A New Interpretation of Dr. Walter Lewin’s Paradox

Much has lately been said regarding this paradox which first appeared in one of W. Lewin’s MIT lecture series on ##{YouTube}^{(1)}##.  This lecture was recently critiqued by C. Mabilde in a second YouTube video and submitted as a  post in a PF ##{thread}^{(2)}##.  The latter cited the third source, that of K. T. McDonald of Princeton University, as support for Mabilde’s ##{presentation}^{(3)} {and}^{(4)}##.  Finally, Charles Link, PF Homework Helper, and Insight Author posted in Advisory Lounge Inner Circle (#1, May 25, 2018) on the same subject.  Furthermore, several other PF individuals (and probably others still)  have been involved in this topic.

I think a key concept, which none of the three sources mentions, is that there are two ##E## fields running around here.  One is the static field ##E_s## which by definition is conservative and the field lines of which begin and end on charges.  The second is the emf-induced field ##E_m## 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.

In the case of Faraday induction its circulation is Faraday’s  ## \frac {-d\phi} {dt} ##. The two fields cancel each other in any loop wire segment but only ##E_s## exists in the resistors.  (This statement assumes negligible resistor body lengths and zero-resistance wires).  The net ##E## field is anywhere and everywhere just ## E = E_s + E_m ##, algebraically summed.

A voltmeter reads the line integral of ##E_s##, not any part of ##E_m##.

For example, in his battery setup  (Fig. 1) Dr. Lewin assumes a net battery ##E## field opposite to the direction of the ##E## field in the resistor.  Yet the battery has two canceling ##E## fields.  ##E_s## points + to – and ##E_m## points – to +. The line integral of ##E_m## over the length of the battery (- to +)  is the emf of the battery.  A voltmeter senses the line integral of ##E_s## only, otherwise the meter would read 0V DC.  The resistor has an ##E_s## field pointing + to -; the circulation of ##E_s## around the loop is zero as required by Kirchhoff’s voltage law.  The circulation of ##E_m##, and thus ##E##, is ##iR##, ##i## being the current and ##R## the resistor.*

Now to address the main topic here, that of the solenoid, the single-turn loop, and voltmeters positioned as shown in Fig. 2. In what follows, loop resistance is again assumed zero.

Let

##\phi## = magnetic flux inside loop,

loop current = ##i##

total loop induced emf = ## \oint \bf E_m \cdot \bf dl ##

then $$E_m = \frac {\frac {-d\phi} {dt}} {2\pi a}$$.

Around the loop with or without the two resistors ##R1## and ##R2## in it, a continuum of ##E_m## field exists throughout the loop, with an ##E_s## field running in the opposite direction.

In the resistors ##E_s## can be very large as ## E_s = iR/d ##  with d the length of each resistor, ##d \ll {2\pi a} ##. All of this follows immediately from the realization that  ##\oint \vec E_s \cdot d \vec l = 0## and the line integral of the two resistors’ ##E_s## fields = ##\mathcal E##, being canceled around the loop by the wire segments’ ##E_s## fields.

Let us next show that a voltmeter as arranged in Fig. 3 and connected at two points a and b along the loop not including a resistor, reads 0V.  That this crucial fact is perhaps paradoxical but entirely logical can be shown as follows:

First, we remind ourselves that ##E_s## and ##E_m## are always equal and opposite in the wire including the shorter section a-b, as well as in the meter leads since a perfect conductor cannot have a net E field.  This is a crucial assumption.

Let

##E_{mw}## = the static field in the meter leads ,

##E_{sw}## = the emf field in the meter leads,

##E_{sv}## = the static field  in the voltmeter,

##E_{mv}## =the emf  field int he voltmeter,

##l_w## = the total meter lead length,

We model the voltmeter as a resistor ##r## of finite physical length ##d##, of arbitrarily high resistance ##r## and  passing correspondingly low current  ##i_v##  with the voltmeter reading ## i_v \cdot r####.

