# B Is there a maximum limit to voltage for a conductor?

1. Oct 29, 2016

### Trance-

This question comes from the equation E = vB (moving conductor in a magnetic field -- E = electric intensity, v= speed of the conductor's movement, and B = magnetic field strength). Say B is constant, so the only thing we have to rely on to vary the electric intensity in the conductor is its speed. The faster you move the conductor, the more potential difference you set up in the conductor -- the magnetic force pushes the charge in the conductor to move to one of its ends and thereby set up the voltage. Now, the question, roughly, is: What if only a finite charge can be moved within a conductor and therefore only a finite voltage can be set up? (if the voltage is finite, then E is finite -- and if E is finite, it means there's a value of v beyond which the above equation won't work)

This question came to my mind from the -- misconception (probably) -- fact that in a step-up transformer, a high p.d is set up with only negligible current flowing because all the charge is used up to set up p.d.

I have a feeling my understanding is wrong on some really fundamental level. Key concepts which I have a feeling I have wrong are of voltage and electric intensity. I'll be grateful if you can help clear my concepts.

2. Oct 29, 2016

### Staff: Mentor

I don't understand your question. But "using up charge" in this context makes no sense.

3. Oct 30, 2016

### Trance-

Not used up as in 'consumed' but like, electrons shifted to one side to set up a p.d. Hard to explain my doubts because they're really going too outside the actual concepts (I can feel it already) so it might just be easier to simply tell me how p.d is actually set up. That might reveal something.

4. Oct 30, 2016

### Staff: Mentor

There was a very long thread about that a while back. Perhaps @jim hardy might find a link to it.

Be warned, it gets hairy.

5. Oct 30, 2016

### cnh1995

Is something like this in your mind?
Say you are moving a metal rod in a magnetic field B with a continuously increasing velocity v. So the motional emf E=Blv will also be continuously increasing. What will be the maximum value of the emf induced physically possible? Is that what you mean by charge being "used up"? Are you asking if there will be a situation when "all" the electrons are piled up on one side of the rod and there are none left to be accelerated by the Blv force?

6. Oct 30, 2016

### jim hardy

A question well stated is halfway answered . I'm a little confused just what is the question... i will take a stab at reducing ambiguity .

The way to keep your thinking straight is to start with the most basic thought experiment , free of unnecessary complications, then add to it.

What is the simplest thought experiment ?
For me it's a single charge , perhaps an electron or perhaps a whole coulomb of conventional positive charge, moving in an electric field not even constrained to the inside of a wire.
It experiences the familiar Lorentz force, f = qv cross B. Force is perpendicular to both v and B . See http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movchg.html
So naturally it'll take a circular path as its course gets deflected by the magnetic field . Spend a few minutes with his guy's video .

Next we will add a complication, constrain your charge to inside of a straight wire so it can only move one direction.
Now as you correctly surmised the charge will be pushed toward one end of the wire . That uneven distribution of charge will result in an electric field that just balances the Lorentz force , trying to redistribute the charge equally .
Force on a charge in an electric field is f = qE
so we have produced by moving the wire in a B field an electric field (E = f/q ) that's making a force equal and opposite to the Lorentz force f=qv crossB
equating the forces
qvcrossB = qE and solve for E i see q on both sides of that equation, dividing both sides by q leaves E = vcrossB for any nonzero value of q.
(at zero q there would be a term q/q=0/0, which is indeterminate)
so if i understood your question to be
My answer would be "Who said that premise is true? It might be true for zero charge but not for any finite charge."

Lorentz force is what makes e=blv true. To get from f = qv cross b to e= blv is a derivation, Many educators use work done a charge moving along your moving conductor to demonstrate, see middle example here
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html

7. Oct 30, 2016

### Staff: Mentor

I might also add that electrical reasoning by thinking of one (or several) electrons at a time is almost always not helpful. I urge the OP to stop doing that. Think equations, not electrons.

8. Oct 30, 2016

### Biker

That is a really interesting question, Never thought of it while studying the topic.

I would presume when you run out of charges that a field will be actually created inside the conductor ( not connected to any circuit), There will be a magnetic force bigger than the electric force created by the charges. But I would imagine if you connect it to a circuit which will probably add a whole lot more of charges to the wire to balance the magnetic force.

I am not really sure how the resistance of a wire will change when you connect it to the rod. Another question that should be discussed because I think this is an interesting topic at least for me :D is by how much does the charge density of the wire connecting the rod to make a full circle will change. Not sure if this actually is true or false.

9. Oct 30, 2016

### Trance-

Yes but it has been clarified by Jim Hardy to some extent.

Thanks for the post. I have only one question remaining: The equation E = blv clearly tells us that if you keep on increasing the speed of the conductor, the electric intensity keeps on increasing -- is that true?

This derivation (E = blv) seems to tell us that charge (q) has no role in building electric intensity inside the conductor and the only factors upon which E depends is the magnetic field strength and the speed (which basically says ''rate of change of flux''). If it is true that E will keep on increasing as we increase v, then I want to know exactly what happens inside the conductor to establish greater and greater E -- because, clear enough, E is basically the time differential of Potential Difference which clearly then depends upon charge too. What I want here is the inner picture: What happens inside the conductor? how is its speed influencing its electric intensity?

I hope I was clear enough.

10. Oct 30, 2016

### jim hardy

As you guys all know by now i am a plodder who has to see a physical analog for every term of an equation. It's a handicap i have.

i think so. While it's not supposed to be possible for an Efield to exist inside a conductor because the charges are free to migrate ,
i have no problem accepting the idea of balanced forces from an Efield and Lorentz forces pushing on the charges in opposite directions. I could call it charging the capacitance of free space between the ends of the wire, but haven't explored that train of thought to see if it leads to some paradox..... so treat it with caution.

