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Magnetic field in matter.

  1. Oct 2, 2005 #1

    quasar987

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    The problem: In the regions of space where [itex]\vec{J_f} = 0[/itex], the curl of [itex]\vec{H}[/itex] vanishes, and hence we can define a scalar potential [itex]V_m[/itex].

    (a) Show that [itex]V_m[/itex] must be continuous at the boundary of material. - Done

    (b) Consider a very long cylinder of radius a made out of a linear magnetic material of relative permeability [itex]\mu_r[/itex]. The axis of the cylinder is oriented along [itex]\hat{z}[/itex] and the cylinder is emerged in a field [itex]\vec{H}[/itex] that is worth [itex]H_0 \hat{x}[/itex] very far from it. From symetrical considerations, [itex]V_m[/itex] must be of the form

    [tex]V_{m_1}=(As+B/s)cos\phi[/tex]
    [tex]V_{m_2}=Cscos\phi[/tex]

    Where [itex]V_{m1}[/itex] is the potential outside the cylinder and [itex]V_{m2}[/itex] the one inside. Find the value of the constant A, B and C in terms of the other parameters.


    My solution: I used the condition of continuity to find B in terms of C, and I used the condition at infinity to find A = -H_0.

    What is the 3rd condition on V that'll let me find the value of the third constant?
     
    Last edited: Oct 2, 2005
  2. jcsd
  3. Oct 3, 2005 #2

    quasar987

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    Nerver mind, I found it. It was crazy! It had to do with the boundary conditions on H, expressed as the gradient of V and noticing that in the cylinder along phi = pi/2, the H field is purely solenoidal, then so is the magnetization, which implied that the dot product of the gradient of V outside in the limit s-->a with the normal unit vector is 0, which allowed to recover a second relation btw B and C. :surprised

    Second Edit: Wrong again! OMG it doesn't end. Finally, I got the right thing. The third relation was lying not to far below the boundary condition on H perpendicular. Much less complicated than what I previcously thought but also much less fun. :frown:
     
    Last edited: Oct 3, 2005
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