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Current in a wire

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data

    If the current density in a wire or radius R is given by J = kr, 0 < r < R , what is the current in the wire?

    2. Relevant equations
    I used j = I/A, the definition of current density: current per unit cross sectional area.
    the formula for the area of a circle and for a cylinder



    3. The attempt at a solution
    First I tried to do the cross sectional area (a circle) times the current density, but I got none of the answer that were displayed, and by just using a circle, I'm only finding the current for a small section of the wire. So I found the area of a cylinder.

    A = 2*pi*R*L (L is the length of the cylinder, not given in the problem)
    j = kr

    so A*j = 2*pi*R*L*k*r
    if you set L and r = to R (no idea why you'd do this), then you get 2*pi*R^3*k, but the correct answer is (2*pi*k*R^3)/3?

    What am I missing here, because this should be a pretty straight forward question... i think.
     
  2. jcsd
  3. Apr 8, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The current density is not constant across the cross section, but is a function of r. Integrate!
     
  4. Apr 8, 2012 #3
    Ok, I understand that it's not constant everywhere, it's an average (from my understanding). So how do you get 2 *(k*pi*R^3)/3??

    I can only get (k*pi*R^3)/3. What i did this time was use the cross sectional area A = pi*R^2 and I multiplied it by current density j = k*r. I then set up the integral:
    I = ∫ [0,R] k*pi*R^2*dr (I substituted r for dr because for the current density function, r is any radius between 0 and R)
    I = (k*pi*R^3)/3

    A circle is the right cross sectional shape, correct? And am I integrating the correct bounds? I'm really confused about this question.
     
  5. Apr 8, 2012 #4

    Doc Al

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    Staff: Mentor

    You can't use a disk, since the current density is not constant over a disk. Instead, break the cross section into rings of thickness dr. What's the area of each ring? Use that to set up your integral.
    See above.
     
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