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Homework Help: Bound currents problem

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A coaxial cable has a linear insulating material of magnetic susceptibility [tex]\chi_m[/tex] separating the conductors. A current I flows down the inner conductor and returns along the outer one. Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents and confirm that they generate the correct field

    2. Relevant equations

    3. The attempt at a solution
    I don't have agreement between the two methods. I think my result for the first part is probably right, so it must be the second part that's wrong.

    Firstly, I have
    [tex]\oint \textbf{H} \cdot d \textbf{L} = I[/tex]
    [tex]\Rightarrow 2 \pi r H = I[/tex]
    [tex]\Rightarrow H = \frac{I}{2 \pi r} [/tex]
    [tex]\Rightarrow B = \frac{\mu I}{2 \pi r} [/tex]
    [tex]=(\chi_m + 1) \frac{\mu_0 I}{2 \pi r} [/tex]

    And for the second part I have
    [tex]\textbf{M}=\chi_m \textbf{H}[/tex]
    [tex]=\chi_m \frac{I}{2 \pi r} \hat{\textbf{\phi}}[/tex]

    [tex]\textbf{j_{bound}} =\nabla \cdot \textbf{M}[/tex]

    [tex]\textbf{k_{bound}} =\textbf{M}\times \hat{\textbf{n}} [/tex]
    [tex]=\textbf{M}\times \hat{\textbf{r}}[/tex]
    [tex]=-\chi_m \frac{I}{2 \pi r} \hat{\textbf{z}}[/tex]

    [tex]\oint \textbf{B} \cdot d \textbf{L} = \mu_0 I (1+\textbf{k}_{bound})[/tex]
    [tex]\Rightarrow B =\mu_0 \frac{I}{2 \pi r} (1-\frac{\chi_m}{2 \pi r} )[/tex]

    Now clearly the bound current should be in the same direction as the current I, so I have messed up the cross product(EDIT: No, wait. It should be in the opposite direction to I, so I had it right the first time...not sure what I've done wrong then). The main thing though is the division of chi by 2 pi r, as I can't see how to get rid of that. Any help here would be appreciated, thanks :) (EDIT2: I just realised that I've neglected a factor of 1/r in computing the cross product, so there should be r^2 in the denominator of k_bound instead of r, but that doesn't solve my problem.)

    P.S Sorry about the weird tex, I'm not sure what I've done. It's meant to read j_bound=div(M) and k_bound=M X n. The rest looks right.

    (EDIT2: no wonder I'm having so much trouble with this problem, I've made so many mistakes. I just realised that the bound current is A/m, and I've been treating it as though it is just A. I also realised that I could r=r_1 for the surface current because we are only dealing with a current that resides at r_1. I now have this
    [tex]B=I/2\pi r(1-\chi_m/r_1)[/tex]
    It's looking much better, but still not right. I'm still hoping someone can help me with this.
    Last edited: Apr 16, 2009
  2. jcsd
  3. Apr 16, 2009 #2
    Okay, I figured it out. Although I'm still not sure if I am using the cross product properly. If anyone wants to tell me how the cross product works in different coordinate systems, I would still love to hear. Scroll to the bottom for an explanation.

    My direction for the bound current was wrong because I was taking n=r, but a the centre the normal points in the NEGATIVE r direction. I could find very little in the way of explanations for how to cross two vectors in cylindrical coordinates. I now think this is because it is exactly the same as in cartesian coordinates. One source, I think a reply to a thread on here, said to use the formula for the curl of a vector in cyl. coordinates, but replace del with the other vector. That is what I was doing before, but it can't be right because then swapping the order of vectors does not just make it negative. It was this realisation that had me try just using the formula for cross product that works in cartesian coordinates, and I came up with the right answer, so I think that must be right.
  4. Apr 16, 2009 #3
    Sorry for bumping, I meant to edit, but accidentally replied instead. Upshot is, I solved the problem but still am not entirely sure of an aspect of the use of cross product in cylindrical coordinates. Look at the reply for an elaboration. Thanks.

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