M_{12} I_1 I_2 + \frac{1}{2} L_{22} I_2^2$, so the formula also gives the force between two circuits by differentiating $M_{12} I_1 I_2$ with respect to their relative position. In numerical work, inductances of printed circuits are often computed by discretising this double line integral.