Induced Voltage in wire due to current

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gar

Senior Member
Location
Ann Arbor, Michigan
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EE
100108-2146 EST

steve66:

Consider this:
An infinitely large square loop conductor closed by a voltmeter.
Thus you have your closed path.
A small 1" square coil 1" from a side of the large coil.
A switch and battery in series with the 1" coil.
Switch open and no magnetic flux exists around either wire.
Switch closes and a magnetic field expands out from the 1" coil linking a small portion of the large coil inducing a voltage pulse.
Reduce the large coil to 1,000,000 miles. The induced voltage will be insignificantly different.

The major voltage distribution along a portion of the large coil will be mostly in a 1" to 2" region along the large coil.

So whether the large coil is closed or not the voltage exists in the small region of the wire close to the small coil.

.
 

mull982

Senior Member
O.K. Lets consider a case where we do not want to to run an analog cable in the same conduit of power wiring. Are we trying to avoid a voltage being induced on the analog cable as a result of capacitance coupling or induced voltage? Or are we trying to avoid a 60Hz frequency from interfering with the analog signal?

Now if we look at the case where the shield on a cable has a voltage induced on it from running along side a current carrying conductor. If there is a voltage induced on this shield then if we ground one end of this shield it will be at the same potential as ground and there will be no current flowing through the connection to ground.

Since this shield is at the same potential as ground with no current flow, then does this prevent induced voltage from effecting the wire it is surrounding or prevent shock when the shield is touched?

What if this shield is grounded at both ends. Will there now be current flow due to the voltage induced on the shield, and the fact that the shield has a resistance? Will this current flow cause additional heating as well as cause interference?

With this current flow I would imagine that this would cause an additional current flow on the origonal current carrying wire causing the induced voltage due to the fact that this would be similar to the primary of a transformer.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100117-2150 EST

mull982:

First, at low frequencies an electrostatic shield has little shielding effect on magnetic fields. Therefore, whether or not the electrostatic shield is present will have virtually no effect on voltages induced into the signal wires inside the electrostatic shield. You want to use twisted pair signal wires to minimized the effect from magnetic fields.

Typically for the currents in a power wire adjacent to a signal wire the induced voltage in the signal wire is small, but not small compared to some signals. For example in some straingage applications I resolve voltages as low as 1 microvolt. So a 10 microvolt induced signal might be a problem.

Now if we look at the case where the shield on a cable has a voltage induced on it from running along side a current carrying conductor. If there is a voltage induced on this shield then if we ground one end of this shield it will be at the same potential as ground and there will be no current flowing through the connection to ground.
True.
Since this shield is at the same potential as ground with no current flow, then does this prevent induced voltage from effecting the wire it is surrounding or prevent shock when the shield is touched?
I already answered part of this. Shock not likely for typical currents.
What if this shield is grounded at both ends. Will there now be current flow due to the voltage induced on the shield, and the fact that the shield has a resistance? Will this current flow cause additional heating as well as cause interference?
Yes. Don't know if it is significant. Depends upon magnitudes.
With this current flow I would imagine that this would cause an additional current flow on the original current carrying wire causing the induced voltage due to the fact that this would be similar to the primary of a transformer.
Not clear on the question.

.
 

mull982

Senior Member
Typically for the currents in a power wire adjacent to a signal wire the induced voltage in the signal wire is small, but not small compared to some signals. For example in some straingage applications I resolve voltages as low as 1 microvolt. So a 10 microvolt induced signal might be a problem.

So in an industrial setting, voltages that are caused by induction will be typically small, maybe in mV range? If there are voltages that appear at 10V, 20V, etc.. it is most likely that these are caused by capacitance coupling?

So heres a question, would it be worse to run a signal cable alongside a 120V wire carrying 20A of current, or a 480V wire carrying 1A of current. One would typically be inclined to say that its better to run it alongside the 120V cable because its less voltage, but if its carrying more current wouldn't this cable be more of a source for induction? Or is capacitive coupling more prevelent and thus the 480V be a higher source for inducing voltage through capacitance coupling even though current is small?

I already answered part of this. Shock not likely for typical currents.

I'm not following what you've stated on this? I guess shock would be more of an issue on high voltge cable shielding. I have also heard that for long distances you need to ground the HV cable shield at both ends to reduce voltage on shield. Is this due to the shielding resistance on long cable runs?

Not clear on the question.
.

