Induced Voltage in wire due to current

[charlie b]

Anyhow, I think there is a similar flaw in your model: If you have a wire, you can't have a voltage between the two ends without current flow in the wire. That wouldn't satisify ohms law, since the wire has a finite resistance. Zero current through X ohms must equal 0 volts.

So there cannot exist such a thing as a battery, since it is alleged to have a voltage between its two terminals, even when it sits unpurchased on a store shelf, and since current cannot be flowing in such a situation? :-?

The math error in your logic is that, absent a complete circuit the resistance between the wire terminals (which essentially constitute a voltage source at that point) is infinite. Zero current through infinite ohms need not result in zero volts.

 

[steve66]

So there cannot exist such a thing as a battery, since it is alleged to have a voltage between its two terminals, even when it sits unpurchased on a store shelf, and since current cannot be flowing in such a situation? :-?

The math error in your logic is that, absent a complete circuit the resistance between the wire terminals (which essentially constitute a voltage source at that point) is infinite. Zero current through infinite ohms need not result in zero volts.

Not at all. A battery and a wire are two totally different things.

However, after looking at Jim's post, I've been looking Faraday's law, and it does look like you are correct about a voltage being induced. All the equations appear to show a voltage as proportional to the di/dt.

And it's also clear that a coil can have a voltage across its terminals, even though it is a short circuit, and even with zero current. But we must still satisfy ohms law - and I'm missing how that works in this case. Maybe it's the reactance component I'm missing.

 

[weressl]

More and more I am seeing new "electricians" who have no idea what a Wiggy (or Knopp) is, let alone why it is different than a digital multimeter. They "see" voltages with their expensive Fluke DMMs and get excited, then I pull out my Wiggy and like magic it's gone! Confuses the heck out of them...

Stupid electrical tricks to confuse the civilians:

• Grabbing the above wire
• pocking with a paperclip into the neutral of a receptacle, claiming to investigate if there is any electricity hiding there...
• having a bunch of folks join a chain, holding hands and grabbing a live wire on each end of the chain. Then removing one person at a time, to see who can withstand the most juice....

 

[charlie b]

But we must still satisfy ohms law - and I'm missing how that works in this case.

I had to look this up, to get it straight. According to Eshbach's "Handbook of Engineering Fundamentals," Ohm's law states, "If a steady difference in potential V (in volts) is impressed across a conductor which (a) is held at constant temperature and in which (b) there is no internal emf, V = rI. . . ." It goes on to explain the terms V, r, and I, and they are what we would expect.

In this situation, we have a wire that has an internal emf. It is produced by the nearby magnetic field. Thus, Ohm's Law does not govern the current (or lack thereof) through the wire.

It is the same situation with the battery. There is a measurable voltage across its terminals, and internal to the battery there is a conductive material that is held at constant temperature. But internal to the battery there is an emf, so a battery on the shelf (no complete circuit) will not have a current, and that fact does not create a violation of Ohm's Law.

 

[LarryFine]

They "see" voltages with their expensive Fluke DMMs and get excited, then I pull out my Wiggy and like magic it's gone! Confuses the heck out of them...

It would be even more effective to have them keep their VM on when you connect the wiggy.

 

[LarryFine]

I think the opposite can be just as valid: Voltage can be the result of current pushing through a resistance.

No, for that to happen, the load would have to "suck" current, rather than "permit current to pass," as loads do. Current is definitely the result.

 

[skeshesh]

sidenote: I was up in the mountains for new years and was craving forum reads like a junkie! hope all had a good one.

In this situation, we have a wire that has an internal emf. It is produced by the nearby magnetic field. Thus, Ohm's Law does not govern the current (or lack thereof) through the wire.

I'm pretty sure that's mostly correct but I don't think it's correct to say ohm's law doesn't "govern" the current since you can use ohm's law and the resistance of the wire to find the induced current (flux = B.A ; B= mag. field, A= Area & induced voltage = dflux/dt if I remember correctly; Induced current = induced V/R). I'm pretty sure I did this is college and had a lab too where we induced voltage, measure then measured a battery source then combined waveform.

