Magnet Myths and Misconceptions

[Qwert]

MileHigh, I believe, your implication(s) on electrons behavior in DC (Direct Current) in solid conductor are scientifically supported. Can you show us a link or any reference on that matter?

Edit:

Oops! MH, you are supported: http://en.wikipedia.org/wiki/Charge_carrier

Then we have a dilemma.

[MileHigh]

MH said,
Oh? What if the end of that wire is connected to the cathode of an electron gun in a CRT? Where do the electrons in the beam come from, if not from out of the wire supplying the cathode?  Or have you gone over to the TA side, where you don't believe that there is a beam of electrons, focussed and directed by changing magnetic fields, in a CRT?

http://en.wikipedia.org/wiki/Cathode_ray

TK:

A CRT is a regular circuit with a current loop.  The electrons leave the hot cathode (using FET lingo we can all that the source), and then are accelerated by the anode plates and then strike the phosphor.  Then there is a wire on the side of the CRT that acts as the drain for the electrons to complete the circuit.  I am assuming that there may be a voltage jump when the electrons flow from the drain wire back to the hot cathode to sustain the current loop also.  Sorry, I haven't looked at a CRT schematic in many years.

I don't see where you imply there is an issue.  There is an electric field making the electrons move through the current loop just like there is in a wire in a conventional circuit.  Note also that the beam of electrons can be induced to change direction by either an external electric field or by an external magnetic field.  Isn't it the yoke that produces the raster scan?  (i.e. "deflecting coils.) So the yoke is bending the electron beam because it's generating an external magnetic field where there are two "ramp" stimuli, one for the horizontal and one for the vertical.  I am assuming that there are CRTs that use horizontal and vertical ramp-function voltage potentials to do the same thing.  So instead of a yoke you have two sets of what look like big parallel plate capacitors, one for the horizontal and one for the vertical.

MileHigh

[Liberty]

TK:

A CRT is a regular circuit with a current loop.  The electrons leave the hot cathode (using FET lingo we can all that the source), and then are accelerated by the anode plates and then strike the phosphor.  Then there is a wire on the side of the CRT that acts as the drain for the electrons to complete the circuit.  I am assuming that there may be a voltage jump when the electrons flow from the drain wire back to the hot cathode to sustain the current loop also.  Sorry, I haven't looked at a CRT schematic in many years.

I don't see where you imply there is an issue.  There is an electric field making the electrons move through the current loop just like there is in a wire in a conventional circuit.  Note also that the beam of electrons can be induced to change direction by either an external electric field or by an external magnetic field.  Isn't it the yoke that produces the raster scan?  (i.e. "deflecting coils.) So the yoke is bending the electron beam because it's generating an external magnetic field where there are two "ramp" stimuli, one for the horizontal and one for the vertical.  I am assuming that there are CRTs that use horizontal and vertical ramp-function voltage potentials to do the same thing.  So instead of a yoke you have two sets of what look like big parallel plate capacitors, one for the horizontal and one for the vertical.

MileHigh

"The electrons leave the hot cathode"

Hello Milehigh and TK,

If I recall correctly, I think that the cathode was usually "painted" with a chemical that had a rich supply of electrons available, that when heated with the filament, the electrons would be free to boil on the cathode.  This allowed the tube to have electrons to flow with the high voltage potential of the grid and screens and eventually the plate.

Liberty

[MileHigh]

MH said,
Charge up a capacitor with DC. The plates of the capacitor and the conductors connected to them have a net charge, equal and opposite since charge is a conserved quantity. Install the capacitor in an AC oscillating circuit and the plates and conductors attached to them will have net charges, alternating polarity as the capacitor charges and discharges and recharges in the opposite polarity. Right?

