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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 cross-sectional area?

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  1. Michele_Laino
    • one year ago
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    I think no, since "n" is the density of free electrons with respect to the volume of the conductor

  2. science0229
    • one year ago
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    Then, would the drift speed be inversely proportional to the cross-sectional area?

  3. Michele_Laino
    • one year ago
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    no, since J is inversely proportional to the cross sectional area too

  4. Michele_Laino
    • one year ago
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    the drift speed is proportional to the distance traveled by our free electrons

  5. science0229
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \sigma is the conductibility of that conductor

  8. Michele_Laino
    • one year ago
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    now we can write J as follows: \[\Large {\mathbf{J}} = Ne{\mathbf{v}}\]

  9. science0229
    • one year ago
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    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
    • one year ago
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    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. science0229
    • one year ago
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    okay

  12. Michele_Laino
    • one year ago
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    I try to solve the first case, namely V is applied between left-right faces

  13. science0229
    • one year ago
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    You meant to say that v is proportional to the length, right?

  14. Michele_Laino
    • one year ago
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    I mean the length of the space traveled by free electrons

  15. science0229
    • one year ago
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    alright

  16. science0229
    • one year ago
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    But, isn't the dimensions for the current density A/m^2?

  17. Michele_Laino
    • one year ago
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    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
    • one year ago
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    as we can see the drift speed is proportional to the space traveled by free electrons inside an unitary interval of time

  19. science0229
    • one year ago
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    okay

  20. Michele_Laino
    • one year ago
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    ok! for your first case, we can write: \[J = \sigma \frac{V}{{3L}}, \Rightarrow nev = \sigma \frac{V}{{3L}}\]

  21. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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. science0229
    • one year ago
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    okay

  24. Michele_Laino
    • one year ago
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    or equivalently: \[\Large v = \sigma \frac{V}{3}\frac{A}{{Ne}},\quad {\text{since}}:A = 2{L^2}\]

  25. Michele_Laino
    • one year ago
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    namely, the speed drift is proportional to the cross sectional area

  26. Michele_Laino
    • one year ago
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    |dw:1434257535131:dw|

  27. science0229
    • one year ago
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    I think you made a little calculation mistake above.

  28. Michele_Laino
    • one year ago
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    please tell me

  29. Michele_Laino
    • one year ago
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    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. science0229
    • one year ago
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    \[v=\sigma \frac{ V }{ 3L }\frac{ 6L^3 }{ Ne }=\sigma \frac{ 2VL^2 }{ Ne }\]

  31. science0229
    • one year ago
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    But still, our conclusion hold; the drift speed is proportional to the cross-sectional area.

  32. science0229
    • one year ago
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    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
    • one year ago
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    that model, namely the water hose model it is not similar to ours

  34. science0229
    • one year ago
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    Then, is there some kind of a "model" that can help me understand this intuitively?

  35. Michele_Laino
    • one year ago
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    I think that a possible model can be this: we have to imagine a large conductor, and the electron density n is constant

  36. science0229
    • one year ago
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    okay

  37. Michele_Laino
    • one year ago
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    now think about a square inside that conductor, like this: |dw:1434258444976:dw|

  38. science0229
    • one year ago
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    got it

  39. Michele_Laino
    • one year ago
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    next we compute how many electrons are flowing into that square, inside a time interval \Delta t

  40. Michele_Laino
    • one year ago
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    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. science0229
    • one year ago
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    okay

  42. Michele_Laino
    • one year ago
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    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
    • one year ago
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    since the volume of our parallelepiped is: \[\Large volume = {L^2}u\Delta t\]

  44. science0229
    • one year ago
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    okay

  45. Michele_Laino
    • one year ago
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    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. science0229
    • one year ago
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    That means that the drift speed is inversely proportional to the cross-sectional area!

  47. Michele_Laino
    • one year ago
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    yes!

  48. science0229
    • one year ago
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    But above, from what we did, wasn't the drift speed proportional to the cross-sectional area? What's going on?

  49. Michele_Laino
    • one year ago
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    since in previous exercise what is constant is the voltage drop V, and as you know, we have: V=R*I

  50. Michele_Laino
    • one year ago
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    in my model above, what is constant is the current I

  51. Michele_Laino
    • one year ago
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    the resiatance R of a conductor depends on the geometrical shape of that conductor

  52. Michele_Laino
    • one year ago
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    resistance*

  53. science0229
    • one year ago
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    Wait.

  54. science0229
    • one year ago
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    Oh. I think I got it! Thank you!

  55. science0229
    • one year ago
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    It seems like I was a bit confused on the definitions.

  56. Michele_Laino
    • one year ago
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    we have to consider each time what are the quantitieswhich are constant and what are the ones which are not constant

  57. science0229
    • one year ago
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    I'll always keep that in mind! Again, thank you for sticking with me for almost an hour!

  58. Michele_Laino
    • one year ago
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    :)

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