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Magnetic field rod

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data
    i have a rod of 35 cm and a current of 10 A. the rod is a paramagnetic aluminum rod. how do i find the magnetic field inside the rod?


    2. Relevant equations
    the fact that the rod is paramagnetic changes anything?


    3. The attempt at a solution
     
  2. jcsd
  3. Nov 3, 2009 #2
    Yes. You now have a magnetization inside the rod that prohibits use of the biot-savart law there.

    Also, I'm not sure what you mean by "35 cm" - are you refering to the length? If so, you should be very careful in applying ampéres law as you do not have a perfect cylindrical symmetry.

    Please clarify
     
  4. Nov 3, 2009 #3
    sorry it is the diameter. and the rod is consider to to infinitely long.

    so what should I use?
     
  5. Nov 3, 2009 #4
    Okay - that eases thing up a bit :)

    Use Ampéres law for the H-field:

    [tex]\oint_{\partial \mathcal{S}}\vec H \cdot d\vec \ell = \int_{\mathcal{S}}\vec J_\mathrm{free}\cdot d\vec a[/tex]

    For points outside the wire, this of course reduces to the familiar form:

    [tex]\oint_{\partial \mathcal{S}}\vec H \cdot d\vec \ell = I_\mathrm{free,encl}[/tex]

    but for points inside the wire, you must be careful to only include the part of the current that is enclosed by the loop - assume that the current density is uniform, if not stated otherwise.

    THen you should find:

    [tex]H\cdot2\pi r = J_0 \pi r^2 \quad\Rightarrow\quad H=\frac{J_0 r}{2}[/tex] for r < 17.5 cm

    Where J0 is the totalt current divided by the crossectional area of the wire

    (take a moment to verify this)

    Now use

    [tex]\vec B = \mu_0(\vec M + \vec H)[/tex]

    together with

    [tex]\vec M = \chi \vec H[/tex]

    to find the required result
     
    Last edited: Nov 3, 2009
  6. Nov 3, 2009 #5
    so once i find the H which is the magnetic field i just plugged in the
    [tex]
    \vec B = \mu_0(\vec M + \vec H)
    [/tex]
    along with
    [tex]
    \vec M = \chi \vec H
    [/tex]

    so is [tex]
    J_0
    [/tex]
    the current they gave me?
     
  7. Nov 3, 2009 #6
    As I stated, it is the current density which is the total current divided by the cross-sectional area of the wire. THat is:

    [tex]
    J_0=\frac{10\,\mathrm{A}}{\pi(0.35 \,\mathrm{m})^2}
    [/tex]
     
  8. Nov 5, 2009 #7
    how will I find the bound current density Jb and Kb.


    Jb= curl M

    Kb= cross product of M and n.

    is this right and how do I do the curl?
     
  9. Nov 5, 2009 #8
    It's right - so just plug-in the magnetization you got from [tex]\vec M = \chi \vec H[/tex]
     
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