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Quick Help!

  1. Feb 7, 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 = ??

    Can someone help me out on the answer for this problem? Thanks!
  2. jcsd
  3. Feb 7, 2005 #2


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    Pay attention with the units... :grumpy:
    [tex] |\vec{j}|=\frac{I}{S}= n |q_{el.}| |\vec{v}_{drift}| [/tex]

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