Microscopic Ohm's Law


by ed2288
Tags: conductivity, field, microscopic, ohm's law
ed2288
ed2288 is offline
#1
Apr16-09, 01:44 PM
P: 25
1. The problem statement, all variables and given/known data
Explain the terms used in the microscopic version of Ohm's Law


2. Relevant equations
j=oE



3. The attempt at a solution
Well I know what we have here is the current density within a conductor is equal to its conductivity multiplied by an 'E-field'. But what is this E-field? Is it an external field that acts upon the wire? But if this is the case will the electrons in the conductor not just move in a few microseconds to go back into equilibrium meaning j=0. You could act on it with an varying field but then how would the calculation work out?
Or is E the field inside the conductor caused by the current density?
Thanks
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LowlyPion
LowlyPion is offline
#2
Apr16-09, 01:57 PM
HW Helper
P: 5,346
Quote Quote by ed2288 View Post
1. The problem statement, all variables and given/known data
Explain the terms used in the microscopic version of Ohm's Law


2. Relevant equations
j=oE

3. The attempt at a solution
Well I know what we have here is the current density within a conductor is equal to its conductivity multiplied by an 'E-field'. But what is this E-field? Is it an external field that acts upon the wire? But if this is the case will the electrons in the conductor not just move in a few microseconds to go back into equilibrium meaning j=0. You could act on it with an varying field but then how would the calculation work out?
Or is E the field inside the conductor caused by the current density?
Thanks
Maybe this helps?
http://hyperphysics.phy-astr.gsu.edu...ic/ohmmic.html
ed2288
ed2288 is offline
#3
Apr16-09, 04:23 PM
P: 25
Hmmm I guess I've spent so long doing electrostatics that I'm struggling a bit with the intuition of currents.
So from the link above, am I right in thinking the E field is from within the wire caused by a potential difference from a power source?


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