Newton cooling of a Joule-heated wire ($T(a)-T_{\mathrm{air}}=I^2/(2{\pi }^2{a}^3\sigma \alpha )$)

The surface flux of a uniformly heated cylinder is $Ha/2$. Equating it to the Newton-cooling flux $\alpha [T(a)-T_{\mathrm{air}}]$ 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.