1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the drift velocity

  1. Apr 8, 2017 #1
    1. The problem statement, all variables and given/known data
    A 5.00-A current runs through a 12 gauge copper wire (diameter 2.05mm) and through a light bulb. Copper has 8.5 x 1028 free electrons per cubic meter.

    c) At what see does a typical electron pass by any given point in the wire?

    2. Relevant equations
    J = nqvd

    3. The attempt at a solution
    The previous parts of the question I solved for current density.

    J = 3.105 x 106 A / m2

    n = 8.5 x 1028 electrons / m3

    vd = J / n|q|

    I substituted all my values in:

    vd = (3.105 x 106 ) / (8.5 x 1028)(1.602 x 10-28)

    vd = (3.105 x 106) / 13.167 m/s

    vd = 2.28 x 105 m/s

    the correct answer is vd = 0.111 mm/s

    I'm sorry this is probably an arithmetic error but i can't see what I did wrong
     
  2. jcsd
  3. Apr 8, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    Can you show us how you arrived at your current density?
     
  4. Apr 8, 2017 #3
    b) What is the current density?

    cross section area = A = πr2 = πd2 / 4 = π(2.05x106) / 4 = 1.61 x 10-6

    I = 5 A

    Then current density = J = I / A = 5 / 1.61 x 10-6 = 3.105 x 106

    ..
    Also the very first part:
    a) How many electrons pass through the lightbulb each second?

    I = 5A = 5 C/s

    5 C/s * 1s = 5C

    5C / (-1.602 x 10-28) = -3.121 x 1028 electrons
     
  5. Apr 8, 2017 #4

    gneill

    User Avatar

    Staff: Mentor

    You didn't square your diameter when you calculated the cross sectional area.

    Use the magnitude of the charge on the electron. The sign of the charge is unimportant if you're counting quantity, which should be a positive number.
     
  6. Apr 8, 2017 #5
    Ok I see what you mean .. so the correct cross sectional area is:

    A = π((2.05)2 x 10-3*2) / 4 = 3.30 x 10-6
    J = 5 / 3.30 x 10-6 = 1.51 x 106 A / m2

    So I know my J and |q| are correct..I also forgot to mention that the problem gave "Copper has 8.5 x 1028 free electrons per cubic meter" which I use as my n value.

    So vd = J / n|q|.
    J = 1.51 x 106
    n = 8.5 x 1028 free electron / m3
    |q| = 1.602 x 10-28 C


    vd = (1.51 x 106) / (8.5)(1.602) = (1.51 x 106) / 13.617 = 110,890 m/s

    never mind I just realized |q| = 1.602 x 10-19.. so 110,890 * 10-9 = .111 x 10-3 which is the answer in the book.

    Thanks for helping me on this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find the drift velocity
  1. Drift velocity ? (Replies: 1)

  2. Drift velocity (Replies: 3)

Loading...