Voltage and Energy

[ed84c]

Hmm Ive been wrestling with this for a while now (hence my return to sfn ).

 

It goes as follows;

 

Imagine a wire of 0 resistance. OK so unless there's a rule of physics im missing then we arn't loosing any energy in the wire.

 

So i decided to wave a flornscent tube around the wire. The magnetic fields around it make it glow. . So wheres the energy coming from? By introducing the florescent tube am i introducing a voltage drop?

[314159]

Well, in short, it probably wouldn't light up, unless, possibly, the field was absolutely vast.

 

Normally that experiment is electric fields, such as waving a flourescent tube near a Van der Graaff. There the electric field provides the energy.

 

If you managed to get any effect with waving a flourescent tube, it would be equivalent to waving a tube next to a strong permanent magnet. I've just double checked by having a go with a flourescent tube and my trusty neodymium magnets, to no effect. But if there were a case where you did, with a far stronger field, it would have to be down to the changing magnetic field, and so the energy would come from you waving the tube. With no field it would be easier to move it.

[swansont]

It would light up, briefly, but that's a load on the system. You run into trouble postulating zero resistance; you can't assume that and solve any standard problems like this. Superconductors have peculiar behaviors.

[YT2095]

it would depend on what the wire`s carrying IMO.

if it was RF as sufficient pawer then the tube would light up and take nothing from the wire that wasn`t alreay being "Given away" in respect of Photons.

at DC levels, then you simply have a Magnetic field, and so any Excitation to raise the tube gas to a different ionisation level must come from You waving it.

 

 

think of this, a Radio station doesn`t "Know" if your listening or tuned in or not, the power given of is just that "Given off" at the transmiting antenna

 

beleive me, the massive Radio Corps would LOVE if it worked that way, THEN they`d get a better and faster update on Viewing figures

[swansont]

it would depend on what the wire`s carrying IMO.

if it was RF as sufficient pawer then the tube would light up and take nothing from the wire that wasn`t alreay being "Given away" in respect of Photons.

at DC levels' date=' then you simply have a Magnetic field, and so any Excitation to raise the tube gas to a different ionisation level must come from You waving it.

 

 

think of this, a Radio station doesn`t "Know" if your listening or tuned in or not, the power given of is just that "Given off" at the transmiting antenna

[/quote']

 

 

But could a power station tell if you planted a transformer near the power lines, and tapped into it? I think yes.

[314159]

But could a power station tell if you planted a transformer near the power lines, and tapped into it? I think yes.

 

Yes, I remember I famous (and possibly apocryhal?) story of someone who had a power line built over his property, so he built a secondary coil and produced useable power from it. He was effectively stealing the power, as that produces an equivalent excess load on the system.

 

Similarly, a radio produces a greater drain on a radio transmitter, but only by an incredibly small amount. In reality it won't be easily measurable. So practically it won't be measurable, although theoretically it would.

 

It's an interesting question how correctly tuning the radio affects the original transmitter. I'll have to think about that one...

[woelen]

When a non-stationary current is flowing through a wire, then you will have magnetic fields, and at the same time electrical fields.

 

The total energy flux through a small surface dN of space equals [(E×B)•e/μ]dN, here E is the electric field strength, B is the magnetic flux strength, e is the unit vector, perpendicular to the surface area dN, and μ is the permeability of the medium, through which the waves are propagating, × and • are the exterior and interior product operators for 3D vectors. Perform a google search on the words "Poynting vector" and things probably may become more clear. What this shows, is that electromagnetic power can propagate through space without the need of any wires. In your setup, with a zero-resistance wire, which is connected to an AC-current source (not voltage source!!), there will be power drain from the source, due to the radiation. For similar reasons, there will be power drain from an AC-voltage source, to which a wire is connected, without closing the circuit. This effect is negligible at 50Hz AC voltage, but it can easily be observed at 100 MHz AC voltage.

 

I once did a very nice experiment. I made a 100 MHz oscillator with a HF-transistor and a simple capacitor/inductor circuit. This oscillator was capable of producing several 10's of mW of power output. Now I did the following experiment.

At the emitter of the output transistor, I connected a small tungsten bulb (2.5V/50 mA or something like that). On the other contact of that tungsten bulb I connected a little copper wire, around 1 m length. The other end of the wire was not connected with anything. Remarkably, the little bulb started glowing. What happens is that energy is lost at the wire (radiation into space), this causes a certain current drain from the oscillator, and this current drain is sufficiently large to make the little lamp glow a little. When I touched the other end of the wire, then the little light started glowing brigher. Then I became part of the irradiating object, through which energy is sent into space.

[5614]

Swansont touched on an interesting point. The only way you can get a wire where R=0 is with a superconductor. However if you apply a magnetic field to a superconductor it will lose it superconducting properties... so how can you apply a magnetic field to a wire with R=0?

[swansont]

Swansont touched on an interesting point. The only way you can get a wire where R=0 is with a superconductor. However if you apply a magnetic field to a superconductor it will lose it superconducting properties... so how can you apply a magnetic field to a wire with R=0?

 

Superconductors actually reject magnetic fields, which is how you can levitate magnets above them (Meissner effect)

[YT2095]

But could a power station tell if you planted a transformer near the power lines, and tapped into it? I think yes.

 

there`s no reason at all that it Shouldn`t be detectable if sufficient power were taken that actualy Registered on a Meter somewhere.

although I doubt you`de need a whole transformer for that, just a correctly placed pick-up coil ought to do the trick.

[5614]

Superconductors actually reject magnetic fields, which is how you can levitate magnets[/url'] above them (Meissner effect)

Yes, but that is not what I meant.

 

Cooper pairs form when electrons have antiparralel spin, however magnetic fields require electrons to have parallel spin. As the forces between Cooper pairs is quite small a magnetic field of suficient size can quite easily force the electrons to spin parallel to each other at which point they are no longer Cooper pairs, hence it will not superconduct.

 

With Type 1 superconductors when a magnetic field exceeds [math]H_c[/math] it loses it's superconducting properties.

 

With a Type 2 superconductor you have [math]H_{c1}[/math] at which point some of the magnetic field penetrates the superconductor and [math]H_{c2}[/math] at which point it loses all it's superconducting properties.

 

Although obviously you could apply a small(er than Hc) magnetic field and successfully induce a current that way.