# Motional EMF and Electrons

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1. Aug 1, 2015

### Gerry Rzeppa

Here is a typical description of motional EMF:

"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 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."

My question is: How do we calculate the number of excess/deficit electrons for a given L, v, and B?

2. Aug 1, 2015

### Staff: Mentor

Calculate the voltage that leads to a force equilibrium, and then the corresponding charge densities (the second part can be messy in the general case).

3. Aug 1, 2015

### Gerry Rzeppa

I'm going to need a little more help with the math (it's a weakness of mine).

That sounds like LvB to me. Yes? If so, let's say we select L and v and B so LvB is 1 volt.

And I'm stuck.

4. Aug 1, 2015

### Staff: Mentor

There are methods to calculate a charge distribution for a given field distribution. But if you want to do that properly, then you have to consider that your conductor has a finite size, and do those calculations in three dimensions.

Where do you plan to use the charge distribution?
If you are just interested in an order of magnitude, assume your rod is a capacitor with large plates (large compared to the gap size), and calculate the charge density there.

5. Aug 1, 2015

### Jeff Rosenbury

It is a complicated problem. You need to know what the charge distribution is in your particular alloy (charge carriers/unit volume). In most materials electrons dominate, but not always.

In addition, once the charges start to move, eddy currents will form. That's not a problem if something is pushing the rod at a constant speed; the push will just push harder. But in the real world there might be slowing due to power loss in the eddy currents, etc.

6. Aug 1, 2015

### Gerry Rzeppa

Rats! I was hoping for something simple like the definition of current: "1 amp is equal to 6.28x10^18 electrons moving past a point every second." Or perhaps a simple equation like E=LvB or F=ma or E=IR.

The ultimate goal is to explain the workings of a basic vacuum tube guitar amp circuit to a ten-year-old in electron-flow terms. Electron-flow terms because the workings of vacuum tubes are invariably described from an electron-flow perspective. (I realize this is an unusual approach to circuit analysis in these modern times, but it seems appropriate to me in this particular case.) Consider, for example, this excerpt from a book explaining how the varying flow of electrons coming out of a vacuum tube is converted to changes in voltage that are passed on to the next stage of the amp:

I think that's something the kid can easily picture: the varying voltage (at the point marked with the blue arrow) is the result of the varying excess/deficit of electrons at that spot (relative to ground). Which reminded me of descriptions of motional EMF from my college days (decades ago): hence this thread. After all, I thought, we're able to define current easily enough in terms of electron flow; perhaps we can similarly define voltage for the kid in terms of electron concentration (eg, excess electrons per proton, or per cubic centimeter, or something like that) and he can thus picture voltage (not as some abstract quantity with units of measure he's never heard of -- like "joules" and "coulombs" -- but) as the natural result of the mutual repulsion of electrons; as a measure of the electron's "unhappiness" or "dissatisfaction" at being unnaturally crammed together; a measure of their desire to redistribute themselves.

7. Aug 1, 2015

### Jeff Rosenbury

That sounds like a fair description. The electron flow model has some serious weaknesses, but as long as you understand that, it's a good intuitive model.

The reason the problem is hard is not conceptual. It's hard because it's hard to know how many free electrons are in a wire. It's hard to know how they self interact. These difficulties are more in measurement than concept.

8. Aug 1, 2015

### Gerry Rzeppa

Well, how about just an order-of-magnitude kind of guess for me? If one amp is a gazillion electrons (6.28x10^18) moving past a point per second, roughly how many "excess" or "displaced" or "accumulated" electrons does it take to get, say, a 1-volt peak out of that vacuum tube?

9. Aug 2, 2015

### Staff: Mentor

You cannot convert voltage to electron density, that does not work. They are completely different things.
For a specific setup they can be proportional to each other, but then the factor between them depends on the setup. As analogy: there is no fixed time you need to travel 100 kilometers. For a given transportation mode (walking, car on a highway, airplane, light...) you can make a good guess how long it will take, but a general relation does not exist, not even as order of magnitude.

10. Aug 2, 2015

### Jeff Rosenbury

I disagree. One can convert electron density to voltage. It is called capacitance. There are other factors involved such as geometry, but a capacitor does just that.

A 1 Farad capacitor contains 1 Coulomb of electrons.

So to answer the OP's question, it depends on the capacitance of the tube element (the grid?). Conveniently 1 Coulomb per second is an Amp. So that gives you an order of magnitude.

You might want to study what affects the capacitance to understand more about how charge turns into voltage.

11. Aug 2, 2015

### Staff: Mentor

Capacitance depends on the setup, that's my point.
Also, charge and charge density are not the same thing either.

