How Does Faraday's Law Apply to a Moving Loop in a Magnetic Field?

[Paulanddiw]

I think some people call it "quasi-static", your example of choosing one loop inside the room and the next outside. There won't be an electric field in either loop because neither is moving.

I agree there's lots of confusion in the books about faraday's law in motion. You may talk about "fixed" loop, but does that mean it doesn't move? or does that mean it moves without changing shape?

 

[pardesi]

no what i say is there should be clear distinction between where farday's laws hold purely and where u need emf due to lorentz force also

 

[Paulanddiw]

They always put that sliding-contact illustration (constant B, find E) in all the E-M books and they, apparently, leave it as a homework problem to show that it is equivalent to the Lorentz force. You have to accept that it’s OK to increase the area by sliding the contacts, however. (Alternately, you could imagine that the lateral, stationary, conductors are stretchy, or maybe install springs in place of the sliding contacts. In the old days they‘d dip the ends of the wires in troughs of mercury.)

If you let the distance slid be equal to vt and the length of the sliding conductor be Dy, then, since the generated voltage will be small, De, the linked magnetic flux will be:
B vt Dy, and De = BvDy. The electric field, E = de/dy, so E = Bv.

Looking at your drawing, you can see that the directions of E, B, and v agree with the vector notation in v´B . The important thing is, that one side of the loop is moving and the other is stationary (For small t, the lateral sides will be negligible.) The stationary side represents the fact that the Lorentz force is observed by stationary observers.

If you were raveling along with the moving side of the loop, the Lorentz force would be difficult to measure because your instruments would be subjected to the same v´B forces as the loop.

If your instrument were a charge on a spring, for instance, at the terminals at the stationary side of the loop, the generated voltage would cause the charge to deflect the spring. But, if the instrument were moving along with the moving side, the magnetic field would create a force on the charge that just cancels that generated by the moving conductor.

But, I do agree, however, that the Lorentz force is much more “user friendly” than flux linking.

 

[Paulanddiw]

I guess some of my symbols got corrupted. I wanted a capital delta instead of a "D" in the equations about B and E and v.

I wanted a vector cross product instead of a ' in the expressions v'B.

 

[lugita15]

No, this is incorrect. Faraday's law handles motional EMF (moving loop) just fine.

I repeat, the integral form of Faraday's law holds for BOTH motional EMF (EMF induced in a moving conductor in a B-field) and transformer EMF (EMF induced in a stationary conductor in a time invariant B-field).

However, the differential form of Faraday's law (curl of E = -dB/dt) only holds for transformer EMF (fixed loop).

I know that Faraday's law handles a motional emf, I'm not disputing that. I'm just saying that strictly speaking, Faraday's Law only refers to a fixed loop. I know that Faraday's Law is valid for a changing loop as well. I was just arguing semantics.

 

[leright]

I am, however, confused about your other assertion. The integral form is mathematically equivalent to the differential form of Faraday's Law. So how could one be valid for motional emf and the other not be?

NO, they are NOT mathematically equivalent. look at the differential form of faraday's law again. What if the B-field is NOT changing with time (static B-field) but the conducting loop is time varying (its size of orientation is changing)? There is still an induced EMF! This is explained by the integral form, but is neglected by the differential form.

When you transform from the integral form to the differential form you bring the time derivative into the flux integral and only differentiate the B-field component but not the dS, but in some cases the B-field is time invariant and the surface area or angle the surface makes with the B-field changes...this is not accounted for in the differential form.

Sorry, this is difficult for me to explain since I am bad with Latex.

