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Question 2

  1. Feb 4, 2005 #1
    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 have considered a wire of cross-sectional area A, with current I. V is the drift velocity of the free charges in the conductor and n is the number of charged particles per unit volume. Then the current is given by I = nqva, where q is the magnitude of the charge on a charge carrier.

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

    I am unsure of how to finish the problem. We are looking for n. But what is q? Are all my other numbers correct?
  2. jcsd
  3. Feb 4, 2005 #2


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    All u need to know (and apparently u don't) is that
    [tex] j=n_{el}e v [/tex]

  4. Feb 4, 2005 #3
    I am unsure of what the equation you gave was. In my book, I was given I = nqva.

    Care to explain on how to do the problem?
  5. Feb 4, 2005 #4


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    Yes,the two formulas are obviously equivalent.It's better to use yours as it has already built in the numbers/variables u need to plug in.
    What is "n" (or as i denoted it,n_{el.}) & what is its SI unit...?

  6. Feb 4, 2005 #5
    The SI unit is m^-3.

    How do I find q? Or do I even need q in this equation?
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