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Current in a Wire

  1. Feb 1, 2008 #1
    [SOLVED] Current in a Wire

    1. The problem statement, all variables and given/known data
    A metallic wire has a diameter of 4.12 mm. When the current in the wire is 8 A, the drift velocity is 5.4*10^-5 m/s.

    What is the density of free electrons in the metal?

    I = 8
    n = density??
    A = (2.06)^2 * 3.14
    v = 5.4 x 10^-5
    q = 1.6x10^-19

    Express your answer numerically in m^-3 to two significant figures.

    2. Relevant equations

    I = ne[tex]v_{d}[/tex]A

    A = [tex]\pi r^2[/tex]

    3. The attempt at a solution

    A = [tex]\pi r^2[/tex] = [tex]\pi (2.06x10^-3)^2[/tex] = 1.3 x 10^-5

    So manipulating the equation, I get n = I / evA

    n = [tex]\frac{8}{(1.6x10^-19)(5.4x10^5)(1.3 x 10^-5)}[/tex] = 6.94 x 10^28

    and since I need it in millimeters, I multiply it by 1000 (since I did the calculations in SI) and get 6.94 x 10 ^31, but it says it's wrong. Any ideas where I went wrong?
    Last edited: Feb 1, 2008
  2. jcsd
  3. Feb 1, 2008 #2


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    The question you've posted says express the answer in m-3. Does the answer need to be in meters cubed or millimeters cubed?
  4. Feb 1, 2008 #3
    it's to the power of -3 so m^-3 is millimetres
  5. Feb 1, 2008 #4


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    it is not 10^-3 m it is 1/(m^3) where m^3 is the unit of volume (cubic meter)
  6. Feb 1, 2008 #5
    so the answer would be 1 / 6.94 x 10^28 ?
  7. Feb 1, 2008 #6


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    The variable n is a number density or the number of electrons per unit volume so it is already expressed in terms of m-3
  8. Feb 1, 2008 #7
    Thank You.
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