Induced Voltage in wire due to current

[mull982]

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?

 

[zog]

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?

Capacitive coupling, here is a nice link about it.
http://www.siemon.com/us/white_papers/02-03-22-emi.asp

 

[charlie b]

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.

Yes it will. But it won't be much.

Current flowing through a wire will create a magnetic field around the wire. This will induce a voltage in any electrically conductive material "within range" of that field. The closer the material is to the source of the field (i.e., the current-carrying wire), the higher the induced voltage will be. Wrapping a second wire in the shape of a coil simply puts more of the conductive material into range of the original field. In addition, the voltage induced in one coil of the second wire is added to the voltage induced in the adjacent coil, with the result being the more the coils the higher the induced voltgage. But if all you have is one straight wire, the induced voltage is very small (not zero, but perhaps too small to measure).

 

[drbond24]

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.

The portion of the magnetic field around the first wire that intersects the second wire will induce a current in the second wire. This quantity will be most directly affected by the strength of the magnetic field (i.e. how much current is flowing in the first wire) and how close the second wire is to the first.

As others have said, however, the induced current won't be much.

 

[Jraef]

I had a project with 24VAC inputs on a PLC which were connected to wires that ran in the same conduit with conductors for a 100HP 460V motor, about 150ft run length. Once the pump turned on, the current in the motor leads induced enough voltage on the control wires that the PLC inputs would remain Hi even when the circuit was opened at the other end (limit switch on a control valve) so the pump would never shut off. It did not induce enough to energize a Lo input, but it was enough to prevent a Hi input from being read Lo when opened. A simple dropping resistor solved it, but it was there to be sure. I interpreted that the PLC had almost zero burden but had there been more, as evidenced by the resistor, it dropped to being immeasurable.

 

[drbond24]

I remember a homework problem we worked in college. It was to calculate the voltage induced on a phone line due to its proximity to a three phase overhead power line. We did the calculations with the three phases of the power line in several different configurations, and we got discover why you see the phone line directly below the center phase conductor (at least around my part of the world, I assume its that way everywhere). At that point, most of the magnetic fields from the three phases are cancelling each other out so there is little net effect on the phone line.

It was also pointed out to us that the high voltage of the power lines created a magnetic field that was, to the low voltage phone line, rather large. This is the same reason that power and control circuits are separated in cable tray. The small amount of voltage induced by the power wiring can be enough to wreak havoc on the control wiring because they are much more sensitive due to their lower voltages. If a 120 V wire induced 1 V onto a second nearby 120 V wire, that is a less than 1% change in voltage. If that same 1 V was induced onto a 24 V control circuit, that is a 4.2% change in voltage. It makes a much bigger deal on the lower voltage wires. It can be enough to fool a control device into thinking there is a voltage at the input when there isn't, as Jraef described.

 

[steve66]

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?

For the most part, I agree with the other replies. Yes, a voltage can be induced in a wire by inductance.

One point I would like to make it that it is actually a current that is induced. And you only get an induced current if the current on the other wire is AC.

The induced current is proportional to the coupling between the two wires. For the most part, the coupling would depend on the spacing between the two wires, the lengths of the two wires, and any shielding, and if the conductors are twisted, etc. So the coupling is basically determined by the physical layout. It's also dependent on the frequency of the applied current - a higher current gives a higher di/dt, and more induced current.

So in most cases, I would agree with Charlie that the induced current would be weak. However, a few exceptions where you could get a high induced voltage would include high frequency signals, a high impedence on the line with the induced current (high impedence makes a small current develop a high voltage), or very long wire lengths.

For your "ghost voltage", the wire getting the induced voltage is usually basically an open circuit. So the high impedence factor comes into play, and you could easily see 30 or 40 volts on a good digital voltmeter, even at 60 hz with typical branch circuit wire lengths.

Steve

 

[TxEngr]

As others have pointed out, although the current is small, the real impact is the voltage 'induced' on the wire. This can wreak havoc on your control system inputs and not just the 24 volt stuff. I have seen as much as 40VAC induced on cables in industrial plants. Where you have solid state inputs and leakage currents on solid state field devices, this is enough to cause a false 'ON' at the input that will never go away. The solution as others pointed out is to load it with something. I've even seen light bulbs placed on inputs to solve the problem of loading. A-B makes interposing relays designed specifically for this problem.

 

[charlie b]

One point I would like to make it that it is actually a current that is induced.

I think not. If the wire were a single length that is not connected to anything, you could not get current (no complete path), but you would still get a voltage induced between the two ends of the wire. This is the same as the voltage induced on the secondary of a transformer, with nothing connected to the secondary terminals.

