Electricity - detailed explanation ?

[JPC]

hey
from what i have been told , in an electric circuit , electrons only move at a few millimeters per sec

but then, how can u produce energy kilometers away from these moving electrons ?

i mean if we look on an atomic level , u have electrons moving very slowly at the beginning of the circuit , and at a more general level , u have heat being produced at the end of circuit (also on all the circuit)

This leads me to my more general question : How does electricity work ? how can we explain it on an atomic level (or why not quantic level) ?

does it have something to do with the magnetic field of each atom of the circuit ?

 

[lpfr]

Take a long rod and begin to push one extremity very slowly. The entire rod will begin to move almost at the same time as the extremity you are pushing (in less than a thousandth of second).

For your other questions, it is not possible to post an electricity book. You can begin consulting htp://en.wikipedia.org; it is not always clear and true, but it can be a beginning.

 

[JPC]

but is it like each electron; at the same time jump from one atom (lets say of copper) to the other ?

 

[lpfr]

You can see it so. If fact, all electrons are pushed by the electric field due to voltage difference. There are a lot of electrons. They do not jump simultaneously but they jump. It is like a disordered march (not a military parade) all people advance but not simultaneously.

 

[rcgldr]

Although the drift velocity of individual electrons is slow, the propagation rate of the electrical wave, or velocity factor, is quite high. 60+% of the speed of light is typical in transmission lines.

is it like each electron; at the same time jump from one atom (lets say of copper) to the other?

Yes, but when one electron is capture by an atom, another is released almost imediately. Electrons travel at near light speed between molecules.

here's a link with a better explanation:

http://www.allaboutcircuits.com/vol_2/chpt_14/4.html

 

[lpfr]

Yes, but when one electron is capture by an atom, another is released almost imediately. Electrons travel at near light speed between molecules.

In fact, in a conductor, the electrons which participate in conduction are free electrons.
That is, they are not attached to any particular atom and they do not jump from atom to atom.

Of course this is a big number statistical process and there are continuously electrons that reattach to atoms and electrons that go free. But this process is independent from conduction.

At ambient temperature the speed of electrons in a conductor is about $$1.17\,\,10^5$$ m/s which is very small compared to speed of light in vacuum.

 

[JPC]

Ok

that clears it up

thanks for ur comments :)

 

[rcgldr]

At ambient temperature the speed of electrons in a conductor is about $$1.17\,\,10^5$$ m/s which is very small compared to speed of light in vacuum.

Make that average velocity. The electrons travel at high speed, but in almost random directions with a small net component of flow related to current.

 

[lpfr]

Make that average velocity. The electrons travel at high speed, but in almost random directions with a small net component of flow related to current.

You are wrong.
With free particles in thermal equilibrium, as electrons atoms or molecules, the kinetic energy is distributed among the degrees of freedom. Each degree of freedom contributes for
$${1\over 2}kT$$
of energy. Here T is the temperature (Kelvin) and $$k=1.38\,10^{-23}$$ is the Boltzmann constant.
Electrons and monatomic gases have 3 degrees of freedom. Then, you can write:
$${1\over 2}mv^2={3\over 2}kT$$
and verify if I didn't make a mistake in my numerical calculus.
Of course, this is just the velocity RMS value. The velocity distribution is a Maxwell-Boltzmann.

As for the velocity due to current, it is indeed very small. If a current of 1 A circulates in a copper wire of 1 mm in diameter, the speed due to current is $$\simeq 10^{-4}$$ m/s.

EDIT: TeX seems to have problems.
first fomula {1\over 2}kT
second: k=1.38\,10^{-23}
third:{1\over 2}mv^2={3\over 2}kT
last: \simeq 10^{-4}