Also, I think the voltage difference between two points results in current flow not the other way around. It seemed that some of the posts were suggesting that the same way an open circuit has zero current a short circuit can have 0 voltage and some current... I have no idea how this conclusion was drawn but I believe this is not correct.

 

[weressl]

sidenote: I was up in the mountains for new years and was craving forum reads like a junkie! hope all had a good one.



I'm pretty sure that's mostly correct but I don't think it's correct to say ohm's law doesn't "govern" the current since you can use ohm's law and the resistance of the wire to find the induced current (flux = B.A ; B= mag. field, A= Area & induced voltage = dflux/dt if I remember correctly; Induced current = induced V/R). I'm pretty sure I did this is college and had a lab too where we induced voltage, measure then measured a battery source then combined waveform.

Also, I think the voltage difference between two points results in current flow not the other way around. It seemed that some of the posts were suggesting that the same way an open circuit has zero current a short circuit can have 0 voltage and some current... I have no idea how this conclusion was drawn but I believe this is not correct.

It?s the same old philosophical dilemma as the question posed: ?Does the fall of the tree in the woods make a sound where there is no-one to hear it?? Is the voltage there if not measured? Or rather what is the nature of the steady state voltage?

 

[steve66]

No, for that to happen, the load would have to "suck" current, rather than "permit current to pass," as loads do. Current is definitely the result.

Larry: The load isn't sucking the current, the current pushes its way through the load.

In electronics and other areas of electricity, current sources are frequently found. Transistors and other electronics sometimes are more like a current source than a voltage source. In these cases, it is frequently helpful to think of the current as producing the voltage. Sometimes that helps to understand what is going on.

Of course, most real current sources start with a voltage to get their power. So I completely understand when you say it's "voltage that produces current". But at the same time, I'm not 100% sure that it isn't just as correct to say a voltage can be produced by a current.

Just something to think about.

Steve

 

[weressl]

No, for that to happen, the load would have to "suck" current, rather than "permit current to pass," as loads do. Current is definitely the result.

Larry,

Logical reasoning, EXCEPT we really do not know exactly how electricity works. There is no way of providing empirical proof to your conclusion. (See my previous post.)

 

[steve66]

I had to look this up, to get it straight. According to Eshbach's "Handbook of Engineering Fundamentals," Ohm's law states, "If a steady difference in potential V (in volts) is impressed across a conductor which (a) is held at constant temperature and in which (b) there is no internal emf, V = rI. . . ." It goes on to explain the terms V, r, and I, and they are what we would expect.

In this situation, we have a wire that has an internal emf. It is produced by the nearby magnetic field. Thus, Ohm's Law does not govern the current (or lack thereof) through the wire.

It is the same situation with the battery. There is a measurable voltage across its terminals, and internal to the battery there is a conductive material that is held at constant temperature. But internal to the battery there is an emf, so a battery on the shelf (no complete circuit) will not have a current, and that fact does not create a violation of Ohm's Law.

Charlie:

I still say the battery is a totally different animal. If I remember correctly, batteries have a "one way valve" in them that keeps current from flowing backwards through the battery. Otherwise all the current would flow inside the battery until the battery died. That one way valve makes the battery very different than the wire we are talking about.


Back to our induced voltage or current in a wire: Faraday's law must be taken over a closed surface. The induced voltage is proportional to the change in flux through the area enclosed by a complete circuit.

With a single straight wire, there is no enlcosed area. Therefore, you can't say that any voltage is induced. I think that was the original question, and ELA's comment hit the nail on the head. It isn't until you hook up a voltmeter that an actual voltage is developed. And just how you run the test leads determines how much voltage is induced. If you run them in a big loop, you get a high voltage. If you run them right against the wire you are testing, you get almost no voltage.

