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Homework Help: Field of long cylindrical shell conductor

  1. Mar 18, 2008 #1
    [SOLVED] Field of long cylindrical shell conductor

    1. The problem statement, all variables and given/known data
    Shown above is the cross-section of a long cylindrical shell conductor of inner radius a=4.00 cm and outer radius b=7.00 cm which carries a current into the page.
    The current density J (current/area) is uniform across the shell from r=a to r=b and has the magnitude J=3.6E3 A/m2 where r is the distance from the axis of the shell. (The magnitude of the total current carried by the conductor is thus given by I=J*π*(b2-a2)=37.3 A.)

    (a) Find the magnitude of the magnetic field at r=2b=14 cm.

    (b) Find the magnitude of the magnetic field at r=(a+b)/2=5.5 cm (halfway between the two shells).

    (Hint: how much current is enclosed by the Amperian loop in each case?)


    2. Relevant equations

    B = (uIr)/(2pi*R^2) (inside conductor, r<R)
    B = (uI)/2pi*r (outside conductor r>R)

    where u = 4*pi*10^-7

    3. The attempt at a solution

    I already solved for part a, getting 5.33E-5 T using the second relevant equation.

    But when I used the first relevant equation for part 1, I got 8.4 E-5 which I found out isn't correct. Here's how I did it:

    (u*.055*37.3)/(2*pi*(.07^2)) = 8.4E-5 T.

    What am I doing wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 18, 2008 #2
    So I don't have a clue what you are plugging in. Most of the people here will be able to follow algebra much easier than seemingly random numbers, just for a future pointer. Anyway, how would you rewrite that first equation in terms of the current density, which is given to you?
     
  4. Mar 18, 2008 #3
    Thanks for your help.

    Here's what I got:

    Since

    [tex]J=\frac{I}{\pi*R^{2}}[/tex]

    Then
    [tex]B=\frac{\mu_{0}Ir}{2\pi*R^{2}}[/tex]

    becomes

    [tex]B=J*\frac{\mu_{0}*r}{2}[/tex]

    and so

    [tex]J=\frac{2B}{\mu_0*r}[/tex]
     
    Last edited: Mar 18, 2008
  5. Mar 22, 2008 #4
    Sorry for the bump. I still haven't figured out the problem. Is there anyone else that could help?
     
  6. Mar 22, 2008 #5
    Sorry about leaving you hanging, didn't realize you responded. That should be right.

    Inside:
    [tex]\mathbf{B}(r) = \frac{\mu_0 J r}{2} \hat{r}[/tex]

    Did you get your units right?
     
    Last edited: Mar 22, 2008
  7. Mar 23, 2008 #6
    Hm, I'm still not getting the right answer. Just to let you know, the answer I should be getting is 5.86E-5 T.

    Using the formula you gave me, I did as followed:

    1) For [tex]r[/tex] I used 5.5 cm = .055 m

    2) and then I plugged in 3.6E3 for J (since it was given in the problem).

    3)[tex]\mu_{0}=4\pi*10^{-7}[/tex]

    [tex]\frac{\mu_{0}*3.6*10^{3}*.055}{2}=1.244*10^{-4} T[/tex]

    Am I using the wrong numbers?
     
  8. Mar 24, 2008 #7
    I figured it out. I had been using the incorrect value for J.
     
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