The tiny batteries model, or: "You don't know the half of it""You keep using that word.
I don't think it means what you think it means."
-- Inigo Montoya
Here is something that KVLers will find... unconceivable. (I will come back to this post for formatting and for adding the pictures in a couple of days)
Let's start from Faraday's law, in its modern integral form
circulation of E = - d/dt flux of B
We already established that there is no particular place on the path where we can localize the inductive EMF represented by the surface integral on the right. Even if we split the circulation of E in the path integrals along several segments we can partition the path into, none of the segment can account for the inductive EMF on the right. If we bring the whole surface integral on the left hand side of the equation we get
circulation of E + d/dt flux of B = 0
But this won't make E conservative.
There is, though, some mathematical trick we can pull to transform the surface integral of B into a line integral of something related to B and then incorporate this something into the integrand of the original path integral of E.
SPOILER ALERT GALORE:
What we will find is the circulation of something that is not E, but something else that has zero circulation and as such admits a scalar potential (which I will call phi to avoid any confusion with voltage).
Spoiler alert part I: it turns out that 'that something else' is the conservative part of the electric field E, a quantity that I will denote with Ecoul.
Spoiler alert part II: the electric field E can be seen as the composition of its conservative (irrotational) part Ecoul with its non-conservative (solenoidal) part Eind. This is the Helmoltz decomposition of a generic vector field and it holds under very general assumption (basically always):
E = Ecoul + Eind
Moreover, to avoid confusion, I will also use the symbol Etot, instead of just E, to the denote the electric field.
Etot = Ecoul + Eind
But it has to be clear that Etot(x,y,z,t) (that is E(x,y,z,t) ) is the one and only electric field an electron, or any test charge, will experience at the point (x,y,z) at time t. Electrons do not have an accountant to tell them how much of the electric field they experience is due to the changing flux and how much to the displaced charge. The just react to the whole shebang.
We can start to massage our relation starting from the the vector potential A. Since div B = 0 we can express the magnetic field B as the curl of another vector field A:
B =
curl AOur Faraday equation will become
circulation of
Etot + d/dt flux of
curl A = 0
We can now use Stokes' theorem to turn the surface integral of the curl of A into the path integral of A along the boundary of the surface
circulation of
Etot + d/dt circulation of
A = 0
If the ring is stationary we can bring the time derivative inside the integral
circulation of
Etot + circulation of d
A/dt = 0
Now we can incorporate the second circulation integral into the first and we obtain the circulation of a field that is conservative
circulation of (
Etot + d
A/dt) = 0
But notice what the integrand is: it is
Ecoul, the conservartive PART of the total electric field
Ecoul =
Etot -
Eind =
Etot - (-d
A/dt) =
Etot + d
A/dt
Ecoul has zero circulation and as such admits a potential function. This potential function is the scalar electric potential phi (some call it V, generating confusion) that obeys what we could call KsPDL (Kirchhoff's scalar Potential Difference Law).
It represents a partial description of the system because phi completely describes
Ecoul, but only PARTIALLY describes
Etot. In order to know
Etot (the actual electric field felt by the charges) we need to know BOTH potentials: the electric scalar potential phi AND the magnetic vector potential
A.
Etot = -
grad phi - d
A/dt
Where does the tiny batteries model come from?The tiny batteries model is not a model of the
complete physical system "ring with resistors" described by
Etot, but only a model of the effects of the partial, conservative part
Ecoul alone. As a matter of fact, the tiny batteries represent the term associated with the negative of the induced field
Eind = -d
A/dt that we need to strip from
Etot to get the conservative field
Ecoul
Ecoul =
Etot -
Eind =
Etot - (-d
A/dt)
This figure show the fields inside the conductor and the resistors of Lewin's ring for the case of zero-resistance wires.
Only the ring on the right expresses a conservative field, and as such it can be represented by tiny lumped components obeying K"V"L (it's KsPDL, actually). I am using tiny batteries because it makes easier to see at first glance the polarity.It would be more accurate to use tiny generators, since in general the EMF contribute is time-varying, but if we freeze the system at a particular instant in time, we can pretend we are dealing with tiny batteries.
fig model with tiny batteries, no resistance in wires
The model can be generalized to the case of resistive wires. In practice we break the ring in tiny segments for which we have two contributes: the tiny resistor that represents the conductivity of the material and the tiny battery that represent the 'EMF eraser' that will recover
Ecoul from
Etot.
fig model with tiny batteries and resistors
This being the model of a mathematical part of the system, it will break the laws of physics obeyed by the actual, complete, system. Case in point: Ohm's law. Consider any portion of copper conductor and we will see that a piece of metal with a resistance that is almost zero will appear to drop a voltage (actually it's a scalar electric potential difference) of, say, 0.25 volts when a current of just 1 mA flows through it.
-buffering-