Steve

 

[charlie b]

. . . I don't think it's correct to say ohm's law doesn't "govern" the current since you can use ohm's law and the resistance of the wire to find the induced current

I was speaking of a wire that is not connected to anything, but was in the presence of a varying magnetic field, and I was speaking about a battery sitting on the shelf. Ohm's Law would not govern either situation. Neither case has current flowing, and in both cases the component under discussion has an internal emf.

I'm pretty sure I did this is college and had a lab too where we induced voltage, measure then measured a battery source then combined waveform.

Then you did that experiment with a complete circuit that had current flowing. Different animal.

 

[charlie b]

If I remember correctly, batteries have a "one way valve" in them that keeps current from flowing backwards through the battery. Otherwise all the current would flow inside the battery until the battery died. That one way valve makes the battery very different than the wire we are talking about.

I don’t think there is such a “one way valve,” but I might be wrong. I think that if you place a 1.5 volt battery in series with a 9 volt battery, with the two in opposite polarity, the higher capacity battery would push current backwards through the other battery. It might eventually destroy the other battery, and that might not take much time, but current would flow.

Faraday's law must be taken over a closed surface. The induced voltage is proportional to the change in flux through the area enclosed by a complete circuit.

I can’t address this without access to my physics book, which I keep at home. I fear that if I tried to look it up on-line, I would be led to (Heaven help us all) wikedpedia!

. . . ELA's comment hit the nail on the head. It isn't until you hook up a voltmeter that an actual voltage is developed.

I like Laszlo’s comment on this. A voltage source with no load attached still has a voltage at its terminals, whether or not you connect a voltmeter to the terminals to detect or measure the voltage.

 

[W=EI]

I quite often hear the term "induction" used to describe phantom voltages on wires. From previous threads I have come to learn that the cause of these phantom voltages is usually a capacitance coupling between wires.

My question is though can a voltage be induced through "induction" between two wires traveling side by side with current in one wire inducing a voltage onto the other wire. I am inclined to say no due to the fact that induction requires a magnetic flux which is defined as a magnetic field through a closed surface. Since these two wires are simply traveling next to each other then there is no closed surface to speak of and thus the magnetic field from the current carrying wire wont produce a magnetic flux onto the other wire.

If this second wire was somehow wrapped several times around the current carrying wire then we would have a closed loop and thus the magnetic field from the current carrying wire would produce a flux through the second wire and induce a voltage. This is also true of what happens when a current carrying wire is run in a metallic conduit, or has a shielding around the cable. Since the conduit or the shielding creates a closed surface, then the magnetic field creates a flux through this surface and thus induceds a voltage.

Is this correct?

Induced means induction. A rotating 60 hertz AC field exists around AC circuit conductors. Often the field is cancelled by return current in the other paired circuit conductor and this is why all of the conductors of a circuit are required to be bundled...(in raceway or cable) per NEC 300.3(B). Hysterisis is a similar problem and is addressed by NEC 300.20. I have several times required rerouting and grouping of conductors where the electrician has routed the grounded conductors thru one chase nipple and routed the phase conductors thru another...looks neat but not permitted...

 

[mull982]

Wow great discussion guys! Great theory.

I am still a little confused as to weather or not a voltage would or would not be induced in this wire since there was no closed loop?

As I mentioned in my OP, I am familiar with capacitance coupling and how when you use a high impedance meter or a wiggy you are essentially creating a voltage divider and the voltage will drop across the higher impedance. This I have seen on many low voltage inputs causing problems as others have mentioned.

However it seems that people that were referencing these phantom voltages causing problems were refereing to these voltages as being a result of induction. I was always under the impression that the "phantom" voltages were striclty a result of capacitance coupling. So that leads me to ask the questions, in an industrial enviornmental are these phantom voltges mostly a result of capacitance coupling or inductance coupling?