As an example lets say there is a current carrying wire of 1A inducing a voltage through induction on a second wire. Now if the second wire is grounded on both ends there will be current flowing in this wire. So if there is current flowing in this secondary wire this is seen as a secondary load is a sense and therefore there will have to be an associated current flowing in the primary or current carrying wire since this arrangement is similar to a transformer. This additional current through the induction will cause the primary current on the wire to be something greater than 1A since it will add to this initial 1A current flowing in this wire.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
So in an industrial setting, voltages that are caused by induction will be typically small, maybe in mV range? If there are voltages that appear at 10V, 20V, etc.. it is most likely that these are caused by capacitance coupling?

So heres a question, would it be worse to run a signal cable alongside a 120V wire carrying 20A of current, or a 480V wire carrying 1A of current. One would typically be inclined to say that its better to run it alongside the 120V cable because its less voltage, but if its carrying more current wouldn't this cable be more of a source for induction? Or is capacitive coupling more prevelent and thus the 480V be a higher source for inducing voltage through capacitance coupling even though current is small?

Each case is going to be different depending on the size of the wires, and the spacing. Induced voltage depends largely on the distance between the conductor routing of the circuit that is getting the induced voltage. Use a twisted pair, and you get very little induced voltage. Run your hot wire overhead, and your return wire under the floor, and you will get a lot more induced voltage.

However, I would guess (its only a guess) that in most building wiring cases, the inductive induced voltage would be much greater than the capacitive induced voltage.

As an example lets say there is a current carrying wire of 1A inducing a voltage through induction on a second wire. Now if the second wire is grounded on both ends there will be current flowing in this wire. So if there is current flowing in this secondary wire this is seen as a secondary load is a sense and therefore there will have to be an associated current flowing in the primary or current carrying wire since this arrangement is similar to a transformer. This additional current through the induction will cause the primary current on the wire to be something greater than 1A since it will add to this initial 1A current flowing in this wire.


Yes, I would agree. If you are inducing current into another circuit, the current flowing in the first circuit must increase. Otherwise you would be violating the law of conservation of energy.

Steve
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
100108-2146 EST

steve66:

Consider this:
An infinitely large square loop conductor closed by a voltmeter.
Thus you have your closed path.
A small 1" square coil 1" from a side of the large coil.
A switch and battery in series with the 1" coil.
Switch open and no magnetic flux exists around either wire.
Switch closes and a magnetic field expands out from the 1" coil linking a small portion of the large coil inducing a voltage pulse.
Reduce the large coil to 1,000,000 miles. The induced voltage will be insignificantly different.

The major voltage distribution along a portion of the large coil will be mostly in a 1" to 2" region along the large coil.

So whether the large coil is closed or not the voltage exists in the small region of the wire close to the small coil.

.

OK, but the whole topic of discussion was about what happens if this "large coil" is not a coil at all. What happens if it is just a straight wire?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100118-1619 EST

mull982:

I have a test coil about 1" in diameter. It is labeled 20*f/60 = MV/Gauss. I have no idea how many turns are on this coil.

Using said test coil close, means touching the handle, to some of my breakers in my main panel I get readings up about 21 MV. Immediately next to my water line about 1.2 MV. Adjacent to an extension cord with 12 A about 1 MV. A 500 turn coil on a 1/2" square cardboard coil form and about 1/2" long has about 1.2 MV directly adjacent to the extension cord.

As soon as you separate the forward and return current paths by a large amount the magnetic flux density increases. Using a single wire about 6' long adjacent to the extension cord and several feet away for the return path produced about 0.1 MV.

A welder might be a major source of magnetically induced voltage because the currents are large and the forward and return current paths are usually separated by many feet.

If you have a floating conductor, high impedance to ground, not touching anything, then the capacitive coupled voltage is most likely much greater than any inductively coupled voltage. Further the inductively coupled voltage will be along the conductor and not to ground.

In some susceptibility tests I ran on my I232 Isolator System using a MIG welder I did not get any data errors until my CAT-5 cable was within about 4 feet of the welding cable. CAT-5 uses twisted pairs for data transfer.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100118-1625 EST

steve66:

My point was that as you make the coil very large it approximates the field around a single straight wire for positions moderately near the wire compared to the size of the coil.

Thus, the magnetic field around a straight wire for a 100 ft from the wire is about the same as if that wire was a portion of a large coil of 1,000,000 miles on a side.

.
 
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