Video: 'How electricity actually works'

[cianfa72]

TL;DR Summary

About the misconception on how electricity actually works

Hi, I found this interesting video about How electricity actually works.

The point he makes (see for example the video at minute 7:23) is that the energy in a light bulb connected to a battery is actually transferred by the electromagnetic field and not by electrons flowing through it. The analysis is done by mean of Poynting vector flux.

What do you think about? Thanks.

 

[Baluncore]

That is all correct.
The conductors are there to guide the electric and magnetic fields to the load.

There were many technicians quickly trained during the 20th Century for service in the armed forces. The hydraulic analogy was employed, with the product of voltage pressure, and current flow, being power, because that was what could be measured, understood, and taught quickly to High School graduates.

Now it is admitted, that when we lead a current into an ammeter, we are guiding the magnetic field into the meter to be "weighed". When we measure the voltage with a voltmeter, we are measuring the electric field between the guide wires. The cross product of the electric and magnetic fields is the Poynting vector, that represents the magnitude and direction of energy flow, the real power.

Electrical contractors can be quickly educated to follow an Electrical Code. That prohibits practices that would cause safety problems with the use of the hydraulic analogy, rather than the more realistic Poynting analysis.

 

[cianfa72]

That is all correct.
The conductors are there to guide the electric and magnetic fields to the load.

The hydraulic analogy was employed, with the product of voltage pressure, and current flow, being power, because that was what could be measured, understood, and taught quickly to High School graduates.

So in the hydraulic analogy, the voltage is the pressure, and the product of voltage times electric current flow (the power i.e. energy for unit of time) is the product of pressure times fluid current flow.

On the other hand, from the analysis is the video, one can deliver electric power to a device without using electric wires at all.

 

[renormalize]

The point he makes (see for example the video at minute 7:23) is that the energy in a light bulb connected to a battery is actually transferred by the electromagnetic field and not by electrons flowing through it. The analysis is done by mean of Poynting vector flux.
What do you think about? Thanks.

There have been previous discussions here on PF about how power flows outside of conductors. One example is this thread that I participated in:
https://www.physicsforums.com/threads/power-flow-outside-a-wire-how-close.1053463/

 

[cianfa72]

The cross product of the electric and magnetic fields is the Poynting vector, that represents the magnitude and direction of energy flow, the real power.

The video at minute 14:00 shows that the Poynting vector in the outer region near a wire segment points towards it.

If one assume infinite conducibility for the wire, the electric field $\vec E$ inside it should be negligible hence $\vec J \cdot \vec E=0$ i.e. there is not energy from the electromagnetic field transformed into heat inside the wire. So if one take a closed surface near the wire accross it, is there a balance between the energy flux entering that closed surface from outside and the flux leaving it ?

 

[Baluncore]

So in the hydraulic analogy, the voltage is the pressure, and the product of voltage times electric current flow (the power i.e. energy for unit of time) is the product of pressure times fluid current flow.

Pressure change (in pascals), multiplied by flow (in cubic metres per second), is the hydraulic power in watts.
Voltage change (in volts), multiplied by current (in coulombs per second, or amps), is the electrical power in watts.

On the other hand, from the analysis is the video, one can deliver electric power to a device without using electric wires at all.

That is true, but it is inefficient if the energy radiates. Efficiency requires energy be guided by some form of conductor. A good conductor is a good mirror of magnetic fields, and a good conductor of electric fields.

So if one take a closed surface near the wire is there a balance between the energy flux entering that closed surface from outside and the flux leaving it ?

Then there is a voltage drop along the wire. The wire does radiate heat. Energy is conserved.

 

[berkeman]

There have been previous discussions here on PF about how power flows outside of conductors. One example is this thread that I participated in:
https://www.physicsforums.com/threads/power-flow-outside-a-wire-how-close.1053463/

I was sent a PM with more similar threads...

https://www.physicsforums.com/threads/an-interesting-question-from-veritasium-on-youtube.1009268/

https://www.physicsforums.com/threads/energy-flux-direction-in-a-conducting-wire.1013582/

[Edit: Checking these links now...]

 

[DaveE]

The scary part of the video! Yep CalTech/Ligo has money and lots of instruments. Your not doing that with my oscilloscope, which is a damn sight cheaper than that one.

