Energy Density of a Magnetic Field — Question 4

Question 4

Internal wire energy remains finite for a uniform current density

A long nonmagnetic solid wire of radius a carries a uniform current density with total current I. Inside, $B(r)=\mu_0Ir/(2\pi a^2)$. Consider only the energy inside the conductor.

1. Integrate the internal energy per unit length and infer internal inductance per unit length.

2. Find the fraction of internal energy lying outside radius a/2.

Question 4 - Solution

Internal wire energy remains finite for a uniform current density

(a) Method and calculation

$$\frac{U_{\mathrm{int}}}\ell=\int_0^a\frac{\mu_0I^2r^2}{8\pi^2a^4}2\pi r\,dr.$$$$\boxed{U_{\mathrm{int}}/\ell=\frac{\mu_0I^2}{16\pi},\quad L_{\mathrm{int}}/\ell=\frac{\mu_0}{8\pi}}.$$

(b) Interpretation and further calculation

$$U(r<c)/U_{\mathrm{int}}=c^4/a^4.$$$$\boxed{U(a/2<r<a)/U_{\mathrm{int}}=1-2^{-4}=15/16}.$$

Physical check

The radius independence assumes a uniform current distribution; it does not apply to current confined to the surface.