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

[pardesi]

well iwas going through the standard example that is there is a uniform magnetic field in aregion going into the palne and ther is a rectangular 'loop' with one side which can slide on metallic rods with constant speed v and the the relation

\epsilon=BLv
well the example is (almost) here
well i could intutively understand that from lorentz' force concept that when the side starts moving so do the electrons so they experience a force and a current is produced.

well i wanted to go about it from farady's alw.well what i couldn't get is farady's law states when the "flux across a circuit...".but here our loop itself is changing so how do i utilize the law

 

[marlon]

The flux (number of magnetic field lines per unit of surface) needs to change with respect to time : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

So, the voltage is created whenever to loop rotates in the magnetic field or when it moves in/out the magnetic field !

marlon

 

[pardesi]

but here the 'loop' itself is not constant

 

[marlon]

but here the 'loop' itself is not constant

What do you mean by the loop is not constant ?

Generally, the flux changes due to :
1) change in the position of the loop with respect to the B-field (rotation or translation)

2) the loop changes its form (rectangle becomes circle eg).

The basic ingredient is the fact that the flux must change with respect to time ! If this does not happen, no potential is created.

marlon

 

[pardesi]

i mean here the rectangle grows in size it does have a change in area but this change is due to addition of 'more conducting element' as oppsoed to whenit changes toa circle in which the lopp is 'constant' but it's shape changes

 

[marlon]

i mean here the rectangle grows in size it does have a change in area but this change is due to addition of 'more conducting element' as oppsoed to whenit changes toa circle in which the lopp is 'constant' but it's shape changes

First of all, if it changes in surface, this means that the flux WILL change and a voltage will be created.

But how does the rectangle increases its surface ? I don't get the "due to more conducting elements" reasoning.


marlon

 

[pardesi]

well waht i meant was consider the rectangular loop PQRS no suppose i start stretching it or change it into a circle in presence of a mganetic field then the flux changes and an 'emf' is induced across the loop PQRS.
but here the loop PQRS not only schanges its' shape but also gets some additional new elements.

 

[marlon]

but here the loop PQRS not only schanges its' shape but also gets some additional new elements.

How does the loop acquire additional new elemnts and what are these elements ? I really don't get it. Besides, what exactly do you want to know other than the consequence of the applicaton of Faraday's law which we covered here.

marlon

 

[pardesi]

well i got thsi ...i read that in griffith where it is explained nicely...well it is sheer coincidence he writes that the motional emf has the same formula as that for the other two cases...
great

 

[marlon]

well i got thsi ...i read that in griffith where it is explained nicely...well it is sheer coincidence he writes that the motional emf has the same formula as that for the other two cases...
great

By motional emf you mean the moving loop right ? What same formula ? You mean Faraday's Law ?

Actually, could you explain the solution to your problem to me because i don't get it.

marlon

 

[pardesi]

yes motional emf means moving coil but stationary fie;d(here).
and the formula refers to faraday's law
if u could get a copy of griffith then he very beutifully denies to accept that the emf induced due to motion of the cil as the one given by faraday's law.infact goes on to say that in some sense it is not the 'true' farady's law that holds here.
and most probably for the reason which i was worrying about
because if u go through faraday's law it starts witha line that "if through a loop..." so the loop is fixed and by loop i mean a collection some points.
but while we consider the motional emf as in my case the rectangle is taken to be the loop whose one side is moving.so the 'points' which make up the loop are not the same.
crudely speaking the loop itself is changing
so there is no point in assuming 'farday's law' for granted...though he discovered for this case also.it is just sheer coincidence taht thsi happened

 

[marlon]

faraday's law it starts witha line that "if through a loop..." so the loop is fixed

That's incorrect. What matters is the variation of the flux with respect to time, inside the loop. The loop may vary in structure as well.

marlon

 

[pardesi]

flux through what

 

[pervect]

If you're hopelessly confused by the area changing (honestly, I don't see why), just perform a line integral of the electric field around some closed curve.

http://www.ucl.ac.uk/Mathematics/geomath/level2/mult/mi3.html

You will find that the intergal over a curve of the electric field component pointing along the curve is non-zero.

