# Prove I=nvqA?

1. Jul 23, 2012

### Scopesys

I've been sitting at my desk trying to prove this and nothing, already tried googling.. Any help guys? I wanna know how to derive it from basic principles.
Thanks

How I came by it: I'm a high school student and was doing my exam then get asked about drift velocity and it's not even in the syllabus and hence in not in any of my books >_<

All I could find on it is===>

This formula is used to calculate the current carried by a conductor in which we know the charge carrier density and the drift velocity of the charge carriers

I: current
n: # of charge carriers per unit volume
q: charge
v: drift velocity
A: area of conductor in which charges are moving

Last edited: Jul 23, 2012
2. Jul 23, 2012

### Latrace

$I$ is the current, so the amount of charge passing through the surface you get when you cut the wire perpendicular to the direction of flow (area $A$), per second. Try to imagine, before starting your watch, which electrons will pass through the surface during one second (say the flow is from left to right, you might want to make a drawing). Of course, all the electrons very close to the surface on the left side will surely pass through (you also see that the precise position of the electrons is not important, only the distance to the surface, because they presumably <a class="inlineAdmedialink" href="#">travel</a> in straight lines). But how far back left can you go, still being assured that these electrons will pass through the surface during that one second? To know this, you have to calculate the distance an electron will travel in one second (I'm sure you can do that). So this is how far back in the wire you can go (electrons further will not reach the surface in time). You can now see that the electrons that will pass through the wire in one second are precisely the electrons in the volume starting from the position you calculated (the perpendicular slice, that is) up to the surface itself. Using the given $n$ you can calculate the amount of charge in that volume, and get the formula you need.

3. Jul 24, 2012

### mikelepore

Wire with a uniform cross sectional area A.
A piece of wire with length x has volume A x.
The wire has n free electrons per unit volume.
The number of free electrons in that volume of wire is n A x.
Each electron has charge e.
The total charge in that volume of the wire is Q = e n A x.
In a time t, the charge in that volume flows past a reference point on the wire.
The rate at which the charge flows is I = Q / t = e n A x / t.
x/t is the drift velocity v,
therefore I = e n A v.