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Current and Avogadro's Number

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data

    The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.20 mm.
    (a) The beam current is 7.75 µA. Find the current density in the beam assuming it is uniform throughout.
    correct check mark A/m2

    (b) The speed of the electrons is so close to the speed of light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam.
    correct check mark m-3

    (c) Over what time interval does Avogadro's number of electrons emerge from the accelerator?
    s



    2. Relevant equations

    [tex]J=I/A[/tex]
    [tex]I_{avg} = nqv_{d}A



    3. The attempt at a solution

    Part a and b are straight forward.

    For part a I have: 1.71 A/m^2

    For part b I have: 3.565 x 10^10 m^-3

    I am having issues with part c. I know I need to figure out how many electrons are leaving the wire per second and then from there it should be a straight division problem using the 6.022 x 10^23 for Avogradro's number.
     
  2. jcsd
  3. Sep 18, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Think of it as a bucket. How long to fill'er up.

    So what's the definition of an ampere?
     
  4. Sep 18, 2008 #3
    Right...an ampere is a coulomb per second. So we take the current which is a coulomb per second and divide it by the elementary charge to get the number of electrons per second. Afterwards it's a simple division of avogadro's number by the aforementioned number...far easier than I anticipated. Sometimes your mind can just be clouded.
     
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