Author Topic: Induction question  (Read 1667 times)

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Offline nix85

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Induction question
« on: July 12, 2019, 07:41:44 pm »
I'd like someone to clear this up. If we look at the square shaped coil below, let's for a moment imagine it's just two vertical parallel wires not connected.

As magnet sweeps across them as shown by arrows obviously voltage is induced in each wire in same direction.

When we connect the two vertical wires into a coil, two opposing voltages should cancel out.. but this is not what happens.

In this configuration voltage is normally induced and no one so far seemed to ever question this.

There must be a simple explanation, so..anyone?

 

Offline T3sl4co1l

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Re: Induction question
« Reply #1 on: July 12, 2019, 10:17:12 pm »
So what, the magnet sweeps past one wire, then the other?  The magnet moving (translationally) in the plane of the loop?  Not through it (axially)?

Note that it's not meaningful to speak of the voltage across some open chunk of wire.  You only get an EMF around a loop.  The loop can be an imaginary one, returned through space itself -- it doesn't need to be 100% conductor -- but in all cases, the loop path must be defined because the induction is path-dependent.

(This gets more nuanced at radio frequencies, where an "open chunk of wire" can behave as an antenna; but unless we're talking about relativistically accelerated magnets in this example, we don't need to worry about this, and the magnetostatic, or quasi-static, case is relevant.)

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Offline nix85

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Re: Induction question
« Reply #2 on: July 12, 2019, 10:34:00 pm »
So what, the magnet sweeps past one wire, then the other?  The magnet moving (translationally) in the plane of the loop?  Not through it (axially)?

Note that it's not meaningful to speak of the voltage across some open chunk of wire.  You only get an EMF around a loop.  The loop can be an imaginary one, returned through space itself -- it doesn't need to be 100% conductor -- but in all cases, the loop path must be defined because the induction is path-dependent.

(This gets more nuanced at radio frequencies, where an "open chunk of wire" can behave as an antenna; but unless we're talking about relativistically accelerated magnets in this example, we don't need to worry about this, and the magnetostatic, or quasi-static, case is relevant.)

Tim

Of course "magnet moving (translationally) in the plane of the loop", as indicated by arrows.

Voltage can be induced in open circuit, current can flow only in closed one.

And your solution "So what, the magnet sweeps past one wire, then the other" does not cut it.

I again point out induced voltage is in same direction on two parallel wires of the coil and should cancel out, but this does not happen. Why.

I could have drawn the two wires of the square coil with 0 distance between them, that would illustrate the point even better and would not change anything as we all know such coils exist and work perfectly well.
« Last Edit: July 12, 2019, 10:37:08 pm by nix85 »
 

Offline T3sl4co1l

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Re: Induction question
« Reply #3 on: July 12, 2019, 10:53:08 pm »
That wasn't a statement of "so what.", but more of a "so let me get this straight", sorry if that was unclear.

The picture is ambiguous, as all 3D-to-2D projections are, so I wanted to clarify more precisely.

Voltage can only be measured at a point, such as at a small gap in a loop; I think you will find it difficult to do this any other way. :-+

If the coil has zero width or height or both, then the induction is indeed the same and the result is zero.  Otherwise, you can imagine the magnet's fringe (where dPhi/dx is greatest) cutting across one wire first, then the other.  It's the area of the coil that generates the voltage, and nonzero area needs nonzero width and height. :)

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Offline nix85

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Re: Induction question
« Reply #4 on: July 12, 2019, 11:12:57 pm »
If the coil has zero width or height or both, then the induction is indeed the same and the result is zero.  Otherwise, you can imagine the magnet's fringe (where dPhi/dx is greatest) cutting across one wire first, then the other.  It's the area of the coil that generates the voltage, and nonzero area needs nonzero width and height. :)

Tim

It seems you claim voltage can not be measured unless wire is looped onto itself. I don't think anything could be farther from the truth.

Zero width or height? I never said anything about such impossible dimensions, but only that distance between two vertical wires can be 0, no horizontal spacing. That is irrelevant.

Again, point is voltage in left and right side of the coil should be in same direction and should accordingly cancel out but dont. Why.

