Inductors and Inductance — Question 2

Question 2

Toroidal inductance retains a logarithmic dependence on radial thickness

An ideal air-core toroid has N turns and rectangular cross section of height h and radii a and b. Its azimuthal field is $B(r)=\mu_0NI/(2\pi r)$. All turns link the same cross-sectional flux.

1. Derive L without assuming the toroid is thin.

2. Recover the thin-toroid expression for b=a+w with w much smaller than a.

Question 2 - Solution

Toroidal inductance retains a logarithmic dependence on radial thickness

(a) Method and calculation

$$\Phi=\int_a^bB(r)h\,dr=\frac{\mu_0NIh}{2\pi}\ln(b/a).$$$$\boxed{L=N\Phi/I=\frac{\mu_0N^2h}{2\pi}\ln(b/a)}.$$

(b) Interpretation and further calculation

$$\ln(1+w/a)=w/a+O(w^2/a^2).$$$$\boxed{L\simeq\frac{\mu_0N^2(hw)}{2\pi a}}.$$

Physical check

The approximation replaces a varying inverse-radius field by its leading thin-cross-section value.