 

[lugita15]

That's what I thought, but haven't had a chance to dig into my books (Jackson et al.) to verify. Thanks for saving me that work. :)

What we call the integral form of Faraday's law,

\varepsilon = - \frac {d \Phi_B} {d t}

includes two kinds of emf \varepsilon: (a) the emf produced by the electric force on the electrons:

\varepsilon_E = \frac{1}{q} \oint {\vec F_{elec} \cdot d \vec \ell} = \oint {\vec E \cdot d \vec \ell}

and (b) the motional emf produced by the magnetic force on the electrons that is caused by the motion of the wire:

\varepsilon_B = \frac{1}{q} \oint {\vec F_{mag} \cdot d \vec \ell} = \oint {(\vec v \times \vec B) \cdot d \vec \ell}[/itex]<br /> <br /> If the loop of wire is fixed (\vec v = 0 all around the loop) , then \varepsilon_B = 0 and \varepsilon= \varepsilon_E. If the loop is changing size or shape, then \varepsilon= \varepsilon_E + \varepsilon_B.<br /> <br /> When you take the differential form of Faraday&#039;s law (without currents)<br /> <br /> \nabla \times E = - \frac {\partial \vec B} {\partial t}<br /> <br /> and apply Stokes&#039;s Theorem to convert it to an integral equation (which requires a fixed loop for the line integral on the right-hand side), the left-hand side gives you only \varepsilon_E. This is not the complete &quot;integral version of Faraday&#039;s law&quot; because it doesn&#039;t include \varepsilon_B. Including \varepsilon_B allows us to relax the requirement that we use a fixed loop when calculating \Phi_B.<br /> <br /> So what we call the &quot;integral version of Faraday&#039;s law&quot; is not purely the integral of the differential version of Faraday&#039;s Law.

<br /> I have read about that before in a paper by Giuseppe Giuliani. However, that is the only place I&#039;ve read about it. Do you know whether any textbook (or any published source other than Giuliani for that matter!) which discusses the &quot;true&quot; integral form of Faraday&#039;s Law? It seems surprising to me that no one else in the literature has even thought about this issue.

 

[lugita15]

NO, they are NOT mathematically equivalent. look at the differential form of faraday's law again. What if the B-field is NOT changing with time (static B-field) but the conducting loop is time varying (its size of orientation is changing)? There is still an induced EMF! This is explained by the integral form, but is neglected by the differential form.

When you transform from the integral form to the differential form you bring the time derivative into the flux integral and only differentiate the B-field component but not the dS, but in some cases the B-field is time invariant and the surface area or angle the surface makes with the B-field changes...this is not accounted for in the differential form.

Sorry, this is difficult for me to explain since I am bad with Latex.

Never mind about my confusion. I just realized why the differential form doesn't hold for motional emf. I retracted that part about two minutes after I said. Unfortunately, you replied to it before that.

 

[leright]

I have read about that before in a paper by Giuseppe Giuliani. However, that is the only place I've read about it. Do you know whether any textbook (or any published source other than Giuliani for that matter!) which discusses the "true" integral form of Faraday's Law? It seems surprising to me that no one else in the literature has even thought about this issue.

actually, jackson, in a cryptic way, describes the derivation of the 'correct' differential form of Faraday's law where he differentiates both the B-field and the dS parts. He introduces a convective derivative.

Here is a thread (that I started a long time ago) addressing this difficult issue.

https://www.physicsforums.com/showthread.php?t=120518

 

[pardesi]

I know that Faraday's law handles a motional emf, I'm not disputing that. I'm just saying that strictly speaking, Faraday's Law only refers to a fixed loop. I know that Faraday's Law is valid for a changing loop as well. I was just arguing semantics.

right exactly what i am was saying.also lugita15 can u post any e-link to the paper by Giuseppe Giuliani.
also griffith has it

 

[lugita15]

right exactly what i am was saying.also lugita15 can u post any e-link to the paper by Giuseppe Giuliani.
also griffith has it

This is the paper by Giuliani:
http://arxiv.org/PS_cache/physics/pdf/0008/0008006v1.pdf

Are you sure that Griffith talks about it? If so, what does Griffith say on the subject?

 

[lugita15]

actually, jackson, in a cryptic way, describes the derivation of the 'correct' differential form of Faraday's law where he differentiates both the B-field and the dS parts. He introduces a convective derivative.

