So what happens if you puy the lamp inside a mu-metal shield?
Draw a diagram or picture of what you mean. I would assume you mean something other than a completely sealed Faraday cage with a disconnected lamp inside. The circuit wires still must be connected to the lamp from the outside so the field disturbances from charge displacement (separation) that carry energy outside the shield follow (surround) the wire through the holes in the shields as energy carrying fields inside the shield where the wiring is connected to the lamp.So what happens if you puy the lamp inside a mu-metal shield?
https://www.physicsforums.com/insights/circuit-analysis-assumptions/Where were you, Mr Poynting, when I met Oersted and Faraday?
I don't recall reading of this before. Why?
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/Edit to add: y sin entender nada aún.
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A recurring problem on PF is raised by students who may have learned about components and CA and wish to go a “little bit” deeper to understand electrical conduction. To make it worse, many are unwilling to go further with serious study and unwilling to deal with math beyond what they already know. Sometimes, the tip-off is when the student mentions electrons.
Smoke and mirrors both are valid communications technologies.Sounds like smoke and mirrors to me, I think he must have got that from the service manual of the Enterprise warp drive.
Great video for the most part but I would say there are many times where the fact the energy is outside the wire matters (even at DC). Any time where the rate of change (wavelength) is a sizable fraction of the circuit dimensions we need to account for the dielectric media (vacuum, air, PCB substrate) the energy passes through as a propagation delay, phase shift and radiation. We can use lump-sum transmission line solutions but IMO that don't give the level of intuition you get by knowing and visualizing the energy flow paths across space. Today we use EM solvers in daily engineering to see the fields so we do care because the tools are much easier to use today with high speed computers on every desk.An Australian response…
But that may be where the "Energy Harvesting" fringe people get some of their ideas from.Circuit theory describes how much energy is transferred, but it never makes any claim about where energy is located within a circuit element nor where energy crosses a lumped element's boundary. There is no conflict here (with energy outside the wire) because circuit theory makes no claim on the question.
Bingo.But that may be where the "Energy Harvesting" fringe people get some of their ideas from.
So how does the field carrying the energy work with a DC current and how does it change with a change in the DC voltage?the fact the energy is outside the wire matters (even at DC).
A simple example is the DC motor (magnetic field) or energy storage in a DC filter capacitor(electric field)? The DC circuit is not static because there is a energy flow from the combination of electric and magnetic field components from a electrical energy source via the wires to physical mechanical load in a DC motor. The magnetic field (component of the EM force) energy is what pushes or pulls the motor round, not the electrons in the wires that stay neatly insulated from each other wire and the motor. Like I said, the electrons are like chain links in a power chain, they don't physical change, or increase then decrease in energy during the transfer of power from sprocket to sprocket but they do move round and round (like current/movement of charge). The analogy to fields is the bike chain tension that changes with the load and level of effort applied to the pedals (energy). There is nothing in circuit theory that says how the energy actually moves or is transferred, Circuit theory provides the solutions to quantities of energy, not solutions to the physical implementation of energy transfer. There is no conflict with the EM field explanation of energy transfer.So how does the field carrying the energy work with a DC current and how does it change with a change in the DC voltage?
scalar quantity equal to the line integral of the electric field strength E along a specific path linking two points a and b: |
As a consequence, even if we have extracted all possible energy from a Fermi gas by cooling it to near absolute zero temperature, the fermions are still moving around at a high speed. The fastest ones are moving at a velocity corresponding to a kinetic energy equal to the Fermi energy. This speed is known as the Fermi velocity. Only when the temperature exceeds the related Fermi temperature, do the electrons begin to move significantly faster than at absolute zero.
This enormous coulomb force is why fusion requires enormous energies to crack the barrier.In 1820, André-Marie Ampère showed that parallel wires having currents in the same direction attract one another. To the electrons, the wire contracts slightly, causing the protons of the opposite wire to be locally denser. As the electrons in the opposite wire are moving as well, they do not contract (as much). This results in an apparent local imbalance between electrons and protons; the moving electrons in one wire are attracted to the extra protons in the other. The reverse can also be considered. To the static proton's frame of reference, the electrons are moving and contracted, resulting in the same imbalance. The electron drift velocity is relatively very slow, on the order of a meter an hour but the force between an electron and proton is so enormous that even at this very slow speed the relativistic contraction causes significant effects.
So if the whole apparatus was dipped in a medium where light slows down, would it then take a longer time for the bulb to light up?