In the meantime I browsed through this thread in a bit more detail. Here are some quotes from a textbook by Antiphon which are plain wrong, and Prof. Lewin's explanation is way better. I think all this has been answered already in the first few postings of this thread, but let's summarize it again.
Antiphon said:
"6.10 The Concept of Voltage and Kirchoff's Laws
[...] Kirchoff's voltage law states that the sum of the branch voltages along any closed path in the circuit (measured in the same direction) must be equal to zero.
That's plain wrong according to Faraday's Law. Kirchoff's Laws are strictly valid only for DC circuits. As soon as you have time-dependent magnetic fields they do not hold anymore. Faraday's Law says (in its local form)
\vec{\nabla} \times \vec{E}=-\partial_t \vec{B},
i.e., es soon as you have a time-varying magnetic field the electric field is not conserved anymore, and the "sum of the voltages" along the circuit is not 0. Of course "sum of the voltages" here means the integral along the closed circuit, and this value is according to the above equation the negative time derivative of the magnetic flux through any surface with the boundary given by this loop.
To make things easy, and I think that's also the case discussed in this thread, let's only discuss circuits without moving parts, i.e., in the following all surfaces and their boundaries are assumed to be at rest.
Now suppose, we have the most simple case of one resistor in a closed loop, and a volt meter measuring the voltage drop across that resistor. What we have then are effectively two loops, namely the resistor loop and the volt-meter loop of this parallel circuit. I assume a simple volt meter which I can treat as another resistor (of high resistance). Then we can use Faraday's Law for these two loops. Let's start with the resistor loop and integrate Faraday's Law over an arbitrary surface with a boundary given by this loop. The left-hand side can be taken as path integral along that path, using Stoke's Law. This gives
U_1=R i_1=-\dot{\Phi}_1,
where \Phi_1 is the magnetic flux through the surface. The integral is independent of the particular choice of this surface due to Faraday's Law. So there's no ambiguity here.
Now by the same argument we can integrate Faraday's Law across the area with the boundary given by the volt-meter loop, giving
-R i_1+R_V i_2=\dot{\Phi}_1+U_V=-\dot{\Phi}_2,
where R_V is the resistance of the volt meter and U_V the corresponding voltage. Thus, what you measure is
U_V=-\dot{\Phi}_1-\dot{\Phi}_2,
and of course the voltage, measured by the volt meter, depends on both fluxes, i.e., the volt-meter reading will change when the volt-meter loop is changed. [This result you can of course also get, if you integrate along the outer loop, containing only the volt meter as a resistance. The total magnetic flux is of course the sum of the fluxes through the resistor and the volt-meter loops.] If you want to measure the magnetic flux through the resistor loop alone, you not only have to make the resistance of the volt meter, R_{V} \gg R (as would be sufficient for DC circuits) but also make sure that the magnetic flux through the volt-meter loop can be neglected (by either arranging it to be outside the relevant time-varying magnetic field or making it as small as possible).
Antiphon said:
This law is the equivalent of Maxwell's first equation, i.e. of Faraday's induction law. This equivalence, however, is not as directly evident as the relation between Kirchoff's current law and the conservation of charge. Indeed, the voltage law depends on how the branch voltages are defined in herms of the electromagnetic field. Although the concept of voltage has already been discussed in Sec. 6.8 in connection with inductive fields, it deserves some further, careful consideration in view of its key role in circuit theory.
To obtain a better feeling for what is involved in in the circuit concept of voltage, it is helpful to consider its definition from an experimental point of view. A little thought will make it obvious that all voltmeters are designed to measure the line intrgral of the electric field along the path formed by the connecting leads. This is evident in the case of electrostatic voltmeters whose operation depends directly on the forces exerted by the electric field. Other more common insturments measure actually the current through a resistor of known value; the current desnity in any such resistor is proportional, by Ohm's law, to the elctric field and, therefore the total current is proportional to the line integral of the electric field between the terminals of the resistor. On the other hand, there are implicit limitations on the use of voltmeters. For instance, nobody in his right mind would wrap the leads of a voltmeter around the core of a transformer in determining the voltage between two points in a circuit. Furthermore, it is understood that the leads of a voltmeter should be kept reasonably short and that little meaning should be attached to an indications which depends on the exact position of the leads. [YOUNGMAN, THIS IS YOUR EXPERIMENT]
This is the same thing with words as I derived for this most simple example above, but it's wrong to call this "voltage". A voltage is a difference of an electric potential. In the case of time-dependent fields, there is no electric potential. In this case, one must use not only a scalar but also a vector potential to describe the electromagnetic field, i.e.,
\vec{E}=-\vec{\nabla} \Phi-\frac{\partial}{\partial t} \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.
For a given electromagnetic field the electromagnetic potentials (relativistically the four-vector potential) is not unique but only determined up to a gauge transformation and have not a clear physical meaning except of giving the fields in a way such that the homogeneous Maxwell equations (i.e. Farday's Law and the absence of magnetic monopoles) is fulfilled, but that's not the point here.
In any case, if you have a time varying magnetic field, \vec{E} is not a conserved vector field, which however is already clear from Faraday's Law in terms of the electromagnetic field itself, without using the potentials. "Voltage" thus doesn't make sense here. Of course sometimes, one calls L \frac{\mathrm{d} i}{\mathrm{d} t} a "voltage", but that's at least misleading and precisely the reason for unnecessary confusion as in this thread. Prof. Lewin is right to stress this point as in
http://ocw.mit.edu/courses/physics/...netism-spring-2002/lecture-notes/lecsup41.pdf (which has already been quoted in #9 of this thread).
Antiphon said:
These limitations on the use of voltmeters indicate that the voltage between two points has meaning only wjen the line integral of the electric field between two points is closely independent of the path of integration for all reasonably short paths. In mathematical terms, this amounts to saying that a voltage can be defined only between between points of a region in which there exists a scalar potential whose negative gradient is closely euqal to the electric field [VIOLATED BY PROFESSOR LEWIN'S EXPERIMENT]. Thus the concept of voltage in the presence of of time-varying currents is strictly an extension of the concept of voltage as defined in electrostatic systems; this extension is valid only when the path of integration used in the computation of the voltage is contained in a region of space in which the electric field behaves approximately as an electrostatic field."[THIS IS WHY A CIRCUIT HAS TO BE OF INFINITESIMAL SIZE;]
No, it's simply a wrong statement, as explained in detail above and in much more detail in the above quoted lecture note by Prof. Lewin. Volt meters simply have to be used in the right way to measure the very quantity you are interested in. Circuit theory can be used for any circuit as long as the quasistationary limit is applicable (i.e. as long as the typical wave length of the em. fields under consideration are much larger than the size of the circuit and thus Maxwell's displacement current can be neglected) and as long as all emf's from time-varying magnetic fluxes are taken into account properly.
