# 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 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 apparently countless others over the decades) have been deeply 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 main loop wire segment but only ##E_s## exists in the resistors.The net ##E## field is 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 only 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, single-turn loop, and voltmeters positioned as shown in Fig. 2. In what follows, loop resistance is again assumed zero. Line integration and circulation are always assumed counter-clockwise in keeping with polar coordinates definition.

Let

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

coil radius = a

coil current = i

total main loop induced emf = ## \mathcal E_0##

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

Around the main loop with or without the two resistors R1 and R2 in it, a continuum of ##E_m## field exists throughout the loop, with ##E_s## fields running in the opposite direction. Letting the coil radius = a, ## \oint \vec E_m \cdot d \vec l = \mathcal E_0 = -\frac {d\phi} {dt} ## in the coil 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 main loop not including a resistor, reads 0V. That this crucial fact is not only perhaps paradoxical but entirely logical can be proven as follows:

Around the meter loop, including the coil segment a-b, ## \oint \vec E_m \cdot d \vec l = 0## since there are no dB/dt fields or other sources of emf inside this loop. Thus, if

##E_{mw}## is the ##E_m## field in the meter leads,

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

##\theta## is the angle subtended by a and b,

##E_{sw}## is the ##E_s## field in the meter leads, and

VM is the voltmeter reading.

So, from a to b and back,

$$ \oint \vec E_m \cdot d \vec l = -\frac {\mathcal E_0\theta} {2\pi} + E_{mw} \cdot l_w = 0 $$.

Also, ##\oint\vec E_s \cdot d \vec l = 0## for any loop so,

and combining the two equations,

$$–VM + (\frac {\mathcal E_0\theta} {2\pi})(1 – \frac{E_{sw}} {E_{mw}}) = 0 $$ after some math.

But ##E_s## and ##E_m## must be equal and opposite in a wire since ##E = E_s + E_m = 0 ## so finally VM = 0.

This of course explains the Lewin conundrum. Since VM = 0 for a wire length by itsef, it follows that the voltage read by a voltmeter probing a length of wire with some resistance R inbetween, 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.

We distinguish between VM readings and the so-called “scalar potential” difference which is here ##\oint \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 a “correct” way of measuring the scalar potential experimentally with his interior voltage measuring setup, arriving at the scalar potential in all cases. I am not wholly convinced that his readings, which are in agreement with Dr. McDonald’s and my scalar potential computations, represent a means of truly measuring the scalar potential. There may be unknown cancelling effects to get to the scalar potential numbers. Still, his demonstration is impressive and I accept its conclusions until someone shows me why I should’t.

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 really meter “lead dress” that matters; any meter loop 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.

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 must above all be aware of two separate electric fields in the loop wiring vs. 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 coil and a meter loop. Spurious coupling must be identified and avoided if one wishes to obtain scalar potential differences; this may not always be an easy task.

Any refusal 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.

References:

- https://www.youtube.com/watch?v=FUUMCT7FjaI
- https://www.physicsforums.com/threads/faradays-law-circular-loop-with-a-triangle.926206/page-4
- Attachment: K. McDonald: “Lewin’s Circuit Paradox”
- Attachment: K. McDonald, “What Does a Voltmeter Measure?”

AB Engineering and Applied Physics

MSEE

Aerospace electronics career

Used to hike; classical music, esp. contemporary; Agatha Christie mysteries.

EMF is somewhat of a mathematical oddity, because the electrostatic ## E_s ## has ## nabla times E_s=0 ##, and thereby ## oint E_s cdot ds=0 ## (it's a conservative field), but that is not the case for ## E_{induced} ##. ## \ ## One comment is that a voltmeter will not be able to distinguish between an EMF generated by a battery/electrochemical cell as opposed to the voltage from an inductor or a capacitor which can both be considered as voltage sources. The equation ## mathcal{E}= L frac{dI}{dt}+IR+frac{Q}{C} ## can be rewritten with the capacitor and/or inductor source on the EMF (left) side of the equation with a minus sign. The ## IR ## term represents any resistance, including that of a voltmeter. ## \## A very interesting article @rude man . Thank you. I have to study the conclusions in more detail before I can say I agree, but in any case, very good reading. :)

@charles link Thx!

I would emhasize that a battery voltage or an inductor voltage are both emf. A capacitor voltage is just a voltage drop. That derives from the definition of emf which is changing energy of another form to electrical energy. The “other” is of course magnetic in the case of an inductor.

Your circuit of a battery in series with an inductor, capacitor and resistor, looked at it that way, would say: initially there is no net emf in the circuit; L di/dt cancels the battery emf. Eventually, loop emf = battery emf since di/dt = 0. Not sure what the significance of that is …

I think you should be able to edit the original. Not for sure, but I was able to make a couple of changes to my Insights after publishing.

Yeah, I looked for that opportunity but couldn’t find one.

OK I finally figured out how & the error and one other have been corrected. Thx for the tip.

@rude man One additional comment which may essentially be contained in your article: When an inductor which is also an ideal conductor contains an induced electric field ## E_{induced} ## it necessarily must develop an electrostatic ## E_s ## that is equal and opposite the ## E_{induced} ## or the localized current density would be infinite, in the ideal case of zero resistance in the conductor. Since ## oint E_s cdot ds =0 ## around the loop, this means ## int E_s cdot ds ## in the other parts of the loop outside the inductor must be equal to ## int E_{induced} cdot ds ## in the inductor. ## \ ## I think I have most likely repeated what is also contained in your paper. When I read it quickly, this idea/concept appears to be what you are referring to. Once again, I found it very good reading. :)

Thx, great explanation why 2 E fields are present in that inductor. BTW I think I managed to get my figures into the blog, clumsy though they be and clumsily inserted as well. I really appreciate your observations.

Could you elaborate a bit on what you said about current density going to infinity unless there’s an electrostatic field inside the inductor to oppose the emf field? As I said, that certainly fits in with my conception of emf generators but I’d like to understand this a bit better. I believe it applies to all emf generators; it certainly applies to a chemical battery.

The Insight has now been updated with diagrams. Thanks @rude man!

@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.

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:

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?

Thanks CL.

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!