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

[rude man]

Greg Bernhardt submitted a new PF Insights post

A New Interpretation of Dr. Walter Lewin's Paradox

Continue reading the Original PF Insights Post.

 

[Charles Link]

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. :)

 

[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 $2pia$.

 

[rude man]

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!

 

[Charles Link]

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 $2pia$.

@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.

 

[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 $2pia$.

@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.

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

 

[Charles Link]

@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. :)

 

[rude man]

@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.

 

[Greg Bernhardt]

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

 

[Charles Link]

@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.

 

[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 $2pia$.

@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.

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]

@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. :)

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.

 

[Charles Link]

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 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.

 

[rude man]

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.

Thanks CL.

 

[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.

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!

 

[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.

 

[rude man]

Oops, missing post? Somebody asked,
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_m is always clockwise but in a resistor the E_s field points + to - so they add.
Line-integrated E_m will be small, especially if the resistor is short, but line -inegrated (E_s + E_m) = iR so you can see that E_s is the dominant field in R.

 

[Charles Link]

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.

 

[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.

 

[Charles Link]

@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?

 

[vanhees71]

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.

 

[Charles Link]

@vanhees71 Consider two circles of radius $R$. One is centered at the origin $(0,0,0)$ in the x-y plane, and the other centered at $(2R, 0, 0)$ in the x-y plane. Consider $\vec{B}=\beta t \hat{z}$. By Stokes theorem, the $E$ from the first circle points in the "clockwise" direction. At the point $(x,y,z)=(R,0,0)$, we have $\vec{E}=-\beta \frac{\pi r^2}{2 \pi r} \hat{j} =-\beta \frac{R}{2} \hat{j}$. $\\$ If symmetry arguments are employed, computing $E$ from the circle on the right, (centered at $(2R,0,0))$, at the same point $(R,0,0)$, $E$ points "clockwise" which is upward at that same point, so that $E=+\beta \frac{R}{2} \hat{j}$. This calculation is not consistent with $E=\frac{1}{2} \vec{r} \times \beta \hat{z}$ where a single origin is used. $\\$ In computing the second circle, it has a shifted origin, so that perhaps $E_{shift} =\frac{1}{2} (2R \hat{i}) \times \beta \hat{z}=-\beta R \hat{j}$ needs to get added to the $E$ that we compute with the shifted origin.. Adding a constant to $E$ does not change the EMF around a closed loop, which is really what our calculation does. $\\$ It seems though, an $E$ that changes depending on the location of the circle is inconsistent with symmetry arguments. $\\$ ..................... $\\$ Edit: Let's see if I can perhaps partly answer my own question: $\\$ Consider the $E$ field inside a long solenoid with changing current in the windings so that $B =\beta t \hat{z}$, we can compute $E$ using Stokes theorem, and we can take advantage of the symmetry. What these results suggest though is it would be incorrect to draw a circle off-axis in the long solenoid, even though $B_z$ is uniform throughout the entire interior, and compute $E$ around the circle using Stokes theorem, and assume that $E$ is uniform around that circle, i.e. points in/opposite the $\hat{a}_{\phi}$ direction w.r.t. the new off-axis origin that we have chosen as the center of our circle. The symmetry exists in the problem because of the entire symmetry of the complete system, and not simply because of the uniformity of $B_z$ inside the region of interest of the off-axis circle that we can draw. To assume symmetry for this latter case is incorrect.

 

[vanhees71]

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).

As I argued, for your setup the electric field reads
$$\vec{E}=-1/2 \vec{r} \times \vec{\beta}-\vec{\nabla} \phi = -\beta/2(x_2,-x_1,0).$$
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
$$\vec{r}=(R \cos \varphi,R \sin \varphi,0)+(2R,0,0).$$
[Error of previous version corrected]
Then we have along the circle
$$\vec{E} \cdot \vec{r}=-\frac{\beta}{2} \pi R^2 [1+2\cos \varphi],$$
and integrating leads to the same result as before.

Contrary to what I wrote before, the electric field is not homogeneous, but its curl!

 

[rude man]

@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_m + E_s the E_s of course does not contribute to the curl. E_s is the -∇φ of post 19 which I think gives a good explanation.

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_m 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.

EDIT I think I came to this party a bit late ...

 

[Charles Link]

@vanhees71 @rude man Please see the "Edit" at the bottom of post 22=that is the solution to this problem that puzzled me.

