Veritasium is wrong about time to light a light bulb in a long wire circuit.
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- เผยแพร่เมื่อ 10 ธ.ค. 2021
- In this video I show why the correct answer for the time to light the lamp in the thought experiment is none of the above. Then I show an alternate topology of the thought experiment where there may not be any electric or magnetic fields to produce a Poynting vector.
Here is a link to the original Veritasium video and some of the responses to it.
• Poynting Vector - วิทยาศาสตร์และเทคโนโลยี
It is nice to see when people discovering high frequency electronics and learn new things.
Wow a punch line with a straight face, love it!
Fantastic counterargument
I have to agree with you for the most part.
Veritasium and his followers are misinterpreting how energy flow or is transferred in a simple DC circuit. I think they can't explain energy flow in your last diagram because of course the currents in opposite directions will cancel each other out. Good example.
Very good video. Regarding the pointing vector: It is defined as S = E x H. If the distance between the two conductors becomes smaller, the magnetic field H becomes smaller, but the electric field between the conductors becomes larger. The product of both remains the same. If both conductors collapse, the magnetic field is 0 and the electric field is infinite.
Thanks for your comment. Actually there is no electric field between these conductors. They are parts of different circuits and connecting a voltmeter between them will show zero voltage.
@@Jimscoolstuff Both circuits have no reference potential, so they can theoretically be at any potential. At the beginning the wires are normally potential-free, they will be charged by connecting them to the battery. The upper wire becomes negative, the lower positive. This applies to both circuits. The potential of the battery (Ub) is between the upper and the lower cable. Assuming the cables have the same line capacity, the potential is distributed equally. + Ub / 2 on the upper wire and -Ub / 2 on the lower one. The same applies to the second circuit. As a result, there is a potential between the lower wire of the first circuit and the upper wire of the second circuit of Ub and, accordingly, an electric field. If the two cables are connected to a voltmeter, even if this has high-resistance, the two charges on the cables are balanced. This means that the measurement is the reason that the wires are at the same potential. This discharge is difficult to measure in the laboratory but a practical example shows the phenomenon. Helicopters are used to maintain high-voltage transmission lines. A worker must not touch the high-voltage line immediately from the helicopter. Because it is on a different potential. First, he has to charge the helicopter by making a cable connection between the helicopter and the high-voltage line. This connection brings the helicopter to the same potential as the high-voltage line so that the worker can safely touch the high-voltage line. Otherwise, the equalizing current would flow through the worker and likely kill him. The helicopter corresponds to the first circuit (it has also an internal voltage supply), the high-voltage line corresponds to the second circuit.
@@axelbeer1000 We are talking apples and oranges. With your AC powerline example there is a capacitance between the powerline and ground. There is a capacitance from the helicopter to ground and a capacitance from the helicopter to the powerline. There is no discharge going on. There would be a steady state AC current thru the capacitors. We can eliminate all this in my DC circuit by replacing the voltmeter with a wire and leaving it connected. This then eliminates any possible field between the wires.
Some commenter asked Veritasium what if the circuit was circular. Good point to make.
What happen if the wires were in a zip zap shape? They are all over-thinking a simple problem. You need current flow in an electric circuit for power or energy to be transferred.
The energy flux for an electromagnetic wave can be obtained integrating the Poynting vector over a surface, and you can get from that which is the energy radiated by the wire.
But the electric power is transmitted by the movement of the electrons. Those little guys don't want to be one near the other, and if you "push" them together, then is very much like when you compress a spring. That "compression energy" is just the Voltage.
The moving electrons at their drift speed will push - by the repulsion of the electric field - the electrons ahead in the wire, and that's how the electric energy flows.
Of course, the electric field force is mediated by photons.
In the circuit pictured the switch is useless, the circuit remains open.
The actual length of the wires is probably less than 200 feet: of course the bulb lights ‘instantly ‘. If wire and bulb are ideal as Derek says 2 seconds is the answer. The physics has the expanding electron crossing over from the subatomic realm to atomic realm at @“c”, moving from high pressure- excess electrons- to low pressure, depleted, ‘positive ‘ terminal.
If the special Veritasium wires have 0 resistance then the path is only in that. All the energy will remain and will not take detours to the stuff 1 meter away.
The actual length of the wires is probably less than 200 feet: of course the bulb lights ‘instantly ‘. If bulb and wires are ideal ( as Derek says) the answer is 2 seconds. The expanding electron crosses over from the subatomic realm to atomic realm at @“c”. That’s the physics.DC produces a magnetic ‘field ‘ which is expanding electrons physically leaving the wire. Not aware that magnetic fields or magnetism can light a bulb.
The actual wire wire is probably less than 200 feet: of course the bulb lights ‘instantly ‘. Time to complete circuit is 2 seconds ( per ideal bulb/ wires). The physics has the expanding electrons crossing over from subatomic realm to atomic realm at @“c”. DC produces a magnetic ‘field ‘- physical electrons- leaving the wire perpendicular and spiraling ( skin effect). Magnetic fields do not light up a bulb.
You could literally test the side by side batteries, rather than saying "I don't know." That bothers me. A lot.
I'm pretty sure he knows.
No matter how close your wires are, as long as they are not touching, there will be magnetic field in between. The Poynting vector in your examples are squeezed in between the wires, and almost zero elsewhere. See Poynting vector on a coax cable in Wikipedia. en.wikipedia.org/wiki/Poynting_vector
Thanks for your comment. In the video I point out that there is a magnetic field between the wires, but I do not believe that there is an electric field. We can make sure that there is no electric field by connecting the wires together a only one point. If we connect the two wires on the right near the midpoint of their length, the circuit will operate the same and this will guarantee that there is no electric field between the wires. A piece of coax could be used in place of the two wires. If the shield is connected to the center conductor at either end there will not be an electric field across the dielectric. The wires are carrying power from the batteries to the lamps but I do not see where any Poynting vector is possible.
@@Jimscoolstuff At #17:42, let's say you connect the vertical wires on the right at their mid points, which is fine. There will be no current through that junction and the wires on the right will have the same potential: no E field.This kills Poynting vector on the right side of the circuit.
But look at the wires on the left now. There are sitting at different potentials: one at +V one at -V, that is 2V difference. That creates an E field in between the wires (along with the magnetic field). There is still Poynting vector in between the lines on the left half of the circuit.
If you connect the wires on the left too, you kill the left side as well, but that means you are shorting two batteries- you won't get any power to the lamps, consistent with having no Poynting vector pointing to the lamps.
@@ramanian Yes you are right. This discussion has made me realize that I did not intend to challenge the Pointing vector. I actually wanted to challenge the idea that the fields carry the energy in a DC circuit.
A field can carry energy only when it is changing in intensity or direction. Energy can be conveyed thru a capacitor or a transformer without electron flow, but only when the field is changing. Here is a demonstration at the 12 minute mark. th-cam.com/video/XZJ2-oF7T9E/w-d-xo.html
In a DC circuit ,after the initial transient, the fields stop changing and thus can not convey energy. The electrons must convey the energy after the fields reach steady state.
Thank you for challenging me and making me think harder.