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Homework Help: Ampere's Law: Cylindrical conducter with varying current

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data

    The current density of a cylindrical conductor of radius R varies as J(r) = J0e−r/R (in the region from zero to R). Express the magnitude of the magnetic field in the regions r < R and r > R. (Use any variable or symbol stated above along with the following as necessary: μ0.)

    Produce a sketch of the radial dependence, B(r).

    2. Relevant equations

    Ampere's Law

    [itex]\oint B \bullet ds [/itex] = μ0ienc

    3. The attempt at a solution

    At r < R:

    B = μ0J(r)A/2∏r

    J(r) = J0e−r/R
    J(r)A = J0e−r/R(∏r2)

    B = (μ0J0e−r/Rr)/2

    This is not correct.
  2. jcsd
  3. Mar 25, 2013 #2
    How current is defined in terms of current density?

    (Hint: The relation between current and current density involves a integral.)
  4. Mar 26, 2013 #3
    I = J(r)A
    dI = J(r)Adr
    dI = J0∏r2e-r/Rdr

    I = J0∏∫r2e-r/Rdr

    I used a computer to solve the integral, which is an integration by parts.

    I = -J0∏R(r2+2rR+2R2)/er/R

    B = μ0I/2∏r

    B = -μ0J0R(r2+2rR+2R2)/2rer/R
  5. Mar 27, 2013 #4
    You can't write that. According to the definition,
    [tex]I=\int \vec{j}\cdot\vec{dA}[/tex]

    What is dA?
  6. Mar 27, 2013 #5
    In that definition, is J constant? It seems to me that it would not be, but I don't know how to integrate

  7. Mar 27, 2013 #6
    No, J is not constant. You have the expression for J and it varies with r. About the differential area dA, see the attachment. We have to find the total current enclosed within the Amperian loop. As we move out away from the axis of cylinder, the current density varies so we select a very small area (differential area, dA) where we can assume that current density is effectively constant. Then we sum up (or integrate) the expression we get for the current passing through that small region. See the attachment. Can you calculate dA now? I have drawn two circles of radius r and the other with r+dr. The shaded region is dA.

    Attached Files:

  8. Mar 27, 2013 #7
    da = 2∏rdr

    2∏J0∫e(-r/R)rdr = -2J0∏R(r+R)/er/R

    B = -μ0J0R(r+R)/rer/R

    That's what I get and it isn't correct.
    Last edited: Mar 28, 2013
  9. Mar 28, 2013 #8
    What were your limits? Do you know how to evaluate a definite integral?
  10. Mar 28, 2013 #9
    limits of integration were from 0 to r, r < R.
  11. Mar 28, 2013 #10
    Right but check your work again. I get a different answer. (##e^0=1##)
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