- #71
jsmith613
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cabraham said:Make sense? I hope I helped.
Claude
How can the polarity suddenly switch direction.. all previous posts seem to suggest the emf CAN exist without the current??
cabraham said:Make sense? I hope I helped.
Claude
jsmith613 said:How can the polarity suddenly switch direction.. all previous posts seem to suggest the emf CAN exist without the current??
cabraham said:...
Claude
I have just re-read this. What did you mean by "in the tube"? The emf will be different at all points on the tube. The Mean EMF (in time and distance) will be zero, of course - or there would be an endless DC component.jsmith613 said:the EMF induced in the tube?
if I drop a magnet into the tube for the entire time the magnet is in the tube, won't the net emf be zero and hence our graph would have a curved ends with a long flat bit in between
a little bit (but not exactly like:)
/\_________________
........\/
Ignore the ... they are there just to position the V
(this is an emf-time graph)
jsmith613 said:it seems you missed the point of the question.
I was not referrering to a radio signal
I was referring to the car moving in a magnetic field where the mag field is CONSTANT
hence the cutting of the flux induces emf...surely in this situation no current would flow but an emf would still exits?
sophiecentaur said:I have just re-read this. What did you mean by "in the tube"? The emf will be different at all points on the tube. The Mean EMF (in time and distance) will be zero, of course - or there would be an endless DC component.
sophiecentaur said:This is correct but does not contradict anything that has been written. d∅/dt is constant in this case (as the car moves steadily through the field) so the EMF will be constant and no current would flow (load = ∞Ω).
IFFF you actually tried to connect between the ends of your car aerial, to measure or get power, you would be introducing another conductor with an EMF exactly the same as the Aerial so no current would flow. The 'line integral' around the circuit would be zero so your arrangement cannot be used as a source of DC.
jsmith613 said:I mean as it is falling through the tube
two currents, in different directions, will be produced
hence cancelling - net emf is zero so graph should look like:
/\_________________
........\/
right??
sophiecentaur said:How are these two currents "cancelling"? They are in different places and at anyone time. It's like saying AC 'cancels' because there are as many positives as negatives. I still don't think you have the right picture of what's going on. Currents are flowing all over this tube in different directions and they are all changing in time. There is no one, single, value of current. You could produce a similar effect if your tube were to be replaced by a series of insulated copper rings, fixed one on top of another.
Have you heard of 'eddy currents' which are induced in iron cores of motors and transformers? They are the same phenomenon as in the tube. By laminating the core, you introduce insulating layers between the leaves of the core and eliminate / reduce these eddy currents. If you milled a slot down one side of the tube, the EMFs would still be induced but currents wouldn't flow so the magnet would fall unimpeded.
jsmith613 said:it seems you missed the point of the question.
I was not referrering to a radio signal
I was referring to the car moving in a magnetic field where the mag field is CONSTANT
hence the cutting of the flux induces emf...surely in this situation no current would flow but an emf would still exits?
sophiecentaur said:You are right in saying that it's all the Laws together that should be considered if you want to cover all cases.
I think that the models being used here are too mixed to come to a conclusion. If we're talking about a wire moving through a field then an EMF, initially, will cause some displacement of charges. (If the resistance of the wire were approaching, the current, however, would be near zero but the EMF would be the same. This emf is the sum of all elemental EMFs along the wire and the amount of current flowing will depend upon the capacity between the 'two halves' of the wire and the resistance. Eventually, current will stop and the PD across the ends will reach a steady state. That is the EMF that is referred to when discussing a transformer, for instance and the EMF is there even though the current has stopped flowing. A battery EMF is specifically defined as the PD with no current being taken.
I think we / you have stated all the facts. It remains to explain just how they apply in each new circumstance. There is a bit of chicken and egg with this topic.
sophiecentaur said:But what do you mean by "net"?
cabraham said:...
Claude
sophiecentaur said:I guess my point is that, even with no wire there, if you move your observation point through the magnetic field then there is an emf. As the resistance is infinite then there is no current but the emf would still be there because the flux is still being cut by your wire probe. Here's the thing, then. If you had two wires of equal dimensions (same capacity) and different resistivities, instantly starting to move at the same speed through the field, what you are implying, I think, is that the rate of change of measured PD wouldn't just be proportional to resistivity (i.e. not proportional to the induced EMF from dΦ/dt). I am implying that it would just be proportional to RC. This all seem like straightforward bookwork and I can't see where the current is necessarily the primary factor. Don't we always say "induced emf" in these problems?
If we're talking in terms of steady state for this wire / car then the EMF / PD will still be there but the current will have stopped flowing long ago. If the EMF were not there, the charge would leak back into place - so it must still be there. (I think that's a QED, actually)
sophiecentaur said:I think this boils down to how we choose to see things. I am happy with an emf existing even though there is no current flowing. You are happy with the current needing to flow before the emf can be measured. But EMF has units of Joules/Coulomb which is independent of the number of coulombs that have actually flowed (Current times time). I get the same answer no matter how many Coulombs have flowed but you need some current to flow and are bringing in the microscopic (electrons) into your reasoning. That seems to be going somewhere that's not necessary. (Did Maxwell use electrons or even charges with mass?) I do follow what you say, though, and it's quite reasonable.
