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Let's say, we have a loop of wire of length \$L>0\$. Let's say, hypothetically, it is a loop of non-ideal wire. That means that the wire has a resistance per unit length \$ρ>0\$. Let's further assume that there is a changing magnetic flux through that loop of wire. Faraday's Law of Induction, as it is widely understood, dictates that a current will be induced in that loop of wire. Let's call that current \$I\$ (where \$I>0\$).

Now, for the sake of argument, let's divide that wire into infinitely many segments of infinitesimal length. Let each of those segments have length \$dl\$. That means that each segment will have resistance \$ρdl\$, and since each segment has the same current \$I\$ passing through it (KCL), the voltage drop across each segment should be \$Iρdl\$ (Ohm's Law).

Starting at any point in the loop of wire, we can sum up the voltage drop across each infinitesimal segment for the entire loop (the segments are in series):

\$\int\limits_{0}^{L}Iρdl=IρL>0V\$

Now, since we start at any point in the loop and come back around to the same point, that means that the voltage at that point (relative to itself) is both \$0V\$ (trivially) and \$IρL\$, which is a contradiction since we assumed that neither of \$I\$, \$ρ\$, and \$L\$ is \$0\$.

This is essentially a long way of saying that KVL finds itself contradicted. But I took this long way to avoid answers such as "KVL simply doesn't work with magnetic fluxes". But how can it not work? What did I do wrong in all those steps? Which assumption was incorrect?

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    \$\begingroup\$ Voltage is induced not current. How many times have I said this over the years..... \$\endgroup\$
    – Andy aka
    Jul 27, 2020 at 17:39
  • \$\begingroup\$ A current is induced as well. Bet? \$\endgroup\$
    – nc404
    Jul 31, 2020 at 0:06
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    \$\begingroup\$ Andy aka you are correct and you nc404 are wrong. If the resistance of the circuit is infinite voltage is induced but current doesn't flow. \$\endgroup\$ Jul 31, 2020 at 0:13
  • \$\begingroup\$ Theoretically speaking yeah. \$\endgroup\$
    – nc404
    Jul 31, 2020 at 23:06

4 Answers 4

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This is essentially a long way of saying that KVL finds itself contradicted.

Kirchhoff's Voltage Law is a law from circuit theory that is valid in the lumped circuit approximation. One of the requirements for the lumped circuit approximation is that there be no significant changing magnetic flux passing through the circuit. If this requirement is violated then we absolutely don't expect KVL to apply.

What did I do wrong in all those steps?

When you said, " since we start at any point in the loop and come back around to the same point, that means that the voltage at that point (relative to itself) is both 0 V (trivially) and ..." you have already assumed that the lumped circuit approximation is valid and that there are no changing magnetic fluxes through the loop.

You assumed KVL would apply in a circuit where it doesn't apply.

Which assumption was incorrect?

The assumption that there was no magnetic flux through the circuit being evaluated

But I took this long way to avoid answers such as "KVL simply doesn't work with magnetic fluxes".

You basically demonstrated the reason why KVL doesn't work in circuits with changing magnetic flux.

If you want a reason why KVL doesn't work in this scenario, then your logic is exactly that reason. It's because in this scenario the integral of the EMF around the loop is not zero but something that depends on the changing magnetic flux through the loop.

If you want to model this circuit within the context of the lumped circuit approximation and KVL, then you should treat each element of the loop not as a small resistor, but as a small section of the secondary of a transformer (with the primary being whatever is generating the magnetic flux through the loop). Then you have a term for the voltage across each element that won't sum to zero as you go around the loop and you won't get a contradiction.

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    \$\begingroup\$ @nc404 It's not really a voltage. A voltage in the strict sense only makes sense in a conservative field where the path you take when integrating over the E-field doesn't change the result as long as start and end point are identical. This is not the case when you have a changing magnetic flux. \$\endgroup\$
    – Felix S
    Jul 27, 2020 at 16:10
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    \$\begingroup\$ At the very least, both your assertion (the voltage at that point (relative to itself) is 0V) and KVL depend on the potential being a well-defined quantity at every point, and that isn't true when there are changing magnetic fluxes around. \$\endgroup\$
    – The Photon
    Jul 27, 2020 at 16:10
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    \$\begingroup\$ @FelixS, be careful. "Voltage" can refer to either EMF or (quasi)electrostatic potential difference. The voltage around a loop cutting a changing magnetic field is an EMF but not a potential difference. (but we still can call it "voltage") \$\endgroup\$
    – The Photon
    Jul 27, 2020 at 16:12
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    \$\begingroup\$ @nc404, yes, that's another way to say it. \$\endgroup\$
    – The Photon
    Jul 27, 2020 at 16:15
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    \$\begingroup\$ @Felix, it's not true that "a voltage is always between two points". A potential difference (sometimes called a voltage) is measured between two points. An electromotive force (EMF) is also sometimes called a voltage and is measured on a specific path between two points, not on the two points alone. In OP's scenario the potential difference (voltage) is not defined, but the EMF (also voltage) is defined (and depends on the path taken around the loop). \$\endgroup\$
    – The Photon
    Jul 27, 2020 at 16:38
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The current in a loop of length Len with resistivity ρ in a changing magnetic field can be successfully calculated with the Faraday's, Kichhoff's, and Ohm's law. The lumped element model for a conductive loop in a changing magnetic field is a voltage source of EMFext of the external magnetic field, a current-controlled voltage source EMFself of the self-induced current, and a resistor R with the resistance of the loop's wire whole length.

