# Motional EMF from an Electron Flow Perspective

Discussion in 'Physics' started by Gerry Rzeppa, Jul 6, 2015.

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1. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
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Here's a typical description of Motional EMF (from http://teacher.nsrl.rochester.edu/phy122/Lecture_Notes/Chapter32/chapter32.html, edited for brevity; there are many similar descriptions on various sites):

The figure below shows a conducting rod of length L being moved with a velocity v in a uniform magnetic field B:

The magnetic force acting on a free electron in the rod will be directed upwards. As a result, electrons (the blue dots) will start to accumulate at the top of the rod. The charge distribution of the rod will therefore change, and the top of the rod will have an excess of electrons (negative charge) while the bottom of the rod will have a deficit of electrons (positive charge). This will result in a potential difference between the ends of the rod equal to LvB.

Now looking at that description, I can easily see how a longer rod will result in a greater potential difference -- we've got more electrons for the force to act on, and thus more electrons will tend to accumulate at the top of the rod:

My problem is that I would also expect a wider rod to have a similar effect:

Yet the formula (LvB) says width has nothing to do with it.

So my question is: Why doesn't the width of the rod affect the induced potential difference?

2. ### BR-549 AAC Fanatic!

Sep 22, 2013
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As you say, the width, i.e. surface area will effect the number of electrons, but it will not effect the ratio.
It's the ratio that sets the potential.

3. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
Okay, but why does the length affect the ratio? Staring at those illustrations some more, I think I may have an idea. Is it because the longer rod crosses over twice as many Xs (lines of force) and thus twice as much force is applied and twice as many electrons get moved? In other words, each X (line of force) has only so much "oomph" or "capability" per swipe of the rod and that oomph is used up on the first electron (or group of electrons) encountered. Which would mean the third drawing should look more like this, with greater effect on the leading side of the rod (making the ratio the same as the first):

Or is it that all the free electrons in the wider rod move, only not as far or fast?

Related question: I've read that copper "normally" has one free electron per atom. When these electrons accumulate as in the drawings above, it seems (since the rest of each atom doesn't move) that the atoms at the bottom must lose their free electrons, and the atoms at the top gain more than one each. Is this so? If so, how many free electrons can each of the "overloaded" atoms at the top accumulate? If not, is the ratio we're talking about above (in copper) simply the ratio of atoms with/without a free electron? In this latter case, it would seem that an "uncharged" rod would have some of its atoms with a free electron, and about an equal number without, all more-or-less equally distributed in the rod. Yes?

4. ### BR-549 AAC Fanatic!

Sep 22, 2013
4,813
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Anytime you have a flux field and a conductor moving relative to each other, you get induction.

This induction will be proportional to the AREA swept by the conductor, or the flux field.

The conductor length and the perpendicular velocity(width) constitutes the area.

If you want to use the conductor as a power source, assuming the flux and velocity is constant, the length will set the voltage and the conductor width(surface area) will control the capacity(amp-hour).

The electron accumulation on the surface of the conductor is always a gradient and symmetrical.

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5. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
That helps a lot; thank you. Let me make sure I've got it. Here's my revised drawing, where darker blue indicates a greater concentration of free electrons:

Assuming the flux and the velocity of all the rods are constant:

1. Rods (A) and (C) will have the same voltage potential because the ratio of electron accumulation at the top vs the bottom is the same. Yes?

2. Rods (B) and (D) will have significantly more voltage potential than (A) and (C) because the ratio of electron accumulation at the top vs the bottom is significantly greater. Yes?

3. Rods (C) and (D) will have greater current capacity than corresponding rods (A) and (B) because more electrons are available per unit length. Yes?

Unanswered questions from my previous post:

4. I've read that copper "normally" has one free electron per atom. When these electrons accumulate as in the drawings above, it seems that the atoms at the bottom must lose their free electrons, and the atoms at the top gain more than one each. Is this so?

4a. If so, how many free electrons can each of the "overloaded" atoms at the top accumulate?

4b. If not, exactly what is going on here, electron-flow-wise?

New question:

5. I notice that most of the formulas for transformers focus on "number of turns" in the coils, rather than the length of the wire. If we drag a coil (rather than a rod) through the field above, will we see (a) a single electron gradient spanning the whole coil? or (b) a separate gradient in each loop of the coil?

6. ### Lool Member

May 8, 2013
116
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I think a simple way to see it is to realize that the Lorentz force from the magnetic field on the moving charges can be thought of as an electric field, since electric field is force per unit charge, by definition. The electric field does not depend on the width, nor the length, but only on the speed of motion of the charge and the magnetic field that the charge is moving through. The electric field is along the direction of the force, and electric field is the negative of the gradient of potential, which (when E field is constant) becomes simply E=-V/L. That is, potential is V=-EL, and hence voltage depends on length, and not width.

