Is the Self Inductance of the Hairpin Loop Calculated Correctly?

[stunner5000pt]

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

Homework Equations

\Phi = LI
L = self inductance and I is the current through the loop
induced emf is
\epsilon = - L \frac{dI}{dt}

The Attempt at a Solution

ok the flux due to straight part on top is (and its legnth is l)
\Phi = \frac{\mu_{0} Il}{2 \pi} \int_{\epsilon}^{d} \frac{dr}{r} = \frac{\mu_{0} Il}{2 \pi} \ln \frac{d}{\epsilon}

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

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

is this correct??

Thanks for the help!

 

[esun]

Almost. You have a magical 1/2 that shows up in your expression for L that shouldn't be there.