# A New Interpretation of Dr. Walter Lewin’s Paradox

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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 a 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 the 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).  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 magnitude of each 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 mostly an ##E_s## field pointing + to -; the circulation of ##E_s## around the loop is zero as required.  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 = ## \mathcal E_0##

then $$E_m = \frac {\frac {-d\phi} {dt}} {2\pi a}$$ and  ## \oint \vec E_m \cdot d \vec l  = \mathcal E_0##.

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.

## \oint \vec E_m \cdot d \vec l = \mathcal E_0 =  -\frac {d\phi} {dt} ##  in the loop wiring; but in the resistors ##E_s## is much larger, being ##iR1##/d and ##iR2##/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 loop 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##  passing current  ##i_v##  with the property that if  ##i_v = 0##,  the VM reading is also 0.

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 ##.

This of course explains the Lewin conundrum.  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 some resistance ##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 \vec E_s \cdot d \vec l.##  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.

On the other hand, I believe I have offered a simpler and more 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.

To summarize, one should above all be aware of two separate electric fields in the loop and meter lead wiring vs. (essentially) one only in the resistors.  Voltmeters are insensitive to any line-integrated ##E_m## fields unless those fields set up ##E_s## fields, which when line-integrated constitute “real”, i.e. scalar, potential differences .  Coupling effects are entirely predictable and can be shown to be nothing more than ##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 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 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.  This may be an eye-opener to some who might previously have exclusively associated the curl of E with  ## -\frac {\partial B} {\partial t} ##.

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64 replies
1. rude man says:
2. Ronald Hamilton says:

If the covenant says Abraham I will make your descendants as many as the stars, and latest research estimates 2 trillion galaxies with each having about 10,000,000 billion stars which is give or take an inch about 10 the 27th, and the earth's carrying capacity is estimated at about 10 billion how much energy and time is required to complete such a task? What speed must we achieve for this?

3. rude man says:

The inductor is a conductor (It is made of conducting wire, with typically very ideal conduction). The current density at any location is given by ## vec{J}=sigma vec{E} ##, where the conductivity ## sigma ## is quite large and essentially nearly infinite. When a conductor experiences an electric field and is part of a loop with a resistor, the resistor will limit the current density and make it quite finite. In order to have the same finite current everywhere in the loop, there will be a redistribution of electrical charges in the conductor, and the redistribution is such as to create a static ## vec{E}_s ## that will make the current and current density finite. In order to do this, this implies ##vec{E}=vec{E}_{total} approx 0=vec{E}_s+vec{E}_{induced} ## in the inductor. ## \ ## In the case of a chemical battery, there normally is an internal resistance, so the full voltage is only measured in a nearly open circuit configuration, with a voltmeter with a large resistance. In the case of a chemical battery, (which because of the internal resistance has a very finite conductivity ## sigma ##), with a smaller load resistor in the loop, ## vec{E}=vec{E}_{total} ## could certainly be non-zero. ## \ ## Perhaps the thing that each of these cases has in common is that the Kirchhoff Voltage Laws (KVL) always apply. To get to the reason behind why KVL works, it does help to treat separately the electric fields ## E_s ## and ## E_{induced} ##, as you @rude man have done in your Insights article. Once again, very good reading. :smile:

Thanks CL.

4. 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!