Motional EMF from an Electron Flow Perspective

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RdAdr

Joined May 19, 2013
214
Who would ever suspect, looking at Ohm's Law, that current is easy to picture and describe in electron terms, and resistance is easy to picture and describe in electron terms, but voltage, in electron terms, is beyond the understanding -- not to mention the imagination -- of the average man?
Everyone with a basic understanding of physics would understand.
Current represents a movement of the electrons themselves, while voltage is a property of the electrons via the electric field that they create. So, there is a difference between something and the property of something.

What you say is the same as: the falling of balls with mass in a gravitational field is easy to picture in terms of balls, while the potential difference (between some given height and the ground), in balls terms, is beyond the understanding of the average man.
 
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Lool

Joined May 8, 2013
116
It sounds like you're saying the purple paragraph should be reworded as follows (your implied additions in bold):
As I said, I would not reword anything. Your references are correct as written.

Your confusion comes from the fact that you have nothing to compare the physics of non-conservative forces with. I'll bet you still do not even know what it means for a force to be non-conservative. This is a key piece of evidence about your laziness. You have been told about the importance of understanding non-conservative forces, many days ago and many posts ago. You said you don't know what that means. Well, did you try to find out what it means? No you did not because you don't want to do any real learning. Instead of asking, "what does that mean", you revert back to your own ideas. This is why no one will here will try to help you now. If you wont help yourself, how can we help you?
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Let me see what you guys think of this. We've got our rod moving through the magnetic field, as before. But in this particular experiment:

1. We select the length, velocity, and field strength so LvB equals 1 volt.
2. We use a rod whose resistance, end to end, is 1 ohm.
3. We limit the experiment to 1 second of movement.

Now Ohm's Law tells us that it takes 1 volt to push 1 amp's worth of current through 1 ohm of resistance. And we know that 1 amp is 6.28x10^18 electrons moving past a point (say, the vertical center of the rod) per second. Seems to me, then, at the end of our 1-second experiment, the critical part of the purple paragraph would read (just in this particular case, my additions in bold):

The charge distribution of the rod will therefore change, and the top of the rod will have an excess of 6.28x10^18 electrons while the bottom of the rod will have a deficit of 6.28x10^18 electrons.

Yes?



 

RdAdr

Joined May 19, 2013
214
Someone correct me if I am wrong, but.


You see the rod as resistance R of 1 ohm being connected to an external LvB=1V source voltage. Thus, the current is 1A. Thus, after one second, 6.28x10^18 electrons would have gone through some cross-section of the rod.

So, you assume that the voltage LvB across the rod is constant during this second.
But when the rod starts moving in the magnetic field at t=0, the voltage is 0. And after some time, it reaches LvB. And you want to find how many electrons pass through the cross-section in this transient time.
 

Lool

Joined May 8, 2013
116
Let me see what you guys think of this. We've got our rod moving through the magnetic field, as before. But in this particular experiment:

1. We select the length, velocity, and field strength so LvB equals 1 volt.
2. We use a rod whose resistance, end to end, is 1 ohm.
3. We limit the experiment to 1 second of movement.

Now Ohm's Law tells us that it takes 1 volt to push 1 amp's worth of current through 1 ohm of resistance. And we know that 1 amp is 6.28x10^18 electrons moving past a point (say, the vertical center of the rod) per second. Seems to me, then, at the end of our 1-second experiment, the critical part of the purple paragraph would read (just in this particular case, my additions in bold):

The charge distribution of the rod will therefore change, and the top of the rod will have an excess of 6.28x10^18 electrons while the bottom of the rod will have a deficit of 6.28x10^18 electrons.

Yes?


Take any of the following answers.

1. No.
2. Total garbage.
3. Are you kidding?
4. Spoken like a true troll.
5. You must have brain damage, and memory failure.
6. Are you a human being or a trained chimpanzee that was taught to use a computer?
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Someone correct me if I am wrong, but. You see the rod as resistance R of 1 ohm being connected to an external LvB=1V source voltage.
Not quite. I see the voltage as the result of the action of the field on the electrons in the rod; it's the LvB mentioned at the very end of the purple paragraph:

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.