As we stated we can put a resistance load in series with a capacitance coupling and either see the phantom voltage or make it dissapear depending on what resistance value (meter) we are using. But what happens when you put an impedance on a wire or point that is coupled through induction? I would then think that this would act like the secondary of a transformer?

So for our given wire wich we are assuming has an induced voltage (and no current as discussed) if we take a wire and connect this wire to ground what will happen? What will be the voltage potential on the wire, will it be the induced voltage value, or the ground value?

 

[LarryFine]

. . . are these phantom voltges mostly a result of capacitance coupling or inductance coupling?

. . . what happens when you put an impedance on a wire or point that is coupled through induction?

I wouldn't think it matters. Inductance and capacitance have the impedance they each have at 60 Hz, and will behave accordingly.

The voltage will be what it will be with the coupling impednace in series with the loading resistance.

A resistor doesn't necessarily zero an induced voltage unless it's low enough. If it stops nuisance behavior, it's low enough.

 

[gar]

100108-1957 EST

From "Electric and Magnetic Fields", by Stephen S. Attwood, 3rd Edition, 1949, John Wiley & Son; page 299:

From this discussion it should be obvious that a voltage may be induced even in an open conductor when the latter is moved across a magnetic field.

Attwood was head of the Electrical Engineering Department of the University of Michigan in the mid 1950s. William G Dow followed him.

.

 

[steve66]

100108-1957 EST

From "Electric and Magnetic Fields", by Stephen S. Attwood, 3rd Edition, 1949, John Wiley & Son; page 299:
Attwood was head of the Electrical Engineering Department of the University of Michigan in the mid 1950s. William G Dow followed him.

.

Gar:

But we are not moving a wire through a magnetic field. We've got a stationary wire with a changing magnetic field. A few days ago I would have said they are the same thing, but there is a discussion on Wikipedia (sorry Charlie - if it makes you feel better, answers.com has the exact same discussion - I think they just copy the text off Wikipedia) about how two different equations are needed to describe the two different physical experiments that seem to be the same principle.

See Feynman's quote on :
http://en.wikipedia.org/wiki/Faraday's_law_of_induction

I'm not saying you are wrong, but your quote doesn't apply to this situation because we need to use the del cross E equation, not the v times B equation.

Steve

 

[gar]

100108-2055 EST

steve66:

Whether you move a wire thru a magnetic field or the field moves relative to a fixed wire the result is the same. When you have a fixed wire and a changing magnetic field from any source you get an induced voltage. The wire has no knowledge of who or what is changing the magnetic field.

.

.

 

[steve66]

Wow great discussion guys! Great theory.

As we stated we can put a resistance load in series with a capacitance coupling and either see the phantom voltage or make it dissapear depending on what resistance value (meter) we are using. But what happens when you put an impedance on a wire or point that is coupled through induction? I would then think that this would act like the secondary of a transformer?

There is one major difference between this and a transformer. In a transformer, almost all the flux is confined inside the core. Almost all the flux that the primary produces induces voltage into the secondary.

Lets start with a straight wire, open circuit on each end, and lets ask how much voltage is induced in this wire by a neighboring wire that has current flowing in it.

I think the answer depends on how much flux is linked. With a single straight wire, I don't see how any flux is linked. (Flux is linked when the magnetic field from one circuit flows through a turn of wire created by another circuit.) So I'm still somewhat skeptical on the whole voltage induced in a single straight wire idea.

However this isn't a problem for most real world applications. As soon as you hook your voltmeter probes to the two ends of our wire, you have created a loop. Now it would be easy to determine how much flux links our circuit, and we can all agree that a voltage is induced.

The magnitude of teh induced voltage depends on how we run our meter leads. If we run them right along the wire, we get a small loop, and we only link a little flux. If we route our meter leads in a big circle, we link more flux, and the induced voltage is higher.

If our straight wire and meter don't have any connection to ground, grounding a single point on the wire won't have any effect on the voltage we measure.

Steve