Energy Transport in Circuits (Electromagnetic waves and electrons)

Question

I've been reading about how energy is actually transported in electric circuits (I first read about it here: http://amasci.com/miscon/ener1.html). The description in this article of why electrons don't move around the circuit carrying energy like buckets, picking up energy from the battery and dumping it at the load, is reasonable to me, some points in particular being that electron drift velocity in circuits is actually very low and that in AC circuits electrons don't even drift along the entire length of the circuit but rather oscillate, so they don't even travel from sources to loads and back. Instead, it's electromagnetic waves that carry energy through a circuit.

I can accept this but now I'm confused about how you would interpret calculations of energy transfer in terms of electrons, calculations which also seem reasonable but that I don't know how to interpret in terms of electromagnetic waves.

For example, suppose you have a 9 V battery and you want to increase its energy by 500 J. Since a volt is in units of J/C, $ \Delta E = V \Delta C \rightarrow \Delta C = \frac{\Delta E}{V} \rightarrow \Delta C = \frac{500 \ \text J}{9 \ \text J/\text C} \approx 55.56\ \text C$. Converting this into electrons, this is about $ 3.47*10^{20}$ electrons. This seems to suggest that you would physically need this number of electrons to pass between the terminals of the battery to transfer this energy. Is this calculation invalid? If not, how do you interpret this calculation in terms of electromagnetic waves?

Answer 1

how do you interpret this calculation in terms of electromagnetic waves?

In electromagnetism there is a lot of interplay between parameters. In particular, with Ampere's law anything where you see current can be rewritten in terms of magnetic fields and electric fields: $$\nabla \times \vec B = \mu_0 \left( \vec J + \epsilon_0 \frac{\partial \vec E}{\partial t}\right)$$

See section 11.3 and the derivation of equation 29 here: https://web.mit.edu/6.013_book/www/book.html

Basically, if you have a "lumped" circuit element where the magnetic flux is contained inside some boundary and the net charge is 0 and the only current is going in through some discrete set of terminals, then from Maxwell's equations you can derive $$-\oint_S \vec E \times \vec H \cdot d\vec a = \Sigma v_i i_i$$ The term on the left is the power flux in the fields from Poynting's theorem. So this states that the quantity given by the fields around the circuit elements can be calculated from the usual formula in circuit theory.

Notice that circuit theory tells you how much power is transferred, but it does not tell you anything about where it is transferred. Indeed, circuit theory simply cannot answer such a question because all of the geometrical and spatial information is abstracted away in circuit theory. If you want to know where the energy goes then you need to use a theory that retains the spatial information: classical electromagnetism. When you do so, you can prove that the power is transferred through the fields surrounding the circuit element and also that the amount of power is given by $P=iv$

Answer 2

Electromagnetic energy from a battery or an AC system transforming into other forms of energy doesn't really require description on the level of electromagnetic waves. The process is usually electrostatic, or in the case of motors, quasi-electro/magneto static (that is, things move but so slowly that wave-like propagation can be neglected).

The linked article correctly points out that EM energy is not transported from the source to the consumer while localized in the moving electrons. Energy is available to the consumer almost instantly, before individual electron (from the standpoint of classical theory) could move from the energy source to the consumer. So the energy gets from the power plant (or a battery) to the consuming device very fast, almost with the speed of light in vacuum. We know it isn't instant, as any connection/disconnection can manifest far away only with some delay, so we assume (and EM theory confirms) the propagation of energy is wavelike. That part of the article is fine.

But the article is also somewhat misleading, because most usual electrical appliances such as bulb or electric motor can be perfectly well understood without adding ideas of EM wave propagation. This is because the EM wave is important only in the points of dramatic change, such as switching power on or off, after that, the field is not wavelike at all. The wave is so fast, we can neglect it takes some time to travel from the source to the consumer, and we can assume that EM energy is available instantly, because the field is present already and maintained by the wires and the power source far away.

Any process of EM energy conversion into mechanical energy or heat is a local one. EM energy at point X transforms into other forms and further energy is supplied from the surrounding region. This conversion happens only when electric charge motion is present, so that's why the electrons have to move. The moving electrons do not carry the energy from the source to consumer, but they are a "handle" by which the consumer can access the EM energy already present at its location.

In your example with charging a battery, you would need a device with somewhat higher voltage that will push the electrons into the 12V battery against the usual discharge direction. This could be lab source or another battery of higher voltage. The energy of the source will be transformed into EM energy and will be present in some small amount throughout all the region where the circuit is, but the most will be near the wires with current where the magnetic field is strongest. This EM energy will flow mostly along the wires into the 12 V battery. This can be imagined as a wave process, but the distance is usually so small we can neglect any delays and just assume that the battery gets its new energy from EM field that surrounds it, but mostly from direction where the wires are coming in. The moving electrons are required as a way to extract this EM energy and store it as chemical energy inside.

Answer 3

Your calculation currently says the following:

If we take a bunch of charge with a potential of 9 V and reduce its potential to 0 V by transferring that potential energy somehow to the battery, then we would need to absorb the potential of $10^{20}$ electrons to increase the energy of the battery by 500 J.

Nowhere does it require that charge actually passes through the battery; converting volts to J/C like you did only really says "there are some charges that initially carried this energy, and now they don't carry this energy." What actually happens to the charge depends on the electrochemistry of the particular battery you're examining.

It doesn't suggest that physical movement of charge between the terminals is necessary.