If you like, you can use a vector identity to convert this line integral into an area integral. This is known as stokes theorem. This is why people are talking about integrating the flux, this is justified by Stokes theorem.

 

[pardesi]

i am not worried about the changing area...and stokes theorem is fine
also when u ask me to calculate the line integral it has to be about a fixed loop
what i am worried about is when we say that flux through a loop changes ...then obviously the flux has to be through a fixed loop,obviously i f i vary my loop with time my flux has to change(in most cases) ,so what i am saying is though the faraday's law holds true in case of motional emf...but that's just because it coincidentally happened to be so...
what i am opposing is 'direct' use of farady's law in getting the induced emf in case of motional emf..though it gives u the right answer but that won't be using faraday's laws in the right sense,as most books do (thankfully griffith didn't) and for the reason i stated above

 

[pardesi]

to clarify things further as in this case the we assume the loop to be the rectangle PQRS then calculate the flux through it at any time t to be \Phi=Blx
then we write E=\frac{d \Phi}{dt}=Blv where v is the speed of the movable side.
but this makes no sense because the flux through the loop is not changing as the field is constant here.what actually is changing is the loop which we are considering so the calculated flux changes.though the flux through PQRS is not changing.
also here the emf induced is best explained by lorentz force.
it is just by sheere coincidence(and i repeat) that what we do is working out right.

 

[marlon]

because the flux through the loop is not changing as the field is constant here.

what actually is changing is the loop which we are considering so the calculated flux changes.

But can't you see the contradictio in terminis in the above two sentences ? First you say the flux is constant, then you say it changes. Again, the flux IS changing because the number of magnetic fieldlines in the loop changes with time. The reason of this change is the varying loop ! The flux can change due to 1) the B field or 2) the loop that encloses the field lines !

marlon

 

[pardesi]

yes what i am saying is the flux through any loop is not changing with time.what is changing is the loop which we have taken into consideration.
say for example there is magnetic field uniform in a room.at time t i take my loop to be a square one inside the room.the next moment 'i' will take my loop to be a square one outside the room.so where will the induced emf be .

again i say the thing is the proof by faraday' law is not all correct(at least to me).lorentz force proof seems right.

 

[marlon]

yes what i am saying is the flux through any loop is not changing with time.what is changing is the loop which we have taken into consideration.

Again, if the loop is changing the flux is also changing. I would advise you to carefully investigate the definition of FLUX. After having done that, you will understand that what you are saying is incorrect.

say for example there is magnetic field uniform in a room.at time t i take my loop to be a square one inside the room.the next moment 'i' will take my loop to be a square one outside the room.so where will the induced emf be .

Yes because the flux has changed. Again, make sure you look at the definition of flux.

marlon

 

[pardesi]

flux \phi = \int_{S} \vec B . \vec dS

yes what i am saying is the flux through any loop is not changing with time.what is changing is the loop which we have taken into consideration.

by loop i mean any fixed loop .consider any loop u want to the flux through it doesn't change with time.what is changing is the loop u are considering .so just because u r changing the loop of your consideration why should an emf be developed.

Yes because the flux has changed. Again, make sure you look at the definition of flux.

when u say the flux has changed can u tell me flux through what has changed?

 

[lugita15]

again i say the thing is the proof by faraday' law is not all correct(at least to me).lorentz force proof seems right.

