Inductors and Inductance — Question 4

Question 4

Internal inductance accounts for only the fraction of current linked by each flux strip

A solid round wire of radius a carries uniform axial current I. Inside, $B(r)=\mu_0Ir/(2\pi a^2)$. For length ell, use the supplied internal-linkage rule $d\Lambda=[I(r)/I]B(r)\ell\,dr$, where $I(r)=Ir^2/a^2$.

1. Derive the internal inductance per unit length.

2. Compare it with a model in which all current is on the wire surface and the internal magnetic field is zero.

Question 4 - Solution

Internal inductance accounts for only the fraction of current linked by each flux strip

(a) Method and calculation

$$\Lambda_{\mathrm{int}}=\int_0^a\frac{r^2}{a^2}\frac{\mu_0Ir}{2\pi a^2}\ell\,dr=\frac{\mu_0I\ell}{2\pi a^4}\frac{a^4}{4}.$$$$\boxed{L_{\mathrm{int}}/\ell=\mu_0/(8\pi)}.$$

(b) Interpretation and further calculation

$$B_{\mathrm{interior}}=0\Rightarrow\Lambda_{\mathrm{int}}=0.$$$$\boxed{(L_{\mathrm{int}}/\ell)_{\mathrm{surface}}=0}.$$

Physical check

This comparison concerns internal inductance only; the external return-path geometry still contributes inductance.