 

[cianfa72]

Then there is a voltage drop along the wire. The wire does radiate heat. Energy is conserved.

If we assume infinite conducibility then the voltage drop along a wire segment is null (basically there is a net current density $\vec J$ inside the wire even though the electric field $\vec E$ driving it inside the wire is negligible). So there is not electromagnetic energy transformed into heat inside the wire segment ($\vec J \cdot \vec E = 0$). Hence I believe the integral of the Poynting vector over a closed surface accross the wire segment (the total energy flux accross that closed surface) must be zero.

If the above statement is true, then does perhaps inside the wire the Poynting vector actually point outside the picked closed surface ?

 

[renormalize]

If the above statement is true, then does perhaps inside the wire the Poynting vector actually point outside the picked closed surface ?

The Poynting vector is $\vec{S}=\mu_{0}^{-1}\vec{E}\times\vec{B}$, and because $\vec E$ is strictly zero inside a perfectly conducting wire, so is $\vec S$. Thus, it can't point in any direction inside the wire.

 

[cianfa72]

The Poynting vector is $\vec{S}=\mu_{0}^{-1}\vec{E}\times\vec{B}$, and because $\vec E$ is strictly zero inside a perfectly conducting wire, so is $\vec S$. Thus, it can't point in any direction inside the wire.

Ok, so that means that, for a perfectly conductive wire, the component of Poynting vector field pointing radially inward into the wire from the surrounding space is actually zero. However, there is a component of Poynting vector field parallel to the wire, hence a flux of electromagnetic energy accross a surface orthogonal to the wire.

On the other hand if we consider a closed surface around a segment of wire then the net electromagnetic energy flux (the integral of the Poynting vector field over such a closed surface) is zero - i.e. the electromagnetic energy entering the "left" wire's cross section is the same as the one coming out from the "right" cross section.

Does the above make sense? Thanks.

 

[renormalize]

On the other hand if we consider a closed surface around a segment of wire then the net electromagnetic energy flux (the integral of the Poynting vector field over such a closed surface) is zero - i.e. the electromagnetic energy entering the "left" wire's cross section is the same as the one coming out from the "right" cross section.

Sorry, but I'm inept at visualizing the scenario from your description. Could you post a simple diagram that depicts the configuration of your wire(s), surface(s) and Poynting vectors? Thanks.

 

[Baluncore]

If the wire is a perfect conductor, then it is a perfect reflector of the magnetic field. The sum of the incident and reflected magnetic fields must then prevent energy entering the wire. Near the wire, the Poynting vector field must be parallel to the axis of the wire.

i.e. the electromagnetic energy entering the "left" wire's cross section is the same as the one coming out from the "right" cross section.

You are saying zero = zero.
There can be no EM field in a perfectly conductive wire. There is no Poynting vector inside the perfect conductor. All energy propagates outside the conductor.

 

[cianfa72]

@renormalize this is my diagram. $\vec P$ is the Poynting vector field pointing parallel to the wire's axis.

If the wire is a perfect conductor, then it is a perfect reflector of the magnetic field. The sum of the incident and reflected magnetic fields must then prevent energy entering the wire. Near the wire, the Poynting vector field must be parallel to the axis of the wire.

In steady-state conditions (stationary current flowing through the wire) why are you considering incident and reflected magnetic fields on at all ?

You are saying zero = zero.
There can be no EM field in a perfectly conductive wire. There is no Poynting vector inside the perfect conductor. All energy propagates outside the conductor.

Sorry I was sloppy. As in the attached diagram, I was considering a cylindrical surface around a wire with bases's width greater than wire cross section. In this case the energy flux integrated over each of the cylinder's bases is not null, however the energy flux entering the right basis is exactly equal to the energy flux coming out from left basis.

 

[renormalize]

@renormalize this is my diagram. $\vec P$ is the Poynting vector field pointing parallel to the wire's axis.

