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Comparing Current in two different wires

  1. Oct 10, 2009 #1
    Problem:
    Suppose wire A and wire B are made of different metals, and are subjected to the same electric field in two different circuits. Wire B has 5 times the cross-sectional area, 1.1 times as many mobile electrons per cubic centimeter, and 2 times the mobility of wire A. In the steady state, 2E18 electrons enter wire A every second. How many electrons enter wire B every second?

    Equation:
    i = nAuE
    n = Electrons/cubic meter
    A = cross-sectional area
    u = mobility of electrons
    E = electric field


    Solution:
    ib=ia * 2*5*1.1E6
    it is 1.1E6 because the mobility factor needs to be in electrons/ cubic meter

    ib = 2.2E25

    Webassign says thats wrong. I can't come up with anything else
     
  2. jcsd
  3. Oct 10, 2009 #2

    Delphi51

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    Homework Helper

    Omit the E6. The whole story is
    i = nAuE = 2*n1*5*A1*1.1*u1*E
    = 2*5*1.1*n1*A1*u1*E
    = 2*5*1.1*i1
    = 2*5*1.1*2E18
    If 1.1 times as many electrons per cm^3, then there will be 1.1 times as many electrons per m^3 too.
     
  4. Oct 10, 2009 #3
    there are a million cubic centimeters in a meter cube. if each cubic cm has 1.1 more electron then the meter cube has 1.1E6 extra electrions
    (Also, I already tried that and webassign said I was wrong).
     
  5. Oct 10, 2009 #4

    Delphi51

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    Homework Helper

    No. It doesn't give a count of the extra electrons. It says the other wire has 10% more per cm^3. That means 10% more per m^3. Say the first one has 100 per cm^3. Then the second will have 110.
    Per m^3 the first one will have 100*1E6 and the second will have
    110*1E6. The ratio will be 110E6/100E6 = 1.1


    Kind of like saying what if every student in the class overeats and comes back to class on Monday 10% heavier. Then the whole class together will be 10% heavier, not 10*the number of students heavier in pounds, kg or tons.
     
  6. Oct 11, 2009 #5
    You were right. Thanks.
     
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