I know that the drift speed can be found by v=J/ne where J is the current density and ne is the carrier charge density. My question is that is ne constant regardless of the cross-sectional area?

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I know that the drift speed can be found by v=J/ne where J is the current density and ne is the carrier charge density. My question is that is ne constant regardless of the cross-sectional area?

Physics
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I think no, since "n" is the density of free electrons with respect to the volume of the conductor
Then, would the drift speed be inversely proportional to the cross-sectional area?
no, since J is inversely proportional to the cross sectional area too

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the drift speed is proportional to the distance traveled by our free electrons
Wait. Since drift speed is proportional to the current density, and the current density is inversely proportional to the cross-sectional area, shouldn't the drift speed be inversely proportional to the cross-sectional area?
I think that the reasoning is as below: inside a conductor, we can write: \[\Large {\mathbf{J}} = \sigma {\mathbf{E}}\] so what we have to consider as constant is the electric field inside the conductor
\sigma is the conductibility of that conductor
now we can write J as follows: \[\Large {\mathbf{J}} = Ne{\mathbf{v}}\]
Then, using this, how can I solve this problem? Figure shows a rectangular solid conductor of edge lengths L, 2L, and 3L. A potential difference V is to be applied uniformly between pairs of opposite faces of the conductor. First, V is applied between the left–right faces, then between the top–bottom faces, and then between the front–back faces. Rank those pairs, greatest first, according to the drift speeds of the electrons.|dw:1434256446254:dw|
from which we can see that J is proportional to a length, since, dimensionally speaking we have: \[v = \frac{J}{{Ne}} = \frac{{Coulomb/{m^2}}}{{\left( {1/{m^3}} \right)Coulomb}} \sim m\]
okay
I try to solve the first case, namely V is applied between left-right faces
You meant to say that v is proportional to the length, right?
I mean the length of the space traveled by free electrons
alright
But, isn't the dimensions for the current density A/m^2?
sorry you are right! \[v = \frac{J}{{Ne}} = \frac{{Coulomb/\left( {s \times {m^2}} \right)}}{{\left( {1/{m^3}} \right)Coulomb}} \sim \frac{m}{s}\]
as we can see the drift speed is proportional to the space traveled by free electrons inside an unitary interval of time
okay
ok! for your first case, we can write: \[J = \sigma \frac{V}{{3L}}, \Rightarrow nev = \sigma \frac{V}{{3L}}\]
now we can suppose to have N free electrons, inside your conductor, so we can write: \[n = \frac{N}{{volume}} = \frac{N}{{L \times 2L \times 3L}}\]
so substituting we have: \[\Large v = \sigma \frac{V}{{3L}}\frac{1}{{ne}} = \sigma \frac{V}{{3L}}\frac{{6{L^3}}}{{Ne}} = \sigma \frac{{2V}}{3}\frac{{{L^2}}}{{Ne}}\]
okay
or equivalently: \[\Large v = \sigma \frac{V}{3}\frac{A}{{Ne}},\quad {\text{since}}:A = 2{L^2}\]
namely, the speed drift is proportional to the cross sectional area
|dw:1434257535131:dw|
I think you made a little calculation mistake above.
please tell me
please note that, we have started from hypothesis that total free electron is constant, which is very reasonable hypothesis, since we have a finite conductor
\[v=\sigma \frac{ V }{ 3L }\frac{ 6L^3 }{ Ne }=\sigma \frac{ 2VL^2 }{ Ne }\]
But still, our conclusion hold; the drift speed is proportional to the cross-sectional area.
Still, intuitively, as the cross-sectional area gets smaller, shouldn't the speed be faster. I'm thinking of it as a water hose; as you squeeze the hose, the cross-sectional area decreases, but the speed of the water increases.
that model, namely the water hose model it is not similar to ours
Then, is there some kind of a "model" that can help me understand this intuitively?
I think that a possible model can be this: we have to imagine a large conductor, and the electron density n is constant
okay
now think about a square inside that conductor, like this: |dw:1434258444976:dw|
got it
next we compute how many electrons are flowing into that square, inside a time interval \Delta t
the answer is all thoise electrons that are inside a parallelepiped whose length is: \[u \times \Delta t\] |dw:1434258654231:dw| being u the common speed of our free electrons
okay
now suppose that the side of our square is L: |dw:1434258774249:dw| so total current I is: \[\Large I = \frac{{en{L^2}u\Delta t}}{{\Delta t}} = en{L^2}u = enAu\]
since the volume of our parallelepiped is: \[\Large volume = {L^2}u\Delta t\]
okay
now, when we study conduction in solids, we are in steady conditions, namely the current I is constant, so we can write: \[\Large u = \frac{I}{{enA}} = \frac{{const}}{A}\]
That means that the drift speed is inversely proportional to the cross-sectional area!
yes!
But above, from what we did, wasn't the drift speed proportional to the cross-sectional area? What's going on?
since in previous exercise what is constant is the voltage drop V, and as you know, we have: V=R*I
in my model above, what is constant is the current I
the resiatance R of a conductor depends on the geometrical shape of that conductor
resistance*
Wait.
Oh. I think I got it! Thank you!
It seems like I was a bit confused on the definitions.
we have to consider each time what are the quantitieswhich are constant and what are the ones which are not constant
I'll always keep that in mind! Again, thank you for sticking with me for almost an hour!
:)

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