Loop Rule

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  • #26
Tide
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rjb,
there is still a concept of how much work...
Indeed! However, with regard to KVL, you're interested in potential differences. In that regard, there can be no difference between the potential at a given point and the potential at the same point!

Think of it as hills and valleys. Their heights and depths may vary in all kinds of strange and fantastic ways but the elevation at a given point at a given instant in time is a single value. The peak of Mt. Everest cannot be both 29,000 feet above sea level and 2,500 feet below sea level at the same time - no matter how you draw a loop on your topographic map.
 
  • #27
rbj
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Tide said:
rjb,Indeed! However, with regard to KVL, you're interested in potential differences.
it's the sum of a bunch of potential differences (that, in this adverse case do not add to zero) that i am interested in.

In that regard, there can be no difference between the potential at a given point and the potential at the same point!
in a non-conservative field, potential (if you're gonna define it, which might not be such a good idea for a non-conservative field) is not fully defined by only the location of that given point. it's the same point. but, in a non-conservative field, getting to that same point requires a different quantity of energy, depending on which route you take.

Think of it as hills and valleys. Their heights and depths may vary in all kinds of strange and fantastic ways but the elevation at a given point at a given instant in time is a single value. The peak of Mt. Everest cannot be both 29,000 feet above sea level and 2,500 feet below sea level at the same time - no matter how you draw a loop on your topographic map.
it's a very good comparison to electrostatic fields. hell, the math is the same in both cases (inverse-square law, additive superposition of forces), just replace charge with mass and Coulomb Force Constant with G. but your analogy is not accurate with a general electromagnetic field where the right hand side of Faraday's Law is non-zero.

there is a gravitational counterpart to magnetic forces in gravitation. since magnetic forces are only a manifestation of electrostatic forces, but with relativity taken into consideration and the same can be said for gravitation - inverse-square law, additive superposition of forces, relativity creating a "GravitoElectroMagnetic" (or GEM) counterpart to Maxwell's Equations for the case of reasonably flat space-time. so we could conceive of a situation where climbing around from the peak of Mt. Everest around some closed loop back to the peak of Mt. Everest would not yeild a net amount of work that is zero, but, from a POV of the human-scale, G is a helluva lot smaller than k, the Coulomb force constant. there would have to be some really nasty gravito-magnetic effects going on in order for the local gravitational field to not be a conservative field (to any degree that we could measure). if there were enough gravito-magnetic fields around to be measurable, i don't think i would be able to keep my lunch from coming up, if you know what i mean.

so, Tide, it's a nice analogy. if fits well with electrostatic fields but it does not apply to the case of Faraday's Law with a measureably significant changing magnetic field.
 
  • #28
Tide
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rbj,

There can be no potential difference between a point and itself - however much work you need to physically traverse a loop.

We can extend the topographical analogy to include friction. Take a global snapshot of the topography/potential. Now take a virtual trip in your temporally frozen world. Obviously, if you leave the peak of Everest by foot, car, or rail and travel whatever path you want and finally return to the peak of Everest you end up at the same potential as where you started - because you're in the same place! You will have expended enormous amounts of energy overcoming friction along the way but the height of Everest (at our instant in time) simply will not have changed. You will be at the same potential.

And that is true whether we're talking gravitational potential, electrostatic potential, vector potential or nuclear potential and it's true whether we're including time varying fields or topography. At a given instant, a point cannot be at two or more different potentials.

Think about this: If the potential (electrostatic and/or electromagnetic) is multivalued, how could you possibly derive fields from them if the corresponding electric and magnetic fields are gradients, curls or time derivatives of those potentials?
 
  • #29
rbj
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Tide said:
There can be no potential difference between a point and itself - however much work you need to physically traverse a loop.
...
And that is true whether we're talking gravitational potential, electrostatic potential, vector potential or nuclear potential and it's true whether we're including time varying fields or topography. At a given instant, a point cannot be at two or more different potentials.
listen, after this, i'm leaving the thread alone. KVL really is simply a consequence of this one of Maxwell's Equations (in integral form), the one originally called Faraday's Law.