We must then also take cognizance of the fact that, for the voltmeter, ##i_v \cdot r  =  (E_{sv} – E_{mv}) \cdot d ## since ##E_{sv} ## and ##E_{mv} ## oppose. Thus, ##d \cdot E_{sv} = i \cdot r   +   d \cdot E_{mv} ##.

With this in mind, performing the circulation of ##E_s## around the meter circuit,

## +E_{sw} \cdot l_w   +  i_v \cdot r + E_{mv} \cdot d ~  –  ~ E_s\theta a  = 0 ##

The circulation of ##E_m## is also zero since there are no sources of emf within or around that contour:

## E_{mw} \cdot l_w + d \cdot E_{mv}  – E_m \theta a = 0 ##

Combining these last  two, with ##E_{sw} = E_{mw} ## and ##E_s = E_m ## as required,

##i_v\cdot r  = 0,  i_v = 0, VM = 0 ##.

Since VM = 0 for an uninterrupted section of the loop, it follows that the voltage read by a voltmeter probing a length of wire with a resistor ##R## in-between, VM = ##iR## and is not dependent on the length of wire segments surrounding ##R##. This argument includes of course Lewin’s famous A and D points which are located at the top and bottom respectively of the loop as in figs. 2 or 3.

We thus distinguish between VM readings and the so-called “scalar potential” difference which is here ##\int \bf E_s \cdot \bf dl.##  McDonald rightly offers that meter wire coupling is accountable for the difference, which is why he argues for accepting scalar potential differences as the “true” potential difference, not subject to measurement detail.  This is also the explanation offered by Mr. Mabilde. The latter demonstrated an apparently valid way of measuring the scalar potential experimentally with his interior voltage measuring setup, arriving at the correct scalar potential in all cases.  However, his demo is simply a simulation of the Es field profile around the ring, set up by the area of his meter leads..

His statement criticizing Lewin’s “Kirchhoff was wrong” is however spot-on if we understand that the Kirchhoff voltage law applies to voltages in the correct sense of the word, which is the line integral of the Es field only and does not apply to Em fields.

I want to emphasize that the voltmeter readings in Lewin’s setup can be entirely accounted for without splitting the E field into Em and Es.  What I think I contribute is more insight into the observed phenomena.   I believe I have offered a precise explanation for the difference between meter readings and scalar potential differences.  It’s not meter lead dress as suggested by Mr. Mabilde that matters; any meter loop in any orientation will give the same results.  It’s simply ##E_m## and ##E_s## sharing between loops.  As McDonald points out, if you want to avoid the consequences of loop coupling then the meter probes must be connected right at a resistor or the scalar potential difference reading will be wrong since those potentials associated with the loop wiring will not be included..

Lewin did not to my recollection place the VM leads in the middle of the solenoid.  I think we can all accept that the reading would be half-way between -0.1V and + 0.9V, i.e +0.4V.  Reflecting on the numerical mismatch between expected and actual VM readings, we see that the ##R1## meter reading was -0.1V – 0.4V = -0.5V i.e. too negative, while for the ##R2## loop it was +0.9V – 0.4V = +0.5V (too positive).  Now, if we integrate the ##E_s## field over the meter loop with the meter positioned half-way, i.e. directly above, the solenoid, we can sum the line integral of ##E_s## as follows:

## \mathcal E##/2 – ##iR2##  +  VM = 0 or VM  = +0.4V.

Or,  –VM + ##\mathcal E##/2 –##iR1## = 0 also giving VM = +0.4V.  Which agrees with the data.  Suspending the voltmeter with its leads directly above the magnetic source gives the correct voltage.

To summarize, one should be aware of two separate electric fields in the loop and meter lead wiring vs. (essentially) one only in the resistors.  Voltmeters give erroneous voltage readings if the meter circuit forms an alternative path for the Em field, as it does in the Lewin set up and readings.    Coupling effects are predictable and can be shown to be ##E_m## and ##E_s## field sharing between the main loop and the meter loop.  Spurious coupling must be identified and avoided if one wishes to obtain actual scalar potential differences; this may not always be an easy task.

