(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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Self inductance Griffiths problem

**Physics Forums | Science Articles, Homework Help, Discussion**