On Poynting vector and electric circuits Consider the following setup.

The answer is $(3)$ - spoiler alert - and it's explained here, using Poynting vector. The thing is though

How can the electromagnetic energy flux even reach the light bulb?

In this picture it's perfectly clear how the energy flux (blue arrows) flows

Around the resistor there are both an electric field and a magnetic field so the energy flux, given by all contributions
$$ {\bf S} = {1 \over \mu_0} \bf E \times B$$
is radially inward.
But in the first image there aren't electric and magnetic fields around the light bulb (within the first second), so what makes it light up almost immediately?
At first glance, I thought it was the energy flux coming out from the battery that finally reaches the light bulb, but it can't be! Otherwise every source of electromagnetic energy reaching the bulb would make it light up.
If I put it next to an alternating dipole nothing happens (and it should, if all that matters were an energy flux).
 A: You are correct.
The video is only partly right. It is true that EM energy flows along the Poynting vector. This means that energy in a circuit is not transported by the electrons but rather by the fields outside the wire. But despite that, it is not correct that the bulb will light after 1 m/c, it will take the the full second.
There are a couple of important issues that were neglected in the video. The first is the amount of energy flux and its direction. The vast majority of the energy flux is in the space just immediately outside the wires and the energy flows parallel to the wires. There is thus very little energy* that goes across the 1 m gap between the wires. The bulb needs a rather large amount of power to light up, so the small amount that goes across the gap will not do it.
The other issue is the one you identified. It is not enough merely for the energy to reach the bulb. It must flow into the bulb and remain instead of just flowing in and out through it. According to Poynting’s theorem that requires a current in the bulb. The bulb is not an antenna, so it is not designed to produce large currents from small passing EM waves. That means that out of the small amount of power that reaches the bulb an even smaller amount remains in it.
These two issues make the conclusion wrong, even though the discussion of Poynting’s theorem and the idea that energy flows outside the wire is correct. It takes the full second for those fields, traveling just immediately outside the wire to carry the energy the long way around. It actually takes a little longer even, because the speed of this signal around the wire is less than c.
*The “very little energy” comment is assuming a typical setup with a normal lightbulb and a normal battery as shown in the video. It is possible to special build an apparatus that would transfer more energy, but such is not shown. I believe that any apparatus that would cause the bulb to light at 1 m/c would cause it to melt at 1 s.
A: An oscillating dipole with a large enough dipole moment would cause the bulb to light up. If the resistor is just a resistor with short conductive leads on each end then what happens is the original dipole induces an induced oscillating dipole moment in the resistor. This induce dipole moment results in an electric field pointing into the resistor, a voltage across the resistor, B field lines around the resistor and current through the resistor. All of the above results in the resistor lighting up if it’s a bulb.
The oscillating dipole acts like an emitting antenna, and the resistor a receiving antenna. Obviously the power transfer coefficient will be super super low and geometry dependent. But, for any geometry the transfer coefficient will technically be non zero. So if you use insane amounts of energy radiated by the dipole it will light up the bulb.
The key insight of the Veritasium video that no one seems to explicitly be pointing out is as follows: Under certain conditions, the lumped circuit model for circuits breaks down. If you look at Veritasium's circuit then it is clear the lumped elements are:

*

*The battery

*The bulb

*The switch

*The wire between the battery and switch, the wire between the battery and bulb and the wire between the switch and the bulb.

If you just look at this circuit on paper you would expect the bulb to light up instantaneously, no matter the size of the wires. The circuit schematic only encodes information about what lumped elements appear in the circuit and their topology. It does not encode any geometric information about the circuit. The point of the Veritasium video is to show that, under certain conditions, geometry is important to accurately predict circuit behavior.
The example in the Veritasium video is contrived, but these exact same physical principles become critically important at higher frequencies and for more sensitive circuits. For example, it a physics lab, you could imagine a 10 A current supply switching on and inducing a 10 mV or more noise voltage on a nearby sensitive photodetector detection line. This level of noise could be problematic for certain applications. The physical principles are the exact same for the two scenarios.
The effect can be explained either by (1) saying the lumped circuit model fails for certain geometric circuits OR (2) saying that a full model of the circuit requires the inclusion of additional, non-obvious, lumped elements such as distributed capacitors and inductors between and along wires.
A: I don't think there is a static electric field between the wires. When the circuit is first closed, there may be some small amount of energy transport via electromagnetic radiation outside the wires due to the acceleration of electrons in the wire. But each wire has no net charge at any time (before the switch is closed, immediately after the switch is closed and the current is increasing, or when the current is established and steady). When a steady current is reached, only the magnetic field exists outside the wires and battery and lamp, so there will be no Poynting vector outside the circuit (outside the wires, battery and lamp). The energy is transported by the electrons.
