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Mutual inductance of a long wire and a triangular loop?
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[QUOTE="TheSodesa, post: 5442117, member: 512094"] Let's try this again: [tex]d\phi = \frac{\mu I}{2 \pi }\frac{dr}{(\frac{a}{r}+1)}[/tex] Then [tex] \phi = \int_{0}^{a}\frac{\mu I}{2 \pi }\frac{dr}{(\frac{a}{r}+1)} [/tex] [tex] \phi = \frac{\mu I}{2 \pi } \int_{0}^{a}\frac{dr}{(\frac{a}{r}+1)} [/tex] [tex] \phi = \frac{\mu I}{2 \pi } \int_{0}^{a}\frac{r}{(a+r)}dr [/tex] Now let's do long division on our rational expression (off screen, 'cause duck doing that in LaTeX): [tex] \frac{r}{(a+r)} = 1 + \frac{-a}{r+a} = 1 - \frac{a}{r+a} [/tex] This looks like something we can work with, so let's replace the integrand with our quotient: [tex] \phi = \frac{\mu I}{2 \pi } \int_{0}^{a}\frac{r}{(a+r)}dr [/tex] [tex] \phi = \frac{\mu I}{2 \pi } \int_{0}^{a}1-\frac{a}{r+a}dr [/tex] [tex] \phi = \frac{\mu I}{2 \pi } [r-a \ ln(r+a)]_{0}^{a} [/tex] [tex] \phi = \frac{\mu I}{2 \pi } (a-a \ ln(2a) - 0 + a \ ln(a)) [/tex] [tex] \phi = \frac{\mu aI}{2 \pi } (1-ln(2a) - 0 + ln(a)) [/tex] [tex] \phi = \frac{\mu aI}{2 \pi } (1+ln(\frac{a}{2a})) [/tex] Then by substituting this for ##\phi \ in \ M##: [tex] M = \frac{\phi}{I} = \frac{\frac{\mu aI}{2 \pi } (1+ln(\frac{1}{2}))}{I} [/tex] [tex] M = \frac{\mu a}{2 \pi } (1+ln(\frac{1}{2})) [/tex] and by plugging in ##a = 10cm## and ##\mu = 4\pi*10^{-7}NA^{-2}## : [tex] M = \frac{(4\pi*10^{-7}NA^{-2})(10cm)}{2 \pi } (1+ln(\frac{1}{2})) = 6.13706... × 10^-9 \frac{Nm}{A^2} [/tex] [tex] M \approx 6.1 nH [/tex] Booyah! [/QUOTE]
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Mutual inductance of a long wire and a triangular loop?
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