Three Magnetically Coupled Coils

From Gauss' Law for Magnetic Fields, we have:

$B_{NM} A = B_{MI} A$

$B_{MI} A = B_{IJ} A$

$B_{NO} A = B_{OK} A$

$B_{OK} A = B_{KJ} A$

$B_{IJ} A + B_{NJ} (2 A) + B_{KJ} A = 0$ [EQUATION 1]

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Thus, we can let:

$B_{1} = B_{NM} = B_{MI} =B_{IJ}$

$B_{2} =B_{NJ}$

$B_{3} = B_{NO} = B_{OK} = B_{KJ}$

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Rewriting and simplifying [EQUATION 1], we get:

$B_{1} + 2 B_{2} + B_{3} = 0$ [EQUATION 2]

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Let $I_{1}$ be the current through coil circle that flows from the marked end to the unmarked end.

Let $I_{2}$ be the current through coil square that flows from the marked end to the unmarked end.

Let $I_{3}$ be the current through coil triangle that flows from the marked end to the unmarked end.

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Applying Ampere's Law to loops $NJIM$ and $OKJN$ :

$N I_{2} - 2 N I_{1} = \frac{B_{2}}{\mu} 4 L - \frac{B_{1}}{\mu} 9 L$ [EQUATION 3]

$\frac{N}{2} I_{3} - N I_{2} = \frac{B_{3}}{\mu} 9 L - \frac{B_{2}}{\mu} 4 L$ [EQUATION 4]

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From [EQUATION 2], [EQUATION 3], and [EQUATION 4], it can be shown that:

$\left[\begin{array}{c} B_{1} \\ B_{2} \\ B_{3} \end{array}\right]$ $=\frac{\mu N}{L}$ $\left[\begin{array}{ccc} \frac{22}{117} & - \frac{1}{13} & - \frac{1}{117} \\ - \frac{1}{13} & \frac{1}{13} & - \frac{1}{52} \\ - \frac{4}{117} & - \frac{1}{13} & \frac{11}{234} \end{array}\right]$ $\left[\begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \end{array}\right]$

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Applying Faraday's Law to each coil:

$V_{1} = \frac{d}{dt} (2 N B_{1} A)$

$V_{2} = \frac{d}{dt} (N B_{2} (2 A))$

$V_{3} = \frac{d}{dt} (\frac{N}{2} B_{3} A)$

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Thus, we get:

$\left[\begin{array}{c} V_{1} \\ V_{2} \\ V_{3} \end{array}\right]$ $=\frac{\mu N^2 A}{L}$ $\left[\begin{array}{ccc} \frac{44}{117} & - \frac{2}{13} & - \frac{2}{117} \\ - \frac{2}{13} & \frac{2}{13} & - \frac{1}{26} \\ - \frac{2}{117} & - \frac{1}{26} & \frac{11}{468} \end{array}\right]$ $\frac{d}{dt}$ $\left[\begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \end{array}\right]$ [EQUATIONS 5]

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The equivalent inductance is defined as:

$V_{terminal} = L_{eq} \frac{d}{dt} I_{terminal}$ [EQUATION 6]

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From Kirchhoff's Loop Rule, we have:

$V_{terminal} = V_{1} = - V_{2} + V_{3}$ [EQUATION 7]

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From Kirchhoff's Junction Rule, we have:

$I_{terminal} = I_{1} - I_{2} = I_{1} + I_{3}$ [EQUATION 8]

$I_{2} = - I_{3}$ [EQUATION 9]

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Applying [EQUATIONS 5] on [EQUATION 7] and noting [EQUATION 9] to relate the currents yields:

$\frac{d}{dt} I_{3} = \frac{112}{55} \frac{d}{dt} I_{1} = - \frac{d}{dt} I_{2}$ [EQUATION 10]

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Applying [EQUATION 10] on [EQUATION 7] yields:

$V_{terminal} = \frac{36}{55} \frac{\mu N^2 A}{L} \frac{d}{dt} I_{1}$ [EQUATION 11]

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Applying [EQUATION 10] on [EQUATION 8] yields:

$\frac{d}{dt} I_{terminal} = \frac{167}{55} \frac{d}{dt} I_{1}$ [EQUATION 12]

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Using [EQUATION 6], [EQUATION 11], and [EQUATION 12]:

$L_{eq} = \frac{36}{167} \frac{\mu N^2 A}{L}$