Failure to recognize the two types of ##E## field is bound to lead to confusion or even violation of physics laws in any circuit containing one or more sources of emf, be they batteries, magnetic induction, or any other form of emf generation.

Cf. Stanford Professor Emeritus H. H. Skilling, Fundamentals of Electric Waves, probably out of print but readily available on the Web.

* Footnote added 09/12/2018: since the circulation of the ##E_m## field = ##iR##, by Stokes’s theorem there must also exist net curl within the contour.  If we set up  x-y-z coordinates in fig. 1 with the contour in the x-y plane, x axis at the neg. terminal of the battery, y axis coincident with just inside the battery on its right-hand side, L = battery length neg. to pos. terminal, then ## iR = \oint \vec E_m \cdot d \vec l = \iint_S (\nabla \times \vec E_m) \cdot d \vec A ##.
A curl that satisfies this equation is ## \nabla\times\vec E_m = \frac {\partial E_m} {\partial x}~ {\mathbf k}  = E_m \delta(x)~ {\mathbf k} ##.  Then ## \iint_S (\nabla\times\vec E_m) \cdot d \vec A = \int_{0-}^{L+} E_mL~ \delta(x) \, dx = E_mL ##  as expected.

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65 replies
1. rude man says:

Charles Link@vanhees_71 What happens if we choose a circle that does not have the origin at the center? Does that mean the symmetry of ## E ## around the circle is no longer applicable?You mean with a circular B field centerd at the origin I assume. I would think the symmetry is then shot. The only guarantee is in the form of the circulation of E, not symmetry of E.. LIke trying to apply Ampere's law to a finite-length wire.

In ##∇xE = – ∂B/∂t ## with ## E = E[SUB][/SUB][SUB]m[/SUB] + E[SUB][/SUB][SUB]s[/SUB] ## the ## E[SUB]s[/SUB] ## of course does not contribute to the curl. ## E[SUB]s[/SUB] ## is the ## -∇φ ## of post 19 ## .
##
BTW about your two circles touching – I offhand would say there is no effect on either ring after contact is made. Each ring has its E[SUB]m[/SUB] field in the same counterclockwise direction. At the point of contact the fields cancel but I see no issue with this. E fields, both emf and static, can exist alongside, linearly augmenting or canceling each other to some extent. 'Bout all I have to say I guess.

2. vanhees71 says:

I still don't get your problem. Obviously your electric field is fulfilling Faraday's Law (maybe I misunderstand your undefined notation again since you don't tell what ##vec{j}## is; I don't see any current density explicitly mentioned in the problem).

$$vec{E}=-1/2 vec{r} times vec{beta}-vec{nabla} phi = -beta/2(x_2,-x_1,0).$$
If I understand you right, we suppose ##vec{beta}=beta vec{e}_3## and
Circle 1 is parametrized by
$$vec{r}(varphi)=(R cos varphi,R sin varphi,0).$$
The gradient part vanishes when integrated over the closed circle (since there shouldn't be any further "potential vortex like" singularities in this part). Thus we have
$$int_{C_1} mathrm{d} vec{r} cdot vec{E}=int_{0}^{2 pi} mathrm{d} varphi frac{mathrm{d} vec{r}}{mathrm{d} varphi} cdot vec{E} = pi R^2 beta.$$
For Circle 2 you have
$$vec{r}=(R cos varphi,R sin varphi,0)+(2R,0,0).$$
Since the electric field is assumed to be homogeneous in the entire range of consideration the evaluation of the EMF is literally the same as for circle 1 and thus of course yields the same result. You can of course also integrate the magnetic field over the corresponding circular disk, showing that everything is consistent also with the integral form of Faraday's Law as it must be thanks to Stokes's integral theorem.