Last edited: Jul 8, 2015
7. ### WBahn Moderator

Mar 31, 2012
24,070
7,471
The electric field (which determines the voltage) is impacted by the charge DENSITY and the length of the path.

An easy way to visualize it would be to imagine taking a wide rod and cutting it lengthwise in two and separating them by a tiny gap. Now run those two rods through the field. By symmetry, both rods will have the same voltage across them from end to end and, at every point along the gap, the voltage on one rod will be equal to the voltage on the other. As a result, as the rods are moved back together, nothing changes and you will, therefore, have the same voltage lengthwise as you did before you put them back together.

8. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
Thanks, folks. Again, very helpful. Especially the part about cutting the rod lengthwise, etc. I'm very visually-oriented on this kind of stuff. But I think I've got the idea, and it seems my illustration in Post #5 above is consistent with what everyone is saying. Now how about my other outstanding questions:

4. I've read that copper "normally" has one free electron per atom. When these electrons accumulate as in the drawings above, it seems that the atoms at the bottom must lose their free electrons, and the atoms at the top gain more than one each. Is this so?

4a. If so, how many free electrons can each of the "overloaded" atoms at the top accumulate?

4b. If not, exactly what is going on here, electron-flow-wise?

5. I notice that most of the formulas for transformers focus on "number of turns" in the coils, rather than the length of the wire. If we drag a coil (rather than a rod) through the field above, will we see (a) a single electron gradient spanning the whole coil? or (b) a separate gradient in each loop of the coil?

9. ### WBahn Moderator

Mar 31, 2012
24,070
7,471
No. It's important to keep in mind the relative numbers we are talking about.

The density of copper is ~9 x 10^3 kg/m^3.

The mole of copper atoms has a mass of ~63 g.

A mole consists of ~6 x 10^23 copper atoms.

One coulomb of charge is ~6 x 10^18 electrons (in magnitude).

This means that the density of copper is ~9 x 10^10 atoms per cubic micron.

So if every copper atom in just the first micron of the surface of a copper rod has one extra (or one fewer) electron you would have a charge density of ~1.5 C/cm^2. This is a HUGE charge density. For instance, if a 1 cm cube had this charge on opposing faces, the voltage differential between the faces (treating as a parallel plate capacitor, which is a stretch) would be about 1.7 x 10^13 volts. So even if only one atom in a billion (10^9) within this one micron thick layer had an extra electron the voltage would still be 17,000 V.

So only a tiny, tiny fraction of copper atoms either give up or accept an electron as part of the conduction process.

How is the coil oriented relative to the field and to the direction of motion?

Once again, use the visual aid of taking a coil and cutting it into pieces and moving the pieces through the field, seeing what happens to each piece, and then putting the pieces back together.

10. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
Okay, but I'm still not clear regarding the distribution of free electrons. Let's start with an uncharged copper rod (voltage from top to bottom and every point in between equal to zero). How are the free electrons distributed? One per atom?

Now here's my picture again, where darker blue indicates a greater concentration of free electrons:

Let's focus on rod (B), assuming it's pure copper: What is the minimum and maximum of number of free electrons per atom at the bottom? What is the minimum and maximum number of free electrons per atom at the top?

Okay, let me try that with something a little simpler than a coil. Here we have three identical rods connected together by red wires (the red wires are somehow arranged so they are not influenced by the field). Darker blue indicates a greater concentration of free electrons.

I'm thinking this is very roughly analogous to a coil, where each loop (rod) has an equal induced voltage; and the overall voltage (from the bottom of (A) to the top of (C) is the sum of the three since the rods are in series). Like a multi-cell battery. Yes?

If so, then it's clear why "number of turns" in a transformer winding makes a difference in the induced voltage. But it's not clear (at least to me) why the length of each turn (or loop = rod) doesn't affect the induced voltage in a transformer winding.

11. ### Lool Member

May 8, 2013
116
12
Keep in mind that you are most likely thinking about the fields and math in a static reference frame. In the static frame there would be no electric field if the rod is removed. In that frame, the electric field is created only by the accumulation of charges that result in order to balance the Lorentz force from the magnetic field.

However, imagine you are in a reference frame that moves with the rod. In that frame there can be no Lorentz force because the charges are stationary. In that frame, the electric field, with no rod would be E=v x B. Then, when the rod is there the charges move to counteract that electric field and the end result is static E fields with zero field in the conducting rod.

In either frame, potential is proportional to length L because E is the negative gradient of potential. The electric field in the conductor does not point in the direction of the width but only along the length. Hence, the width does not affect potential.

12. ### WBahn Moderator

Mar 31, 2012
24,070
7,471
This is like saying that there are a bunch of cars going down the highway connected by ropes that somehow aren't moving at all. The connecting wires ARE influenced by the field. Just use imagine a square coil where you have one rod in front, a rod that goes from front to back, a rod that goes down in back, a rod the comes back to the front but at a slight angle to connect to the next rod in front.

