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| "Veritasium" (YT) - "The Big Misconception About Electricity" ? |
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| Sredni:
--- Quote from: SandyCox on January 07, 2022, 10:28:55 am ---For now, I would like to point out that the rotating electric and magnetic dipoles do not represent static conditions. The static fields may exert a force on the dipoles. There is no work being done if there is force without motion. As soon as they start rotating the time derivates of the magnetic and/or electric fields are no longer zero. --- End quote --- Yes. You put energy in the field when you push a positive charge in space towards the positive wire. The Poynting vector will appear briefly during the (ideally quasi-static) movement and will disappear when you are in the new steady-state condition. Where did the energy you put into pushing the charge go? In the system (comprising the charge). Now, if you let the charge go by itself it will accelerate, so let's attach it to a damper in order to make it move (pushed by the field) slowly away from the positive wire. Poynting vector will appear again, showing energy going out of the system. Where did it go? Into the mechanical system (EDIT: yes it will eventually be dissipated as heat in the damper) necessary to assure the charge did not significantly accelerate. If you let the charge accelerate, it will radiate and I expect the Poynting vectors to show energy leaving the system in the form of EM radiation. If I wiggle the charge slowly enough not to have appreciable acceleration, I will see energy getting in and out of the system during movement - ideally I will balance to zero (EDIT: but no, the damper will take energy out by dissipating it, we need a magical quasi-static movement over an infinite time). If I wiggle it fast enough, in addition to energy going in and out of the system I will find that some energy is permanently leaving the system. I will not balance to zero for the system alone, because of radiation. I would love to see an animation showing the Poynting vector field (and its time average) when a charge is wiggling from very slowly to very fast. EDIT: I had forgotten that the damper would take energy out of the system. To make the addition and subtraction of energy reversible we need veeeeeeeeery slow motion. |
| SandyCox:
--- Quote from: Sredni on January 07, 2022, 11:42:09 am --- --- Quote from: SandyCox on January 07, 2022, 10:28:55 am ---For now, I would like to point out that the rotating electric and magnetic dipoles do not represent static conditions. The static fields may exert a force on the dipoles. There is no work being done if there is force without motion. As soon as they start rotating the time derivates of the magnetic and/or electric fields are no longer zero. --- End quote --- The Poynting vector will appear briefly during the (ideally quasi-static) movement and will disappear when you are in the new steady-state condition. --- End quote --- This implies time varying electric and/or magnetic fields. This not a static system. The conundrum with the non-unique definition of the voltage is only a problem when we try to calculate the power that is transferred through a single wire. We always have to think of two wires, the current and its return path. We know that electromagnetic energy can be transferred in a quasistatic system. The capacitor and transformer are examples. However, none of them will work in a static system. The question is if electromagnetic energy can be transferred in a purely static system by means other than the flow of charge? Here are two questions to which I do not yet know the answers: 1. Will a fluorescent lightbulb glow in a static electric field? Is the field still static once it starts glowing? 2. Is the following problem static: Chapter 7.5.2 (demo only): Rotation of an Insulating Rod in a Steady Current - YouTube |
| SandyCox:
--- Quote ---EDIT: I had forgotten that the damper would take energy out of the system. To make the addition and subtraction of energy reversible we need veeeeeeeeery slow motion. --- End quote --- Unfortunately this a not allowed. We are looking at the problem from the static point of view. We can split the solution to Maxwell's equations into static and transient solutions. We are only looking at the static solution. |
| Sredni:
--- Quote from: SandyCox on January 07, 2022, 12:27:36 pm --- --- Quote from: Sredni on January 07, 2022, 11:42:09 am --- --- Quote from: SandyCox on January 07, 2022, 10:28:55 am ---For now, I would like to point out that the rotating electric and magnetic dipoles do not represent static conditions. The static fields may exert a force on the dipoles. There is no work being done if there is force without motion. As soon as they start rotating the time derivates of the magnetic and/or electric fields are no longer zero. --- End quote --- The Poynting vector will appear briefly during the (ideally quasi-static) movement and will disappear when you are in the new steady-state condition. --- End quote --- This implies time varying electric and/or magnetic fields. This not a static system. --- End quote --- How do you expect to put or extract energy from an EM system if you do not allow for charges to change position (and moving when doing so)? The circuit with a current flowing is not a static system. It has charges moving. It's quasi-static. --- Quote ---We know that electromagnetic energy can be transferred in a quasistatic system. The capacitor and transformer are examples. However, none of them will work in a static system. --- End quote --- Not even a resistor will 'work' in a static system. DC current is quasistatics. Charges are moving. They are not static. What can be considered static is the pattern of the Poynting vector field, that will show energy coming out of the battery and getting into the resistor, to be converted to thermal energy. Instead of having a single charge losing energy to a miniature damper, I have all the charges moving in the wire losing energy to the lattice inside the resistor (in the classical view). If you want to see what happens in static conditions, wait till the battery is depleted. (and no, subtracting the Poynting vector from itself won't help) |
| snarkysparky:
charges moving along at the same rate produce constant fields. So yes the charges are moving the fields involved are constants. So the partial derivatives wrt time can be zeroed. |
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