12. Aug 2, 2015

### Gerry Rzeppa

1. It seems to me that these two devices are conceptually similar:

The air molecules on the left are analogous to the electrons on the right, and the upward force of the piston is analogous to the upward force of the magnetic field. The air molecules like it less and less as they are pressed closer and closer together, and will immediately separate and distribute themselves more evenly given the opportunity; ditto for the electrons. It seems to me that we ought to be able to somehow relate electrons per cubic inch (or per proton, or something) to voltage (electrical pressure) in the same way that we intuitively relate "molecules per cubic inch" to air pressure.

2. Regarding capacitance: I'm sorry, but I don't see how capacitance applies in this case. The rod is conductive through and through -- it has no insulating dielectric that is being stretched or deformed due to the difference in electron concentration at the ends, and so it just doesn't look like a capacitor to me.

3. It seems odd to me that we can't get a relationship between voltage and a specific number of electrons even though we know that E=IR and that I is defined as 6.28x10^18 electrons per second past a given point. I don't understand the units of measure well enough, and my math isn't good enough, to get me further, but it seems like we ought to be able to relate the two based on that formula and that definition of current.

4. Thought experiment: Let's select L and v and B in the above drawing so LvB equals 1 volt. Then, while the rod is moving, let's connect a 1-ohm resistor across the ends of it. When we do (assuming we keep moving the rod at velocity v) we'll be pushing 1 amp's worth of electrons past any given point in the circuit every second: 6.28x10^18, to be specific. Now, without stopping the rod, we disconnect the resistor. The electrons will continue to "flow" upward until -- until what? Until, I would think, the force exerted by the mutual repulsion of the accumulated electrons at the top of the rod equals the upward force of the field. My guess is that this would be 1 amp's worth (or 6.28x10^18) since this is the number we were able to move, continuously, when we had a 1-ohm path all the way around. But of course that's just an intuitive guess from someone who is more of an artist than a mathematician. What would you guys estimate?

13. Aug 2, 2015

### Staff: Mentor

Even if we fully simulate the problem, the result cannot be a single number of electrons. You will get a large charge density at the top, gradually decreasing as you go down, crossing zero somewhere in the middle, increasing (but with opposite charge) again until you reach the bottom. That is not a number, that is a distribution - and, as mentioned, it is not easy to calculate it.

You'll certainly not move 1 C of charges around. That would be a ridiculously huge amount of charge separation. And there is no reason to expect anything special about that arbitrary, man-made definition of a unit.

14. Aug 2, 2015

### Jeff Rosenbury

Gauss's Law relates charge density to voltage. It states that the electric field over a closed surface is equal to the charge enclosed by that surface.

E ⋅ ds = (1/εº) ∫ ρ dv

Notice this equation depends on geometry as well as the permittivity.

Your question seemed to be about a tube. Why is that not like a capacitor? There are conductors separated by a dielectric (vacuum).

As for the rod, I seem to recall mentioning eddy currents. You are the one putting charges on the ends due to your analogy. I tend to think much the same way, but I recognize it's just an analogy. The electrons don't pile up at the ends because it's a conductor and they move around in the magnetic field.

15. Aug 2, 2015

### Gerry Rzeppa

I understand that it's nearly impossible to track individual electrons (as my earlier drawings might have suggested). So let's use these drawings, where I've added conductive rails for the rod to ride on, and where the concentration of electrons is represented as a gradient, darker blue meaning more electrons in that area:

Now assume that L and v and B have been chosen so LvB is 1 volt, and that the rod is moving steadily to the right.

1. If the overall resistance of the left loop is 1 ohm, then by Ohm's Law (I=E/R) the flow of free electrons must be 1 amp's worth and the concentration of free electrons equal throughout the circuit (as indicated by the uniform blue color). Which means 6.28x10^18 electrons must be moving past any point in that circuit every second. Yes?

2. Now let's increase the resistance of the loop to say, 2 ohms, as depicted in the drawing in the center. Since LvB is still 1 volt, Ohm's Law now gives us 1/2 amp of flow or 3.14x10^18 electrons per second past any point in the circuit. But it seems to me there's still a potential (no pun intended) for greater flow since L and v and B haven't changed. In other words, there will be an excess of electrons at the top and a corresponding deficit of electrons at the bottom of the circuit in this case I've indicated this with the revised gradient -- darker than the center drawing at the top, and lighter at the bottom. Yes?

3. Finally, let's make the resistance of the loop ridiculously large by opening the circuit as in the drawing on the right. Current, in this case, will be negligible. But the potential (LvB) will still be 1 volt. Which means, according to the description of motional EMF at the very top of this thread, that the number of excess electrons at the top (and the corresponding deficit of electrons at the bottom) will be the maximum for this configuration and the selected L, v, and B variables. We thus get very dark blue at the top and a correspondingly very light blue at the bottom. Yes?

