Energy Density of a Magnetic Field — Question 2

Question 2

Integrating toroidal field energy reproduces its inductance

A vacuum toroid of rectangular cross section has inner radius a, outer radius b, height h, and N turns carrying I. Inside, $B(r)=\mu_0NI/(2\pi r)$; neglect leakage.

1. Integrate magnetic energy over the toroidal volume.

2. Infer L and the fraction of energy inside radius $c=\sqrt{ab}$.

Question 2 - Solution

Integrating toroidal field energy reproduces its inductance

(a) Method and calculation

$$d\mathcal V=2\pi rh\,dr,\quad U=\int_a^b\frac{B(r)^2}{2\mu_0}2\pi rh\,dr.$$$$\boxed{U=\frac{\mu_0N^2I^2h}{4\pi}\ln(b/a)}.$$

(b) Interpretation and further calculation

$$\boxed{L=2U/I^2=\frac{\mu_0N^2h}{2\pi}\ln(b/a)}.$$$$\boxed{\frac{U(a<r<c)}{U}=\frac{\ln(c/a)}{\ln(b/a)}=\tfrac12}.$$

Physical check

Equal logarithmic radial intervals contain equal energy, despite unequal geometric volumes.