Electrons can be considered particles, and as such there is a finite but large number of them. In any circuit this number will likely not be nearly as large as you think it is.

Current doesn't care about the number of electrons. It cares about the speed of change in charge through a surface I = dQ/dt. So you can have a finite number of electrons and still have high currents.

 

Fila I think you would confuse Walter Louin

 

Current doesn't care about the number of electrons. It cares about the speed of change in charge through a surface I = dQ/dt. So you can have a finite number of electrons and still have high currents.

Electrons and Protons are the charge carriers. Charges do not exist by themselves. I was not talking about high currents, I was talking about infinite currents. A major difference.

Also, look up the definition of current sometime. It is fundamentally a measurement of the number of electrons or protons passing a static point in a unit of time. It is usually electrons, since humans use wires.

http://en.wikipedia.org/wiki/Ampere

In practical terms, the ampere is a measure of the amount of electric charge passing a point per unit time. Around 6.241 × 1018 electrons, or one coulomb, passing a given point each second constitutes one ampere.[3]

I am a math brick, I took calculus at one time and only retain the most rudimentary basics, which is why I haven't spoken up before and will likely keep quite hereafter. I am a very experience technician however.

 

I know the definition of current and understand that there is no such thing as infinite current. (smiley because I am friendly)

Thank you guys (especially steveb). I will go through your posts a few times and try to grasp all this. I will come back (maybe tomorrow) with more questions!

 

Not trying to say you can have infinite current (I'm not that naive), but for what it's worth, http://en.wikipedia.org/wiki/L'Hôpital's_rule
basically, when you divide by zero you get infinity and vica versa.

also for what it's worth, I figured that out in 8th grade

 

Infinity in the real world is what you get when you misuse math. Just because you have 0 ohms (a superconductor) does not mean you can get infinite current. Like I said, the numbers of electrons (being particles) is finite, therefore you can only get a large but measurable current.

Superconductors also have a maximum magnetic flux they can handle, exceed it and they become conductors (or stop conducting entirely, depending on the material).

A large part of engineering is defining limits.

 

Like I said, I'm not trying to argue that you can ever get infinite anything or anything like that, I was just pointing that out because many consider dividing by zero to be "undefined", I've had multiple arguments on the issue and won them all with the help of an advanced calculus student, who pointed out that rule.
Like you said, if you get it in a real world application, youre doing something wrong.

Also for what it's worth, attempting to measure any voltage across a superconductor will result in zero volts, following ohms law.

 

All good points Bill.

Here, the issue is not really infinity, which we all agree is a nonphysical mathematic abstraction.

Here, the thing to be cautious about is "zero". As we've learned with the discovery of superconductivity, superfluidity etc., science can reveal many unexpected surprises. Turns out zero can be physically achieved. But, it is achieved only within limits, as you point out.

However, even without these limits on superconductivity, the issues with an ideal coil have nothing to do with "infinity". The coil equation says that a perfect inductor will have current increasing as a ramp. So, current gets large, but not infinite. Eventually superconductivity stops, then resistance increases as temperature goes up. Then, assuming the wire does not melt and the power supply does not blow out, the current times resistance equals the applied emf, and the process stops. Hence, every inductor is an LR circuit.

Also, let's not forget that no battery is ideal either. They always have a nonzero source resistance. So, even if you could make an ideal inductor, the circuit would still be an LR curcuit.

 

Me, I'm looking forward to superconducting super caps and superconducting coils. It will be fun to watch.

I'm not trying to be a troll, but whenever you exuberant with math possibilities it is a good time to stop and do a reality check. Things like natural limits and where the power (or electrons) are coming from are good starts. Math is a truly great descriptive tool, but in many cases people using it tend to oversimplify, and a galloping Gertie is born.

 

Charge conservation in its mathematical form is dQ/dt = 0, but if we have some junctions charging up that means that at some time dQ/dt isn't zero. But maybe that happens in a few nanoseconds and after that dQ/dt = 0. So we can live with that. Nothing is simple.

I just wanted to make a clarification here. Charge conservation, as a physical law, in its mathematical form, is not really expressed by dQ/dt=0, but by the continuity equation in field theory.

The continuity equation says that the rate of change of charge in any volume must equal the flow of charge out of that region.

The concept of dQ/dt being zero is too nebulous and unrestrictive to be useful because it requires that the universe be our laboratory and it requires us to monitor charge at every point in the universe to even be sure the law is correct.

