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Jackson Eq. 5.33 (3rd Ed.)

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    I cant seem to figure out how he writes down this equation. Specifically:
    a. Isn't Theta' = 90 degrees. Then why doesn't he write it out explicitly.
    b. Whats the use of adding the Sin(Theta') if he is going to use a delta function using the Cos
    c. What is the radius 'a' doing there. If the wire has no thickness, shouldn't the magnitude of J equal I.
    d. Shouldn't J=I/Area^2 in terms of units.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 8, 2008 #2

    Ben Niehoff

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    Gold Member

    Because it's more convenient to write the measure as

    [tex]\delta(\cos \theta') d(\cos \theta')[/tex]

    rather than

    [tex]\sin \theta' \delta(\theta') d\theta'[/tex]

    It saves a lot of time, trust me.

    The current loop is in the x-y plane, where

    [tex]\sin \theta' = 1[/tex]

    which is the same as

    [tex]\cos \theta' = 0[/tex]

    I suppose he could have technically left out the sin(theta) and just put in 1 (just as he puts in 1/a instead of 1/r). However, perhaps he put it in so that the formula would be easier to generalize.

    No. The integral of J over its cross-sectional area should equal I. That is:

    [tex]I = \iint_R r \; dr \; d(\cos \theta) \; I \delta(\cos \theta) \frac{\delta(r-a)}{a}[/tex]

    where R is some vertical region that cuts across part of the wire. So you see, the 1/a is essential to give the correct value of I.

    Yes, and it is. Or rather, J = I/Area. Remember that delta functions carry reciprocal units. So you get 1/L from the factor of 1/a, and another 1/L from the factor of [itex]\delta(r' - a)[/itex].
     
  4. Dec 8, 2008 #3
    Thanks a lot :)
     
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