 

[LarryFine]

When I was in summer camp years ago, I was in the electronics "club". There was a standby generator system that supplied separate receptacles.

I measured a full 120v on the generator receptacles with a high-impedance voltmeter, which dropped to zero with a 4w night-light plugged in.

I think not. If the wire were a single length that is not connected to anything, you could not get current (no complete path), but you would still get a voltage induced between the two ends of the wire. This is the same as the voltage induced on the secondary of a transformer, with nothing connected to the secondary terminals.

Agreed. You can have a voltage with zero current, but not the other way around. High-voltage transmission lines can develop lethal voltages on "deenergized" lines.

It ain't dead unless it's grounded.

 

[mivey]

Even though it might be a trivial point, I might note that induction requires a changing current, not necessarily an AC current.

 

[weressl]

Even though it might be a trivial point, I might note that induction requires a changing current, not necessarily an AC current.

To be more precise it is the changing magnitude of flux that generates the voltage. As Jraef described it it can create undesireable results on low level signal circuits.

Had a 200HP SSRV starter driving an incinerator blower that would randomly 'decide' to start up. This was way back at the dawn of such things and before the in SSLV inputs to such drives were opto-isolated. It had taken a little time before I figured out what happened. The start signal wire would randomly read 50-95V. Mind you a Wiggy would not indicate voltage since the solenoid within is sufficently low impedance to drop the voltage to practically nothing. Nor would it light up a bulb.

 

[steve66]

I think not. If the wire were a single length that is not connected to anything, you could not get current (no complete path), but you would still get a voltage induced between the two ends of the wire. This is the same as the voltage induced on the secondary of a transformer, with nothing connected to the secondary terminals.

Hmmm...good point. But I'm pretty sure (but not 100% sure) that when a magnetic field cuts a wire, it induces a current, and the current causes a voltage that depends on what impedence the current sees.

Consider the opposite example: Take your open circuit wire, and connect the ends with a short circuit. Now we have an induced current flow, but no induced voltage.

Both seem to be equally vaild examples. But they also seem to be mutually exclusive - it doesn't seem like they can both be right.

Steve

 

[steve66]

Even though it might be a trivial point, I might note that induction requires a changing current, not necessarily an AC current.

Agreed, a changing current is more correct way to phrase it than saying an AC current is required.

 

[charlie b]

But I'm pretty sure (but not 100% sure) that when a magnetic field cuts a wire, it induces a current, and the current causes a voltage that depends on what impedance the current sees.

I prefer to think of it in the following terms: Voltage is what does the pushing; current is the result of the push.

Consider the opposite example: Take your open circuit wire, and connect the ends with a short circuit. Now we have an induced current flow, but no induced voltage.

We have current flow, and we have voltage as well. The voltage depends on the impedance of the wire you use to create the short circuit. No such wires are perfectly without resistance.

 

[Jraef]

... Mind you a Wiggy would not indicate voltage since the solenoid within is sufficently low impedance to drop the voltage to practically nothing...

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...

 

[ELA]

Is there a voltage present on this wire -the one with an induced current/voltage debate-
if you do not connect a voltmeter in order to read it ? ... which draws a current in order to do perform the measurement

 

[rbalex]

Laszlo's statement is the more accurate; we are essentially discussing Faraday's Law:

E=dB/dt (crude form without benefit of "symbols" to show things like closed integrals)

Electromotive Force = derivative of magnetic flux over/time.

Technically, since the magnetic flux lines MUST close, it is impossible for for a voltage to not be induced - even with an open circuit and no current flow. All that needs happen is the conductor sees a change in flux, either by direct variation or movement of the conductor with respect to the flux lines, or both.

 

[steve66]

I prefer to think of it in the following terms: Voltage is what does the pushing; current is the result of the push.

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

The voltage depends on the impedance of the wire you use to create the short circuit. No such wires are perfectly without resistance.

But in theory we can have a short circuit. And with superconductors, we can have a true short circuit.

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.

Steve

 

[steve66]

One point I would like to make it that it is actually a current that is induced.

I think not. If the wire were a single length that is not connected to anything, you could not get current (no complete path), but you would still get a voltage induced between the two ends of the wire. This is the same as the voltage induced on the secondary of a transformer, with nothing connected to the secondary terminals.

I'm not 100% sure if the current induces a voltage or another current. But the point I really wanted to make is that the induced voltage depends on the resistance (or impedence) of the circuit the wire feeds.

If the circuit only supplies a digital multimeter that has several megaohms of input impedence, you might measure 30 or 40 volts. Similar problem with PLC inputs as Jraef mentioned.

Use an older analog meter with several hundred Kohms, and you might only read 10 volts.

Use a wiggy, and the induced voltage goes to zero, as Jraef mentioned.

Steve