= Solution
Replacing the fixed surface <temperature> by <Newton's law of cooling> changes the <boundary condition> but not the radial <derivative> in steady state. At the surface,
$$
-\kappa T'(a)=\frac{Ha}{2}=\gamma\,[T(a)-T_{\rm air}].
$$
Hence
$$
\boxed{T(a)=T_{\rm air}+\frac{Ha}{2\gamma}
=T_{\rm air}+\frac{I^2}{2\pi^2\sigma\gamma a^3}.}
$$
Here $\gamma$ is a surface <heat transfer> coefficient in $\mathrm{W\,m^{-2}K^{-1}}$, unrelated to the <surface tension> symbol used in C10. An independent energy-balance check uses a length $\ell$: the electrical input $I^2\ell/(\sigma\pi a^2)$ must equal the surface loss $2\pi a\ell\gamma[T(a)-T_{\rm air}]$, which gives the same answer.
The complete <temperature> profile is
$$
T(\rho)=T_{\rm air}+\frac{Ha}{2\gamma}+\frac{H(a^2-\rho^2)}{4\kappa}.
$$
<Thermal conductivity> controls the internal gradient, but cancels from the surface-temperature rise because all generated <power> must eventually cross the prescribed surface. This assumes positive constant $\gamma$, uniform <electrical conductivity> and negligible end or additional radiative losses; temperature-dependent material properties could make the heating and boundary problem nonlinear.
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