Insights A New Interpretation of Dr. Walter Lewin's Paradox - Comments

[Charles Link]

@vanhees71 makes a very good point, you can no longer use the term "voltage" when you include a circuit loop that has a changing magnetic field inside of it. $\\$ The "voltmeter" does not measure a "voltage" difference in this case, between the two points on the circuit that it probes.$\\$ Instead, the voltmeter needs to be considered for what it actually is=a couple of wires with a large resistor through which a small current flows. In this case, the voltmeter really doesn't "measure". Instead, it gives a reading which is the (multiplicative) product of the small current times the large resistor. The placement of the wires that form the leads of the voltmeter can yield different results depending on whether the circuit loop that they form encloses the changing magnetic field, in which case there is an EMF around that circuit loop.

 

[vanhees71]

Another equivalent view is that the EMF drives the electrons in the wires making up the volt meter (think of an old-fashioned galvanometer for simplicity), leading to the current @Charles Link mentioned in the previous posting.

This becomes clear if one uses the complete (!) integral form of Faraday's Law of induction. Its fundamental form is, as anything in electromagnetism, the local form in terms of derivatives (SI units):
$$-\partial_t \vec{B}=\mathrm{\nabla} \times \vec{E}.$$
Now if you integrate this over an arbitrarily moving area $A$ with boundary $\partial A$ you can first use Stokes's theorem. The only correct version of this simple treatment is
$$-\int_{A} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B} = \int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}.$$
Now one likes to express in terms of the magnetic flux through the area
$$\Phi_{\vec{B}}=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
Now, if the area and its boundary are moving, you cannot take the partial time derivative out of the integral in the previous equation but you get an additional line integral along the boundary curve of the surface, which you can lump to the integral on the right-hand side. Taking Gauß's Law for the magnetic field, $\vec{\nabla} \cdot \vec{B}=0$ into account the resulting equation gets
$$-\dot \Phi_{\vec{B}} = \int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=:\text{EMF},$$
where $\vec{v}$ is the velocity field along the moving boundary of the area we've integrated over.

Now if you choose the area such that its boundary is along the wires connecting the volt meter with the rest of the circuit, what it measures is in fact the electromotive force, i.e., the line integral along the closed (!) boundary. It's obviously the line integral over the force on a unit charge $\vec{E}+\vec{v} \times \vec{B}$, and this shows that indeed that's the physical picture on what's measured given above: The force on the charges inside the wires connecting the volt meter with the rest of the circuit (including the wires making up the coil in the volt meter, if you take the model of a old-fashioned galvanometer setup).

This considerations also explain why the reading of the volt meter is beyond the simple Kirchhoff circuit theory: It's reading cannot be understood without taking into account the correct geometry of the connection of the volt meter with the rest of the circuit since this you need to calculate the line integral defining the EMF, which is what the volt meter measures. The Kirchhoff theory becomes applicable only if you make the wires connecting the volt meter very short, so that the magnetic flux through the corresponding current loop becomes negligibly small. Then the reading is what you expect according to the Kirchhoff circuit theory, i.e., the EMF through the element of the circuit you want to measure (which may be an Ohmic resistor, a capacitor, or coil).

Note: Another source of confusion is the very name "EMF" for the line integral: Here force is obviously not the modern notion of "force" (which is represented by the Lorentz force per unit charge, $\vec{E}+\vec{v} \times \vec{B}$) but its meaning is more in the sense of "energy". Indeed the EMF is a line integral of the force along a closed loop. The very fact that the quantity is a line integral along a closed loop shows that it is NOT a "voltage". If there'd be a potential for the force integrated over, the integral over any closed loop is 0 (modulo the caveat that the region under consideration is simply connected!).

 

[Henryk]

There is no paradox whatsoever!. I admit, it took me a while to understand what is going on.
First thing first, the loop with two resistor is a red hearing. So, let's remove it and we get a circuit like that:

Now, we have a loop containing two voltmeters encircling flux change of 1 Wb/s. Obviously, the induced EMF is 1 Volt and the direction is indicated by the circular arrow. With the way the voltmeters are connected, the one on the right would show a positive voltage, the other negative voltage, just like in the video.
How much will each of them show?. That would depend on the internal resistance of the voltmeters. Portable meters have resistance of 10 Mohm, if both have this value, one will show 0.5 V, the other -0.5 V. Change the internal resistance of the left voltmeter to 1 Mohm and the other one to 9 Mohm and you will get -0.1 v and 0.9 V. No paradox, just a red herring.
However, Dr Lewin makes a statement in his video that I would disagree. He says that the Kirchhoff (second) law is not valid. The way I was thought physics, it is still valid. I understand that the Kirchhoff law says that for a loop $\sum I_k R_k = \sum EMF$ and that actually agrees with the Faraday law.
Now, I would also like to point out that the supposed tutorial contains some false statements. One of the false statement is

"which is non-conservative in the sense that its circulation is non-zero. E_m can be created by a chemical battery, magnetic induction, the Seebeck effect, and others."

This statement is not correct. The non-conservative electric field can only be created by changing magnetic flux. The field inside a battery is conservative. How is it created.

The key to understand operation of a battery is thermodynamics and equilibrium condition for particle exchange. Thermodynamics tells us that a system is at equilibrium with a reservoir with respect to particle exchange if the chemical potentials are equal. Let's take, for example, an alkaline battery. It consist of a zinc cathode, MnO anode and KOH electrolyte. KOH in solution dissociates into K⁺ and OH^- ions. At the cathode, the following reaction takes place ( see https://en.wikipedia.org/wiki/Alkaline_battery )

Zn(s) + 2OH⁻_(aq) → ZnO(s) + H₂O + 2e⁻

The reaction of solid Zn with OH^- ions produces ZnO, water and free electrons. Where do the free electrons go? they go to the Zn metal charging it up negatively, i.e. increasing the chemical potential of electrons in the metal. The reaction stops when the chemical potential of electrons in the Zn metal become equal to the chemical potential of the electrons attached to OH^- ions. The net result is formation of a potential difference at the electrode/electrolyte interface. This is not unlike creation of the depletion layer in a p-n junction of a semiconductor.
Similarly, there is a potential step created at the anode. The total voltage of an (open circuit) battery is algebraic sum of the two voltage steps.
Seebeck effect, photovoltaic cell EMF can also be understood considering the thermodynamics, that is, EMF is created by a gradient of chemical potential of electrons and the field is conservative.

 

[rude man]

@vanhees71 makes a very good point, you can no longer use the term "voltage" when you include a circuit loop that has a changing magnetic field inside of it. $\\$ The "voltmeter" does not measure a "voltage" difference in this case, between the two points on the circuit that it probes.$\\$ Instead, the voltmeter needs to be considered for what it actually is=a couple of wires with a large resistor through which a small current flows. In this case, the voltmeter really doesn't "measure". Instead, it gives a reading which is the (multiplicative) product of the small current times the large resistor. The placement of the wires that form the leads of the voltmeter can yield different results depending on whether the circuit loop that they form encloses the changing magnetic field, in which case there is an EMF around that circuit loop.

Yes. Your resistive voltmeter shows the field equivalent of Ohm's law which is ir = d(E_s + E_m) with d the length of r. This reduces to ir = dE_s if d << voltmeter wire lead lengths. In my various posts I had made this assumption.

 

[rude man]

BTW Contrary to the statement in the above cited Insight article, of course there are both electric and magnetic fields in stationary circuits (in fact there's only one electromagnetic field in nature, but that's another story)..

E_m is also an electric, not a magnetic field. Two E fields: E_s and E_m. One begins and ends on charges; the other does not.