Veritasium is wrong about The Big Misconception About Electricity

I needed to answer to this Veritasium video, because I think is just wrong. I do this with no ill intend, just some friendly disagreement.
With hum do you agree? Leave it down in the comments. ;)

00:00 - Reviewing all the claims
8:30 - Veritasium's answer
9:24 - My answer

[johnnolen]

Veritasium is wrong. His explanation is correct in part but he's missing a crucial insight into the nature of the Poynting vector. The flux of the Poynting vector through a surface is the electrical power passing through said surface; that is the Integral of S dA = Power through the surface A. If we apply Gauss' divergence theorem to Integral S dA then Integral S dA = Integral div S dV, where dV is the differential volume where div S lives. S = E x H, when you calculate the divergence of a cross product such as this you get div S = - E dot curl H + H dot curl E. The first term is minus the E - field dotted with Ampere's law, and the second term is the H - field dotted with Faraday's law. The result of which expressed explicitly is div S = - E dot (J + epsilon dE/dt) + H dot (- mu dH/dt). This expression can be simplified to div S = - E dot J - d/dt(1/2 epsilon E^2 + 1/2 mu H^2). The expression inside the parentheses is the energy density of the electromagnetic field. Notice that it has to change with respect to time to produce any effect on the power density, div S. Once the current pulse achieves steady state it no longer contains any oscillatory energy, it doesn't change with respect to time. At this point div S reduces to the simple expression div S = - E dot J. There is no time dependence in this expression, and J is the current in the wire over the wire's cross section. (J = sigma E is Ohm's law written in terms of current density and electric field rather than current and voltage.) - E dot J is the ohmic loss that one would expect to find in any long transmission line. Veritasium missed the mark by being unaware of this effect. Electricity does in fact flow in wires!

[radiowqa]

The correct answer comes from a two part analysis: 1) transient pulse in the immediate time around switch closure 2) followed by steady state DC conditions. When the switch is closed, a pulse of energy rich in harmonics (I.e. AC energy -- not DC) energizes the wires...AC creates electromagnetic energy, which can propagate through space....exactly how radio waves propagate. This transient pulse lasts a VERY SHORT period of time, until DC steady state is reached. (We can calculate the pulse energy and duration if enough info about the wires was given...Oliver Heaviside figured this out over 100 years ago...). The bulb is not going to remain illuminated until steady state conditions are reached!

[stephenjackson]

There is some energy that goes from the switch to the bulb in 1m/c seconds. But it is not the amount of energy that can travel in the wires. It's more like the bulb is experiencing a small magnetic force when the switch is closed (but not exactly a magnet, just sort of like that in intensity...) The real power to fully brighten the bulb to intensity must still traverse the entire wire. If the full electrical power traversed the gap, then you'd be shocked just as badly putting your finger between the bulb and the switch as you are if you grab the live bare wires when electricity is flowing. That decidedly does not happen.

[gerrynagel]

Reposting most of this:

(1) The circuit that can be influenced in 1/c time is 1m in length, which means it is effectively an open circuit consisting of ~1 meter lengths of wire connected to the battery/switch and ~1 meter lengths of wire connected to the bulb. These lengths are effectively disconnected within 1/c time, as time goes on the effective lengths grow approximately at the speed of light, until it reaches the end of the loop when the circuit will be effectively closed. Prior to closure, the resistance of the (open) circuit keeps the current to insignificant amounts

(2) If zero volts is across the bulb, then there is no power. Without power, there is no light.
(3) if >zero volts is across the bulb, due to it's ideal nature it will produce infinite current, hence, infinite light.
(2) & (3) => There can be no voltage or current across an ideal bulb without infinite light.

(4) Such an ideal bulb would illuminate by any dynamically changing field, even from sources billions of light years away. To suggest that the fantasy conditions only apply when the switch is closed is ridiculous. The bulb would be be emitting infinite light at all times. The switch wouldn't change anything.

The only way that Veritasium can be taken honestly and/or seriously is if we assume a universe in which the laws of physics operates in some fantastical way in which his statements are the laws of physics themselves. In that case, he is disingenuous when he claims this proves that people have misconceptions about reality.

[dxk]

At 7:50 when you ask the 1st question: The current doesn't need to change. A constant current still produces a magnetic field. There just has to be a moving electric field. The moving electrons produce it.

[rsmt]

Ampere's Law shows that a magnetic field is present around any electrical current. Not necessary for the current to change.

A changing electrical field also produces a magnetic field, though. That does require change; a constant electrical field won't produce a magnetic field.

[ItsEverythingElse]

We all know what he meant by 1/c was that c was just 300, 000, 000, units be damned. I don't know why so many are focusing on that when the entire presentation was just stupid.

[discoveringthegardenofeden]

The answer is 'instantaneous': What is the potential at the location of the lightbulb? And at the breaker? An open circuit with a battery has shared the potential across the wire. Irrelevant is: length of cable, location of lightbulb, magnetic fields, 'flow', signal pulse, induction, speed of light. All these terms do not matter here. As soon as a load exists, allowing for the potential to have an escape point, the potential 'drops' all across the circuit. In an analogy: the entire open circuit is a plastic bucket of water, the switch is an instant needle-prick and the escape point for the potential, for energy potential to leave the circuit, is the lamp in the form of heat and light. The only electron that has to shift position is the electron that sits immediately next to the lightbulb, hence the speed of light is irrelevant as it is instantaneous (same in water sitting next to the hole in the bucket, which flows out immediately as the gravity potential reacts instantaneously on the water mass). Because it is instantaneous, 1/c is equally incorrect.

