Disclaimer: I am purposefully avoiding the mathematics of this theorem when posing my question.

I understand that Poynting's theorem is basically the work-energy theorem of electromagnetics, and I understand how it's derived, but I'm having difficulty wrapping my mind around the intuition.

The theorem is usually expressed in the following terms: "The amount of work done per unit time on charges by the electromagnetic force is equal to the decrease in energy stored in the magnetic and electric fields in some charge and current configuration within some volume, minus the amount of energy that flows out through a surface bounding the volume" My confusion lies in the latter half of the explanation...the bit about electromagnetic energy flowing out through the surface.

This refers only to electromagnetic radiation, does it not? I realize the Poynting vector can be applied to any magnetostatic configuration to describe the direction of the flow of power, but lets say for the example of a current carrying loop of wire (for simplicity lets pretend its a superconducting wire and no energy is being lost to any sort of internal resistance and the circuit bears no load) , where the net flux of the E and B fields through the surface are zero, this cannot possibly transport energy out of the bounding surface, can it?

  • 2
    $\begingroup$ I generally dislike requests for intuition, because yours is not the same as mine. $\endgroup$
    – Jon Custer
    Commented Sep 21, 2016 at 16:00
  • $\begingroup$ In the situation you describe there is no work done on charges, the fields (and therefore the energy density) are constant and there is no Poynting flux. The net flux of the E- and B-fields through the surface is not relevant. $\endgroup$
    – ProfRob
    Commented Sep 21, 2016 at 16:18
  • $\begingroup$ Related http://physics.stackexchange.com/q/109469/59023 and possibly useful http://physics.stackexchange.com/q/234819/59023. $\endgroup$ Commented Sep 23, 2016 at 16:43

2 Answers 2


Apologies if you have read this Wikipedia Poynings Vector article, but it does contains lots of examples of the fields involved.

enter image description here

A DC circuit consisting of a battery (V) and resistor (R), showing the direction of the Poynting vector (S, blue) in the space surrounding it, along with the fields it is derived from; the electric field (E, red) and the magnetic field (H, green). In the region around the battery the Poynting vector is directed outward, indicating power flowing out of the battery into the fields; in the region around the resistor the vector is directed inward, indicating field power flowing into the resistor. Across any plane P between the battery and resistor, the Poynting flux is in the direction of the resistor.

A coaxial cable, as an example.

enter image description here

Poynting vector in a coaxial cable, shown in red.

The Poynting vector contained inside the dielectric insulator of a coaxial cable is almost pointed parallel to the inner core wire axis (assuming no fields outside the cable and a wavelength longer than the diameter of the cable, including DC). Electrical energy delivered to the load is flowing entirely through the dielectric between the conductors. The electric field strength is almost nil, implying that only a small amount of energy flows through the conductors themselves, since the electric field strength is nearly zero. The energy flowing in the conductors flows radially into the conductors and accounts for energy lost to resistive heating of the conductor. Beyond the cable boundary, the magnetic fields of the outer and inner conductor negate each other, resulting in no overall energy flows outside the conductor.


This refers only to electromagnetic radiation, does it not?

Not only EM radiation. If there is some work done on the charges of a circuit (Joule effect for example), there is a flow of energy from the source. In a static configuration like a DC circuit, where the EM energy is constant everywhere, the flow equals the work done.

For a constant current flowing in a superconductor loop, the EM energy comes from the (static) magnetic field, but there is no work anywhere and no energy flow.

  • $\begingroup$ I assume you have come to this conclusion based on E = 0 and thus B×0 = 0, but what about fields caused by surfaces charges. Or better yet, just a current loop that has a net charge? The poynting vector wouldn't be zero in this case? Yet I don't see the supply for this energy? ( obviously if the charge density is moving like a point charge ofcourse there is flow, but not in this case) $\endgroup$ Commented May 8, 2022 at 21:28
  • $\begingroup$ Or would this situation have a poynting vector as the distribution wants to fly apart and re-distribute field energy, I would assume so. $\endgroup$ Commented May 8, 2022 at 21:30
  • $\begingroup$ Or I guess energy circulated back into the same location actually . As div S = 0 so there isn't a constant supply of energy needed to be radiated. $\endgroup$ Commented May 8, 2022 at 21:46
  • $\begingroup$ Yes, I assume no E field, because the potential is the same everywhere for the OP situation, and E is minus the gradient of V for static condition. $\endgroup$ Commented May 8, 2022 at 22:17

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