Veritasium - The big misconception about electricity [video] | Hacker News Reader

Veritasium - The big misconception about electricity [video]

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31 pointssidcool15 comments

While there would be some current inducted into the bulb, I think it doesn't quite tell the whole story.

If you had another bulb and wire, that was strung parallel, but not connected on the (switch side) far end, flipping the switch ON would, by laws of relativity, for entire 0.5s necessarily have exactly the same effect on both bulbs, regardless whether or which one forms a circuit. All power transfer would be inductive/capacitative and one needs to lookup those formulas if one wants to know what will happen during the first 0.5s and some time after that.

Logically then, if the lines were perpendicular rather than parallel, only turning back into a loop far away, the switch's effect on the bulb would be exactly¹ zero until the fields themselves are folded by the wire far away.

¹ counting imperfections, reflections etc - not too close to 0

I had never thought about using superscripts on HN for footnotes. Thanks for showing that. :-)

I had never thought about using footnotes at all; for decades now, and it seems to work just fine. This seems overly pretentious, but that's just me.

  ץ𝓞𝕌'𝐫𝒆 ʷＥＬ𝒸Ｏᵐ乇  ♧

Saw the title in my feed and assumed this would be about Poynting Vectors.

Shaking my head in wonderment that I once routinely solved Jackson problems.

I liked the explanation, but I find the bit at the end confusing. Don’t the electric fields carrying energy need current in the wire to form? How are they carrying energy before the current propagates through the wire?

In the video itself, he stated that electricity propagates from the battery at the speed of light when he describes the diagram.

I think what he has missed describing is that the wire folds carry capacitance (and inductance). The capacitance leads to current flow paths shorter than resistive paths (where resistance is assumed zero).

Some commentors have asked what happens if the wire at the extreme end is cut.

If it is not cut, then intially some current goes via capacitance, making the bulb light up sooner. Ultimately, assuming DC battery output, capacitive current decays down. However, the time-consuming resistive path establishes.

If the wire is cut, capatance makes the bulb light up at first, however, the capacitive current soon decays without establishing of the resistive path, hence the bulb glow would decay.

The 'electrical' model above does not contradict the 'energy flow via fields' description, as the two are one and the same thing at the basic level.

Adding more:

It's a distributed R-C circuit, along with discrete R of the bulb.

C per unit distance depends on the positioning of the wires. How fast current ramps up depends on the distributed C and the bulb's R (as wire's distributed R is ignored in the problem statement).

L too.

Distributed L and C, discrete R.

Consider a point on a wire. As the current gets going, the current at that point will go from zero up to the final value over some (small) period of time. The magnetic field around the wire at that point must also grow from zero to it’s final strength over that period of time. The growing magnetic field causes a current in the wire flowing in the opposite direction as the first current. This slows the flow of current in the wire (so that it doesn’t happen instantaneously), and it’s also the source of the energy in the field; it’s called the self–inductance of the wire. An inductor in a circuit is just a coil of wire so that the magnetic fields from many parts of the wire all reinforce each other and cause a larger opposing current in the wire.

Y'all are too kind in saying that the video "doesn't give the complete picture." Their answer is wrong, the right answer is 1 second.

To see why, consider the same problem with the wire cut at the far end...

Looking at the injection end of an infinitly-long ideal transmission line, you can replace the effect of that in a circuit (as a two-terminal element) with a normal resistor of the same characteristic impedance.

Given a shorted end at some distance in an ideal transmission line, after some number of reflections, in a DC circuit the equivalent two-terminal element will be a zero-ohm resistor.

If you use a perfect matched terminator at the end of an ideal line, at all frequencies you can replace with a normal resistor of the same characteristic impedance.

Basic signal integrity stuff. Am I wrong?

It is all basic stuff indeed. The video, while it is good at explaining energy flow, is not giving the complete picture. Several comments there and here are providing the correct solutions.

There's something in this explanation, or else FTL communication would be possible. Place synchronized atomic clocks at receiver and at the loop of the wire, send bit by connecting or disconnecting the wire at the loop, receive bit by checking whether bulb lights up at the same time.

It's not FTL communication but the 1/c answer is misleading, it's 1m/c and that's because the bulb is 1m away from the switch, so the information propagates at the speed of light, not faster.

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