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About fem produced by coil

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Poldo

Computer
Jun 3, 2013
12
Morning.
I would like to speak about the EMF induced by a magnetic field in a coil.
I refer to the formula fem = d(NAB)/dt
In all texts is always something about an A or a B variable with the time, but nothing about N.
I would like to know if above mentioned formula is still valid in this evenience: fem= AB dN/dt.
I mean: If the area of coils and magnetic field are constant and only the number of coils is time variant, the formula is still valid?
Thanks for the replay
 
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"there is not inversion of flux on it"

Which you can therefore accomplish with a split secondary. The windings are already done for you and using a AC source gives you a better signal to work with.


Why is your company requiring you to do this? This is all proven physics.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529
 
Hi IRstuff...
How I said at the beginning, this is only to make toy to my granddaughter [bigsmile]
I started with normal dinamo (coils and magnet tourning inside... easy to do and easy to see on the oscilloscope) (B variant) and then I have moved to change area... and that was the problem and the discussion originator!!!
The game is over and the curiosity starts !!!

 
My exercise is:
we have a spiral S of N loop of area A (head of wire called 1 and 2) and an other spiral S1 of N1 loop of same area (head of wire called 3 and 4).
Then in each spiral we have B*N*A and B*N1*A flux (we can call them F and F1)
We start with a voltmeter connected to the 1 and 4 points.
If we connect and disconnect, by a switch, this two spirals in serie (connecting 2 and 3 points) obviously we don't measure emf (the total flux doesn't change Ft=F+F1 at any time)

But if we interpose between the 2,3,4 terminals a device (inverter switch, call it D and its terminal D1 and D2) we can obtain:
( the "->" means short-cut)
First position of switch:
Point 1 - coil - Point 2 -> D1 -> Point 3 - coil - Point 4 -> D2

Other position of switch:
Point 1 - coil - Point 2 -> D1 -> Point 4 - coil - Point 3 -> D2

Because the flux has a sign, in the first position we have Ft'=F+F1 but in the second we have Ft"=F+(-|F1|)

Then we obtain a Delta of flux... then I aspect an emf due to the dF/dt.
And because to the step variation I aspect a pulse emf (positive or negative) that is what seems what I see on the low-cost oscilloscope.
You haven’t said the source of your flux. Because you’re later referring to figure 6.5, I’m going to ASSUME assume it’s a permament magnet and the flux density B is constant.

Then your experiment is analogous to Figure 6-23(a) on page 433. The result is stated above that on the same page in paragraph (b) ”Changing the cofiguration of a circuit using a swUtch does not result in an electromotive forces unless the magnetic flux itself changes”.

With reference to the fig.6.5 (pag.400) we can see that, in constant B, moving one side of the loop we increase the flux (B*A). Right? Then we have emf generated. The formula emf= B*dA(t)/dt is working!
Yes. You’ll notice that different parts of the circuit are in different reference frames. It we want to sum the voltages around the loop we must account for that by the logic of paragraph 6-3-1 on page 417 (Galilean transformation). A similar example is worked out in figure 6-14 (page 418) and you see the sum of integral E dot dl in different parts of the circuit and different reference frames.

Where is my conceptual mistake?
You must start with equation (1) on page 395
Integral-along-Closed-Contour-C {E dot dL} = -d/dt ( Integral-Over-Survace-S {B – dA} )
where C is the permiter of S with specified polarity

You have invented a fiction that the right hand side is d/dt (N(t) * B(t) *A(t)) where N(t) changes with change in configuration of the system.

You must define your contour and integral in a way such that rules of math can be applied. If you have constant velocity moving contour, we can transform is to a different reference frame as in Fig 6-14 and calculate a result. If you have a contour which spontaneously changes to something completely different (switching), it is not a change in the contour, it is two completely different contours. We can analyse before and after the switching, but we should not assign the significance d/dt N(t) * B * A that you have assigned during switching. There is nothing in reference frame theory to suggest that...everything remains in the same reference frame with conductors are remaining at rest compared to the magnet.

I can agree it is a challenge to visualize. You are welcome to ponder and experiment how you like. One thing you should not have any doubt is that Zahn knows his subject much better than either of us and is certainly correct in the portion quoted above.



=====================================
(2B)+(2B)' ?
 
Hi!!!!
Then.... reassuming....
If I well understood, the emf will be 0 beacuase there are not moving electrons using switch (instead, moving side of the circuit there are!)... I have electrons inside the moving wire (they are subjected to a force deriving from B field e v of the side) that I don't have in the switch... Right?

