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Self inductance Griffiths problem

  1. Dec 25, 2006 #1
    1. The problem statement, all variables and given/known data
    Griffiths problem 7.23
    Copute the self inducatnce of the hairpin loop shown in the figure. (neglect contribution from ends since mosto f the flux comes form the straight section) To get a definite answer , assume the wire has a tiny radius epsilon, and ignore any flux through the wire itself

    2. Relevant equations
    [tex] \Phi = LI [/tex]
    L = self inductance and I is the current through the loop
    induced emf is
    [tex] \epsilon = - L \frac{dI}{dt} [/tex]

    3. The attempt at a solution
    ok the flux due to straight part on top is (and its legnth is l)
    [tex] \Phi = \frac{\mu_{0} Il}{2 \pi} \int_{\epsilon}^{d} \frac{dr}{r} = \frac{\mu_{0} Il}{2 \pi} \ln \frac{d}{\epsilon} [/tex]

    due to the bottom wire has the same value
    so the total flux is
    [tex] \Phi = \frac{\mu_{0} Il}{\pi} \ln \frac{d}{\epsilon} [/tex]

    now the self indutance is then
    [tex] L= \frac{\mu_{0} l}{2 \pi} \ln \frac{d}{\epsilon} [/tex]

    is this correct??

    Thanks for the help!

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  3. Dec 26, 2006 #2
    Almost. You have a magical 1/2 that shows up in your expression for L that shouldn't be there.
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