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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

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

\sigma is the conductibility of that conductor

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

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?

okay

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

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

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

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?

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

okay

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

okay

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

yes!

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.

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