I'm not so convinced about the different contour that implies different surfaces and then different systems for the flux calculation. Also in the sliding side there is different contour...

From these, we have
emf = 0 ====> 0 = d(B*A)/dt ...
but, because the flux is changing (B constant, A varying), which component kills the non-zero derivative of the flux?

 
If I well understood, the emf will be 0 beacuase there are not moving electrons using switch (instead, moving side of the circuit there are!)... I have electrons inside the moving wire (they are subjected to a force deriving from B field e v of the side) that I don't have in the switch... Right?
Correct

From these, we have
emf = 0 ====> 0 = d(B*A)/dt ...
but, because the flux is changing (B constant, A varying), which component kills the non-zero derivative of the flux?
Refering to the problem of Figure 6-24a (right hand side).

There is no change in the area of any loop.

Let At be the area of one turn and Vt to be the voltage induced in one turn.

Vt(t) = d/dt(B*At) = 0 (because At is constant)

Now sum up the contributions of all the individual turns
Vsum = N(t) * Vt(t) = N(t) * 0 = 0

What I think you're suggesting is that the sum of the turn voltages becomes non-zero simply because we're adding in more turns which individually have zero voltage. That would not be logical.

Note that a "turn" is slightly different than a closed loop, but voltage induced in each is the same. i.e. start with a closed loop where wire is in circumferential direction and flux flows in the axial direction. Now cut that loop and displace the ends axially by small amount to form a turn. The relevant associated projected surface (for surface integral) does not change and induced voltage in the turn is same as induced voltage in the closed loop.


=====================================
(2B)+(2B)' ?
 
Sorry... but this time I don't agree...
I don't speak about voltage in one or more turns...
The Area is changing beacuase the formula and the experience say that if in one turn of area A (omissis...) the flux is BxA, if we have N turns of the same area, the flux is NxAxB...

Your explanation denies this sentence... or I'm wrong understanding....
 
If turns are constant, the expression flux = N* Phi(t) * A can be differentiated with respect to time to find voltage.
In this case, the physical fact that it is N* single turn voltage is still true, but not particular important to us....just plug into formula.

If we change our interest to constant flux, variable turns, we cannot go to expression N K Phi (relevant for constant turns) and expect it to have the same meaning. We have to return to more general law (equation 1) and use the physical understanding (volts per turn) to help us proceed.

If in doubt, refer to bolded statement by Zahn quoted above...surely you don't think he us mistaken?


=====================================
(2B)+(2B)' ?
 
I have some (obvious?) corrections;
(excuse: it was very early and I was typing on my phone)
electricpete corrected said:
If turns are constant, the expression flux = N* Phi(t) B(t) * A can be differentiated with respect to time to find voltage.
In this case, the physical fact that it is N* single turn voltage is still true, but not particular important to us....just plug into formula.

If we change our interest to constant flux, variable turns, we cannot go to expression N K Phi B A (relevant for constant turns) and expect it to have the same meaning. We have to return to more general law (equation 1) and use the physical understanding (volts per turn) to help us proceed.

If in doubt, refer to bolded statement by Zahn quoted above...surely you don't think he us mistaken?

Speaking of "denying"... would you deny the following?
Vcoil(t) = N(t) Vturn(t)
for coil with N(t) identical turns


=====================================
(2B)+(2B)' ?
 
Hi electricpete!
First of all, I want apologize if my English is not very good or it seems not polite... I'm italian!!!
Thank you all for your comments and I apologize for my insistence... but I like to understand and be convinced.

Far from me to think that Zhan (but also you) may be mistaken, but.... I'm hard to convince!!!! :)

About the last formula Vcoil(t), I agree and in fact, I have set to zero the resulting emf
I think that is undeniable that F= B*A1 and if A=N*A1... F=B*N*A1. This F exist even if there are not electron around!!! The flux is line of force of Magnetic field concatenated with an area A (contour or surface delimited by...).

Well...for t=0 we have F'=x; for t1=1ms we have F"=x+y (where y can be positive or negative). Please consider that my contour is always the same (I don't move the measurement point). What I do is reversing the second turns direction (refer to explanation of my experiment in previous post).



 
Most people would probably be bored with this converation by now. Lucky for you I like to hear myself talk, at least when it comes to certain subjects.

I can think of three ways to attack the problem, each spelled out below.


First and easiest – consult an expert. We have the expert words of Zahn and there is no doubt what he says on this exact question. Even if you’re not sure of the details of how to prove it to yourself, you must know that Zahn didn’t just make a mistake when he specifically said (paraphrasing) lots of people get confused about implications of switching, but there is no voltage induced when switching with stationary leads and static flux pattern.

