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Mutual inductance of a long wire and a triangular loop?

  1. Apr 13, 2016 #1

    I originally posted this to the advanced homework -section, but maybe this isn't advanced enough to warrant being posted there :oops:. If a mod sees this, feel free to delete the original on the other forum. I'd do it myself, but I can't.


    1. The problem statement, all variables and given/known data



    Calculate the mutual inductance for the long, straight conductor and a conducting loop shaped like an isosceles triangle in the picture below, when a = 10cm.
    H11_5.JPG

    2. Relevant equations

    Mutual inductance: [itex]M = \frac{N_1\phi _1}{I_2} = \frac{N_2\phi _2}{I_1}[/itex]

    Magnetic flux: [itex]\phi = BA[/itex], where B is the magnetic flux density through a loop and A the area of the loop.

    Magnetic flux density for a long straight wire(Ampere): [itex]B=\mu H = \frac{\mu I}{2 \pi R}[/itex], where R is the orthogonal distance from the wire.

    3. The attempt at a solution

    In order to calculate M, I need to find out the total flux through the triangular loop. We already know the expression for B, however dA is a bit of a mystery.

    Looking at the picture, if at a distance r from the vertex of the triangle we mark:
    \begin{cases}
    \text{Height of top half of triangle} = y(r)\\
    \text{Distance from the vertex} = r
    \end{cases}
    Then by similarity:
    [tex]
    \frac{1/2 a}{a} = \frac{y(r)}{r} \iff y(r) = \frac{r}{2}
    [/tex]
    and if we slice the triangle into small slices with height 2y and width dr:
    [tex]
    dA=2ydr = r dr
    [/tex]
    Now [tex]d\phi = d(B(r)A(r)) = B(r)dA(r)[/tex]

    Therefore
    \begin{array}{ll}
    d\phi
    &= \frac{\mu I r dr}{2 \pi (a+r)}\\
    &= \frac{\mu I r dr}{2r \pi (\frac{a}{r}+1)}\\
    &= \frac{\mu I dr}{2 \pi (\frac{a}{r}+1)}\\
    &= \frac{\mu I dr}{2 \pi (\frac{a}{r}+1)}\\
    &= \frac{\mu I}{2 \pi }\frac{dr}{(\frac{a}{r}+1)}
    \end{array}
    Looking at this it doesn't seem to be possible to integrate this expression with respect to r from 0 to a. My guess is my derivation of dA is wrong. Any help on that front?
     
    Last edited: Apr 13, 2016
  2. jcsd
  3. Apr 13, 2016 #2
    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!
     
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