## anonymous 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 cross-sectional area?

1. Michele_Laino

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

2. anonymous

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

3. Michele_Laino

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

4. Michele_Laino

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

5. anonymous

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?

6. 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

7. Michele_Laino

\sigma is the conductibility of that conductor

8. Michele_Laino

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

9. anonymous

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|

10. Michele_Laino

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

11. anonymous

okay

12. Michele_Laino

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

13. anonymous

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

14. Michele_Laino

I mean the length of the space traveled by free electrons

15. anonymous

alright

16. anonymous

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

17. Michele_Laino

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

18. Michele_Laino

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

19. anonymous

okay

20. Michele_Laino

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

21. Michele_Laino

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

22. Michele_Laino

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

23. anonymous

okay

24. Michele_Laino

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

25. Michele_Laino

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

26. Michele_Laino

|dw:1434257535131:dw|

27. anonymous

I think you made a little calculation mistake above.

28. Michele_Laino

29. 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

30. anonymous

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

31. anonymous

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

32. anonymous

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.

33. Michele_Laino

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

34. anonymous

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

35. 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

36. anonymous

okay

37. Michele_Laino

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

38. anonymous

got it

39. Michele_Laino

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

40. 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

41. anonymous

okay

42. Michele_Laino

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

43. Michele_Laino

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

44. anonymous

okay

45. Michele_Laino

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

46. anonymous

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

47. Michele_Laino

yes!

48. anonymous

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

49. Michele_Laino

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

50. Michele_Laino

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

51. Michele_Laino

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

52. Michele_Laino

resistance*

53. anonymous

Wait.

54. anonymous

Oh. I think I got it! Thank you!

55. anonymous

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

56. Michele_Laino

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

57. anonymous

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

58. Michele_Laino

:)