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E-Field Through Copper Wire given drift velocity

  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data
    If the magnitude of the drift velocity of free electrons in a copper wire is 7.84×10-4 m/s, what is the electric field in the conductor? (Also gives chart that states that the resistivity of Cu is 1.7×10-8 Ωm)

    2. Relevant equations

    vd=(qEτ)/me (where τ is avg. time between collisions)

    ρ=1/σ=me/(nq2τ)


    3. The attempt at a solution

    first I solved for τ in the second equation:
    τ=me/(nq2ρ)

    Then I put this value in for τ in the first equation:
    vd=(qE(me/(nq2ρ))/me
    Simplify:
    vd=E/(qnp)

    When I put in values:
    7.84×10-4=E/(1.6×10-19)(1.7×10-8)(n)
    I didn't know how to get n (number of e- per m3), so I googled the density of copper (8.96 g/ml) and did some conversions to get n=8.475×1028, which gave me a correct answer of E=.18V/m, but I don't think I was supposed to use google, and no density chart is given in the book. Is there another way to do this that I overlooked?
     
  2. jcsd
  3. Nov 2, 2016 #2

    kuruman

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    Homework Helper
    Gold Member

    There is a more direct way that does not involve τ. The current density is related to the electric field by J = (1/ρ) E and to the drift velocity by J = n e vd. You still need n which you cannot get from what is given unless you look things up.
     
  4. Nov 2, 2016 #3
    Thanks kuruman. Just wanted to make sure I wasn't missing anything important.
     
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