[tex] \oint_{s} \mathbf{E} \cdot d\mathbf{s} = - \frac {d\Phi_{\mathbf{B}}} {dt} [/tex]

if the path is taken to go through the electrical components in the loop, the line integral on the left is actually the sum of all of the voltages of these components. those voltages add to the complete, closed, line integral by definition. Faraday's Law says they add to

[tex] - \frac {d\Phi_{\mathbf{B}}} {dt} [/tex]

that is a more general expression of KVL. in practice we usually say that

[tex] \frac {d\Phi_{\mathbf{B}}} {dt} = 0 [/tex]

then, as a consequence

[tex] \oint_{s} \mathbf{E} \cdot d\mathbf{s} = 0 [/tex]

and all the voltages add to zero (and we have a conservative electrostatic field, the concept of scaler potential is meaningful and blah, blah, blah). all of that stuff.

in the case that

[tex] \frac {d\Phi_{\mathbf{B}}} {dt} [/tex]

is not zero, then we normally add that term as a "virtual voltage source", or a "phantom voltage source" or whatever you want to call it. that is the same as saying

[tex] \oint_{s} \mathbf{E} \cdot d\mathbf{s} + \frac {d\Phi_{\mathbf{B}}} {dt} = 0 [/tex]

now, if you add up all of the voltages, plus the phantom induced voltage due to some changing magnetic field in the loop, they still add to zero. we still get to use KVL, but in a slightly modified form.

THAT'S IT. if you still don't get it, there is nothing more than i can do other than repeat the physics (which i choose not to do anymore). if you're still unsatisfied, i would suggest that you take it up with Faraday or Maxwell.

Think about this: If the potential (electrostatic and/or electromagnetic) is multivalued, how could you possibly derive fields from them if the corresponding electric and magnetic fields are gradients, curls or time derivatives of those potentials?
i think you are reversing concepts pedgogically. fields come first. they are there and have the the physical laws that describe how they are created and what they do. fundamentally, that's all you need. concepts like potential comes later and is purely a human construction. the potential of some point in space is the amount of work per unit charge that is required to bring a test charge from a point at infinity to that particular point in space. this concept of potential (as a scaler) is well-defined, makes a lot of sense, and is very useful in conservative fields because then that amount of work per unit charge that is required to bring a test charge from a point at infinity to that particular point in space is only a function of the coordinates of that point in space and it doesn't matter what route is taken. however, it makes much less sense and is much less meaningful in the non conservative fields. i would not choose to draw any concept of "field" from the concept of "potential" because it won't always work.
 
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  • #30
Tide
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rjb,

fields come first
Of course they do. That's why I mentioned them.

the line integral on the left is actually the sum of all of the voltages of these components
No. The sum of the voltages of all the components is

[tex]-\oint \left( \nabla \phi + \frac {1}{c} \frac {\partial \vec A}{\partial t} \right)\cdot d\vec s[/tex]

which, in the absence of charge within or current through the cross section of the loop, adds to zero - i.e. there is no change in potential in traversing a loop.
 
  • #31
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What if you simplify the problem?

Imagine a square wire loop, ABCD, of some resistance R moving at constant velocity, parallel to its plane and perpendicular to a constant B field. Label the leading corners A and B, and the trailing corners C and D.

There will be a constant current in the loop (let it be in the ABCD direction). The current will be the same in all four legs (AB, BC, CD, DA) of the loop.

Along legs AB and CD (excluding the corners!) the voltage is constant, call it zero. The current is caused exclusively by Faraday's Law. If there were an electric potential, you'd get a greater current than Faraday's Law says you should, and you don't.

Along legs BC and DA the magnetic force on the electrons is perpendicular to the wire, so Faraday's Law plays no part. The only other thing that can make electrons move is Coulomb's Law. So from B to C and from D to A (excluding the corners) there has to be a continuous rise in electric potential (conventional current).

So going around the loop, the voltage is zero along AB. Then there's a discontinuous drop at B to some value less than zero. Then there's a continuous rise from B to C to a value greater than zero. Then the voltage drops discontinuously to zero at C and so on.... At the end of DA, the voltage drops back to zero just as you go around the corner; you're back to where you started, and the voltage has the single value of zero.