In other words voltage on one of two sides of the coil is opposite than it would be if we had two disconnected parallel wires instead.
« Last Edit: July 12, 2019, 11:20:52 pm by nix85 »
 

Offline T3sl4co1l

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Re: Induction question
« Reply #5 on: July 12, 2019, 11:21:43 pm »
It seems you claim voltage can not be measured unless wire is looped onto itself. I don't think anything could be farther from the truth.

Produce a diagram showing how it can be done. I'll wait. :D

Quote
Zero width or height? I never said anything about such impossible dimensions, but only that distance between two vertical wires can be 0, no horizontal spacing. That is irrelevant.

If the distance between vertical wires can be zero, then if the vertical wires are cross-connected with horizontal members as well, forming a loop, does that loop not have zero enclosed area?

If the answer is no, then you will have to specify what geometry you are working in, because it's not Euclidian 3D space!

Quote
Again, point is voltage in left and right side of the coil should be in same direction and should accordingly cancel out but dont. Why.

Again, because the field is different for the two wires.  This isn't hard?

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Offline nix85

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Re: Induction question
« Reply #6 on: July 12, 2019, 11:29:53 pm »
Produce a diagram showing how it can be done. I'll wait. :D

No diagram is needed to measure voltage across a piece of wire.

Quote
If the distance between vertical wires can be zero, then if the vertical wires are cross-connected with horizontal members as well, forming a loop, does that loop not have zero enclosed area?

If the answer is no, then you will have to specify what geometry you are working in, because it's not Euclidian 3D space!

You are compliating things in totally useless direction i never pointed to. All i said is that vertical wires can be right next to each other.

Quote
Again, because the field is different for the two wires.  This isn't hard?

Tim

That's not a satisfactory answer. Field surely is slightly different cause wires occupy slightly different position but that's totally irrelevant. Both wires see flux of same polarity approaching them, thus induced voltage should be in same direction in both wires and cancel out. Simple, right?
 

Offline amyk

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Re: Induction question
« Reply #7 on: July 13, 2019, 03:33:59 am »
That's not a satisfactory answer. Field surely is slightly different cause wires occupy slightly different position but that's totally irrelevant. Both wires see flux of same polarity approaching them, thus induced voltage should be in same direction in both wires and cancel out. Simple, right?
I already posted an answer in your other thread (please don't make multiple threads for the same question) --- the one-phrase answer is "flux density", but you can see my full answer there.
 

Offline nix85

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Re: Induction question
« Reply #8 on: July 13, 2019, 08:16:20 am »
That's not a satisfactory answer. Field surely is slightly different cause wires occupy slightly different position but that's totally irrelevant. Both wires see flux of same polarity approaching them, thus induced voltage should be in same direction in both wires and cancel out. Simple, right?
I already posted an answer in your other thread (please don't make multiple threads for the same question) --- the one-phrase answer is "flux density", but you can see my full answer there.

The field strength is increasing on one side of the coil, and decreasing on the other. In the exact middle, the sum of the induced currents cancels, but at any position other than the middle, they do not, and you will see that the midpoint is where the current goes to zero and then reverses direction.

That is not a satisfactory answer either. I already thought about it and if it were true overall voltage would be but a remanant left over from two sides "fighting" each other, but that is NOT the case.

Also, wires can be right next to each other, seeing same flux in same direction approaching them at same time and yet voltage is induced altho according to all known rules it should cancel out.
« Last Edit: July 13, 2019, 08:18:08 am by nix85 »
 

Offline nix85

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Re: Induction question
« Reply #9 on: July 13, 2019, 09:27:27 am »
If we look at this axial generator, magnet first sweeps across one side of the coil inducing + phase and then another for - phase.



Those who make these wind axial gens say if same magent passes across both sides of the coil at the same time it would cancel out and no voltage is induced - as i assumed should be.

I suppose they are right, altho my testing did show voltage induced.