Here is a thread (that I started a long time ago) addressing this difficult issue.

https://www.physicsforums.com/showthread.php?t=120518

After having skimmed that other thread, I have a question for you: if a term concerning velocity is introduced in Faraday's Law, then doesn't it change the whole fundamental character of the Maxwell equations? Maxwell's equations are supposed to the relate the Electric Field, the Magnetic Field, the charge, and the current. If velocity is introduced, then won't that change everything? Won't even the solution of the wave equation be different?
This problem seems important enough that it shouldn't just get a "cryptic" reference from Jackson and an obscure paper by Giuliani. It seems like it should be discussed in more depth in textbooks.

 

[pardesi]

This is the paper by Giuliani:
http://arxiv.org/PS_cache/physics/pdf/0008/0008006v1.pdf

Are you sure that Griffith talks about it? If so, what does Griffith say on the subject?

thanks for that.

griffith as usual is perfect and great .he denies to accept the fact that motional emf as a part of 'true' farady'a laws ang gives a proof of it using lorentz force

 

[lugita15]

thanks for that.

griffith as usual is perfect and great .he denies to accept the fact that motional emf as a part of 'true' farady'a laws ang gives a proof of it using lorentz force

What does Griffith exactly say on the subject of modifying Faraday's Law?

 

[pardesi]

do u want me to write the entire literature here?

 

[lugita15]

do u want me to write the entire literature here?

No. What page number is it on in the third edition?

 

[lugita15]

Never mind, I found the proof that the flux rule is valid for motional emf. Now when I was reading a few pages in Griffith after the proof, I found a statement that Einstein was very surprised about this coincidence, that the same law works in two completely different situations. This is what made him devise the special theory of relativity.

In that case, I believe that the resolution to this dilemma lies in special relativity. I just don't know what that resolution is. It does seem too much of a coincidence that both motional emf and transformer emf are given by the flux rule. Perhaps there is some relativistic link between these phenomena.

 

[pardesi]

yes too much of a co-incidence not well appreciated in most books.

 

[jtbell]

In that case, I believe that the resolution to this dilemma lies in special relativity. I just don't know what that resolution is.

Start with a wire moving perpendicularly with constant velocity across a purely magnetic field, which produces a motional emf. In the reference frame in which the wire is stationary, there is (besides a magnetic field) an electric field which produces an emf in the wire, that has the same net effect as in the original reference frame.

Mathematically you formulate this by constructing the electromagnetic field tensor F_{\mu\nu} in the first reference frame, using the components of \vec B and \vec E. Of course in this frame \vec E = 0. Then you apply the Lorentz transformation to this tensor, which gives you the components of \vec B and \vec E in the other reference frame. It turns out that \vec E \ne 0 in that frame. Griffiths discusses this transformation later in the book, but I don't have my copy handy to look up a page reference.

 

[lugita15]

Start with a wire moving perpendicularly with constant velocity across a purely magnetic field, which produces a motional emf. In the reference frame in which the wire is stationary, there is (besides a magnetic field) an electric field which produces an emf in the wire, that has the same net effect as in the original reference frame.

Mathematically you formulate this by constructing the electromagnetic field tensor F_{\mu\nu} in the first reference frame, using the components of \vec B and \vec E. Of course in this frame \vec E = 0. Then you apply the Lorentz transformation to this tensor, which gives you the components of \vec B and \vec E in the other reference frame. It turns out that \vec E \ne 0 in that frame. Griffiths discusses this transformation later in the book, but I don't have my copy handy to look up a page reference.