Thus, the current is 1A.
Yes. If we've got one volt of push, and one ohm of resistance, we should be able to move one amp's worth of current.

Thus, after one second, 6.28x10^18 electrons would have gone through some cross-section of the rod.
That's how I see it.

So, you assume that the voltage LvB across the rod is constant during this second.
Yes; I'm assuming the velocity is constant and is accomplished instantaneously. See below.

But when the rod starts moving in the magnetic field at t=0, the voltage is 0.
I would say, "before the rod starts moving..."

And after some time, it reaches LvB.
I would say, "And it immediately (or after some amount of time so small it is inconsequential compared to the second-long duration of the experiment) reaches velocity v." If you don't like that, we can get the rod going (say, for 1 second) and then start timing the experiment. I don't see that it matters. During that latter second, 1 amp's worth of electrons will be pushed upward in the rod. Which means during that latter second, 6.28x10^18 electrons will be displaced (from bottom to top). Yes?
 

Lool

Joined May 8, 2013
116
I would say, "And it immediately (or after some amount of time so small it is inconsequential compared to the second-long duration of the experiment) reaches velocity v." If you don't like that, we can get the rod going (say, for 1 second) and then start timing the experiment. I don't see that it matters. During that latter second, 1 amp's worth of electrons will be pushed upward in the rod. Which means during that latter second, 6.28x10^18 electrons will be displaced (from bottom to top). Yes?
No !

You describe nothing even remotely close to what happens. The concept is wrong, the picture is wrong, and the answer is wrong.

You ask again, we'll answer again. The answer does not change just because you ignore it.
 

#12

Joined Nov 30, 2010
18,224
You ask again, we'll answer again.
If I may be allowed to offer some advice...
Among the definitions of a Troll are, He never learns and he never quits asking. If he did, the game would end.
The only answer needed is, "Don't feed the Troll".
 

Lool

Joined May 8, 2013
116
If the troll likes to eat "no's" then I can feed him as many as he wants. :) It's easy to write no and if he really derives some thrill from it, why should I deny him.

But, seriously, this is the first time I ever wondered why a forum member was not banned for stupidity.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Help me understand where my error is. Here's the description:

A conducting rod of length L is moved with velocity v through a uniform magnetic field B such that the magnetic force acting on a free electron in the rod is directed upwards. As a result, electrons 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 while the bottom of the rod will have a deficit of electrons. This will result in a potential difference between the ends of the rod equal to LvB.

And here's my reasoning:

1. The selected rod presents 1 ohm of resistance.

2. L, v, and B are selected so the calculated potential (LvB) is 1 volt.

3. 1 volt can move 1 amp's worth of electrons through a 1-ohm resistance in one second.

4. Therefore, once the rod is up to speed, one amp's worth of electrons (6.28x10^18) should be moving past the vertical center of the rod every second.

5. But of course this movement of electrons cannot go on indefinitely. Eventually, we'll either (a) run out of electrons, which is unlikely, or (b) the counter-force generated by the accumulated electrons will equal the upward force being applied by the field and further accumulation of electrons will cease. I would think this would happen, given the parameters set above, when one amp's worth of electrons have been moved from the bottom to the top of the rod. In other words, the system will stabilize in 1 second or less, with 6.28x10^18 electrons having been displaced, and further movement of the rod at the given velocity will only serve to keep the electrons where they are.

Help me understand where my error is. Keep in mind that name-calling doesn't help, but copying my description up to the first error in fact or logic, explaining what when wrong and what should have been said at that point, does.

I thank you all again for your time a patience.
 

Lool

Joined May 8, 2013
116
Help me understand where my error is.
I already did. Just read my previous posts. It is all there. Just to give one obvious problem, I already told you now much force is implied by 6.28x10^18 electrons. It's over a trillion pounds. Obviously you can't be correct.

So, do you just enjoy making me repeat myself over and over and over again? Or, do you have a short term memory problem? Or, are you off your meds?

Keep in mind that name-calling doesn't help,
Actually it does help by providing a useful service. It provides a flag that newbies and novices might take note of and hence not be mislead by the ravings of a lunatic.
 
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