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1. Jun 25, 2018

rude man

2. Jun 25, 2018

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. :)

3. Jun 25, 2018

rude man

There is an error near the start of the blog.
$cintvec E cdot d vec l = matcal E_0 = -frac dphi dt$ , not divided by $2pia$.

4. Jun 25, 2018

rude man

5. Jun 25, 2018

@rude man 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.

6. Jun 25, 2018

rude man

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

7. Jun 25, 2018

@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. :)

8. Jun 26, 2018

rude man

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.

9. Jun 26, 2018

Greg Bernhardt

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

10. Jun 26, 2018

@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.

Last edited: Jun 26, 2018
11. Jun 27, 2018

rude man

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

12. Jun 27, 2018

rude man

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.

13. Jun 27, 2018

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.

Last edited: Jun 27, 2018
14. Jul 4, 2018

rude man

Thanks CL.

15. Jul 4, 2018

rude man

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!

16. Sep 3, 2018

rude man

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.

17. Sep 5, 2018

rude man

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. Em is always clockwise but in a resistor the Es field points + to - so they add.
Line-integrated Em will be small, especially if the resistor is short, but line -inegrated (Es + Em) = iR so you can see that Es is the dominant field in R.

18. Oct 12, 2018 at 9:46 AM

There is a real puzzle that appears with the Faraday EMF. Suppose we have a region of magnetic field that is changing linearly with time that points into the paper. This will cause an EMF in the counterclockwise direction around a circular loop, and very straightforward calculations allow for the computation of the induced electric field $E_{induced}$ over a circular path. $\mathcal{E}=\oint E_{induced} \cdot dl=-\frac{d \Phi}{dt}$. By symmetry, $E_{induced} \, 2 \pi \, r=-\frac{d \Phi}{dt} =-\pi \, r^2 \, \frac{dB}{dt}$. $\\$ If we consider a circle to the right of the first circle (with the same radius) that makes contact with first circle at one point, we see that the $E_{induced}$ for that path at the point of contact will actually point opposite the direction that it does for the circle on the left. What this means is the $E_{induced}$ that results from the changing magnetic field is a function of the path that is traveled, (or the path that a loop of a circuit takes), rather than being inherently part of an electric field that results from the changing magnetic field. $\\$ In computing the EMF in an inductor, this calculation is very straightforward because the path is well defined. It appears though, without including the path, the computation of $E_{induced}$ has little meaning. We can write the equation for $\nabla \times E=-\frac{\partial{B}}{\partial{t}}$ , but we can't solve for $E$, without knowing the path. The above paradox seems to indicate though that $E$ is not even a well defined function. It would be nice to be able to write $E=E(\vec{r},t)$, but with the above paradox, there is some difficulty encountered in doing this. $\\$ Putting in a conducting loop essentially applies some boundary conditions to the problem. But what about the case of a free electron moving in the changing magnetic field? Is it necessary to consider the path that the electron will follow in order to compute any accelerations from the $E_{induced}$ that it might experience? Perhaps this is where the Lienard-Wiechert solution is required.

Last edited: Oct 12, 2018 at 10:50 AM
19. Oct 13, 2018 at 6:17 AM

vanhees71

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.

20. Oct 13, 2018 at 8:30 AM

@vanhees_71 Your solution for $E$ for this problem is interesting. (I have seen this before for $\nabla \times A=B_o \hat{k}$), but what is incorrect with the following: $\int \nabla \times E \cdot \hat{n} dA=\oint E \cdot dl =-\beta A$ by Stokes theorem, so that $E(2 \pi r)=-\beta \pi r^2$? $\\$ 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?