(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Self inductance Griffiths problem

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