Thank you for the diagram.
You may have a misconception regarding the EM Poynting power flow. While it's true that there is some small flow parallel to a perfectly conducting wire just outside it, for separated conductors as you show in your circuit diagram, the vast majority of the power flows well away from the wires directly through space from the battery (power source) to the resistor (power sink). This is evident from a diagram in the thread I referenced above in post #4, which illustrates a 2D model computation of the Poynting flow between a battery and two resistors:

(from Morris & Styer-Visualizing Poynting vector energy flow in electric circuits)
In 3D, the flow field will extend to the exterior of the pair of conductors, but for the usual case of closely-spaced wires, as is typically used for electric power transmission, the bulk of the Poynting flow is concentrated in the gap between the wires. So pairs of wires are to EM power flow much like the inside-surface of a pipe is to fluid flow.

 

[Baluncore]

In steady-state conditions (stationary current flowing through the wire) why are you considering incident and reflected magnetic fields on at all ?

You are being sloppy again. There is no "(stationary current flowing through the wire)". The current on the surface of the wire is due to the reflection of the magnetic field from the surface. The wire guides the electric and magnetic field around the bends. How else could the EM fields, and the Poynting vector, follow the wire. The Poynting vector field may be steady, but energy still flows. Steady images are still reflected by mirrors.

The perfectly conductive wire, has an equipotential surface, that simply guides the electric field. The magnetic field is guided by reflection.

The incident magnetic field, generates a 90° rotated current on the surface. That 90° rotated current generates the 180° rotated magnetic field, cancelling the incident field that would otherwise have passed into the wire.
Turning 90° left, twice, generates the equal and opposite magnetic field, that then appears to come out of the mirror as the reflection.

 

[cianfa72]

While it's true that there is some small flow parallel to a perfectly conducting wire just outside it, for separated conductors as you show in your circuit diagram, the vast majority of the power flows well away from the wires directly through space from the battery (power source) to the resistor (power sink). This is evident from a diagram in the thread I referenced above in post #4, which illustrates a 2D model computation of the Poynting flow between a battery and two resistors:

Ah ok, so in that diagram there are not lines of Poynting vector field's flux that enter towards the wire.

You are being sloppy again. There is no "(stationary current flowing through the wire)". The current on the surface of the wire is due to the reflection of the magnetic field from the surface. The wire guides the electric and magnetic field around the bends. How else could the EM fields, and the Poynting vector, follow the wire. The Poynting vector field may be steady, but energy still flows. Steady images are still reflected by mirrors.

So, there is not a current flowing inside the wire but only a current on the surface of the wire. What do you mean with "steady images are still reflected by mirrors" ?

The perfectly conductive wire, has an equipotential surface, that simply guides the electric field. The magnetic field is guided by reflection.

The incident magnetic field, generates a 90° rotated current on the surface. That 90° rotated current generates the 180° rotated magnetic field, cancelling the incident field that would otherwise have passed into the wire.
Turning 90° left, twice, generates the equal and opposite magnetic field, that then appears to come out of the mirror as the reflection.

To me it seems a chicken-and-egg problem. Where the magnetic field incident on the wire's surface come from ?

 

[Baluncore]

What do you mean with "steady images are still reflected by mirrors" ?

Stand still in front of a mirror, you do not disappear.

Where the magnetic field incident on the wire's surface come from ?

From the energy source, a current generator, via the space around the wires.

 

[cianfa72]

From the energy source, a current generator, via the space around the wires.

Therefore the magnetic field incident on a wire segment (from the energy source/generator) generates a current flow on the surface on the wire itself (does it follow from Maxwell equation $\nabla \times B = \mu_0 J$ ?). Such a current in turn generates a "mirrored" magnetic field from the wire surface. After all, the total magnetic field around the wire surface in not null. Its cross product with electric field $E$ (in the region outside the wire) is the Poynting vector field which is parallel along the wire outside it (for a perfectly conductive wire).

 

[cianfa72]

BTW, I'm a bit confused about the "sources" involved in Maxwell equations -- see for instance Overdetermination_of_Maxwell's_equations. It claims that

The currents and charges are not unknowns, being freely specifiable subject to charge conservation.

So, in the case at hand, it seems like a chicken-and-egg problem: which are the sources of fields $E$ and $B$ ?

 

[Baluncore]

So, in the case at hand, it seems like a chicken-and-egg problem: which are the sources of fields E and B ?

Why does either E or B have to come first, can they not appear together?

The answer will depend on the type of generator, and how far back into the generator you trace the wires.

