There are magnetic and electric fields outside the DC circuit wire. The fields are static and separate. Energy is stored in the fields. There is a low amount surface charges, and to detect the surface charges requires high voltages. The magnetic fields can be used for application. If the circuit is in a vacuum then neither field is being used and the energy is potential.

The current inside the wire is delivering the energy to the resistor. Interaction between the current electrons and the resistor results in the transfer of energy.

The surface electrons result from partially blocking the current with resistance. The surface charges provide feedback to regulate the current and a local electric field through the resistor. The magnetic field results from the forward progress of the current electrons.

This is the problem I find with the Poynting vector in DC circuits:

Neither field is delivering energy outside the wire. The Poynting vector represents potential energy outside the circuit formed by, and an extension of, the current.

But, interacting with either field outside the wire, with an external electric or magnetic field not part of the circuit, does interact with the delivery of energy inside the wire by interacting with the current in some small or large way.

This is, I think, the problem. The Poynting vector is mathematically correct, but not properly related to the entire circuit.

The stored potential energy in the magnetic field outside the wire can be used for useful work by forming a solenoid. But, if the DC solenoid circuit is in a vacuum then the magnetic field remains potential energy.

A wide carbon resistor has surface charges at the interface between the wire and resistor. The cross product of the Poynting vector appears to be zero. But, there is now a clear mechanism of energy transfer.

enter image description here

Cite against Poynting vector using experiments: enter image description here

Cite describing the depth of the Poynting vector: enter image description here

  • 1
    $\begingroup$ Consider a coaxial cable with a small thickness of the dielectric. Calculate the DC electrical and magnetic field for some load $R$, and the Poynting vector. It is the energy dissipated in the resistor. $\endgroup$
    – user137289
    Commented Jun 23, 2020 at 23:16

3 Answers 3


TLDR: the brief answer to your title question is: No, the Poynting vector represents the flow of energy in all cases including a DC circuit. There is nothing magical about DC that changes the meaning of the Poynting vector.

Neither field is delivering energy outside the wire. The Poynting vector represents potential energy outside the circuit

Unfortunately, this is incorrect, and seems to be a result of a misunderstanding of Poynting’s theorem. Poynting’s theorem can be written: $$\frac{\partial}{\partial t} u + \nabla \cdot \vec S + \vec E \cdot \vec J=0$$ where the Poynting vector is $\vec S = \vec E \times \vec H$. In this equation $u$ is energy density with SI units of $\text{J/m^3}$ and $\vec S$ is an energy flux in SI units of $\text{W/m^2}$

In Poynting’s theorem the amount of energy in the fields at any point is given by $u$ and the work done on matter at any point is $\vec E \cdot \vec J$. Both of those quantities deal with energy at a given point, either energy in the field or energy leaving the field as work on matter. The Poynting vector is the only term that represents the movement of energy through the fields from one point to another. It does not even have the right units to represent a storage of potential energy. It is a flow of energy.

The current inside the wire is delivering the energy to the resistor. Interaction between the current electrons and the resistor results in the transfer of energy.

This is correct but incomplete. The $\vec E \cdot \vec J$ term does result in the transfer of energy from the EM fields to matter, but it does not provide a movement of energy from one location to another. In particular, like $u$, it is a scalar not a vector, so it cannot define a direction for energy to flow. So the question arises, where does that energy come from. That is the role of the Poynting vector. In Poynting’s theorem it is the only term that is both a vector and has the right units for energy flow.

So, the correct understanding of Poynting’s theorem is that the potential energy density in the field at each point is given by $u$ and the interaction with matter is given by $\vec E \cdot \vec J$ at each point while the flow of energy from one place to another is given by $\vec S$

Note that Poynting’s theorem is derived from Maxwell’s equations. Any system which follows Maxwell’s equations follows Poynting’s theorem. The frequency does not matter. Specifically for this question, it applies for static fields and DC circuits which do obey Maxwell's equations.

Edit responding to your newly posted references and comments:

Note that your first reference is from a known predatory publisher AASCIT. You should be highly skeptical of anything found in such journals. They do not actually do peer-review. They simply publish anything submitted, provided the author pays their publication fee. Their business model is based on deception and preys on unsuspecting young researchers and unsuspecting readers, both of whom are tricked into believing that their peer review process is real. In any case, Maxwell's equations trump this article, and Poynting's theorem follows directly from Maxwell's equations.

The reason that this paper was published in a predatory journal instead of a real peer-reviewed journal is that the analysis of their experimental setup is vastly insufficient for supporting their claim. Extraordinary claims require extraordinary evidence. When you seek to experimentally disprove a theory you MUST calculate the exact prediction of the theory for your experimental setup according to the rules of that theory. Such a calculation using the Poynting vector for his "shielded" setup was not even attempted. Without that calculation it is impossible to compare the experimental results to the Poynting vector's theoretical prediction in order to assert a failure of the theory. There is simply no way to make their conclusion based on their analysis.

