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?

- science0229

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Michele_Laino

I think no, since "n" is the density of free electrons with respect to the volume of the conductor

- science0229

Then, would the drift speed be inversely proportional to the cross-sectional area?

- Michele_Laino

no, since J is inversely proportional to the cross sectional area too

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Michele_Laino

the drift speed is proportional to the distance traveled by our free electrons

- science0229

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?

- Michele_Laino

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

- Michele_Laino

\sigma is the conductibility of that conductor

- Michele_Laino

now we can write J as follows:
\[\Large {\mathbf{J}} = Ne{\mathbf{v}}\]

- science0229

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|

- Michele_Laino

from which we can see that J is proportional to a length, since, dimensionally speaking we have:
\[v = \frac{J}{{Ne}} = \frac{{Coulomb/{m^2}}}{{\left( {1/{m^3}} \right)Coulomb}} \sim m\]

- science0229

okay

- Michele_Laino

I try to solve the first case, namely V is applied between left-right faces

- science0229

You meant to say that v is proportional to the length, right?

- Michele_Laino

I mean the length of the space traveled by free electrons

- science0229

alright

- science0229

But, isn't the dimensions for the current density A/m^2?

- Michele_Laino

sorry you are right!
\[v = \frac{J}{{Ne}} = \frac{{Coulomb/\left( {s \times {m^2}} \right)}}{{\left( {1/{m^3}} \right)Coulomb}} \sim \frac{m}{s}\]

- Michele_Laino

as we can see the drift speed is proportional to the space traveled by free electrons inside an unitary interval of time

- science0229

okay

- Michele_Laino

ok! for your first case, we can write:
\[J = \sigma \frac{V}{{3L}}, \Rightarrow nev = \sigma \frac{V}{{3L}}\]

- Michele_Laino

now we can suppose to have N free electrons, inside your conductor, so we can write:
\[n = \frac{N}{{volume}} = \frac{N}{{L \times 2L \times 3L}}\]

- Michele_Laino

so substituting we have:
\[\Large v = \sigma \frac{V}{{3L}}\frac{1}{{ne}} = \sigma \frac{V}{{3L}}\frac{{6{L^3}}}{{Ne}} = \sigma \frac{{2V}}{3}\frac{{{L^2}}}{{Ne}}\]

- science0229

okay

- Michele_Laino

or equivalently:
\[\Large v = \sigma \frac{V}{3}\frac{A}{{Ne}},\quad {\text{since}}:A = 2{L^2}\]

- Michele_Laino

namely, the speed drift is proportional to the cross sectional area

- Michele_Laino

|dw:1434257535131:dw|

- science0229

I think you made a little calculation mistake above.

- Michele_Laino

please tell me

- Michele_Laino

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

- science0229

\[v=\sigma \frac{ V }{ 3L }\frac{ 6L^3 }{ Ne }=\sigma \frac{ 2VL^2 }{ Ne }\]

- science0229

But still, our conclusion hold; the drift speed is proportional to the cross-sectional area.

- science0229

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.

- Michele_Laino

that model, namely the water hose model it is not similar to ours

- science0229

Then, is there some kind of a "model" that can help me understand this intuitively?

- Michele_Laino

I think that a possible model can be this:
we have to imagine a large conductor, and the electron density n is constant

- science0229

okay

- Michele_Laino

now think about a square inside that conductor, like this:
|dw:1434258444976:dw|

- science0229

got it

- Michele_Laino

next we compute how many electrons are flowing into that square, inside a time interval \Delta t

- Michele_Laino

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

- science0229

okay

- Michele_Laino

now suppose that the side of our square is L:
|dw:1434258774249:dw|
so total current I is:
\[\Large I = \frac{{en{L^2}u\Delta t}}{{\Delta t}} = en{L^2}u = enAu\]

- Michele_Laino

since the volume of our parallelepiped is:
\[\Large volume = {L^2}u\Delta t\]

- science0229

okay

- Michele_Laino

now, when we study conduction in solids, we are in steady conditions, namely the current I is constant, so we can write:
\[\Large u = \frac{I}{{enA}} = \frac{{const}}{A}\]

- science0229

That means that the drift speed is inversely proportional to the cross-sectional area!

- Michele_Laino

yes!

- science0229

But above, from what we did, wasn't the drift speed proportional to the cross-sectional area?
What's going on?

- Michele_Laino

since in previous exercise what is constant is the voltage drop V, and as you know, we have:
V=R*I

- Michele_Laino

in my model above, what is constant is the current I

- Michele_Laino

the resiatance R of a conductor depends on the geometrical shape of that conductor

- Michele_Laino

resistance*

- science0229

Wait.

- science0229

Oh. I think I got it!
Thank you!

- science0229

It seems like I was a bit confused on the definitions.

- Michele_Laino

we have to consider each time what are the quantitieswhich are constant and what are the ones which are not constant

- science0229

I'll always keep that in mind!
Again, thank you for sticking with me for almost an hour!

- Michele_Laino

:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.