[victortitov]

you just completely ignored capacitive coupling (i.e., currents induced in a wire when potential of a nearby wire changes). That, together with inductances (voltages induced by changing magnetic fields) makes such a pair of parallel long wires a transmission line, like twisted pair (just not twisted).

BTW, did you check out the expert comments linked in Ve's video? they did tell Ve that the full current will be reached within a lot linger than 1/c, a few times longer than 1 second even, and that only a small fraction of the current will start after 1/c. It bothers me that Ve interprets it as the bulb is on, but if so, his answer is correct.

[noproblem]

all this fuzz comes from the apparently controversy of the net flow of electrons so slow ( few meters/sec) against the speed the light is"lit up", as always this can be explained with the hydraulic analogy : when you open a valve in a pipe FULL of water, it will come out "instantly" (at the speed of sound ) while water close to the valve barely started mooing... we'd say: water energy does not flow with water, but around the pipe as the difference of pressure between in and out, which travels at speed of sound, and makes an effect at the end of the pipe pipe, same would happen if instead of a valve you have a piston back and forth, you wont need a net water flow for energy to travel

[rsmt]

Good point at 25:20. Information can only propagate from the switch to the bulb at speed c or less. So when the switch is located 1 m from the bulb (as at the beginning), it takes 1 m/c for the first wavefront to reach the bulb. When the switch is located 1/2 light-seconds from the bulb, it will take 1/2 second.

[shahid]

Veritasium is wrong on this one, actually measurements can be made using a storage oscilloscope and by making the wire length much smaller, for example 100 meters length would do. If tested it would be evident that the bulb will light up only when the electrons complete the loop. There would be no or extremely negligent transient effect due to the capacitance of the wires.

[joinedupjon]

Agree, I think he went wrong with the chain and wheel - when the wheel is turning backwards it doesn't mean the lightbulb is sending energy back to the grid and it can be shown with the concepts that'd be appropriate to pupils seeing that demo. P=I^2R and I^2 is always positive whether I is positive or negative.
Would have been a lot better if he'd shown a plausible reason why the Maxwell Heaviside stuff gives us a better model rather than giving an incorrect reason to reject chain and wheel.

[robdev]

Thanks for the video. But I don't think you're right. When considering long power lines, or telegraphic lines, the best way to view the equivalent electrical circuit is using the 'Transmission Line Equivalent Circuit' model; If ignoring losses in the wires and the surrounding medium, the Transmission Line is modelled by inductors arranged in series along incremental lengths of the wire, interspersed with capacitors bridging the gap between the wires attached to the positive and negative terminals. In the circuit applicable to the one described, the Transmission Line is terminated with a short circuit. When the switch is closed a Heaviside Step Function propagates along the the circuit. The speed at which the Step Function propagates is not the speed of an EM wave in a vacuum, but instead determined by the inductance (L) and capacitance (C) of the system, which is itself dictated by the geometry of the circuit and the properties of the medium in which the circuit is located. The change in potential due to the leading edge of the Step Function (the high frequency components of the EM field) allows the capacitance of the system to in essence bridge the gap between the wires connecting the bulb to the positive and negative terminals and an electrical current flows through the bulb, well in advance of the time taken for the the Step Function to propagate along the whole length of the circuit and back (i.e. for the potential across the bulb to acquire the same potential as across the battery terminals).

[davidrandell]

The effects might be the same but the cause is unappreciated/ unknown. No energy, charge, photons, waves, spin, fields etc. The expanding electrons push out of surplus- negative- side at @c as “ skin effect “ to depleted- positive- reaching bulb at 1 second and positive terminal yet another 1 second ( the circuit is complete or closed) then the bulb will light up- assuming ( as Derek says) no resistance in wires and the same for ‘ideal ‘ bulb. The expanding electron moves at @c in the atom or outside the atom. Electricity is expanding electrons flowing ( pushing each other) along the wire. DC’s magnetic ‘field ‘ is physical electrons leaving the wire perpendicular yet 360 degrees. The actual wire length in Veritasium in not mentioned- definitely not very long - so we see the bulb light full bright “ instantly”: not necessary or even possibly by the magnetic ‘field ‘. DC does not produce electric ‘fields’? Only AC produces electric ‘fields’ - again expanding electrons- in a capacitor? Enough.

[michaelduke]

He just said that the power to light the bulb comes from the electromagnetic field---not the cable. For a while, I lived in a house where the kitchen lights would flash every time the nearby radar antenna would rotate (2 miles away). At sea, I held a Florence bulb that would glow in a thunderstorm. I held it in my hand. No wires. I think this was the 'free energy' Tesla was trying to broadcast.

[ZhangyXD]

There's no pointing vector before current is established. Put a battery next to a bulb, does it light?

[gupt_uv]

at @8:08 i think a magnetic field is produce by a moving charges in which @Vertassium is correct @DrBernon u mixed Ampere's law here i guess

[petersheldonharris]

YES! this is exactly what I was bothered by. Induced voltage in parallel lines requires a CHANGE in electric field. The only change here is the fraction of a second when the switch is closed.

[fredsalter]

What if the copper wire was only 1 atom thick? Would there still be a skin effect? Thanks! I love tremendous amount of discussion that Derek's video has created!