 

[rude man]

@vanhees71 @rude man Please see the "Edit" at the bottom of post 22=that is the solution to this problem that puzzled me.

@Charles, that is what I responding to. I think you're right, you can't assume symmetry of the E field around the off-axis solenoid; all you can assume is the circulation of E = -d(phi)/dt. And as I said it's like trying to apply Ampere's law to a finite wire: the circulation of H = I always, but that H is not uniform around any circular path. I remember getting faked out (temorarily only of course ha ha) just this way.

As far as vanHees' post is concerned, there is no E_s field anywhere around your path so the gradient term -∇Φ is zero so that does not seem to offer any further enlightenment. I can only think that you have to solve ∇xE = ∇xE_m at every point along your chosen path which is probably prohibitively difficult.

In your case, going with polar coordinates, ∇xE = [∂E_φ/∂r+ E_φ/r - (1/r) ∂E_r/∂φ] k = -∂B/∂t. With the third term on the LHS non-zero plus the boundary values that would be more than the feeble math knowledge of a dumb EE like myself could handle!

EDIT :even if your path doesn't include any B so that ∂B/∂t = 0 the problem is probably not much easier!

 

[Charles Link]

@Charles, that is what I responding to. I think you're right, you can't assume symmetry of the E field around the off-axis solenoid; all you can assume is the circulation of E = -d(phi)/dt. And as I said it's like trying to apply Ampere's law to a finite wire: the circulation of H = I always, but that H is not uniform around any circular path. I remember getting faked out (temorarily only of course ha ha) just this way.

As far as vanHees' post is concerned, there is no E_s field anywhere around your path so the gradient term -∇Φ is zero so that does not seem to offer any further enlightenment. I can only think that you have to solve ∇xE = ∇xE_m at every point along your chosen path which is probably prohibitively difficult.

In your case, going with polar coordinates, ∇xE = [∂E_φ/∂r+ E_φ/r - (1/r) ∂E_r/∂φ] k = -∂B/∂t. With the third term on the LHS non-zero plus the boundary values that would be more than the feeble math knowledge of a dumb EE like myself could handle!

EDIT :even if your path doesn't include any B so that ∂B/∂t = 0 the problem is probably not much easier!

Yes, thank you, I have completely solved this one. I do believe I have seen textbooks that present the problem of a uniform $\frac{dB}{dt}=\beta$ into the plane of the paper and ask you to compute $E$ around an arbitrary circle. $\\$ The long solenoid with current $I(t)=\alpha t$ does have uniform $\frac{dB}{dt}=\beta$, (in the $\hat{z}$ direction), throughout its interior. The surprising thing is it is incorrect to pick an arbitrary circle to compute $E$ and assume uniformity of $E$. The circle must be centered on the axis of the solenoid, or $E$ is not constant, (in the $\hat{a}_{\phi}$ direction), around the circle. The computation of the EMF $\mathcal{E}=\oint E \cdot dl$ works, but a uniform $\frac{dB}{dt}$, surprisingly, doesn't have sufficient symmetry to compute $E$ from the uniform $\frac{dB}{dt}$ inside the circle. $\\$ The $E$ gets computed everywhere by drawing circles of varying radii, that are all centered on the axis of the solenoid. For these circles, $\mathcal{E}=\oint E \cdot dl=2 \pi \, r \, E(r) =-\pi r^2 \frac{dB}{dt}=-\beta \pi r^2$.

 

[rude man]

Great idea, using varying radii about the B field center to find the E field anywhere around any contour! But that still assumes a circular B field, doesn't it? Of course, that's what you get with your long solenoid. Anyhow, you thereby answered your question and why does the lack of symmetry surprise you? Varying radii will give varying E field magnitudes and directions as you go around your contour, right? Nothing symmetrical there.

 

[Charles Link]

Great idea, using varying radii about the B field center to find the E field anywhere around any contour! But that still assumes a circular B field, doesn't it? Of course, that's what you get with your long solenoid. Anyhow, you thereby answered your question and why does the lack of symmetry surprise you? Varying radii will give varying E field magnitudes and directions as you go around your contour, right? Nothing symmetrical there.