I say tomayto and you say tomarto.
going back to the EMF qustion could you quickly look at post #85 and tell me if the diagram there seems correctsophiecentaur said:...
jsmith613 said:going back to the EMF qustion could you quickly look at post #85 and tell me if the diagram there seems correct
sophiecentaur said:Sorry- I though I'd already replied. Must have killed that window without actually posting. The diagram shows what happens for a coil (not tube) and that the peak EMF is greater when the magnet is falling faster. The area under both of those humps would be the same aamof. There is no current (scope is high impedance) so virtually no braking effect in this case.
Are you confusing the two situations or is this just 'for interest'?
jsmith613 said:most probably confusing the two!
i would have thought that it would be the same in both so simple said tube!
in the tube I would therefore presume the copper circlets are independent of each other. therefore the currents coult not cancel
here however it is one long coil so they are all interlinked and the two opposite currents DO cancel
right?
sophiecentaur said:I really don't understand what picture you have in your head. What currents "cancel". You are not using terms that I can understand. Do you mean that the mean current is zero?
When the magnet falls through a copper tube the currents are very high and vary, as I have already said, over time and distance.
sophiecentaur said:I dug out my old textbook (Panofski and Phillips) and found this passage, which, to me, suggests that the effect is there with or without a wire being present. That confirms my opinion that the emf is there without any current being needed.
BruceW said:hmm. If we define emf as the closed-loop integral of E*dL then of course it is possible for an emf to exist even though there are no charges or currents.
cabraham said:I disagree. The author presents his claim that E and B are there with or without a conductor. I agree with him so far. Then he writes the closed path line integral of E*dl, which is voltage, and relates it to B and area. But the integral E*dl, requires a specific path to have a value. It is a closed loop integral.
If the voltage is there in an imaginary closed loop in space, will charges in free space circulate in said loop? I don't think so. If a voltmeter, VM, were to have its probes placed in 2 points in empty space, would it read a non-zero voltage in the presence of a non-zero E field? What do you think?
The author correctly points out that E & B are there even in empty space. He then concludes that emf which is the integral of E*dl, must also be non-zero, which makes me wonder. In a physical conductor immersed in said fields, charges would move. In space they move, but not in a closed loop like they would in a conductor. Lorentz force is there, and the charges in free space do indeed move, but the path changes. The emf due to varying fields is path dependent. Since the path taken by free electrons in space differs from that taken in a conductor, the voltages are not equal.
To say that a voltage exists in free space can be supported by Maxwell et al. But I don't think it is the same value as the case w/ a conductor because electrons would move along a different path, and voltage value is path dependent. A CRT is an example. Two parallel plates have a charge, and an electron beam is projected between the plates. The electrons get attracted towards the positive plate and away from the negative plate.
In free space, at every point between the plates, it is safe to say there is indeed a potential. But the E field here is static. I've already stated that under static conditions, current can exist w/o voltage and vice-versa. This does not negate my earlier statement involving dynamic conditions.
Same problem, but the field between the plates is ac, sinusoidal for example. The sine curve plate voltage results in a sine curve E field. There is indeed a potential in between the plates, sine curve in nature. But the plates carry a current to maintain the sine E field. Free electrons in between the plates move back and forth, which is ac current.
Dynamic conditions, i.e. time-chaning, are different than static. No ac voltage exists w/o ac current. That is my point. In free space there can be a voltage w/o current, but only static, not dynamic.
As far as joules per coulomb goes, the voltage across the car, bumperto bumper, is the joules per coulomb of charge transported from bumper to bumper. If a resistor were connected across the bumpers, large in value so that its own current generates a B field too small to cancel the external B field, then the voltage is equal to the joules of energy per coulombs transported through said resistor.
Is this clear? Do I need further clarification? BR.
Claude
sophiecentaur said:I think it would because there's a field there. A probe (infinite imdedance voltmeter) would register the appropriate voltage for the field strength and its length.
Because they have mass they would not follow any resulting curved field. I think introducing electrons is really a bad idea for this reason. It's only when in a metal (with almost zero speed,) that electrons will follow curved E field lines.
But DC is only the limit of decreasing frequency there can hardly be a step change in what happens when the current is not changing (in any case, there is no such thing as real DC because it was switched on at some time and will be switched off)
A resistor would also have an equal and opposite emf induced in it so no current would flow. I made this point before in a different context.
yessophiecentaur said:I think we are going round in circles here and I wonder how you have arrived at some of your opinions. I am very rusty about a lot of this but, when I re-read textbooks, it usually makes sense.
Regarding my high impedance probe. It takes the form of the shortest 'electric dipole' you can get away with and the smallest meter, in situ. Lead lengths are not considered here, although, connecting to a larger meter with a twisted wire feed at right angles to the conductor shouldn't affect things.
But I'm afraid we have hijacked this thread, which was initially about a more mundane problem, if I remember right.
jsmith613 said:see diagram