I know that part, about how the Lorentz transformations of E and B explains why the flux rule works whether a magnetic field moves or the circuit moves.
The thing that troubles me is the coincidence that the flux rule works in ALL cases. If a loop rotates at a constant angular velocity in the presence of a uniform constant magnetic field, the emf in the loop is given by the flux rule. If the magnitude of a uniform magnetic field perpendicular to a loop increases at a constant rate, the emf is still given by the flux rule. What is the connection between these four cases, that the flux rule works in all four?:
1. Circuit is stationary, flux increases due to region of magnetic field moving into the the area of the circuit.
2. Region of magnetic field is the same is stationary, flux increases due to (part of) circuit moving moving into region of magnetic field.
3. Constant, uniform magnetic field, flux is sinusoidal due to rotating loop with constant angular velocity.
4. Stationary circuit, flux increases due to increase in the magnitude of uniform magnetic field directed perpendicular to the circuit.
In all cases, emf=-(time derivative of magnetic flux)
The first two cases are related to each other by the Lorentz transformations of E and B. How are the third and fourth cases related to each other and to the first two cases?

 

[lugita15]

I just wanted to throw this out there because it is related to the problems we're having with the Flux rule. It is an "exception to the Flux rule" pointed out by Feynman in his Lectures on Physics:
Consider a copper disk rotating with constant angular velocity, as shown in the figure. A bar magnet is directed normal to the surface of the disk, as shown in the following figure:

If a galvanometer is used to measure the induced current in the outer rim of the disk, a nonzero induced current is detected. Using the Lorentz force law, this is easily explainable. Consider a small charge dq on the outer rim of the disk. At any given time, it has a nonzero velocity because of the rotation of the disk. For this reason, the bar magnet exerts a magnetic force on the charge. This makes the charge move differently than the disk itself. Some of these charge will go through the galvanometer, and thus the galvanometer will indicate the existence of a current.

But wait a minute. Let's try applying the integral form of Faraday's Law: the emf along a closed loop is equal to the rate of change of the magnetic flux through the loop. Take the loop to be the outer rim of the copper disk. But the magnetic flux through surface of the copper disk is constant, and thus the rate of change of magnetic flux is zero. This is because both the magnetic field and the area are both constant. So we have a strange situation in which the induced emf is nonzero even though the rate of change of magnetic flux is zero.
So why doesn't the integral form work in this case? What modification of the integral form would allow us to calculate the induced emf?

 

[pardesi]

great example.
yes this proves the point. what is changing in other cases is the area.so it is good that formulaes coincided but fact remains the proof and the assumptions that go behind it should be well appreciated.

 

[Paulanddiw]

great example.
yes this proves the point. what is changing in other cases is the area.so it is good that formulaes coincided but fact remains the proof and the assumptions that go behind it should be well appreciated.

As in your original post, this problem is solved with a loop of changing area--one side moves and the other is stationary. If you have access to R. Becker, "Electromagnetic Fields and Interactions", Vol 1, (1964) look on page 378.

 

[lugita15]

As in your original post, this problem is solved with a loop of changing area--one side moves and the other is stationary. If you have access to R. Becker, "Electromagnetic Fields and Interactions", Vol 1, (1964) look on page 378.

Yes, but what is wrong with using as the loop the outer rim of the disk? The integral form of
Faraday's law applies to any loop.

 

[lugita15]

Does anyone have any answers to the questions I asked in post #50?

 

[Paulanddiw]

Yes, but what is wrong with using as the loop the outer rim of the disk? The integral form of
Faraday's law applies to any loop.

I'm beginning to understand your questions. Some loops work, some don't. I don't remember any of the books explaining how to chose which one.

 

[lugita15]

I'm beginning to understand your questions. Some loops work, some don't. I don't remember any of the books explaining how to chose which one.

Apparently, there are books explaining how to choose the loop over which the integration is done for Faraday's. I'm not sure what textbooks discuss this, but Giuliani mentions these books. The point is, however, that we shouldn't need to be restricted in our choice of loops. All loops are supposed to work.

 

[Paulanddiw]

Yeah but, some loops give different answers than others.

 

[pardesi]

but that shouldn't be the case.so hoe do u resolve it probably an example would help

 

[Paulanddiw]

My immediate, and incomplete, reaction is that if v=0 all the loops give the same answer, but, maybe not for motion.