$E(r)=E_{\phi}(r)$ is computed as a function of $r$. $\\$ The surprising thing with this problem is that stating that $B$ is uniform everywhere across the page and satisfies $B=\beta t \hat{z}$ is insufficient to compute the complete $E$ for this case across that page. Basically, it requires knowing what $B$ is doing outside the page as well. You can't just pick a circle on that page and compute $E$ from Faraday's law. $\\$ (It is possible to compute the EMF $\mathcal{E}$ over any circular path, but not the electric field $E$ without additional information ).$\\$ The long solenoid with current $I(t)=\alpha t$ has sufficient symmetry and has a uniform $B=n \mu_o I(t) \hat{z}=n \mu_o \alpha \, t \, \hat{z}=\beta \, t \, \hat{z}$, (so that $B=\beta \, t \, \hat{z}$), plus enough additional symmetry besides the uniform $B$, that such a calculation for $E$ is possible. In computing $E(r)$ in the solenoid, the circular path that is selected must have its center on the axis of the solenoid.

 

[rude man]

Hope this isn't a duplicate, thought I had posted already, but:
So you assume a B field with finite extent and a circular path somewhere within that field?

 

[Charles Link]

Hope this isn't a duplicate, thought I had posted already, but:
So you assume a B field with finite extent and a circular path somewhere within that field?

Yes. A long solenoid creates the necessary uniform field in the z-direction, and by varying the current linearly with time, we have precisely the magnetic field we are looking for. If the only information given is that the magnetic field is uniform and into the paper, with $\frac{dB}{dt}=$ constant, you cannot solve for $E$. You need to know the location, (i.e. where the center is, etc.), inside the solenoid that is creating that field. There may be alternative ways to create it, but the long solenoid is the simplest and most readily available. $\\$ You can still compute the EMF around any circle without knowing the location, using Faraday's law: $\mathcal{E}=-\frac{d \Phi}{dt}$, $\\$ which is Maxwell's equation integrated over an area with Stokes theorem: $\nabla \times E=-\frac{dB}{dt}$, so that $\int \nabla \times E \cdot \hat{n} \, dA=\oint E \cdot dl=\mathcal{E}=-\int \frac{dB}{dt} \cdot \hat{n}\, dA=-\frac{d \Phi}{dt}$.

 

[rude man]

Anyway, my equation in post 26 must apply and is valid irrespective of surroundings. The thing is you know dB/dt everywhere along the closed circular path but you'll never find the boundary values on E_φ and E_r. Or so I think.

 

[Charles Link]

Anyway, my equation in post 26 must apply and is valid irrespective of surroundings. The thing is you know dB/dt everywhere along the closed circular path but you'll never find the boundary values on E_φ and E_r. Or so I think.

For this case, it seems to be very non-local effects that are occurring. To compute $E$, there must be some knowledge of the magnetic field outside of the circle of interest in order to establish sufficient symmetry conditions for the simple solution. That is the point that I have been trying to make. $\\$ Knowing $\frac{dB}{dt}$ on and inside the circle is insufficient to compute $E$. That is all that is necessary to compute the EMF around the circle, (and then the EMF is the same for any circle that has the same radius as another), but we don't know if $E$ is uniform or not, without knowing the location of the circle relative to the system that is creating the field.

 

[rude man]

I agree with what you say. I still don't see why it's surprising. Interesting but not surprising. I don't think I would ever have assumed otherwise even before your posts but who knows. I cite again the analogy with Ampere's law (mis)applied to a finite-length wire. Circulation of H: check. Detailed knowledge of H: X. In the ampere case the finite length of wire must produce a B field from the returning wiring which distorts the H field. And so with an external B field generating separate E fields which will distort the symmetry of the local E field. Quite complementary, looks like.

Given an arbitrary configuration of the B field covering and surrounding your closed path would mean solving for E_r and E_φ in Maxwell's equation which would be hard or impossible in closed form. In any case I fail to see what's surprising about the non-uniform E field.

 

[Charles Link]

I think we are very fortunate in this case to have a long solenoid, with current $I(t)=\alpha \, t$, that is able to generate exactly what we need in terms of a uniform B with a $\frac{dB}{dt}=$ constant, into the paper, so that we have a practical apparatus to make such a magnetic field. Otherwise it becomes a case where the EMF can be computed from Faraday's law, but not the electric field $E$. $\\$ I do think it is likely the solenoidal geometry proved very important for Faraday and others in coming up with the understanding of magnetism that we presently have. $\\$ It is not immediately obvious from Biot-Savart or Ampere's law, but detailed calculations do show that $B$ is completely uniform inside a long solenoid with $B=\mu_o \, n \, I \, \hat{z}$.