# Transport of energy in DC current

In alternating current (AC), electrons barely move a few centimeters along the conductor, with the usual frequencies and intensities in electric lines. That fact, together with Maxwell's equations, shows that it is actually impossible for it to be the kinetic energy of electrons that provides electrical energy to household appliances. In fact, it is the electromagnetic field surrounding the electrical wires that does carry energy. The Poynting vector calculation shows that there is a net flow from the electrical generator to the electrical devices that consume energy (additional details are discussed here and here).

But, in direct current (DC) once the stationary regime is reached, the electric and magnetic fields supposedly do not change, so the equation for the eletromagnetic energy density variation:

$$\frac{\partial \mathcal{E}_{EM}}{\partial t} = \frac{1}{2}\frac{\partial}{\partial t}\left(\varepsilon_0\mathbf{E}^2+\frac{\mathbf{B}^2}{\mu_0}\right) = -\mathbf{E}\cdot\mathbf{j} - \boldsymbol\nabla \cdot \left( \frac{\mathbf{E}\times\mathbf{B}}{\mu_0} \right)$$

If the electric and magnetic fields are constant then it seems that electromagnetic energy cannot "flow" and then we would have only that:

$$0 = -\mathbf{E}\cdot\mathbf{j} - \boldsymbol\nabla \cdot \left( \frac{\mathbf{E}\times\mathbf{B}}{\mu_0} \right)$$

and, therefore, the Poynting vector would be moving the charges. So it is not clear to me if it is correct to say that in DC the energy consumed by the electrical devices is indeed carried by the electrical charges.

• of course, it can "flow", it is a stationary, ie., time-independent flow Jul 13, 2022 at 23:32

If the electric and magnetic fields are constant then it seems that electromagnetic energy cannot "flow"

It appears that you are misidentifying the terms. The energy flow is given by the $$\mathbf{E}\times \mathbf{B}$$ term. The $$E^2$$ and $$B^2$$ terms are energy density, not energy flow.

So if the fields are constant then the energy density cannot change over time. Energy may still flow, as long as it does so in a way that the energy density doesn’t change.

So the equation you derived says that if net energy is flowing in to a volume then work is being done inside that volume in the form of $$\mathbf{E \cdot j}$$. This makes sense because if field energy is flowing in then either the field energy must increase (which contradicts the DC assumption) or the field must do work to transfer the energy out of the fields.

• Yes, it was deliberate, I posed the question in terms of energy density because it is simpler, integrating over the volume around the cables would naturally provide energy. Jul 13, 2022 at 22:54