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TrojanPoem

  • one year ago

If I is the electric current, A the area of the conductor, the number of electrons for the unit of mass is n_o the speed of this electrons is v so the equation I = 0.25 n_oeva is

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  1. TrojanPoem
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    the relòationship between the current density \(j\) and the number of electrons, is: \(j=nev\), where \(e=1.6 \cdot 10^{-19} coulombs\)

  3. Michele_Laino
    • one year ago
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    relationship*

  4. Michele_Laino
    • one year ago
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    \(n \) is the number of electrons over volume

  5. Michele_Laino
    • one year ago
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    and \(j=I/A\)

  6. Michele_Laino
    • one year ago
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    I don't understand your coefficient 0.25

  7. TrojanPoem
    • one year ago
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    It's written like this in the question

  8. TrojanPoem
    • one year ago
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    so the unit of n_o is 1/kg in (SI) ?

  9. Michele_Laino
    • one year ago
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    yes!

  10. Michele_Laino
    • one year ago
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    from my steps, i can write this: \(I=n_oevA\)

  11. TrojanPoem
    • one year ago
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    M.L^2T^-2* M^-1* L * T^-2 * L^2 L^4 T^-4 ?

  12. TrojanPoem
    • one year ago
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    I can't be right in anyway.

  13. Michele_Laino
    • one year ago
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    I'm thinking at your formula, and it can not be correct if \(a\) is the area and \(e\) is the electron charge since we have this: right side= \[\frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2}\]

  14. Michele_Laino
    • one year ago
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    which is not a current

  15. TrojanPoem
    • one year ago
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    coulombs ?

  16. Michele_Laino
    • one year ago
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    yes! I measure \(e\) in coulombs

  17. TrojanPoem
    • one year ago
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    I meant, why don't you replace it with A * S

  18. TrojanPoem
    • one year ago
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    A * T

  19. Michele_Laino
    • one year ago
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    A is the cross sectional area of your conductor

  20. TrojanPoem
    • one year ago
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    Yeah.

  21. Michele_Laino
    • one year ago
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    \(v\) is the average speed of conduction electrons, so we have: \(I=0.25n_0\;e\;v\;A=\) \[0.25 \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2}\]

  22. Michele_Laino
    • one year ago
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    is 0.25 a quantity with a specific unit of measure?

  23. Michele_Laino
    • one year ago
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    for example, can 0.25 be a density?

  24. TrojanPoem
    • one year ago
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    In dimension problems, It's not necessary to bring to us an equation which we know. Today was the first time to see this. but I think 0.25 is just a constant like last 2 pi

  25. TrojanPoem
    • one year ago
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    Do you suspect that 0.25 has a unit ?

  26. Michele_Laino
    • one year ago
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    yes! since the right side of your formula, is not a current, whereas if 0.25 is a density, then its unit of measure is \(Kg/m^3\), so we can write this: \[\begin{gathered} 0.25 \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2} = \frac{{Kg}}{{{m^3}}} \cdot \frac{{coulombs}}{{Kg}} \cdot \frac{m}{{\sec }} \cdot {m^2} = \hfill \\ \hfill \\ = \frac{{coulombs}}{{\sec }} = amperes \hfill \\ \end{gathered} \]

  27. Michele_Laino
    • one year ago
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    furthermore if \(\rho=m/V\), then we can write this: \[\frac{{dn}}{{dm}} = \frac{{dn}}{{\rho dV}} = \frac{1}{\rho }\frac{{dn}}{{dV}}\]

  28. TrojanPoem
    • one year ago
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    What does dn/dm mean?

  29. Michele_Laino
    • one year ago
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    it is the number of electrons over mass of conductor

  30. TrojanPoem
    • one year ago
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    I looks like a derivative xD

  31. Michele_Laino
    • one year ago
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    yes! whereas \(dn/dV\) is my density of electrons, namely number of electrons over volume of conductor

  32. TrojanPoem
    • one year ago
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    You said: "since the right side of your formula, is not a current, whereas if 0.25 is a density, then its unit of measure is Kg/m3" Well that's convincing but how to ensure so ? the question is as written here.

  33. Michele_Laino
    • one year ago
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    It was my hypothesis, since it is the first time that I see your formula

  34. Michele_Laino
    • one year ago
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    when I was at university, I have always used this one: \[j = {n_0}ev\]

  35. TrojanPoem
    • one year ago
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    choices: a) correct,b) correct according to dimensions c) wrong d)wrong " " "

  36. Michele_Laino
    • one year ago
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    I think it is wrong!

  37. TrojanPoem
    • one year ago
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    So it's wrong :) thanks again again

  38. Michele_Laino
    • one year ago
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    :)

  39. Michele_Laino
    • one year ago
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    I understand the presence of 0.25

  40. Michele_Laino
    • one year ago
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    if we have this formula: \[I = {n_0}evA\] with \(n_0\) as number of electrons over mass, we can say that it is wrong according to the dimensions. nevertheless, due to the presence of the coefficient 0.25, we have to say that your formula is wrong, since using dimensional analysis we are not able to write numerical coefficients

  41. Michele_Laino
    • one year ago
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    here is why there is the coefficient 0.25

  42. TrojanPoem
    • one year ago
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    So 0.25 makes the formula wrong not wrong according to the dimensions ?

  43. Michele_Laino
    • one year ago
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    yes!

  44. TrojanPoem
    • one year ago
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    I tried googling this formula and no results

  45. Michele_Laino
    • one year ago
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    please wait

  46. Michele_Laino
    • one year ago
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    I'm sorry, as I wrote before, using units of measure, the right side is not a current, so I think, that your formula is wrong according to dimensions

  47. TrojanPoem
    • one year ago
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    so when to say wrong, when to say wrong according to dimensions

  48. Michele_Laino
    • one year ago
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    I say wrong according dimensions when, as in our case, the right has the wrong dimensions

  49. Michele_Laino
    • one year ago
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    right side*

  50. TrojanPoem
    • one year ago
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    so wrong when the constant is wrong ?

  51. Michele_Laino
    • one year ago
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    yes I think so!

  52. TrojanPoem
    • one year ago
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    But there is no way to know that In the last question you said correct as you knew that T= 2 pi sqrt(L/g) you know the formula but I don't.

  53. Michele_Laino
    • one year ago
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    for example if I write this formula: \[T = 3\sqrt {\frac{l}{g}} \] what can we conclude?

  54. TrojanPoem
    • one year ago
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    Wrong, but if It's the first time to see it ?

  55. TrojanPoem
    • one year ago
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    wait wait, in Problem #21 , Prove that I = n_oeva according to dimensions

  56. TrojanPoem
    • one year ago
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    See the 6th formula in this table: https://quizlet.com/3348670/physics-formulae-and-constants-flash-cards/

  57. Michele_Laino
    • one year ago
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    it is wrong according to dimensions, since N is the number of free electrons/m^3, whereas A is the cross sectional area so m^2

  58. TrojanPoem
    • one year ago
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    I made a mistake :o

  59. TrojanPoem
    • one year ago
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    I have combined two parts of two questions

  60. Michele_Laino
    • one year ago
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    no worries! :)

  61. TrojanPoem
    • one year ago
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    n is the number of electrons for the unit of volume , SORRY

  62. Michele_Laino
    • one year ago
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    and A is the cross sectional area whose unit of measure is m^2

  63. TrojanPoem
    • one year ago
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    so it's I = neva L^-3 * A * T * L * T^-1 * L^2 = A

  64. TrojanPoem
    • one year ago
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    so as you said it's wrong

  65. TrojanPoem
    • one year ago
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    What's the difference between G , g ?

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