Why do you suggest the magnetic force is bigger?
If there's no current they must be equal, if there is current then there is EMF left over to do external work.

I've been speaking of DC. If we swap to AC then we get into another math entirely.
I've been musing what happens when we approach a limit on Trance's speed - ie a Bfield passes by a conductor at c.
We have now an antenna .
Do Individual electrons gain mass because of relativistic effects , accelerating differently than for the DC case ?
Is an electron's q/m different from that of an ideal Coulomb?

^^ There's a real world example of @anorlunda 's admonition , we must be rigorously attentive to detail when making up physical analogs for our equations. That's the sort of thing i do while driving or trying to wake up in that 'twilight time' before the alarm goes off.

Interesting questions guys with interesting ramifications. It's fun to pick apart equations and figure out from whence each term arises. But i have to start from the most basic equations, not ones that have been through many steps of derivation.
So while i'd never make it in grad school i sure respect those who do.

old jim

11. Oct 30, 2016

### jim hardy

I absolutely believe that it is. If there's no charge there is no Lorentz force to push on the (nonexistent) charges .

Now there's a limit to speed, c. I suppose an electron's mass would approach infinity as speed approaches c hence it would be less inclined to move toward end of the wire
but that's reducto ad absurdum because any real electric machine would have long since been torn to pieces by centrifugal force.

It is a fact that voltage is linear with velocity.
For DC motors it's the basic equation, open circuit voltage = KΦRPM with K being a constant for that particular motor. and Φ being flux.
At least until the armature conductors get flung out of their slots as above.

hope this helps.
I feel awkward here, my math is not so good as that of most PF members. I do this to help beginners over the same stumbling blocks i struggled with. One can contribute meaningfully in industry if he learns to apply basics to real world problems.

old jim

12. Oct 30, 2016

### Biker

Trust me I have that disease too. I had to understand drude model in order to grasp the concepts of electronics(Which tbh they introduced it badly in my books) while it is not required in our high school level.

You agree that there will be a field but not force?
If you put a single positive test charge lets say in the middle of the conductor and lets assume that this is the maximum voltage that the conductor can provide, Then if you increase the speed just a bit more the magnetic force will increase and the electric force will stay constant. Which means there is a field inside.

There is an unbalanced force inside but nothing moves inside the conductor because everything is at it maximum place.( In my point of view)

Thank you jim for the picture, It was really interesting to see this. Just a question about it, So this thing happened because the ions core wanted to move in a circular motion inside the wire but the wire is stuck to both ending of the machine... ?

13. Oct 30, 2016

### cnh1995

I agree. But what if we kept the velocity constant at some finite value and went on increasing the strength of the B-field? Will there ever be a point when the charges will not be able to redistribute themselves to create a larger E-field with increasing B-field, like the E-field is "saturated". If this were true, then I think this "saturated" electric field would generate several hundred kV potential difference, and by that time, the air (or whatever medium) surrounding the conductor would be broken down. (This could of course be non-sense because it's purely based on my intuition and not a single equation.)

14. Oct 30, 2016

### jim hardy

you mean something like Vanderwal's forces?
hmmm ideal charge has zero volume, i don know about charge carrier like an electron.

15. Oct 30, 2016

### jim hardy

That damage happened because the motor spun too fast and centrifugal force threw the copper conductors out of the spinning rotor into the airgap space between rotor and stator where they got wedged and chewed up . Purely a mechanical not electrical or magnetic phenomenon.

That happens when a series wound motor gets disconnected from its load with full voltage applied.
That one was in an electric car, the guy floored the accelerator while transmission was in neutral.

16. Oct 30, 2016

### jim hardy

i'm having difficulty parsing that paragraph.
Can you use paint to make a primitive picture ?
Moving charge inside the conductor leaves one end with deficient and the other wih surplus charge, so the electric field doesn't stay constant until the charge stops moving..
If no external current the E and qvcross b forces are equal and opposite. Mother Nature loves a balance.

Last edited: Oct 30, 2016
17. Oct 30, 2016

### Delta Kilo

It ceases to be a conductor then.
It is worth noting just how much free charge is floating inside a conductor. If I'm not grossly mistaken (which happens every now and then), there is about 1022 atoms in 1 g of copper which makes about 1.5 kilocoulombs worth of free electrons. If you pull them all to one side at a distance of 1cm, we are talking about 1015 volts and 1020 newtons of force.

18. Oct 31, 2016

### Trance-

Oh boy, I'm glad to have started this topic. Thanks for the great answers everyone. I've gotten the clarification I wanted.

19. Oct 31, 2016

### jartsa

Let us build a generator that produces a high voltage. As we know that generator will have a large internal resistance, because we are using long and thin wire to make the coils.

Then let's use a voltmeter to measure the EMF of the generator.

There is a voltage drop $I*R$ in the generator coils. If we want to measure the real EMF correctly, $I$ must be small.

EMF is always $l*v*B$. Sometimes part of the EMF is used to heat the wire were the EMF is generated, but that does not happen when we are measuring the EMF correctly, using a voltmeter with high enough resistance.

Now let us move a metal rod in a magnetic field, let's call this thing a generator, and let's say its internal resistance is infinite. Because of the motion and the magnetic field there is a EMF $l*v*B$, and charges feel a force $q*v*B$, which force must be caused by an electric field, the strength of which is $v*B$.

Last edited: Oct 31, 2016