13. ### Lool Member

May 8, 2013
116
12
Yes, from a simplistic point of view, that is basically correct. However, a bulk piece of conductor might better be viewed as a "sea" of free electrons. These outer 1 or 2 electrons per atom of Cu are free to move around under the influence of forces.

A lot more than you can usually create in practical situations. It does not take a very significant fraction of electrons to create very large electric effects.
To understand exactly what is going on requires a quantum mechanical analysis. This is usually taught in a Solid State Physics class/book or Semiconductor Physics class/book.

Think of one loop to start. The single rod you started with can be combined with another rod in parallel with didstance D between them (rods have length L). The two rods can be connected with perpendicular rods, and you will have a rectangular loop with length L and width D.

If you drag the entire loop, the the full loop will have no EMF generated because the two rods will have equal potential, but they are connected ++ and --, so they sum to zero. This makes sense if you think of Faraday's law because the area of the loop is not changing and the B field is not changing, so there is no time rate of change of flux and EMF should be zero for the closed loop. However, there is still the same potential difference from top to bottom of the loop.

Now imagine that the loop has 3 sides not moving and one bar is sliding across the perpendicular connecting bars. Here, only the moving rod generates a potential, and the closed loop EMF is not zero any more, but will be the negative of time rate of change of flux, due to v x B in the moving rod. This is consistent with Faraday's law because the area is changing, and hence the flux is changing.

These simple problems let you see directly how Faraday's law works.

14. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
Imagining... but having a little trouble...

I don't know what color to make the verticals (where darker blue means more electrons). Which one is correct?

I see that. In both drawings, points A and B are the roughly the same color.

So how, then, do we get anything out of a coil in a transformer (ie, any difference in the concentration of electrons between points A and B)?

15. ### Lool Member

May 8, 2013
116
12
Transformers usually use changing magnetic field from AC. There needs to be a change in flux within the coil. Since flux is the cross product of area and field (B A sin(theta)) then the B field must change, the area must change or the angle must change so that there is a flux change in the loop.

16. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
Since you didn't specifically say (or you did and I missed it), I'm going to assume that the illustration on the right better represents what's happening with the electrons in a coil moving in a uniform field (darker blue representing more free electrons in that area):

After all, if the electrons get pushed to the back on the horizontal legs, I would think the vertical legs in the back would experience the same kind of concentration of electrons as the back ends of the horizontals -- not because of the action of the field on the verticals, which is essentially nil, but simply because the electrons at the back ends of those rods naturally want to spread out as evenly as possible.

Now regarding the transformer, it seems to me that at the two extreme points of the fluctuating field (first in one direction, then in the other) we'd get something like this regarding the flow/concentration of electrons:

In other words, the electrons are first pushed to the back of the coil, then to the front, and so on. My problem is I still don't see how we get anything out of the coil since the electron concentrations at points A and B appear to be the same at any point in time.

17. ### WBahn Moderator

Mar 31, 2012
24,070
7,471
I think where you might be getting tripped up is that when you have a changing magnetic field at play, you do not have a conservative electric field, meaning that the voltage difference between two points is no longer path independent. Your thought experiment seems to be premised on the electric fields in the conductors being conservative.

18. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
Frankly, I'm not sure what that means. But I think it's clear that the free electron distribution in a coil in a transformer must alternate between what we see on the right below (darker blue means more free electrons), and its inverse, as the field goes first one way and then the other:

What's not clear is how to get from the electron distribution in the simple rod on the left to the distribution on the right. Though it seems reasonable (to me) to sum it all up in rather sloppy and colloquial terms, like this:

1. When we move a wire through a magnetic field the free electrons tend to get pushed to one end -- whether the wire is straight, coiled, or configured in some other way.

2. Depending on the various angles between the wire and the field, this movement may be from A to B, from B to A, or in some less effective direction, perhaps resulting in no concerted movement of electrons at all.

3. But all other things being equal, the longer the wire, the greater the resulting potential difference between the two ends of the wire because more of the field will exert influence on more of the free electrons in the wire.

Is that anywhere close to the truth?

19. ### Lool Member

May 8, 2013
116
12
No not even close. It seems you didn't follow my progress explanation above about a moving loop having zero potential around the loop.the figure you put on the right is not correct.

Actually I'm not sure that I am not losing track of what you are saying . this is too hard to explain over the internet. We all need to be standing at a blackboard. One thing seems clear , you are probably missing a couple of key concepts and we are having trouble identifying the sticking point.

Last edited: Jul 9, 2015
20. ### Gerry Rzeppa Thread Starter Member

Jun 17, 2015
126
1
I'm pretty sure we could wrap this up in just a couple of posts if you could supply a diagram showing the distribution of electrons in the secondary coil of a transformer when the field is "maxed out" in one direction. Something like this would do (darker blue indicating a greater concentration of free electrons):

That, plus a couple of plain English remarks on why those electrons accumulate when and where they do, I think, would do the trick.

Last edited: Jul 10, 2015