EMFext is a given value; EMFself = -L·dI/dt, where L is the loop inductance; R = ρ·Len. KVL for this circuit: $$ EMF_{ext} - L·dI/dt + R·I = 0 $$ The solution is $$ I = EMF_{ext}·(exp(-(R/L)·t)-1)/R $$ To arrive at a correct answer, you only need to apply Faraday's, Kichhoff's, and Ohm's in a straitforward manner to the lumped element model of your setup.

TL;DR

I can only guess why you come up with this slicing of the entire loop into "infinitely many segments of infinitesimal length". I will attempt to "reverse engineer" your thinking, OK?

Let us formulate your electromagnetic problem with a more specific configuration. Now, it is a radius Len/2π circular wire loop of a constant resistivity ρ in a changing uniform magnetic field $$ B_z = -(EMF_{ext}/S)·t = -(EMF_{ext}/(Len^2/4π))·t $$ Rather than a three-component series network of my solution, with your slicing you would have a series network with 3·Len/dl elements, each of Len/dl elementary sections consisting of a voltage source (EMFext/Len)·dl, a current controlled voltage source (EMFself/Len)·dl, and a resistor (ρ·Len·I/Len)·dl serially connected in any order you like. I am ready to admit that this construction appease your doubts about an ambiguity of electric potential accumulating ρ·Len·I volts with every revolution along the loop. It makes evident that you simply missed a contribution of EMFs.

I evolve a bit on your construction. Consider EMFext/Len and EMFself/Len terms: because of the setup's axial symmetry, those are external and self-induced electric fields. You proved an equivalence of EM and lumped-element solutions.

You may be inspired by a distributed-element model of coax in your construction. Still, while the derivation of telegrapher's equation for coaxial cable requires splicing into infinitesimal fragments, the lumped-element model gives an exact solution for a conducting loop in a changing magnetic field, if the magnetic field change rate is constant. If this is the case, the electric field generated by changing magnetic field is constant and does not generate a second-order magnetic field correction.

The distributed-element modeling of a circuit with changing magnetic flux passing through the circuit becomes necessary in not-so-symmetrical setups unlike that considered above. In fact, constructing a 2D and 3D grids of distributed-element grid cells is a well-established technique used by electromagnetic solvers. In this approach, the Maxwell grid equations of finite difference time domain scheme and boundary conditions are derived from Kirchhoff's, Ohm's, electric and magnetic fluxes, and material equations for charges and currents.

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  • \$\begingroup\$ First, you do not need Kirchhoff if you use Faraday. Second, the transmission line distributed model is based on the transmission of energy along the line and requires two types of energy storage. This is not the case. \$\endgroup\$ Jul 28, 2020 at 17:37
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You will have EDDY CURRENTS that produce a counter_vailing (opposing) field, thus the net induced voltage will be reduced by from what you expect. IMHO

Thus sheets of metal, without any slits, can circulate currents on the surface that oppose the external field, using whatever 2_D shape of circulation is needed.

But a circuit, with some wires and some components that serve to limit the current, will not produce much of an opposing force.

I use this equation a lot, in predicting vulnerabilities to external "magnetic fields" :

(*) V_induce = (MUo * MUr * LoopArea/(2 * PI * Distance)] * dI/dT

For MUo = 4 * PI * 1e-7 henry/meter, and MUr = 1 for copper or air or FR-4, becomes

V_induce = (2e-7 * Area/Distance) * dI/dT

I've consulted on various "magnetic interference" problems.

One was a speed controller for an electric 10,000 horsepower train.

Using dI/dT = 1,000 amps per microseconds,

LoopArea = 10cm by 10cm

Distance from Wire to Loop (call it BUSSBAR to PCB loop) 4cm, we have

V_induce = [2e-7 * (10cm * 10cm)/4cm] * 1e+9 amp/second

V_induce = 2e-7 henry/meter * (25cm * 1meter/100cm) * 1e+9

V_induce = 0.25 * 2e-7 * 1e+9 == 0.5 * 1e+2 = 50 volts

Thus into the GROUND PLANE of this speed controller, located 4cm from the 2,000 amp high_current buss, was induced 50 volts.

The 10,000,000 watt system sought to impose 50 volts into the Ground Plane.

Lots of bad things happened, including violation of logic_levels in various 5 volt logic signals.

(*) I found this in EDN Magazine in 1990, as a combination of Biot_Savart and Faraday Law of Induction.

The equation assumes there is a planar loop with Area, and current_carrying wire at Distance from the loop. For exactness, you can impose some natural_log maths, but for normal ratios the equation is excellent approximation.

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    \$\begingroup\$ I don't see what this has to do with OP's question, which is about the apparent contradiction between Faraday's Law and KVL. \$\endgroup\$
    – The Photon
    Jul 27, 2020 at 16:45
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Basically, voltage is not well-defined. That means it doesn't have to be a single value relative to a certain node. It can both be \$0V\$ and \$IρL\$ at the same time. The electric field is no longer a conservative force when changing magnetic fluxes interfere.

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