Home exercise 4.2.4: Copper wire

The cross-section is $A=\pi D^2/4=\pi \times 10^{-6}=3.142\times 10^{-6}{\mathrm{m}}^2$, so

both along the wire.

$V=El=54.9\times 10^{-3}\times 100=5.49\mathrm{V}$, and

$J$ and $E$ are uniform and parallel, so

and $I^{2}R=100\times 0.549=54.9\mathrm{W}$, the same.

Units: $J/\sigma$ is (A/m${}^2$)/(S/m) $=$ V/m, and $JE$ is (A/m${}^2$)(V/m) $=$ W/m${}^3$. The power density is $p=J\cdot E=J^2/\sigma =1.75\times 10^{5}W/{\mathrm{m}}^3$, half a watt per metre of wire, spread over its whole volume. The circuit-theory $I^{2}R$ is the Joule phenomenon integrated over the conductor. At the operating temperature of a loaded cable (about $70{}^{\circ }\mathrm{C}$) $\sigma$ is about $20\%$ lower and the loss correspondingly higher.