Does energy transfer from the battery through the load through B-fields, i.e. "electromagnetic current" and not really through any other means?
The vast majority of energy transfer is from the electromagnetic fields (E & B-Fields which are coupled). There might be some fringe situations where ion or electron (-e) transfer contribute to total energy transfer, but I can't think of any off the top of my head. That's the primary point of the page you've linked, and the POYNTING diagram in figure 7 should show you the energy transfer (that's what they do). Notice that it shows all arrows pointing to the load and away from the source.
Is "Electricity" simply heavily concentrated electromagnetic fields when flowing through a conductor? For example, an arc between two high voltage electrodes is the electromagnetic energy jumping the gap, the heat and light of the arc are just by-products of this?
Well, this question isn't well phrased, so I'm going to assume that you are asking what is the principal method of energy transfer, which is the same as your first question.
Directly speaking, electricity is the flow of charged particles. The other point to make here is that the electromagnetic fields flow AROUND a conductor, not through it. Someone might nitpick about this, but it's essentially true.
As to the arc, it is caused by the electromagnetic fields exciting atoms of air to the point where you see a reaction. This is akin to asking, "What is fire? Is it just a byproduct of Oxygen combining with Carbon?". Yes, it is, but what you see is an excitation of gas molecules around that combustion.
What if the path were something like a wire surrounded by a magnetic insulator, would the e-field be pointless, or not "care" that anything is blocking it just existing on either side to create a difference in potential?
Would there be less energy available? I am greatly interested in how energy is transferred through the wire, at least correctly.
Here you need to understand that E and B fields are COUPLED. So, no, the E-Field wouldn't not (double negative, sorry) "care", it would be affected and so would the flow of electrons in response. What you've mentioned here would actually be a passive inductive element in the circuit! Look up the theory around inductors and you'll get a good idea of what's happening here.
Does the insulation of wiring (i.e. plastic) also help in reflecting some of the energy so that it does not drop over a distance? This could sense, being that an exposed wire may pick up energy in the air, and an insulated keep it out.
I'm not exactly sure how to respond to that one. The real job of the insulation is to keep the circuit in isolation from the environment, so your second sentence makes some sense, but reflecting energy isn't the appropriate way to think of what it is accomplishing.
What you are really getting at is: "What is the impedance of a wire in free space". That's simple: 376.73031 ohms (approximately).
To put it differently, a single wire (no ground wire or shield) of infinite length will appear as a 376-ish ohm load to whatever is driving the signal down that wire. This assumes that the copper in the wire has a 0 ohm resistance.
Here is the Wikipedial page for Impedance of Free Space.
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Best Answer
The rule that says a wire has the same potential at every point is a rule for lumped circuits. The lumped circuit approximation applies when the dimensions of the circuit are much smaller than the wavelength associated with the frequencies of the signals present. An antenna, on the other hand, typically must have a dimension that is a substantial fraction of the wavelength, for example 1/4 or 1/2 wavelength. Therefore an antenna is, practically by definition, not a lumped circuit, so that rule doesn't apply.
In fact there will be a potential difference (disregarding the fact that you have to be careful about even using the concept of potential in a distributed circuit) between the points along the antenna, which pushes the charge around and produces the radiation.
The red curve in the animation shows how the potential varies along the antenna. It shows there absolutely is a potential difference between different locations along the antenna.
It is true that the potential difference goes to zero at the instant when the currents are at a maximum, and vice versa. This phase difference comes about because the antenna has both inductance and capacitance, and again because the dimensions of the antenna are on the same order of magnitude as the wavelength of the signal.
The current flowing to you is why you reach the same potential as the wire. Before you touched the wire you were at some other potential. When you touched it, some charge flows in to (or out of) your body to the wire, and that is why you achieve the same potential as the wire.