============

Second way – intuition based on the physics. I suggest maybe if you study the various incarnations of Faraday’s law you’ll see they all involve some motion or time variation and in fact are interrelated. This is something you should do on your own.

But I’ll throw out some thoughts for consideration which I freely admit are not rigorous proof but I hope may provide some intuition. Consider two cases of coil moving next to magnet and magnet moving next to coil. In the case of loop moving next to magnet, we explain the force using F = q V x B. But in the case of magnet moving next to loop, velocity is zero and we explain the force as induced voltage due to rate of change of flux linked by the loop. The two problems can be identical except for different choice of inertial reference frame viewed in and by necessity the predicted force is the same. This gives us some idea that motion and time rate of change are the basic common ingredients and they are related to each other. Do a third experiment where there is no motion, but time rate of change of the flux in the plane of the loop. Here we have no obvious motion. But we are looking only at one plane and that’s all the loop knows about. We could probably create that same time varying voltage in the place of the loop by setting up (with permanent magnets) a 3-d spatially-varying flux pattern moving which is static in its own reference frame but moving with respect to the loop. It’s somewhat loose and not exact but imo goes to extend the intuition of the unifying factors underlying all induced voltage: motion or time rate of change.


Third Way: Attempt at math application of Maxwell’s law of induction.
I would challenge you to find a calculus textbook that will tell you how to you find the derivative of a surface integral whose defining perimeter contour spontaneously morphs to something completely different. Go get your calculus books, I’ll wait....

Didn’t find anything? That’s because it’s not there. There is no way to solve that because it’s an invalid changing contour. I don’t know how to talk about a proof that doesn’t exist. Let’s instead talk about some similar but realistic time varying contours leading to a scenario that resembles your switching and see how they act.

See attachment figure.

We have in Case 1 a voltmeter in circuit with insulated wire which is roughly broken into two loops by a rubber band that keeps it physically together (but not electrically contacting) in the middle. The wires are located in a plane just above a permanent magnet which produces flux density B over area A in the plane of the wire, in a location that links loop 1 but not loop 2. (if you want a more realistic situation to explain why no flux in loop 2, then consider instead that the wires are in the plane of the airgap of a C-shaped permanent magnet / core so that there is flux in the airgap but no significant flux elsewhere in the plane).

The flux linked in the circuit for case 1 is 1*BA (since loop 2 contributes nothing).

Now see case 2. To get from case 1 to case 2 we grab the right side of loop 2 and pull it straight left. Now we have two turns with opposite polarity, the flux linked by the circuit is zero. Note that in getting from case 1 to case 2, loop 2 had to cross over the area of B (there would have been voltage induced in the process).

Now see case 3. To get from case 1 to case 2, we grab the right side of loop 2 and twist it and pull it to the left. The result is two turns of same polarity. The flux linked by the circuit is 2*A* B. Note that in getting from case 1 to case 3, loop 2 had to cross over the area of B (there would have been voltage induced in the process).

Let’s go from case 2 to case 3. We have to grab the left side of loop 2 and flip it. Two sides of loop 2 will cross through B and there will be voltage induced in the process.

What you are proposing is putting a switch on the right side of the figure and transitioning direclty from 2 to 3 by use of switch. No wires cross over the area of B and no voltage is induced. Is it conflict with Faraday? Nope, this is not an evolution of the contour, it involves opening and closing circuits to switch between two completely different contours in an unnatural way which is not a simple evolution of a single contour.

By the way, I should have used “turn” vs “loop” in this figure, but I’m not going back to change it.

About the last formula Vcoil(t), I agree and in fact, I have set to zero the resulting emf
That should be the answer. Zero times any finite number of turns will still be zero.

I think that is undeniable that F= B*A1 and if A=N*A1... F=B*N*A1. This F exist even if there are not electron around!!! The flux is line of force of Magnetic field concatenated with an area A (contour or surface delimited by...).

Well...for t=0 we have F'=x; for t1=1ms we have F"=x+y (where y can be positive or negative). Please consider that my contour is always the same (I don't move the measurement point). What I do is reversing the second turns direction (refer to explanation of my experiment in previous post).
Your contour is not always the same. You have disconnected and reconnected to new wires. It is a spontenaous change in circuit topology, it is not a continous evolution of a given topology.


=====================================
(2B)+(2B)' ?
 
 http://files.engineering.com/getfile.aspx?folder=da47c19f-5894-4b7f-967d-ad9b7ee2435c&file=2013-06-15_14.00.22.jpg
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