MODELLING CURRENT AND ELECTRODYNAMICS

satan 29

Dec 14, 2020 11:35

suppose we draw a simple circuit: EMF E, connected to a resistance R. All wires are ideal conductors.

John Rennie

OK ... ?

satan 29

suppose we select two points on a piece of wire a and b, such that the battery and the resistor both are NOT between a and b.

now. The battery sets up an electric field within the wires, yes?

John Rennie

I think the easiest way to understand what happens is using the hydraulic analogy. I know it's cliched, but it works really well.

satan 29

but sir my doubt is specifically concerned with electromagnetism.

John Rennie

OK. The battery only sets up a field inside the battery. Once outside the battery any field that exists is due to a very, very small difference in the density of the conduction electrons.

The simplest way to understand this is to start with a battery that has nothing attached.

satan 29

Dec 14, 2020 11:42

ok

John Rennie

Inside the battery a chemical reaction absorbs electrons at the +ve terminal and creates electrons at the negative terminal. The net effect is that the battery acts like a pump and pumps electrons from the +ve to the -ve.

So we get a net -ve charge at the -ve terminal and a net +ve charge at the positive terminal.

The battery has some capacitance, C, so the charge separation creates a potential difference given by the usual V = Q/C.

satan 29

hmm

John Rennie

This potential difference is of course just the battery voltage.

satan 29

yes

John Rennie

The point is that now we have a slightly greater electron density at the -ve terminal than at the +ve terminal because electrons have been umped through the battery. It's almost like a gas that has a higher pressure at one terminal than the other.

satan 29

Dec 14, 2020 11:47

yes

John Rennie

If you know connect the terminals with a wire then the electrons will flow out of the high density region near the -ve, through the circuit and back into the battery at the +ve terminal.

satan 29

yes

John Rennie

At any point in the circuit the electron flow is driven by the gradient in the density of the electrons.

satan 29

alright

John Rennie

That gradient is what creates the field in the resistor, wire, or whatever the electrons are flowing through.

Now, your question is about the electric field in the wire, and I guess you're assuming this is an ideal zero resistance wire. Yes?

satan 29

Dec 14, 2020 11:50

yes.

John Rennie

If there is no resistance then there is no obstruction to the electron flow. Well, in principle the electrons have a mass so they can't be accelerated infinitely fast, but in practice the electron mass has little effect.

So what happens is that the electron density in the wire is effectively constant. And it's constant because any density gradient would create a field that accelerates the electrons, and with no resistance to motion that acceleration is effectively infinitely large. So any density differences get damped out instantly.

OK so far?

satan 29

yes

John Rennie

So in an ideal, zero resistance wire there is no field because there is no density gradient.

satan 29

Ah, so no density gradient----->no electrostatic fields

John Rennie

Yes, though of course any real wire has a resistance so there is a field.

satan 29

Dec 14, 2020 11:55

and inside a resistor, there will be field?

John Rennie

Yes.

satan 29

but how is there an electron density gradient inside the resistance?

John Rennie

In your ideal case there is no potential difference in the wires so the potential drop across the resistor is equal to the potential across the battery.

@satan29 a resistor works by restricting the motion of electrons. The field accelerates electrons in the resistor so their speed increases, but the resistor is designed so that electrons flowing through it collide inelastically with atoms in the resistor. The KE of the electrons gets converted to lattice vibrations of the atoms in the resistor.

satan 29

@JohnRennie "the field accelerates" but sir, my question is that how does the field establish inside the resisitor in the first place.

John Rennie

We've agreed the electron density is higher at the battery -ve terminal than at the +ve terminal, and since the wires are ideal the density at the -ve end of the resistor is higher than the density at the +ve end of the resistor.

And that density difference creates an electric field.

satan 29

Dec 14, 2020 12:01

ahhhhhhh ofcourse

whats the hydraulic equivalent of this situation?

John Rennie

Electrons in conductors behave remarkably like a gas, though it's a very incompressible gas - almost a liquid.

satan 29

drude model, right? moving on..

John Rennie

So the battery "pumps" the gas and creates a pressure difference between the two ends of the pump. This pressure difference then makes the gas flow round the circuit.

satan 29

right.

John Rennie

The analogy is remarkably good even down to a surprising level of detail.

For example you know that in gases the speeds of the gas molecules are hundreds of metres per second.

satan 29

Dec 14, 2020 12:05

yes

John Rennie

When we say a gas is flowing at, for example, one metre per second the gas molecules are moving a hundred times faster than this and the 1 m/s is a difference in the average velocities of the gas molecules.

Well in metals the conduction electrons have momenta that is much higher than you'd calculate from the drift velocity. The drift velocity is a difference in the average velocities of the conduction electrons in exactly the same way as for gas flow.

satan 29

how can it have average velocitie S i.e plural

John Rennie

You can calculate the average velocities in the two directions along the wire, and you'd find they were unequal.

But yes, at the end of the day there is only one overall average.

satan 29

Ah yes, I see

how do you account for energy losses in resistors?

John Rennie

In anything that conducts the conduction electrons have wavefunctions that are approximately plane waves. To a good approximation they can be treated as a free particle.

If this was really the case then the electron wavefunctions would have no interaction with the conducting material and they could never exchange energy with it. In fact this is what happens in superconductors, and it's why superconductors have no resistance.

satan 29

Dec 14, 2020 12:13

okay

John Rennie

But in real conductors the electron wavefunctions can couple to the thermal vibrations of the atoms in the conductor, and this coupling means the two can exchange energy.

You'll often see this described as a moving electron crashing head on into an atom and making it vibrate, though this is a very crude description since the electrons are delocalised over distances much larger than the atomic spacing and they don't behave like little balls.

satan 29

aight

John Rennie

But anyhow the result is that the electrons can transfer their KE into lattice vibrations in the material that the resistor is made from.

satan 29

right. Although my original question (perhaps poorly phrased) was that how do we account for energy losses using the hydraulic analogy

John Rennie

I guess it depends on how you model a resistor in the hydraulic analogy.

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