# Cause of the Electromotive Force.

1. Jul 22, 2008

### understand.

In an electric current there must be forces that push electrons in a general direction (current). My question is, what creates this force (electromotive force)?

I assume that this force, that moves electrons, is created by a surplus of electrons in one area. Perhaps they move because the electrons get too close together and push each other away. In a wire, those electrons get pushed through because it's the most accessible path.

If that is the case, then a positive terminal is not required for a current to flow. The positive terminal can't move, so it can't be a factor in the movement of electrons at the other end anyway. All that is needed is a bunch of electrons in one spot.

Am I on the right track?

2. Jul 22, 2008

### Staff: Mentor

The electromotive force is simply the electric force $\vec F = q \vec E$ on the electrons, exerted by the electric field inside the wire. In a circuit with a battery, the electric field is produced by the potential difference that the battery creates between its terminals.

3. Jul 22, 2008

### cabraham

Electric fields are not produced by potential difference. The potential difference is defined as the line integral of the electric field along a specified path. The electric field has an energy associated with it. By definition the work done moving a charge through this field per unit charge is the potential difference, aka voltage. What produces the electric field is the inherent nature of charge. Every charged particle has an electric field, and exerts a force on other charges.

To create a non-zero electric field in a circuit, work must be done. Since E fields carry energy, any change in that energy involves work. It takes power to change the value of the E field, and since work cannot change instantly, but over a finite time span, the work per unit time is power. A non-zero power requires current and voltage both.

To set up an E field, you need amp-seconds and volts. To set up an H field, you need volt-seconds and amps. Either way, you must do work.

I hope this helps.

4. Aug 8, 2008

### nealh149

cabraham, you're analysis is slightly incorrect. Electric fields do not have to have any charge associated with them-they can be created by a time-variant magnetic field. The beauty of electromagnetism is that we do not know have to know the location of the charges to analyze the fields. That being said, I think jtbell's choice to say that the potential "creates" the electric field is justified because potential is really just a scalar representation of the electric field-they're essentially equivalent.

5. Aug 9, 2008

### cabraham

I fully understand that E fields can be due to charge, or due to time-varying magnetic fields. I've stated that many times. Search my post history and you will confirm the same. Sorry to oppose you, but you are incorrect when you say that potential creates the field. If they are indeed equivalent, how can either be the cause of the other? One could just as well argue that the E field creates the potential. Since neither exists independently of the other, they simply coexist. No causality here at all. To say that they co-exist is perfectly correct. To say that potential "causes" the field has no basis in science at all. Also, if E is created by a time varying magnetic field, then E is not created by potential. But it was previously suggested that E is created by potential. A paradigm that contradicts its own self cannot be valid. Peace and best regards.

Claude

6. Aug 9, 2008

### Defennder

7. Aug 11, 2008

### Phrak

The OP was interested in local cause of electron flow in a conductor, I believe.

8. Aug 12, 2008

### Defennder

Yes but apparently even if what he says is true (and I think it sounds plausible) you will still need a positive terminal to decrease the concentration of electrons at one point so that those behind it along the wire can "move along".

9. Aug 12, 2008

### h0dgey84bc

Imagine taking a long straight length of conductor and placing this in a uniform E field (let's say to the right). What would happen? Initially electrons would begin to move in the opposing direction of the E field, and as a result the leftmost side of the bar would gain a net negative charge and the right hand side would gain a net positive charge (due a lack of electrons). This process cannot continue forever, at some point the electric field due to the net positive and negative charges at each end of the bar, will cancel out the external E field, restoring equilibrium and ceasing motion of the electrons. Thus, no more current flows.

With respect to a negative test charge, the -ve end is at a higher potential than the +ve end. If you turn off the external E field, electrons from the -ve end will move to the right to the lower potential (the voltage has been "dropped" across the bar). Unfortunatley in this example the current will be transistory, not steady, and therefore useless for an application.

Batteries (or other simpler sources of emf like moving a closed wire through a uniform B field) work by causing electrons to move against the potential (i.e. go from where there is already a surples of positive to where there is a surplus of negative). Infact the simplest source of emf is to mechanically (on a kind of treadmill) move the negative charges to the higher potential (w.r.t. to the neg charges). This is what supplies the energy (in the example above of the bar, emf could be generated by me "picking up" electrons and moving the from the right to the left.

A good analogy is that of the water fountain, water flows out and falls to lower gravitational potential (this is the electrons flowing naturally from the -ve end to the +ve end, when the external field is switched off). But if this was all there was to the fountain, it would be a very short display (transitory current). What is needed is a way to take the water from the lower potential (ground level) to a higher potential (top of pump), against the gradient. This is were the pump comes in (analogue of the emf). The pump mechanically does work to move the water "up the potential hill", so that this potential can then be "dropped", i.e. the water releases that potential in the form of kinetic energy and flows to bottom of fountain.

One thing to note is that real sources of emf have an internal resistance, this is the resistance encountered for example within the battery, when moving the electrons from +ve to -ve (in the unnatural direction). Therefore the potential that can ultimatley be dropped accross the circuit is slightly less than the emf (note emf is a misnomer, as its not a force but expresses the amount of work required to climb the potential hill). V=emf-Ir, where r is the internal resistance. Noting from ohms law that V=IR (R is the usual circuitory resistance), we arrive at the expression IR=emf-Ir => emf=I(R+r).

In the fountain analogy the internal resistance could be thought of as something like friction, by which I mean the pump does x amount of work to raise the water to the top of the fountain, but not all of that work will go into raising the water's P.E. some will be dissappated because of friction etc. So the amount of potential that can be dropped by the water will be slightly less than x. V=emf(of pump)-energy_dissappated.

10. Aug 12, 2008

### Gear300

The electromotive force is actually electric potential (voltage). Electrons move through a circuit because of an electric field, which propagates much more easily through conductors than insulators. The electric field exists in relation to the existence of a voltage.

If you want to know how this voltage is produced, you can look up electrochemical cells or voltage generated through induction (the latter one is better understood quantum mechanically).

11. Aug 12, 2008

### Phrak

Yeah, that's the point that needed to be made. Of course, the OP is long gone, but no matter.

I had thought the local effects could be explained by an electron density difference to either side of any electron in question that produces an electric field, but it's only valid for a straight, infinitely long wire.

For a wire that varies in cross-section or is bent back on itself in places, you have to say that the overall electric field in any section of the conductor always points along the wire from the positive to the negative terminal. But, it's still not an answer unless you can show that the electric charge arranges itself such that the field will do this. After all, how do we show that some tangled arangement of the wire won't leave a spot where the field points in the wrong direction, and current flow stops?

Basically, I don't know how a wire works.

Last edited: Aug 12, 2008
12. Aug 12, 2008

### Defennder

I'd think along the lines of a potential drop along the wire. The wire has a small but finite resistance, and clearly the potential at any point along the wire decreases with distance the further it trails from the positive terminal. You can model the local effect of a large positive potential over the length of the wire by imagining there are various small emf sources along the length. Now does this make sense? I'm hoping someone could address this.

13. Aug 13, 2008

### Phrak

Yeah, I think you've got half of it. It has to be for a uniform cross-section wire, no matter how it's shaped. The current is constant through any cross-section. The resistance, dR is constant over a length dl. And so the voltage, dV is constant over a length dl.

But somehow all the resultant charge densities throughout the length of the wire must conspire to obtain a electric fields, E such that dE/dl=dV for any cross-section along the length of the wire. The wire can have any path through space that forms a closed loop with an emf located somewhere in the loop.