Real charge conservation is more restrictive than this ideal form of conservation because it says that in addition to charge being conserved, it must also flow in a continuous way. This is fortunate because now we need only have a small laboratory with monitors inside and on all the walls, in order to prove the law.

 

I was trying to find a good link to a description of the continuity equation, and why it's important to make the distinction between rigid conservation and conservation/continuity. In the process, I ran across this nice write-up that I think is worth sharing because it is more intuitive than the usual field derivation of the continuity equation, and it is applied to circuits which maybe makes it more interesting for the members in this forum.

 

Nice attachment. The charge is conserved and everybody is happy.
About the ideal coil. You said that the current would rise sky high, but if the source has a sinusoidal EMF the current will have its limit. For a certain inductance it would even be harmless.

EDIT: I've read that in superconducting wires the induced EMF is not proportional to dB/dt, it is only proportional to B. The current rises to a value to keep the flux the way it was. The equation was rot(J) = kappa * B , J and B being vectors and kappa a value for conductivity. J is current density and B magnetic induction. I don't care much about fancy formulas. Is that really how the current is induced in superconducting material?

 

About the ideal coil. You said that the current would rise sky high, but if the source has a sinusoidal EMF the current will have its limit. For a certain inductance it would even be harmless.?

Yes, absolutely. I was implying the application of a constant emf from something like a battery, but it's important to be clear. A sinusoidal signal with absolutely no DC offset will keep the current under control to a harmless (or even useful) level. However, even a small DC voltage offset will allow current to get integrated to high levels. So, the residual resistance is useful in this regard. Often a feedback loop is used to prevent a problem when DC offsets are not small (e.g. DC/DC converters).

EDIT: I've read that in superconducting wires the induced EMF is not proportional to dB/dt, it is only proportional to B. The current rises to a value to keep the flux the way it was. The equation was rot(J) = kappa * B , J and B being vectors and kappa a value for conductivity. J is current density and B magnetic induction. I don't care much about fancy formulas. Is that really how the current is induced in superconducting material?

I'm not really up on all the details of superconductors, so I hesitate to answer definatively. The key thing in this case is that the superconductor expels the magnetic field from the interior, which does not happen in standard conductors. From a "classical fields" point of view, this can only happen if the internal currents in the superconductor flow in a way that creates zero field inside. This implies unusual eddy currents. The flow of these eddy currents is not prohibited because the internal resistance is zero.

However, it's not clear to me why emf would not be proportional to the external flux change through and with inductance L. I say this because one of the applciations of superconductors is to make very effective lossless coils. I'd expect a usual inductive formula, but obviously zero resistance has consequences to how you would drive the device. Once the flux is established, no emf is needed to maintain the current, unless you try to do work by harnessing the energy stored in the external magnetic field (which would cause the flux to change).

I'll look into this more. It's certainly an interesting question.

 

I've read that in superconducting wires the induced EMF is not proportional to dB/dt, it is only proportional to B. The current rises to a value to keep the flux the way it was. The equation was rot(J) = kappa * B , J and B being vectors and kappa a value for conductivity. J is current density and B magnetic induction. I don't care much about fancy formulas. Is that really how the current is induced in superconducting material?

OK, so looking this up, I see you are talking about the London equations. These equations relate the current density to the internal electric and magnetic fields inside the superconductor. The nice thing about these equations is that they successfully describe the Meissner effect which is the expulsion of the fields at the transition to superconductivity.

http://en.wikipedia.org/wiki/London_equations

So this verifies your description that rot(J) = kappa * B. However, it's not clear how you get to the conclusion that "in superconducting wires the induced EMF is not proportional to dB/dt, it is only proportional to B". According to the London equations, the curl of the current density is proportional to the internal magnetic field B, but you seem to also be saying that emf is proportional to B, which I don't understand how you arrive at that conclusion.

 

According to the London equations, the curl of the current density is proportional to the internal magnetic field B, but you seem to also be saying that emf is proportional to B, which I don't understand how you arrive at that conclusion.

If you write J = kappa * E than rot(E) is proportional to B. I think. I just saw this equation in a book which I used in a first semester to learn about magnetism. There was a small chapter about superconductors.

 

steveb just to conclude the discussion I would like to ask you to verify the conclusions in my post.