"If both radial legs were over magnets with the same polarity – i.e. both north poles, the generated currents would cancel one another out: the left side of the coil generating a current in the clockwise direction, and the right side of the coil generating a current in the anticlockwise direction resulting in no current flowing around the coil."



http://www.reuk.co.uk/wordpress/wind/wind-turbine-alternator-basics/
« Last Edit: July 13, 2019, 09:30:19 am by nix85 »
 

Offline nix85

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Re: Induction question
« Reply #10 on: July 13, 2019, 11:36:03 am »
This video shows what happens when both sides of the coil pass through same flux, no current.

 

Offline DannyTheGhost

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Re: Induction question
« Reply #11 on: July 14, 2019, 09:54:46 am »
I think you miss the main point in all of this.
What 'sees' flux is not the wires
It's the surface of 'wire current loop' what sees it
Therefore  there is voltage induced in the coil until flux is changing inside loop surface
And your example clearly shows it
 

Offline TimFox

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Re: Induction question
« Reply #12 on: July 14, 2019, 01:52:31 pm »
The technical term for the flux that induces emf in a loop is the "linked flux", i.e., the flux that passes through the loop-shaped wire.  That is why a larger emf is induced in a multi-turn coil, where the flux links each turn and the total emf adds up.
 

Offline nix85

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Re: Induction question
« Reply #13 on: July 15, 2019, 08:50:41 am »
I think you miss the main point in all of this.
What 'sees' flux is not the wires
It's the surface of 'wire current loop' what sees it
Therefore  there is voltage induced in the coil until flux is changing inside loop surface
And your example clearly shows it

I think you are the one who misses it. What you are referring to is amperian loop and i am well aware of Amper's laws.

And yes, it IS the wires that see the flux, be it single piece of wire or coil made of many turns.
 

Offline TimFox

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Re: Induction question
« Reply #14 on: July 15, 2019, 11:54:43 am »
If you connect a voltmeter by two probe wires to different points along a wire, then you have formed a loop:  (Wire end A) to (Wire end B) to (Probe B) to (Voltmeter Box) to (Probe A) to (Wire end A).  If the distance from A to B is non-trivial, then the area of the loop is also non-trivial, especially if the wire is not straight.
The “point” at which the voltage is measured is the input terminal of the voltmeter, ideally a coaxial connector.
« Last Edit: July 15, 2019, 07:28:51 pm by TimFox »
 

Offline SiliconWizard

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Re: Induction question
« Reply #15 on: July 15, 2019, 12:05:10 pm »
Time to call Dr Lewin. :popcorn:
 

Offline T3sl4co1l

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Re: Induction question
« Reply #16 on: July 15, 2019, 12:29:01 pm »
I think you miss the main point in all of this.
What 'sees' flux is not the wires
It's the surface of 'wire current loop' what sees it
Therefore  there is voltage induced in the coil until flux is changing inside loop surface
And your example clearly shows it

I think you are the one who misses it. What you are referring to is amperian loop and i am well aware of Amper's laws.

And yes, it IS the wires that see the flux, be it single piece of wire or coil made of many turns.

The distinction is subtle.  For a sufficiently rapid AC flux, the voltage induced in the surface of the wire causes a current flow which opposes the external magnetic field.  Consequently, the magnetic field is attenuated into the depth of the wire -- skin effect.  It is this surface which Danny is referring to.

The true underlying mechanisms, and the math which describes them, are very specific, very well known, and very very accurate (indeed perfect, as far as we can tell).  Of particular note, when working a problem of this type, one may use a loop integral, which is perfectly equivalent to an area integral, for the loop (perimeter) and area being of the same continuous region.  This is Green's theorem.

This may seem surprising -- the loop integral does not actually need to sample the field inside the region.  We can construct arbitrary fields very easily which violate this; but as it happens, such fields cannot be magnetic or electric fields!  It is a consequence of this property -- that these real fields must be continuous and analytic -- that allow such shortcuts as this to exist. :)

Also note that the loop must be closed.  You can't enclose a region with an open path.  That's meaningless.  The closed loop path is exactly the path taken by the current flow -- as I said, and TimFox desmonstrates, voltage cannot be measured without forming a loop.