The square circuit diagram that separates the wires is rarely seen in the real world. The circuit is usually long and thin, with the wires close together and reasonably parallel.

One wire is insufficient for the analysis. Efficient electrical power transfer requires a closed circuit. There are two or more wires, making a transmission line, from the generator to the load. A transmission line has a characteristic impedance, that couples the electric and magnetic fields.

The AC power distribution grid employs transformers that magnetically couple the energy into a local transmission line.

An electrochemical battery provides an electric field that is switched into a transmission line or circuit. A magnet, spinning near the windings in an alternator, generates a voltage. The current that flows is dependent on the transmission line load.

 

[cianfa72]

Sorry, maybe I misinterpreted your point. For the the circuit in post#14 (in steady-state DC condition) is the magnetic field $B$ null near/on the surface of the perfectly conductive wire ?

Edit: for the previous question in #20 about the "sources" I believe the complete solution is given by using Maxwell equations + the Lorentz force equation + Newton 2nd law applied to charges.

 

[Baluncore]

Sorry, maybe I misinterpreted a point. For the the circuit in post#14 (in steady-state DC condition) is the magnetic field null near/on the surface of the perfectly conductive wire ?

The incident and reflected fields cancel into the conductive surface, but sum outwards. So B is not zero outside the wire.

If you embed math in text, it will not appear in a reply quote. Use letters, symbols, and formatting from the toolbar if you want it quoted.

 

[cianfa72]

The incident and reflected fields cancel into the conductive surface, but sum outwards. So B is not zero outside the wire.

Ok, on the other hand, as you pointed out, the surface of perfectly conductive wires are equipotentials hence the Electric field $E$ on the surface is orthogonal to the surface itself. Therefore there is Poynting vector field not null outside near the wires that is parallel to the wires itself.

 

[Baluncore]

Now would be a good time to consider how the magnetic fields of the two equal and opposite currents of a return circuit, sum between the wires, but cancel away from the wires. That, combined with the electric field between the wires, keeps the majority of the Poynting vector field in close proximity to the two wires.

 

[cianfa72]

Now would be a good time to consider how the magnetic fields of the two equal and opposite currents of a return circuit, sum between the wires, but cancel away from the wires. That, combined with the electric field between the wires, keeps the majority of the Poynting vector field in close proximity to the two wires.

Ok, you mean that the majority of Poynting vector field (hence its flux) is actually "confined/bound" into the external region between the two wires.

 

[Baluncore]

Ok, you mean that ...

No, I meant what I wrote.

I did not use the terms "confined/bound".
The phrase "the external region between the two wires" is confusing, and open to misinterpretation.

 

[renormalize]

Ok, you mean that the majority of Poynting vector field (hence its flux) is actually "confined/bound" into the external region between the two wires.

Here is an illustration of the actual Poynting vector field arising from direct current flowing in two parallel round wires:

(from Marc Boulé-DC Power Transported by Two Infinite Parallel Wires)
Power flow is overwhelmingly concentrated in the space between the wires, but there is nonetheless some flow above, below, and outside the pair.

 

[cianfa72]

(from Marc Boulé-DC Power Transported by Two Infinite Parallel Wires)
Power flow is overwhelmingly concentrated in the space between the wires, but there is nonetheless some flow above, below, and outside the pair.

Very interesting paper, thanks . In the conclusion he claims:

Currents and free charges are located on the surfaces of the wires, and because the fields are equal to zero in the perfect conductors, no power flows within them.

I believe the currents he talks about (for perfectly conductive wires) are actually due to the "drift" of free charges over the surfaces of the wires.

 

[cianfa72]

Btw, I found this interesting discussion about energy flux into capacitors and wires from Feynman's lectures.

 

[sophiecentaur]

Electrical contractors can be quickly educated to follow an Electrical Code.

The appropriate word is "trained".

Someone who asks a question on PF should surely be helped past that point where misconceptions are generated at every twist and turn by using the water flow analogy.

Using the term "Pressure" for Voltage can be unhelpful. FWIW, the quantity 'emf' is at least called a sort of force per unit charge and not a pressure. This is a perfect example where people bend over backwards to maintain the equivalence between two totally different sides of an analogy. Use the water analogy very briefly and very early on in EM education and then throw it away.