Faraday's Law, in integral form, states that at any given time, if a fixed closed loop is taken, the line integral of E around that closed loop is equal to the negative of the derivative of the magnetic flux through any surface bounded by the loop. The loop, as referred to in Faraday's law, MUST be fixed. I agree with you completely, in that respect.
In my view, this is how to resolve your question:
In order to find the emf at any time t, consider the fixed loop with dimensions l and vt. From time 0 to time t, the negative of the rate of change of magnetic flux through this fixed loop is -Blv. Even before the moving conducting rod reaches the position of one side of the fixed loop, this is still the rate of change of magnetic flux through the loop. When the conducting rod finally reaches position vt, the voltage of the physical circuit becomes -Blv. Although the physical circuit has changed, we are only concerning ourselves with a fixed loop when we calculate the emf at an arbitrary time t.

 

[pardesi]

u can also go through the proof using the lorentz force.it seems really good and fundamental.

 

[marlon]

flux \phi = \int_{S} \vec B . \vec dS


by loop i mean any fixed loop .consider any loop u want to the flux through it doesn't change with time.what is changing is the loop u are considering .so just because u r changing the loop of your consideration why should an emf be developed.

But if the loop changes, the above integral will have a different outcome as well and thus, your calculated flux has changed !

marlon

 

[pardesi]

yes that's what i am telling our loop itself is changing and farady's law talks of a fixed loop.
anyway when u change the loop then on which loop will the emf be developed or mathematically which loop's line integration...will be the rate of cahnge of flux.
take the very sam example i gava couple of posts back.
just because u have taken a loop once inside the room and the next moment outside another completely different from nowhere why should there be an induced electric field?

 

[lugita15]

just because u have taken a loop once inside the room and the next moment outside another completely different from nowhere why should there be an induced electric field?

Yes, that's a very good illustration of the fact that Faraday's Law refers integration around a fixed path.

 

[leright]

yes that's what i am telling our loop itself is changing and farady's law talks of a fixed loop.
anyway when u change the loop then on which loop will the emf be developed or mathematically which loop's line integration...will be the rate of cahnge of flux.
take the very sam example i gava couple of posts back.
just because u have taken a loop once inside the room and the next moment outside another completely different from nowhere why should there be an induced electric field?

You still need to carefully consider the concept of flux. The magnetic flux through the loop is a function of the magnetic field distribution, the area of the loop and the angle at which the vector defining the plane of the loop (the vector normal to the loop), dS, makes with the B-field at any given point in the loop. The key issue here is the DOT PRODUCT in the definition of the flux.

oops...for some reason I thought this was a problem dealing with a rotating loop...well, Faraday's law STILL applies to your situation. But instead of the angle the loop makes with the B-field changing, the AREA of the loop is changing.

 

[leright]

The loop, as referred to in Faraday's law, MUST be fixed. I agree with you completely, in that respect.

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).

 

[Paulanddiw]

I read in Jackson that faraday’s law works for any area whether it is changing because of change of size, rotation, stationary, etc.

It seems to me you are nervous about the sliding contacts of the moving conductor. As it slides, more of the stationary part of the mechanism is added to the circuit. Instead of sliding contacts, imagine that the wire is folded up at the intersection between the moving and stationary part of the circuit. As the moving section moves, the wire unfolds. Now all parts of the circuit are always in the same magnetic field, and the generated voltage will be Blv.

 

[jtbell]

I read in Jackson that faraday’s law works for any area whether it is changing because of change of size, rotation, stationary, etc.

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.

 

[pardesi]

i never said farday's law was wrong..what is said is that faraday's law although it holds for motional case as u can see above can only be in some sense 'extended' by including the lorentz force term.
now the question is why?:
because when the loop itself is changing talking of an 'induced emf' is somewhat confusing and meaningless because our loop itself is not fixed.further for a time-independent magnetic field flux through any loop is time independentand a proof as given in some books without acknowledging lorentz force is not that correct.
again why?
go to my previous example where u physically had no loop.but just a collection of points(u thought of it) inside a room with uniform magnetic field .then u thought of another loop outside the room where there is no magnetic field...and then u say flux is changing hence u have an induced field well that's meaningless ?

 

[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?