So, Faraday and Kirchoff are both in tact and operating compatibly!

PS I had to think about this one for a while: What makes the electrons go down the potential discontinuity at B and D?
 
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  • #32
rbj
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i'm gonna regret getting back into this.

i posed a similar question about this way back (that was never taken on, until now):

rbj said:
what happens when you have a single circular closed loop of wire in the presence of a non-zero and changing magnetic field? what happens when you apply

[tex] \oint_{s} \mathbf{E} \cdot d\mathbf{s} = - \frac {d\Phi_{\mathbf{B}}} {dt} [/tex] ?

that first integral, on the left hand side, is a voltage. do you get zero?
jdavel, i think you have a point, but there are some problems.

jdavel said:
The current is caused exclusively by Faraday's Law.
first, it's a voltage that gets caused by Faraday's Law. and

jdavel said:
The only other thing that can make electrons move is Coulomb's Law.
is not true in general (perhaps you mean it only in the context of legs BC and DA).

the third problem is, i am not sure if the loop is moving parallel to the plane with a constant B field coming out that there is any net

[tex] \frac {d\Phi_{\mathbf{B}}} {dt} [/tex] .

fourth, if there is a current in the loop, even legs BC and DA will have a voltage drop (and in the same direction around the loop).

however, how about a loop of resistors (in a square or a triangle or octagon or whatever symmetrical shape you like) that is placed right beside a big solenoid with a changing current in the solenoid causing a changing B field (this is essentially a transformer)? the axis of the loop and solenoid are in-line.

we know there will be an infinite number of little (infinitesimal) phantom voltage sources distributed around the loop (these come from the [itex] \frac{dA}{dt} [/itex] in Tide's version), all oriented in the same direction (say, clockwize) with their voltages teaming up (not cancelling). this and Ohm's Law will determine some non-zero current going around the loop.

now, here is where physicists (likeTide) and engineers (like me) look at this differently:

i say the circuit of resistors, as drawn on a piece of paper, is not adhering to KVL, but if the circuit is modified, to have all of those little voltages sources tossed in (usually as one big lumped source), then KVL is satisfied but those voltage sources are not components of the circuit that we started with. for that reason, it is a deviation.

i'll let Tide spell out his version.
 
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  • #33
rbj
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i dunno why, but i thought i would consult the "authoritatve" wikipedia to see what it had to say on the subject. a lot of language is similar to what i was saying all the time (Faraday's Law, "fixing" KVL) with one important exception: perhaps i should have used the term "EMF" instead of "potential difference". perhaps this might help Tide and i to sing the same tune.

from http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws#Kirchhoff.27s_voltage_law

The directed sum of the electrical potential differences around a circuit must be zero.

(Otherwise, it would be possible to build a perpetual motion machine that passed a current in a circle around the circuit.)

This law has a subtlety in its interpretation, because in the presence of a changing magnetic field the electric field is not conservative and it cannot therefore define a pure scalar potential—the line integral of the electric field around the circuit is not zero. Equivalently, energy is being transferred from the magnetic field to the current (or vice versa). In order to "fix" Kirchhoff's voltage law for this case, an effective potential drop, or electromotive force (emf), is associated with the inductance of the circuit, exactly equal to the amount by which the line integral of the electric field is not zero by Faraday's law of induction.
from http://en.wikipedia.org/wiki/Electromotive_force

Motional emf is ultimately due to the electrical effect of a changing magnetic field. In the presence of a changing magnetic field, the electric potential and hence the potential difference (commonly known as voltage) is undefined (see the former) — hence the need for distinct concepts of emf and potential difference. Technically, the emf is an effective potential difference included in a circuit to make Kirchhoff's voltage law valid: it is exactly the amount from Faraday's law of induction by which the line integral of the electric field around the circuit is not zero. The emf is then given by L di/dt, where i is the current and L is the inductance of the circuit.

Given this emf and the resistance of the circuit, the instantaneous current can be computed with Ohm's Law, for example, or more generally by solving the differential equations that arise out of Kirchhoff's laws.
 
  • #34
Tide
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rbj,

perhaps this might help Tide and i to sing the same tune.
Way to go, rb! :)
 

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