In my attached circuit with ideal coil and a sinusoidal emf source the process would go like this. When we connect the terminals of the emf source with the terminals of the coil we create a path for the current to flow. But the current must be in a specific form (with specific values) to obey the Faraday's Law. Or to be more precise - if we take our circuit as a loop then by applying the Faraday's Law in integral form we get

net EMF (in the loop) = source EMF = L*di/dt .

Or we can look at it like this. The current must have a value so that

L*di/dt = source EMF for every t.

Fixed (defined) value of EMF source determines the value of the current.
Solving the above equation for sinusoidal EMF we get a sinusoidal current (finite).

Fixed (defined) value of EMF source also determines the value of the induced EMF. For this specific case

induced EMF = - source EMF because induced EMF = - L*di/dt.

But I was puzzled with the induced EMF and electric fields (induced by the coil) and I posted ''When the coil induces emf there must be an induced electric field, but if the wires are ideal induced electric field is zero so there is no induced emf. But that is contradictory! '' and than you posted

So, what about the mystery of no fields in an ideal inductor? Since the ideal inductor has zero resistance, we conclude that the induced electric field is outside the inductor, and the terminal voltage is the line integral of electric field for any external path that closes the loop at the inductor terminals.
Remember that Faraday's Law only talks about the emf developed around a fully closed path. An inductor, by itself from terminal to terminal, is not a closed path. So, to apply Faraday's law, you must close the path into a loop. Faraday's Law does not tell you where the electric field is around that closed path. It only tells you the net accumulation (line integral) of the field, which it the total emf.

That sounds reasonable to me.
So if we were to connect a voltmeter to the terminals of the coil (+ with + and - with -) we would measure V = L*di/dt. And that measured voltage is what we call the ''voltage drop'' across the inductor (coil). We use that term because of simplicity and because it can be integrated in KVL rule for practical use.

 

If you write J = kappa * E than rot(E) is proportional to B. I think. I just saw this equation in a book which I used in a first semester to learn about magnetism. There was a small chapter about superconductors.

Well, J=kappa*E is the point form of Ohms law, so I'm OK with that relation. However, conductivity (kappa) is infinity for a superconductor (resistance is zero). This is the thing that i find confusing. How do we use the relation J=infinity*E ... or E=0*J?

However, I want to be clear that I'm not saying you're wrong because my knowledge of superconductors is not too deep. I'm just saying it's not clear to me how to arrive at the conclusion that emf is proportional to B in a superconductor. I also realize that you aren't making absolute definitive statements either, but are posing reasonable questions. Unfortunately, I can't answer that particular question definitively right now.

If you have any references, such as which book you found that small chapter in, I'd be interesting to learn more about this topic.

 

steveb just to conclude the discussion I would like to ask you to verify the conclusions in my post.

In my attached circuit with ideal coil and a sinusoidal emf source the process would go like this. When we connect the terminals of the emf source with the terminals of the coil we create a path for the current to flow. But the current must be in a specific form (with specific values) to obey the Faraday's Law. Or to be more precise - if we take our circuit as a loop then by applying the Faraday's Law in integral form we get

net EMF (in the loop) = source EMF = L*di/dt .

Or we can look at it like this. The current must have a value so that

L*di/dt = source EMF for every t.

Fixed (defined) value of EMF source determines the value of the current.
Solving the above equation for sinusoidal EMF we get a sinusoidal current (finite).

Fixed (defined) value of EMF source also determines the value of the induced EMF. For this specific case

induced EMF = - source EMF because induced EMF = - L*di/dt.

But I was puzzled with the induced EMF and electric fields (induced by the coil) and I posted ''When the coil induces emf there must be an induced electric field, but if the wires are ideal induced electric field is zero so there is no induced emf. But that is contradictory! '' and than you posted "... a response ..." That sounds reasonable to me.
So if we were to connect a voltmeter to the terminals of the coil (+ with + and - with -) we would measure V = L*di/dt. And that measured voltage is what we call the ''voltage drop'' across the inductor (coil). We use that term because of simplicity and because it can be integrated in KVL rule for practical use.

I agree with what you say here, if I'm interpreting you correctly.

Note that I made a correction to my previous post above. Resistance (not conductivity) is zero in a superconductor.

 

steveb you don't have to bother with superconductivity.
I just want to say thanks for the discussion. I learned something new.

 

steveb you don't have to bother with superconductivity.
I just want to say thanks for the discussion. I learned something new.

You are welcome. I learned something new too. I didn't know about the London equations. So thank you also.

I do plan to look into superconductivity more soon. It's been one of those things on my list of subjects to explore in more detail.