# Does direct current exist in an infinite straight thin wire?

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?

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.

• This seems like the magnetic analogue of this question Sep 27, 2022 at 13:06
• My orginal question was to have a physical picture of the infinite energy of line current, but it was too specific. So I ask it in a stronger but general way, that is, even for the even field component (among the Fourier spectra) it is infinite.
– XjX
Sep 27, 2022 at 16:55
• Do infinite straight wires exist? No. Do we need them? No. What we may need are better textbooks that teach students how to make physical approximations on finite systems instead of introducing lazy infinite ones to avoid teaching proper approximation techniques for realistic systems. That's just a pet opinion of mine, of course. Sep 27, 2022 at 22:57
• > "The left-hand side is mathematically zero" Why do you think this? The Fourier transform has nothing to do with this. It is not zero but equals total current flowing through the surface $z=0$. Mar 13 at 2:37
• @XjX that makes no sense. Fourier concepts do not imply anything about the system, they just describe that system in a particular way. If the premise is that there is current, then there is current and its magnetic field. Apr 12 at 0:16

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.

• Hmmm... but transmission lines basically have a pair of wires, with counter-flowing currents. So the total flux (RHS) can be zero.
– XjX
Sep 27, 2022 at 17:11
• Do you have any references on "The characteristic impedance is infinite"? I may go check it and then continue the discussion.
– XjX
Sep 27, 2022 at 17:14
• @XjX See edit above. Sep 27, 2022 at 22:08
• Coaxial cable theory is based on the proper treatment of a realistic system, while infinite single wire approximations are "lazy theory" that does not want to think about how much of an effect the necessary return current has on the fields near the wire. Somebody should ask the publishers to have the best selling textbooks updated with proper approximation techniques, especially in the electrodynamics department where the lazy assumptions are all over the place. Sep 27, 2022 at 23:01

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.

• If I use the Stokes theorem, then the LHS will be two line integrals of H, one enclosing the infnite point (which is zero), and the other enclosing x=y=0 point (which is the current flux). Is that what you mean?
– XjX
Sep 27, 2022 at 16:37