You can't cheat it with quirky physics, either.  For example, suppose we put an open wire in an AC field, thus presumably inducing a voltage along its length, then shoot UV light at it, causing it to emit photoelectrons from whatever voltage the surface was at at that instant.  Well, two things must happen before we can detect the emitted electrons: one, they must flow along some definite path* to the detector; and two, they must also flow in free space through the field, which as an AC field, has electric as well as magnetic components to it, and these act on the path taken (which, if we know the field, we can solve backwards for, of course).  The only thing we've done here, is make the loop harder to figure out, than simply connecting wires to the experiment!

*Definite by the rules of quantum mechanical paths: the integral over all possible paths, times the probability of each path.

Tim
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Offline nix85

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Re: Induction question
« Reply #17 on: July 15, 2019, 08:47:08 pm »
Quote
For example, suppose we put an open wire in an AC field, thus presumably inducing a voltage along its length

Well, when we place the probes on the wire we close the loop and voltage is measured.

This does not mean voltage was not induced before the loop was closed.

Maybe i'm wrong, but i find it hard to believe voltage is not induced in open piece of wire by changing flux.
« Last Edit: July 15, 2019, 08:48:57 pm by nix85 »
 

Offline T3sl4co1l

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Re: Induction question
« Reply #18 on: July 16, 2019, 10:31:53 am »
I didn't say voltage is not induced, I said the statement itself is not meaningful.  Voltage can only be measured at a point, not between points!

The issue is not about what the number is; it's also not about whether a number even exists; it's that you're asking what's the state capitol of cheese.  It's the mathematical equivalent of a type mismatch. :D

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Offline nix85

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Re: Induction question
« Reply #19 on: July 16, 2019, 05:03:18 pm »
I didn't say voltage is not induced, I said the statement itself is not meaningful.  Voltage can only be measured at a point, not between points!

The issue is not about what the number is; it's also not about whether a number even exists; it's that you're asking what's the state capitol of cheese.  It's the mathematical equivalent of a type mismatch. :D

Tim

Your reasoning is full of holes like Swiss cheese. :D If you want to get into deeper, implicate order of things, then whichever path you take, you will in the end realize whole creation is unreal, just a lucid dream projected, inhabited and experienced by the Spirit in the eternal now.
 

Offline T3sl4co1l

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Re: Induction question
« Reply #20 on: July 16, 2019, 10:09:03 pm »
You are welcome to identify the holes.  Meanwhile, I will continue constructing meaningful systems and calculating meaningful numbers with our knowledge of the complete and correct theory of electromagnetism. :-+

Tim
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Offline nix85

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Re: Induction question
« Reply #21 on: July 16, 2019, 11:11:58 pm »
You are welcome to identify the holes.  Meanwhile, I will continue constructing meaningful systems and calculating meaningful numbers with our knowledge of the complete and correct theory of electromagnetism. :-+

Tim

Sure, right after you identify the capitol of cheese. Meanwhile, don't delude yourself you are
calculating or constructing anything meaningful. Your understanding of EM laws is like a broken vase with 5/6 pieces missing glued together by a blind man with one hand.

But don't give up, you'll get there some faaaar day off.  :-+
 

Offline TimFox

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Re: Induction question
« Reply #22 on: July 17, 2019, 01:51:07 pm »
This reminds me of one of the classic questions on the famous University of Chicago Candidacy Exam (an important hurdle for PhD candidates in the Physics department).  A circular loop is formed around a region containing a time-varying flux. One side of the loop contains a 20 ohm resistor (at 9:00) and diametrically opposite (at 3:00) is a 10 ohm resistor.  Two AC voltmeters connect to the loop at 12:00 and 6:00, with each voltmeter sitting on the same side as the nearby resistor, further out from center.  What is the ratio of the two voltages?
When one student gave the correct answer (2:1), he wrote that he didn't believe it.
The professor brought in a simple cart-top experiment with two Simipson 260 VOMs to demonstrate.  This is a good example of the relevance of complete loops to the question of induced emf.  In the system, each voltmeter and its nearby resistor form a loop linked by no flux, so the voltage corresponds to the current through that resistor.  The current through the two resistors is the same.
 


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