Your second reference agrees with what I have been describing. Additionally, here is another reference that I like, which quantitatively shows the flow of energy in a coaxial cable: Manoj K. Harbola, "Energy flow from a battery to other circuit elements: Role of surface charges" American Journal of Physics 78, 1203 (2010). Note in particular figures 4, 5, and 6 showing respectively the E-field, the direction of the Poynting vector, and the magnitude of the longitudinal component of the Poynting vector. Figure 6 directly supports your second reference which qualitatively found "that electromagnetic energy flows ... mainly in the vicinity of the wires (and not inside them)".

Regarding the electric field in the wire. In an ideal conductor it is identically 0, and it increases as the resistance of the conductor increases. An increased conductor resistance is associated with increased energy dissipation and reduced energy transmission. So it is immediately clear without even looking at Poynting's theorem that the E-field inside the wire is not associated with energy transfer through the wire but rather with energy dissipation in the wire.

If we use Poynting's theorem inside the wire we gain further insight: the direction of this E-field is longitudinally along the wire, and the magnetic field is circumferential even inside the wire. This results, per Poynting's theorem, in an energy flux into the wire, not along the wire. This energy flux is exactly equal to the $\vec E \cdot \vec J$ energy dissipation in the wire.

  • $\begingroup$ There is a lot about the Poynting vector. The voltage source appears to be a black box providing current electrons that are driven through the circuit by surface charges. Where is the electric field in the wire from the voltage source itself? This electric field should be accelerating the current electrons in addition to surface charges which push against this field to limit the current and provide an additional acceleration through a resistor. There maybe be a problem with the direction of the Poynting vector in wide carbon resistor. $\endgroup$
    – MarkJanus1
    Commented Jun 24, 2020 at 6:19
  • $\begingroup$ Here the surface charges are at the interface between the wire and resistor and parallel to the surface of the wire. In this location, the surface charge electric field, as a separate static field from the magnetic field, is changing potential energy into the kinetic energy and the Poynting vector cross product appears to be zero. $\endgroup$
    – MarkJanus1
    Commented Jun 24, 2020 at 6:27
  • $\begingroup$ Edit: and parallel to the cross sectional area of the the wire. $\endgroup$
    – MarkJanus1
    Commented Jun 24, 2020 at 6:59
  • $\begingroup$ @MarkJanus1 I have responded both to your comments and your question edits in my answer. Please, do not edit questions in ways that invalidate previously received answers. $\endgroup$
    – Dale
    Commented Jun 24, 2020 at 13:29
  • 1
    $\begingroup$ Thank you for bad journal information. $\endgroup$
    – MarkJanus1
    Commented Jun 24, 2020 at 21:04

Think of momentum conservation. Joule effect in a resistor means infrared radiation emmited to the surroundings.

That EM radiation has a momentum, proportional to its Poynting vector. For a typical cylindrical resistor, that vector comes from it in a cylindrical symmetry.

But if we think only of electrons being scattering by the nucleous as a source for the Joule effect and radiation, the change of momentum should be in the wire direction. How a EM wave with a diretion of propagation (and momentum) transversal to the wire could be generated?

The fact is that we can not think of movement of charges forgetting the fields that are involved in this movements. Otherwise momentum for example is not conserved.

From the fields perspective, there is a flow of energy (and momentum) to the wire/resistor proportional to $\mathbf E \times \mathbf B$ during the flow of current.

And there is a flow of momentum from the wire/resistor in the form of EM infrared radiation so that the total momentum is conserved.

  • $\begingroup$ The Poynting vector has depth. But, going to try a follow up question. The current electrons are zig zagging locally and the radiation from the Joule effect can be transversal to the wire. $\endgroup$
    – MarkJanus1
    Commented Jun 24, 2020 at 5:35
  • $\begingroup$ Momentum, I think, is conserved between the de-accerlation of current electrons and the vibration of the resistor lattice which emits heat. I have looked at one paper that experimental shows the Poynting vector is not correct and another that describes the rich depth of the concept. I will place in an edit. Question: what is the mechanism that E x B uses to interact with the interior of the wire? These are external static fields. $\endgroup$
    – MarkJanus1
    Commented Jun 24, 2020 at 5:51
  • $\begingroup$ Basically when we think of electron as particles moving, bouncing (at the lattice) we forget the magnetic field. The interaction of charges doesn't follow the $3^{rd}$ Newton law, and momentum are not conserved if the fields are not included. I think that is the role of E x B. $\endgroup$ Commented Jun 24, 2020 at 16:50

Nope. I think the fallacy all comes from the idea that there will be both a magnetic and electric field around the wires, including in the space between the two wires. There is only a magnetic field in the regions around the wires and between the wires. This misconception is ALL OVER the web. The electric field is entirely within the wires.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.