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science0229
 one year ago
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 crosssectional area?
science0229
 one year ago
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 crosssectional area?

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think no, since "n" is the density of free electrons with respect to the volume of the conductor

science0229
 one year ago
Best ResponseYou've already chosen the best response.0Then, would the drift speed be inversely proportional to the crosssectional area?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, since J is inversely proportional to the cross sectional area too

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the drift speed is proportional to the distance traveled by our free electrons

science0229
 one year ago
Best ResponseYou've already chosen the best response.0Wait. Since drift speed is proportional to the current density, and the current density is inversely proportional to the crosssectional area, shouldn't the drift speed be inversely proportional to the crosssectional area?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\sigma is the conductibility of that conductor

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we can write J as follows: \[\Large {\mathbf{J}} = Ne{\mathbf{v}}\]

science0229
 one year ago
Best ResponseYou've already chosen the best response.0Then, 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1from which we can see that J is proportional to a length, since, dimensionally speaking we have: \[v = \frac{J}{{Ne}} = \frac{{Coulomb/{m^2}}}{{\left( {1/{m^3}} \right)Coulomb}} \sim m\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I try to solve the first case, namely V is applied between leftright faces

science0229
 one year ago
Best ResponseYou've already chosen the best response.0You meant to say that v is proportional to the length, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I mean the length of the space traveled by free electrons

science0229
 one year ago
Best ResponseYou've already chosen the best response.0But, isn't the dimensions for the current density A/m^2?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry you are right! \[v = \frac{J}{{Ne}} = \frac{{Coulomb/\left( {s \times {m^2}} \right)}}{{\left( {1/{m^3}} \right)Coulomb}} \sim \frac{m}{s}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1as we can see the drift speed is proportional to the space traveled by free electrons inside an unitary interval of time

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! for your first case, we can write: \[J = \sigma \frac{V}{{3L}}, \Rightarrow nev = \sigma \frac{V}{{3L}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we can suppose to have N free electrons, inside your conductor, so we can write: \[n = \frac{N}{{volume}} = \frac{N}{{L \times 2L \times 3L}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so substituting we have: \[\Large v = \sigma \frac{V}{{3L}}\frac{1}{{ne}} = \sigma \frac{V}{{3L}}\frac{{6{L^3}}}{{Ne}} = \sigma \frac{{2V}}{3}\frac{{{L^2}}}{{Ne}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1or equivalently: \[\Large v = \sigma \frac{V}{3}\frac{A}{{Ne}},\quad {\text{since}}:A = 2{L^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1namely, the speed drift is proportional to the cross sectional area

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434257535131:dw

science0229
 one year ago
Best ResponseYou've already chosen the best response.0I think you made a little calculation mistake above.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please note that, we have started from hypothesis that total free electron is constant, which is very reasonable hypothesis, since we have a finite conductor

science0229
 one year ago
Best ResponseYou've already chosen the best response.0\[v=\sigma \frac{ V }{ 3L }\frac{ 6L^3 }{ Ne }=\sigma \frac{ 2VL^2 }{ Ne }\]

science0229
 one year ago
Best ResponseYou've already chosen the best response.0But still, our conclusion hold; the drift speed is proportional to the crosssectional area.

science0229
 one year ago
Best ResponseYou've already chosen the best response.0Still, intuitively, as the crosssectional area gets smaller, shouldn't the speed be faster. I'm thinking of it as a water hose; as you squeeze the hose, the crosssectional area decreases, but the speed of the water increases.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that model, namely the water hose model it is not similar to ours

science0229
 one year ago
Best ResponseYou've already chosen the best response.0Then, is there some kind of a "model" that can help me understand this intuitively?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that a possible model can be this: we have to imagine a large conductor, and the electron density n is constant

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now think about a square inside that conductor, like this: dw:1434258444976:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1next we compute how many electrons are flowing into that square, inside a time interval \Delta t

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now suppose that the side of our square is L: dw:1434258774249:dw so total current I is: \[\Large I = \frac{{en{L^2}u\Delta t}}{{\Delta t}} = en{L^2}u = enAu\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since the volume of our parallelepiped is: \[\Large volume = {L^2}u\Delta t\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, when we study conduction in solids, we are in steady conditions, namely the current I is constant, so we can write: \[\Large u = \frac{I}{{enA}} = \frac{{const}}{A}\]

science0229
 one year ago
Best ResponseYou've already chosen the best response.0That means that the drift speed is inversely proportional to the crosssectional area!

science0229
 one year ago
Best ResponseYou've already chosen the best response.0But above, from what we did, wasn't the drift speed proportional to the crosssectional area? What's going on?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since in previous exercise what is constant is the voltage drop V, and as you know, we have: V=R*I

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in my model above, what is constant is the current I

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the resiatance R of a conductor depends on the geometrical shape of that conductor

science0229
 one year ago
Best ResponseYou've already chosen the best response.0Oh. I think I got it! Thank you!

science0229
 one year ago
Best ResponseYou've already chosen the best response.0It seems like I was a bit confused on the definitions.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to consider each time what are the quantitieswhich are constant and what are the ones which are not constant

science0229
 one year ago
Best ResponseYou've already chosen the best response.0I'll always keep that in mind! Again, thank you for sticking with me for almost an hour!
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