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1. Homework Statement
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. Homework 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!
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. Homework 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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