Steady temperature of a uniformly heated cylindrical wire ($T(\rho )=T_{\mathrm{air}}+\frac{Ha}{2\gamma }+\frac{H(a^2-{\rho }^2)}{4\kappa }$)

For a long homogeneous circular wire of radius $a$, uniform volumetric heating $H$, constant thermal conductivity $\kappa$ and negligible end losses, the steady heat equation is $\kappa {\rho }^{-1}(\rho T^{\prime }{)}^{\prime }+H=0$. Regularity at the axis excludes a logarithm term, giving $T=T_s+H(a^2-{\rho }^2)/(4\kappa )$. If the surface obeys Newton's law of cooling, $-\kappa T^{\prime }(a)=\gamma (T_s-T_{\mathrm{air}})$, then $T_s=T_{\mathrm{air}}+Ha/(2\gamma )$. Here $\gamma$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ and $H$ has units $\mathrm{W}{\mathrm{m}}^{-3}$. For uniform electric current $I$ and constant electrical conductivity $\sigma$, Joule heating gives $H=I^2/(\sigma {\pi }^2{a}^4)$.