Current flow in a conductor is basically a process of equalizing charge pressure between more positive and more negative unbalanced regions. Only when current stops flowing is charge equalized; conversely, no current flows unless there is a charge imbalance between the ends of the wire. So if you look at a wire carrying current with a very sensitive instrument you will see a voltage drop along the wire, because the wire has a finite resistance. This means that there is a charge imbalance between the ends of the wire, that exists and that can be measured as long as current is flowing in the wire.

The capacitor will have charge and absence of charge on the two plates.  But the wires that supply the current to the two plates of the capacitor will not have any kind of charge imbalance or charge pressure in them.

"Charge imbalance at opposite ends of the wire to induce current flow" is the wrong way of looking it it.  A better way of looking at it is that electrons at some point in a circuit are at some potential level difference compared to some other point in the circuit.  We typically use "ground" as the reference point.  So some electrons can be at a high potential relative to ground but that does not imply some kind of "imbalance" where there are more bunched up electrons on one side of a wire and less bunched up electrons on the other side of a wire.  Yes that happens in capacitors, but they are a different animal.  Capacitors are energy storage devices.

Here is a simple example:

Circuit A is a 10-volt battery connected to a 1-ohm resistor.   Circuit B is a 1-volt battery connected to a 1-ohm resistor.

Are the electrons more densely bunched or imbalanced in Circuit A as compared to Circuit B?

The answer is no, the electrons are evenly distributed in both cases.  However, there are real differences in the relative potential of the electrons in the two circuits.

Let me just switch to conventional current for the rest of this discussion so I don't have to rework everything in my head.

What's the difference between the two circuits?

When the current enters the negative terminal of the battery in Circuit A, it's "takes an elevator ride up by 10 volts in potential" by the time it exits the battery at the positive terminal.  Then when the current hits the resistor it takes a "steep drop" and convert the potential energy into heat.

For Circuit B, the "elevator ride up" from the battery is only one volt, and the drop is a "not so steep drop" with less heat conversion.

Besides that, the current flow and the electron charge density is all the same in both circuits.

All the battery is doing is giving the current a voltage boost from the chemical reactions taking place.

.... see part two... ->

[MileHigh]

Continued...

So here is a thought experiment:   You have two batteries, one is 12 volts, the other one is one million volts.   There is no load on either battery.

When you look at the positive terminals of either battery, does the million-volt battery have more densely packed electrons on it?   (we will ignore the parasitic capacitance between the two terminals that will cause extra charge to appear on the terminals because we are not talking about that aspect.)

So, in my opinion, ignoring the parasitic capacitive effects, you will not observe any difference between the open-circuit positive terminals of each battery.  Both of the positive terminals, being made of metal, will be electrically neutral.   However, the potential of the electrons on the million-volt battery will be much higher that that of the 12-volt battery.

This is pretty "hard core" and I know my limits and all that stuff so I could be wrong in certain aspects.  By in general sense I am pretty confident that I am right.

Almost all circuits are driven by a voltage source.  That means the electric field is king.  The electric field snakes its way through all of the conductors in a circuit.   Some parts of the circuit, and some wires in the circuit may be at very high potential.  In cases like this you have a very very weak electric field inside the high-potential wires.  At the same time, the relative potential of the overall wire itself can be very high.   So you have a very weak electric field at a very high potential.  That may sound contradictory but in fact it's not.

Where you can get a very high electric field is in a resistor.  In wires the electric field strength is very very low, but in resistors the electric field strength can be very high (when you have a large voltage drop).  Sitting on top of all of this is the potential of any point in the circuit with respect to ground.

So you have two concepts of potential going on at the same time.  The first is the concept of relative potential to ground, and the second concept is the local differential potential.  In a wire the local differential potential is almost always very low.

And driving the whole thing is the electric field snaking its way through the wires.   The electrons are just along for the ride as all of this happens.  They don't get more closely bunched up at high voltage potentials.  If all of the electrons in a place in a circuit are at low potential, or if all of the electrons in a place in a circuit are at high potential, there is no difference in local electron density.

MileHigh