3. vanhees71 says:

There's nothing wrong with your calculation. If the field is homogeneous in the region you get of course always ##-beta A##, provided you integrate over a plane area perpendicular to ##vec{beta}##. So I don't understand your statement that you get something else for another circle. As long as you are in the region where the fields are homogeneous, there's no difference where you put the origin of the circle (it only must be entirely in the region where the fields are homogeneous).

Of course, this is just math. It's hard to imagine how to produce such fields in reality.

4. vanhees71 says:

Of course only knowing ##vec{nabla} times vec{E}## is not sufficient to calculate the field. It determines the field only up to a gradient. I'm not sure, whether this is a sound example, because it's not clear to me whether your setup fulfills all of Maxwell's equations. Only solutions that fulfill all Maxwell equations are consistent in describing a situation.

I also don't understand the very beginning of your argument.

Suppose (without thinking much about the physicality of the assumptions) there's a region, where
$$vec{B}=vec{beta} t$$
with ##vec{beta}=text{const}##. You get
$$vec{nabla} times vec{E}=-vec{beta}=text{const}.$$
This implies
$$vec{E}=frac{1}{2} vec{r} times vec{beta} -vec{nabla} phi$$
for an arbitrary scalar field ##phi##.

Up to this gradient the electric field is unique, and thus also the EMF is uniquely defined for any closed loop, giving by construction the ##-dot{Phi}## with the flux according to this loop. Maybe I simply misunderstand your description. Perhaps you can give your concrete calculation to look into the issue further.

5. rude man says:

Hi
In figure 2 inside the resistors, do the Em and Es fields add and help rather than oppose like they do in the connecting wire?

The answer is yes, they are both clockwise in the figure. E[SUB]m[/SUB] is always clockwise but in a resistor the E[SUB]s[/SUB] field points + to – so they add.
Line-integrated E[SUB]m[/SUB] will be small, especially if the resistor is short, but line -inegrated (E[SUB]s[/SUB] + E[SUB]m[/SUB]) = iR so you can see that E[SUB]s[/SUB] is the dominant field in R.

6. rude man says:

I have made some emendations to the original blog. Aside from a few typo corrections etc. I have simplified the math and removed the assumption of finite resistor and meter physical dimensions which were implicit in the original version.

7. rude man says:

@rude man The EMF from a battery seems to be of a slightly different nature than the EMF of Faraday's law. I was trying to come up with an analogy that might describe the type of mechanism involved where the chemical reactions create a potential difference resulting in an electrostatic field: One perhaps related mechanism would be a spring system that pushes apart positive and negative charges: e.g. You could have two capacitor plates, initially at ## d=0 ## with one having positive charge and the other negative charge. The spring system could supply energy to push them apart and create a voltage. In this case, electrostatic fields would be generated having ## oint E_s cdot ds =0 ##. The force from the spring is quite localized and is essentially in the form of an EMF. The external loop could be completed between the two plates with a large resistor that could essentially be the resistance of a voltmeter. ## \ ## Once again, the equations are quite similar, and agree with the concept your Insights article promotes, that the voltmeter actually measures the integral of the electrostatic field ## E_s ## external to the battery. Sounds good.
Somewhere I mentioned another analogy I liked, given by Prof. Shankar of Yale. He likened the process to a ski lift; you need force (Em) to overcome gravity (Es) to get from the bottom to the top, then you ski down the slope (current thru the resistor) but you bump into trees along the way (heat dissipation) so when you get to the bottom you have zero k.e. Then you repeat the process. He mentions this analogy again later when he discusses induction so I still think the two are very much the same thing except as you point out a real battery has internal resistance in which case the Es field has to be reduced from |Em| to allow for the excess Em to push the charges thru the internal R.

I highly recommend Prof. Shankar's youtube lectures. I have picked up a lot from them and am still absorbing.

Thanks for continuing the discussion!

8. rude man says: