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Find drift velocity of the current carriers in the copper strip

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    ABCD is a plane rectangular strip of conducting material of uniform thickness, with a steady current flowing uniformly from AD to BC. The potential difference between E and F, the mid-points respectively of AB and CD, is zero, but when a magnetic field is set up at right angles to ABCD, into the plane of the diagram, a small, steady potential difference appears between E and F. Explain these observations and briefly describe a practical application of the phenomenon.

    e09f448fd306.jpg

    When the rectangle is made of copper, of density 9 * 103 kg m-3, relative atomic mass 63, with thickness 1 mm, breadth 20 mm, carrying a current of 10 A and with an applied field of flux density 1.67 T, the potential difference between E and F is found to be 1 μV, F being positive with respect to E. Taking the electronic charge to be 1.6 * 10-19 C and the Avogadro constant to be 6 * 1023 mol-1, find:

    (a) the drift velocity of the current carriers in the copper;

    (b) the number of charge carriers per unit volume in the copper and the sign of their charge;

    (c) the mobility (drift velocity per unit electric field) of these carriers, given the resistivity of copper to be 1.7 * 10-8 Ω m;

    (d) the ratio of the number of atoms per unit volume to the number of charge carriers per unit volume in the copper.

    Answers: (a) 3.0 * 10-5 m s-1, (b) 1.0(4) * 1029 m-3, (c) 3.5 * 10-3 m2 s-1 V-1, (d) 0.82.

    2. The attempt at a solution
    (a) I = n A v e → v = I / n A e where: I = 10 A, n = 9 * 103 kg m-3, A = (1 / 10 / 100) * (20 / 10 / 100) = 2 * 10-5 m2, e = 1.6 * 10-19 C.

    v = 10 / (9 * 103) * (2 * 10-5) * (1.6 * 10-19) = 3.47 * 1020 m s-1.

    Doesn't fit the answer. What am I missing?
     
  2. jcsd
  3. Oct 15, 2016 #2
    hall effect !!!!
     
  4. Oct 16, 2016 #3
    I used the formula, I've put the numbers into it, still I get a wrong result. What's wrong?
     
  5. Oct 17, 2016 #4
    Any help please?
     
  6. Oct 17, 2016 #5
    I = n A v e

    n = I / A v e = 10 / 2 * 10-5 * 3 * 10-5 * 1.6 * 10-19 = 1.042 * 1029 m-3.

    But no idea how to approach (a), (c) or (d).
     
  7. Oct 18, 2016 #6
    Any help please?
     
  8. Oct 18, 2016 #7

    gneill

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    Staff: Mentor

    There are two useful formulas that pertain to the Hall effect. You've been using only one of them. @lychette gave you an excellent hint. Did you do a bit of research on the Hall Effect?

    Take a look at: Hyperphysics Hall Effect then list the two useful formulas.
     
  9. Oct 18, 2016 #8
    So, VH = B v D, v = VH / B d = 10-6 / (1.67 * 0.02) = 3 * 10-5.

    I saw that formula before, but I didn't think that
    is the Hall Voltage.

    What is mobility? VH = BI / net, where net = mobility?

    Update: vd = μ E and σ = n e μe, where μ is mobility, σ is conductivity (or resistivity?), e is charge and n is number density.
     
    Last edited: Oct 18, 2016
  10. Oct 18, 2016 #9

    gneill

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    Staff: Mentor

    Mobility is the drift velocity per unit of electric field that is causing that velocity. This one's a bit tricky because you need to first determine the electric field. This is not the field established by the Hall voltage which is transverse to the current flow, but the field due to the potential difference caused by the current and the resistance of the copper. So, very much like the potential drop across a resistor.

    Note that they give you the resistivity of copper for this question. You know the current flowing, and you know the cross sectional area of the conductor... so what's the potential drop per unit length of the conductor?
     
  11. Oct 18, 2016 #10
    I think I'll only need to determine n now and then calculate mobility.

    Update: ρ = R A / L → R = ρ L / A = 1.7 * 10-8 * 0.02 / 2 * 10-5 = 1.7 * 10-5 Ω. V = I R = 10 * 1.7 * 10-5 = 1.7 * 10-4 V. vd = μ E → μ = vd / E = 3 * 10-5 / 1.7 * 10-4 = 0.176 m2 / s V.
     
    Last edited: Oct 18, 2016
  12. Oct 18, 2016 #11

    gneill

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    Staff: Mentor

    That will work. It amounts to the same thing. ##\sigma## is conductivity.
     
  13. Oct 18, 2016 #12
    What's missing here? I am mostly unsure in the L figure (length).
     
  14. Oct 18, 2016 #13

    gneill

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    Staff: Mentor

    You don't need the length (or at least not a particular value of length). You have the resistivity. Conductivity is the reciprocal of resistivity.
     
  15. Oct 18, 2016 #14
    σ = 1 / ρ

    σ = n e μ → μ = σ / n e = (1 / (1.7 * 10-8)) / (1.042 * 1029) * ( 1.6 * 10-19) = 3.5 * 10-3 m2 V-1 s-1.

    How shall we address this last one?
     
  16. Oct 18, 2016 #15

    gneill

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    Staff: Mentor

    Well, what do you know already and what remains to found?
     
  17. Oct 18, 2016 #16
    Number of atoms per unit volute and number of charge carriers per unit volume are unknown.

    We have density 9 * 103 kg m-3, relative atomic mass 63 and NA = 6 * 1023 mol-1.
     
  18. Oct 18, 2016 #17

    gneill

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    What did you calculate for part (b)?
    Yes. So work out the material's number density (atoms per unit volume).
     
  19. Oct 18, 2016 #18
    So we only need the number of atoms per unit volume. Using this formula: (NA * density) / atomic mass = (6 * 1023) * (9 * 103) / 63 = 8.57 * 1025 m-3.

    8.57 * 1025 / 1.0(4) * 1029 = 8.24 * 10-4.
     
  20. Oct 18, 2016 #19

    gneill

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    Make sure that your units match between given values.
     
  21. Oct 18, 2016 #20
    Indeed, 63 is g and not kg. The answer is (NA * density) / atomic mass = (6 *1023) * (9 * 103) / 0.063 = 8.57 * 1028 m-3.

    8.57 * 1028 / 1.0(4) * 1029 = 0.82.

    Thank you!
     
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