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Finding the resistance of the halved rod

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1.50 m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x from the left end and obeys the formula ρ(x) = a + bx^2, where a and b are constants. At the left end, the resistivity is 2.25x10^-8 Ω m, while at the right end it is 8.50x10^-8 Ωm.

    A) What is the resistance of this rod?

    B) What is the electric field at its midpoint if it carries a 1.75 A current?

    C) If we cut the rod into two 75.0 m halves, what is the resistance of each half?

    2. Relevant equations




    3. The attempt at a solution

    I have solved parts A and B by doing the following:

    A) R = ρL/A

    First I found the constants a and b by solving for them, which I found to be:

    a = 2.25x10^-8 Ωm
    b = 2.78x10^-8 Ωm

    A = πr^2
    A = 3.8013x10-4 m2

    Taking the integral of the Resistance equation gave me:

    1/A*(ax + bx3/3)

    I then took the integral from 0 to 1.50m, thereby giving me:

    (1/3.801299x10-4)*3.38x10-8*1.50 + 2.78x10-8*1.53/3
    = 1.71x10-4

    This I know to be correct.

    B)

    ∫E.dl = V
    dV/dx = d/dx(IR)

    1.75/A*(a + bx2)

    Plugging in 0.75 for x gave me

    E = 1.76x10^-4 V/m

    This is also correct

    C) This is where I'm stuck. I had thought that if I used the integral in part A to find the integral from 0 to 0.75 then that would be my resistance in the left hand side. Then I thought that I could subtract the left hand value from the value I found in part A to give me the right hand value, but this hasn't worked. My workings are as follow:

    R = ρL/A

    A = 3.8x10^-4 m2

    Rleft hand1/A*(2.25x10^-8)(0.75) + (2.78x10^-8)(0.75^3/3)
    = 4.43966x10^-5

    Then Rright hand = 1.71x10^-4 - 4.43966x10^-5 = 1.266x10^-4

    Unfortunately, this did not work and I' not sure what else to try.
     
  2. jcsd
  3. Oct 3, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    I'm not sure how you've ended up with the value you did for the left had side resistance; The numbers in the formula look okay, but the calculated result does not. Calculator finger problems?
     
  4. Oct 3, 2012 #3
    Yes, I've just gotten the right answer. The correct calculation is as follows:

    I know that the Resistance of the entire (1.50m) rod is 1.71x10^-4

    Thus I used the integral form part A to calculate the value of R from 0 to 0.75, which gave me 5.4677x10^-5. This is the Resistance of the left hand piece of rod. I then subtracted this value from the resistance of the entire rod, thereby giving me a resistance of 1.16323x10^-4 for the right hand rod.

    Thanks for your help. Sorry to waste your time.
     
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