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Free Current

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data
    I am looking at an example problem in the text and they skipped some steps. I think I am missing somthing obvious but none-the-less I don't know what is going on.

    We have a long copper rod o radius R which carries a uniformly distributed (free) current I. Find H, the auxillary field inside and outside the rod.


    2. Relevant equations

    [itex]\oint H dot dl[/itex] = Free current enclosed

    3. The attempt at a solution

    so the LHS would be (magnitude only) H*2pis for some amperian loop. Now I don't know how to get the Free current enclosed inside and outside the rod.

    Thanks for your help!
     
  2. jcsd
  3. Dec 14, 2013 #2
    The Amperian loops are always closed curves (that's because the line integral at LHS is closed), so they are always the circumference of a surface. In our case this is the surface of a circle with radius [itex]\displaystyle{s}[/itex]. The current at RHS is the total current that goes through this surface. When you are outside of the cylinder it's easy to find this current. When you are inside, you just have to take into account that the current is uniformly distributed.
     
  4. Dec 14, 2013 #3
    I mostly understand that. The answer the book gives for I free is = (I*pi*s^2)/(pi*R^2) which makes sense, but I am looking for what definition they used exactly to get that. What equation?
     
  5. Dec 14, 2013 #4
    You can get this by using the current density vector [itex]\displaystyle{\vec{J}}[/itex]. This is the current that flows through a unit area of the wire crossection (or for uniform distributions the ratio current/area). For this case the current flows only in a specific direction, so it's not necessary to use vectors. We can just say that:
    [tex]\displaystyle{J=\frac{I}{\pi R^2}}[/tex]
    because the current is uniformly distributed. The current we need is given by a surface integral:
    [tex]\displaystyle{I_{f,enc}=\int _{S}JdS}[/tex]
    where [itex]\displaystyle{S}[/itex] is the circle's surface. Actually the integral is an overkill, because [itex]\displaystyle{J}[/itex] is constant here. If you have seen any applications of Gauss's law, it's like using the charge density in electrostatics.
     
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