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Magnetic Field Using Amperian Loops inside a wire

  1. Mar 12, 2008 #1
    1. The problem statement, all variables and given/known data
    There is a cross-section wire or radius R with a uniform current density j flowing through it. There is a hole in the wire's cross section (a vacuum) a distance b away from the centre of the cross section. The hole has radius a. What is the magnetic field B in the hole as a function of r?

    2. Relevant equations
    B(integral of ds) = (mu)(current)

    3. The attempt at a solution
    If we draw an amperian loop just outside the hole (so r = b + a) we get:
    B(1) = (mu)jr/2
    and B(2) = (mu)j(a^2)/2r
    and B = B(1) - B(2) = (mu)j(r^2 - a^2)/2r

    If we do the same just before the hole (so r = b - a) we get:

    B = (mu)jr/2 [because the hole does not affect the magnetic field in this case)

    But how can we find B in other places inside the hole (in the middle of the hole B = (mu)jb/2 ). My prof says that the magnetic field is uniformly distributed inside the hole is the same (or maybe he said something different but similar), but how can this be? As far as I understand, we can get the B in the hole by visualizing a current in the wire without the hole and then adding a current in the opposite direction (the question says we can assume it's the same density) in order to form the hole. From my understanding, the B of the wire without the hole gets bigger as we get further from the center, right? But if this is true, then wouldn't B be bigger when r = b + a than when r = b, and even bigger than when r = b - a? Can someone please explain this to me? How is B distributed inside the hole? Thx.

    Also, if we were to try to find B inside the hole (besides the centre, or r=b+a, or r=b-a)
    wouldn't we need to find the area of the hole thats beneath r? So if r was just a little bigger than b, wouldn't the area be just about half of (pi)a^2 ?
  2. jcsd
  3. Mar 12, 2008 #2

    Doc Al

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    Staff: Mentor

    Treat the field as the vector sum of the fields of: (1) A uniform current of radius R, and (2) A uniform current (oppositely directed) of radius a. Use Ampere's law to solve for the field of each as a function of distance from their respective centers. Note that (1) is centered at r = 0, while (2) is centered at the center of the hole.
  4. Mar 12, 2008 #3
    What if r is inside the hole though? Wouldn't i need to calculate the B due to a uniform current of radius r, and then subtract the B due to the current of part of the hole? I know B(1) will be (mu)jr/2, but wouldn't B(2) be (mu)j(area of part of hole)/2(pi)r ?

    Did u read my long question above? Maybe it didn't make sense but I already stated what you told me to do.
  5. Mar 12, 2008 #4

    Doc Al

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    Staff: Mentor

    What about it?
    Not sure I understand what you're saying here. To repeat what I already said, the field at any point, including inside the hole, is the vector sum of the two field contributions already defined.

    I read your post and question about how the field is distributed in the hole. And I thought I answered it accurately, but maybe I misunderstood you. :wink:
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