Magnetic field in matter.

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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?
 
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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:
 
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