Cube Of Magnetically Coupled Coils

The vertices of the cube are labeled according to this figure:

Symmetry

The system exhibits a symmetry across the plane $1207$ . Also, the upper half and the lower half of the cube are symmetric.

Gauss Law for Magnetic Fields

Knowing that the total magnetic flux entering each vertex is zero, and combining that with the symmetries of the problem:

$\mu A H_{25} = \mu A H_{23} = \mu A H_{61} = \mu A H_{41} = \dfrac{\mu A H_{12}}{2}$

$\mu A H_{50} = \mu A H_{30} = \mu A H_{76} = \mu A H_{74} = \dfrac{\mu A H_{07}}{2}$

$\mu A H_{56} = \mu A H_{34} = \dfrac{\mu A (H_x - H_y)}{2}$

where $H_x = H_{12}$ and $H_y = H_{07}$ .

Kirchoff's Current Law

Analogously, we can write the following equations for the currents:

$I_{25} = I_{23} = I_{61} = I_{41} = \dfrac{I_{12}}{2}$

$I_{50} = I_{30} = I_{76} = I_{74} = \dfrac{I_{07}}{2}$

$I_{56} = I_{34} = \dfrac{I_x - I_y}{2}$

where $I_x = I_{12}$ and $I_y = I_{07}$ .

Figure 2 and Ampere's Law

Suppose that the left end of the arm in Figure 2 is labeled as $J$ and the right end $K$ . Suppose that as part of our Amperean loop, we traverse this arm from the left end to the right end. Thus, in our Ampere's Law equation, we add $+ s H_{JK}$ to the left side of the equation, and $+ N I_{JK}$ to the right side of the equation.

Ampere's Law

We apply Ampere's Law to loops $1256$ and $7056$ :

$s \left( \dfrac{H_x + H_x}{2} + \dfrac{H_x - H_y}{2} + \dfrac{H_x}{2}\right) = N \left( \dfrac{I_x + I_x}{2} + \dfrac{I_x - I_y}{2} + \dfrac{I_x}{2}\right)$

$s \left( \dfrac{H_y + H_y}{2} + \dfrac{H_y - H_x}{2} + \dfrac{Hy}{2} \right) = N \left( \dfrac{I_y + I_y}{2} + \dfrac{I_y - I_x}{2} + \dfrac{I_y}{2}\right)$

Solving for $H_x$ and $H_y$ in these two equations yields $H_x = \dfrac{N I_x}{s}$ and $H_y = \dfrac{N I_y}{s}$ .

Self Inductances and Mutual Inductances

Notice the parallelism between the equations derived from Gauss Law and the equations derived from KCL. Using the equations derived from Ampere's Law, it can be easily shown that, for all arms of the cubic frame, the magnetic field $H$ of an arm and its corresponding coil current $I$ are related by $H = \frac{NI}{s}$ . Notice that the magnetic field in an arm depends only on its corresponding coil current and does not depend on other coil currents. This means that we can effectively simplify the circuit of Figure 3 with all coils having the same effective self inductance and without worrying about mutual inductance.

Magnetic Flux Linkage and Equivalent Inductance

The magnetic flux linkage through a coil is $N A \mu \frac{NI}{s}$ . Thus, the effective self inductance of each coil is $\frac{u N^2 A}{s}$ . Simplifying the circuit of Figure 3:

$\frac{12}{5} \frac{\mu N^2 A}{s}$