So (assuming all that is corrent) we know both the flow (6.28x10^18 electrons per second past any spot) and the relative concentration (equal everywhere) in the left case. And we know the flow (3.14x10^18 electrons per second past any spot) in the center case, and the flow (zero) in the right case. What we don't know is the relative (or absolute) concentration of electrons in the latter two cases.

It seems to me we ought to be able to calculate at least the relative concentrations in these latter cases by drawing a horizontal line through the middle of the circuit and averaging what's above and what's below.

And it seems to me that we ought to be able to calculate absolute concentrations once we figure out what to put in the denominator of those figures (ie, per proton, or per cubic centimeter, or something like that). After all, the energy is stored by pressing the electrons closer together than they want to be, which strikes me like compressing a spring (and where the stored potential energy is a simple 1/2kx^2).

Too bad calculus is out of the reach of my ten-year-old. And me (since I haven't used it much since I graduated from college decades ago).

Yes, but I wasn't asking about the relative concentration of electrons on the plate versus the cathode. I was asking about the relative concentration of electrons at a point in the circuit where only resistive elements are involved. We get the same kind of "pile up" of electrons at the pickup end of the guitar amp where no tubes are involved:

As you can see, I've attempted to illustrate the relatively high-voltage (2-volt) / low current (it's nanoamps) behavior of the circuit using the same kind of picture we get with motivational EMF above. Specificially, (a) uniform current throughout the circuit (the darker blue electrons), but (b) not uniform concentration of electrons (dark and light together). It is my understanding that it is this varying concentration of electrons that gives us the voltage swing, both here and in that junction between plate and plate resistor discussed earlier.

What bothers me is that I can tell the kid exactly how many electrons are flowing in that circuit at any point in time (since the current is known), but I can't tell him even what ratio of electrons it takes to account for those varying voltages. Seems like I ought to be able to do that without the need for higher mathematics.

Last edited: Aug 2, 2015
16. Aug 2, 2015

### Staff: Mentor

The concentration won't be the same everywhere.
There is no fundamental difference between the 1 Ohm and 2 Ohm case here. In both cases the electron density will be non-uniform.

There is nothing special about "1 [Unit]" in general. We could have completely different unit systems, and physics would not change at all. We could have defined 1 Ampere as 1 electron per second, or 1 electron per year. Would you expect 1 electron separation (in some way) then?

Right. But we don't get the amount from that consideration. Also, "very large" is probably of the order of pico- to nanocoulombs for typical lab setups.

17. Aug 2, 2015

### Gerry Rzeppa

It sounds like you want to re-word the description of motional EMF at the top of this thread to read (implied changes in bold):

"The magnetic force acting on a free electron in the rod will be directed upwards. As a result, an insignificant number of electrons will start to accumulate at the top of the rod. The charge distribution of the rod will therefore change, but not in any significant way, and the top of the rod will have an insignificant excess of electrons (negative charge) while the bottom of the rod will have an insignificant deficit of electrons (positive charge). This excess/deficit of electrons will not result in a potential difference between the ends of the rod equal to LvB, though the potential difference will be equal to LvB due to other unspecified considerations."

At least that's what I'm hearing.

18. Aug 2, 2015

### Jeff Rosenbury

The relationship between voltage and charge carriers is the capacitance. The fact that something doesn't look like a capacitor has nothing to do with it.

Capacitance exists nearly everywhere, even in solid copper. The only requirement is a separation of charges. Most circuits have stray capacitance which engineers work on reducing for high frequency work. It's why component sizes keep getting smaller as computers get faster.

In your diagrams, there is an electric field (voltage) between the horizontal bars. There is a capacitance as well. To know how many electrons are there, find the capacitance. One Farad equals one Coulomb per Volt. It is a simple equation. Most capacitors are well under 1 F. Stray capacitance is usually in the nF or pF range. Tubes are likely on the high end of that.

As an example, here's the datasheet for the popular 12AX7 tube. It lists all the inter-electrode capacitances.

19. Aug 2, 2015

### Gerry Rzeppa

Again, it seems to me that the capacitance of the three devices below is not pertinent: none of them is able to store any significant potential when the rod is not moving -- the electrons too quickly redistribute themselves evenly -- so none of them behaves like a capacitor.

But let's say you're right: how do I find the capacitance in these three cases? (Six if we count "rod at rest" and "rod at velocity v".) I know the voltage (because I've selected L and v and B to equal 1 volt). But the other two variables are unknown.

I need to see a worked-out example. Preferably three (or six examples) corresponding to the three diagrams above (at rest and in action). Make whatever additional assumptions you need to. Thanks.

Last edited: Aug 2, 2015
20. Aug 2, 2015

### Jeff Rosenbury

Fine.

