 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/15d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Substituting the wire potential from part (b) into the flux formula from part (a) gives Neumann's mutual-inductance formula

L_{12} = \frac{Φ _{12}}{I _{2}} = \frac{μ _{0}}{4 π} \oint_{C_{1}} \oint_{C_{2}} \frac{d x _{1} \cdot d x _{2}}{∣ x _{1} - x _{2} ∣} .

Interchanging the two curves proves

L_{12} = L_{21}

Parametrize the coaxial circles by

x_{1} = (a cos ϕ, a sin ϕ, 0), x_{2} = (b cos ψ, b sin ψ, c) .

Then, with

θ = ϕ - ψ

and

R = a^{2} + b^{2} + c^{2}

d x_{1} \cdot d x_{2} = ab cos θ d ϕ d ψ, ∣ x_{1} - x_{2} ∣ = R 1 - q cos θ,

where

q = 2 ab / R^{2}

. One angular integration contributes

2 π

, so

L_{12} = \frac{μ _{0} ab}{2 R} \int_{0}^{2 π} \frac{c o s θ d θ}{1 - q c o s θ} .

Since

ab = q R^{2} /2

, this is

L_{12} = \frac{μ _{0} R}{4} f (q) = \frac{μ _{0}}{4} a^{2} + b^{2} + c^{2} f (q) .

 New to topics? Read the docs here!