Newton cooling of a Joule-heated wire
= Newton cooling of a Joule-heated wire
{c}
{title2=$T(a)-T_{\rm air}=I^2/(2\pi^2a^3\sigma\alpha)$}
The surface flux of a uniformly heated cylinder is $Ha/2$. Equating it to the Newton-cooling flux $\alpha[T(a)-T_{\rm air}]$ gives the displayed surface <temperature>. The full <temperature> profile adds the internal parabolic excess to this surface value. Electrical power per length equals the surface <heat> loss per length in steady state.