Does direct current exist in an infinite straight thin wire?

Question

Suppose an infinite-long thin wire is placed along $z$ axis in 3D space, with current density $\textbf{J}$ and static magnetic field $\textbf{H}$ satisfying the Ampère's law: $\nabla\times\textbf{H}=\textbf{J}$. By integrating both sides of the equation over surface $z=0$, we have \begin{equation} \int_{z=0}dxdy~\nabla\times\textbf{H}=\int_{z=0}dxdy~\textbf{J}. \end{equation} With finite magnetic field, the left-hand side of the equation is mathematically zero (think about the Fourier transform), leading to the right-hand side—the current flux—be zero as well. In this sense, there should be no current in this thin wire.

One interesting point is the infinite inductance of the wire: \begin{equation} L=\frac{2~\text{magnetic energy}}{\text{current flux}^2}=\frac{\int_\text{3D space} dV~\mu_0 |\textbf{H}|^2}{|\int_{z=0}dxdy~\textbf{J}|^2}=\frac{\text{non-zero value}}{0}=+\infty \end{equation} Perhaps explaining $L$ can help understanding.

Afterall, it is merely a thought experiment to transport electrons endlessly through the universe. But I am still wondering which setting is unphysical in this scenario: is it the infinite long wire, infinite large magnetic field, or others?

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Supplementary:

The $\textbf{k}$-component of 2D Fourier transform of $\textbf{H}(\textbf{x})$ is $\tilde{\textbf{H}}(\textbf{k})=\int dxdy~\textbf{H}(\textbf{x}) e^{i\textbf{k}\cdot\textbf{r}}$. As $\textbf{k}=\text{0}$, $\tilde{\textbf{H}}(\textbf{0})=\int dxdy~\textbf{H}(\textbf{x})$. Fourier transform has the rule: $\frac{d}{dx}f(x) \rightarrow ik\tilde{f}(k)$, so $\nabla\times\textbf{H}(\textbf{x}) \rightarrow i\textbf{k}\times \tilde{\textbf{H}}(\textbf{k})$. Substitute $\textbf{k}=\textbf{0}$, then LHS of the first equation is $\textbf{0}\times\tilde{\textbf{H}}(\textbf{0})$. If $\tilde{\textbf{H}}(\textbf{0})$ is a bounded value, then LHS = 0.

Answer 1

Mathematically, the electromagnetic energy per unit length is divergent for an infinite wire supporting a finite electromagnetic field. The characteristic impedance is infinite. Thus, in principle, you cannot have current on an infinite wire. However, the divergence is logarithmic, so even in the case of an extremely long wire with the return current far away, the wire can support a substantial current.

Practical single-wire transmission lines have characteristic impedances of ~600Ω, not terribly high.

Edit in response to comment:

I don't know of a reference, but it's implicit in the usual calculation of coaxial cable impedance, as $D\rightarrow\infty$. Or, you may notice that the electric field of a linear charge is proportional to $1/r$. Integrating the with respect to $r$ yields the potential, and that diverges. So, any finite charge density on the wire should, theoretically, yield infinite potential. The puzzle to me in my student days was that I knew that real wires don't behave like that, even if they are very long. The solution is that real wires aren't infinite, and they aren't infinitely far from other conductors. The idealization that a long wire suspended in space may be assumed to be infinitely long in an empty vacuum fails.

Answer 2

The integral on the LHS is not zero. Use Stokes theorem, and realize that because $|{\bf H}|\sim 1/r$, the boundary contribution is independent of the distance from the wire.