Suppose the rods are 10cm square and 1m long. Let's say they are 1m apart.

We plug the numbers in our handy capacitance calculator and get 0.88 pF. That doesn't count the "leads" leading to the resistor or the cross bar (which is just a small value resistor). This will cause some fringing and the like. But I'm too lazy to figure that. I'll stick to the plate calculator.

Another example, suppose the rods are 10m by 10m and 1mm apart; that's 0.88µF.

Let's get wild and include a titanium dioxide dielectric. Not we get 70µF.

BTW, I messed up the link to the tube datasheet.

21. Aug 3, 2015

### Gerry Rzeppa

1. Okay, so 0.88 pf which is 8.8e-13 farads. And let's say V=LvB= 1 volt. Then q=CV=8.8e-13*1 coulombs or about 5.5 million excess (or deficit) electrons. Yes?

Now let me try to wrap my head around what's happening here.

2. I find when I make the rails 10 times longer, the capacitance is multiplied by 10 (as we'd expect due to the larger plate area), and the number of electrons also increases by a factor of 10 (55 million now, instead of just 5.5 million). This makes sense since we have 10 times the space ("plate area") to work with -- the mutual repulsion of the electrons on the top plate (and the attraction to protons with a deficit of electrons on the bottom plate) will not be felt as strongly as before until we have 10 times as many electrons as before. Yes?

3. I also find that when I leave the length as it was but make the space between the rails 10 times smaller, the capacitance is again multiplied by 10 and the number of electrons goes up by a corresponding factor of 10. This also makes sense to me, at least capacitance-wise: we can now have 10 times the number of electrons on the top plate because the mutual repulsion of electrons there is offset by the stronger attraction of the much closer positively charged bottom plate. Yes?

Unfortunately, I'm having trouble picturing point (3) above, circuit wise. Consider these three circuits (each with just a battery, a bulb, and some wire):

Let's say we've adjusted everything so the bulb will light with 1 volt, and the capacitance between that top wire and the bottom wire in the left circuit is about 0.88 pf, as above. That would mean we need an excess of about 5.5 million electrons on the top rail to make the thing work. So far, so good.

Now the center circuit is the same except the top and bottom rails are five times as long. Which means the capacitance is now five times greater and that means we need five times more electrons, 27.5 million, to make it work. Which is nevertheless understandable: we've got five times the "space" in those wires, so to get the same "pressure differential" we're going to need five times the number of excess/deficit electrons. Still good.

But now comes the problem. The right circuit is the same as the first except those top and bottom plates are now five times closer to each other, which again gives us five times the capacitance -- and thus the need for five times the number of electrons to make the circuit work. That makes no sense to me. I can see how this circuit makes a better capacitor than the other two; but I can't see why it should take five times the number of electrons to maintain a 1-volt differential between the top and bottom legs of the circuit.

Frankly, this anomaly makes me think the whole "capacitance approach" to the problem is not right. And not only for the reason just stated. It also seems odd to me that the current through that bulb might be as large as 6.2e+18 electrons per second, yet mere millions of electrons are involved in determining the voltage. Seems way out of proportion.

Last edited: Aug 3, 2015
22. Aug 3, 2015

### Jeff Rosenbury

It makes sense to me.

The electromagnetic force is very strong. Electrons don't weigh much. It takes far fewer of them to make charge than it takes for that charge to move them.

But it really doesn't matter what we think. The universe acts as it wants. It pays no attention to our opinions.

23. Aug 3, 2015

### Staff: Mentor

Those two statements are exactly the same, by the definition of capacitance. How can one be unclear while the other one is clear?

You don't need any surplus of electrons anywhere for motion. Many electrons (something like 1026 per kilogram of conductor) contribute to the current, having a few millions more or less does not make any difference.
If you calculate the drift velocity of electrons in conductors, you'll typically get something of the order of millimeters per second.

24. Aug 3, 2015

### Gerry Rzeppa

Then help me to make sense of it. Why should the circuit on the right require five times the number of excess/deficit electrons than the circuit on the left to maintain the same voltage differential across the bulb?

25. Aug 3, 2015

### Gerry Rzeppa

Do I need surplus electrons anywhere for voltage? Or are you still implying that the description of motional EMF at the top of this thread should be modified to read more like this (implied modifications in bold):

"The magnetic force acting on a free electron in the rod will be directed upwards. As a result, an insignificant number of electrons will start to accumulate at the top of the rod. The charge distribution of the rod will therefore change, but not in any significant way, and the top of the rod will have an insignificant excess of electrons (negative charge) while the bottom of the rod will have an insignificant deficit of electrons (positive charge). This excess/deficit of electrons will not result in a potential difference between the ends of the rod equal to LvB, though the potential difference will be equal to LvB due to other unspecified considerations."