1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ampere's Law: Determining magnetic fields of a shell conductor

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

    A current of constant density, J0, flows through a very long cylindrical conducting shell with inner radius a and outer radius b. What is the magnetic field in the regions r < a, a < r < b, and r > b? (Use any variable or symbol stated above along with the following as necessary: μ0.)

    2. Relevant equations

    Ampere's Law
    [itex]\oint B \bullet ds[/itex] = μ0ienc

    3. The attempt at a solution

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

    Solving for B:

    B[itex]\oint ds[/itex] = μ0ienc
    B2∏r = μ0ienc

    B = μ0ienc/2∏r

    At r < a:
    B = 0 because ienc at this point = 0

    At a < r < b:

    B = μ0ienc/2∏r

    I don't know how to get ienc. I know it has something to do with the current density.

    At r > b:

    ienc = itotal, but I would need an expression for the volume of the cylinder.

    V = ∏b2h - ∏a2h
    V = ∏h(b2-a2)
    Ienc = J0V [itex]\Rightarrow[/itex] Ienc = J0∏h(b2-a2)

    B = μ0ienc/2∏r

    B = μ0J0h(b2-a2)/2r

    However, by including h, I am introducing variables that the problem hasn't allowed me to use.
    Last edited: Mar 25, 2013
  2. jcsd
  3. Mar 25, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Current = current density times cross-sectional area. What is the cross-sectional area of your cylinder at a < r < b?
    No. Volume does not enter the picture. Cross-section does.
  4. Mar 25, 2013 #3
    That makes sense.

    So at a < r < b:

    B = μ0J0(r2-a2)/2r

    At r > b:

    B = μ0J0(b2-a2)/2r

    That worked. Thanks for your help.
    Last edited: Mar 25, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted