Does an Electric Field Outside of a Wire Supply Energy to a Resistor?

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GrizzlyBat
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Hey, my friend started asking questions about how energy works with circuits, and he linked me this page. I have never heard of poynting fields before.

The idea I find most interesting is where the "Electromagnetic energy flows out of the battery and into the empty space around the circuit". Is this implying that there is an electric field outside of the wire as well, and it is supplying energy to the resistor some how?
 
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Realistically, the energy is always contained in the fields and waves. Even in a DC circuit, we still have electromagnetic waves because every circuit has a point that it is turned on which prevents it from being truly DC. The Poynting vector is a vector the represents the flow of power of an electromagnetic wave. However, I have seen that you can make a similar treatment even of an idealized DC circuit. I believe that you can show that the fields of a DC current in (well, if it's an ideal conductor the currents are on, not in) the wire have the energy of the circuit. That is, you can show that the energy/power contained in the fields is the same as the ohmic energy, IV.

So the fields and waves contain the actual energy. This is transferred to devices by the fact that the fields induce currents in the wires/transmission lines. These currents are physical charges (electrons) and the charges interact with the circuit devices, causing energy to be dissipated. For a resistor, this is done by what is known as the Drude model. That is, the electrons move through the resistor and have a high rate of collisions with the lattice of atoms that make up the material. Each collision causes the lattice to vibrate which is heat energy. So the collisions of the charges bleed off energy from the fields that accelerate the charges into heat.

For a DC circuit, there is no electric field outside the wire, but there is a magnetic field. But any AC signal is going to have both electric and magnetic fields in an electromagnetic wave.