Exercise 4.3.4: Two coils on a three-limb core (linear)

${\mu }_0{\mu }_{r}S=1.2566\times 10^{-6}\times 1500\times 3\times 10^{-4}=5.655\times 10^{-7}\mathrm{H}\mathrm{m}$, so

Take ${\Phi }_1$ and ${\Phi }_3$ upward in the outer limbs and ${\Phi }_2$ downward in the centre limb. At node A, ${\Phi }_2={\Phi }_1+{\Phi }_3$. The mmfs are $F_1=N_1{I}_1=600\mathrm{At}$ and $F_2=N_2{I}_2=200\mathrm{At}$. Loop 1 (left limb and centre limb) and loop 2 (right limb and centre limb):

In numbers ($\times 10^5{\mathrm{H}}^{-1}$): $7.07{\Phi }_1+1.77{\Phi }_3=600$ and $1.77{\Phi }_1+7.07{\Phi }_3=200$. Subtracting the second from the first, $5.30({\Phi }_1-{\Phi }_3)=400$, so ${\Phi }_1-{\Phi }_3=7.54\times 10^{-4}\mathrm{Wb}$. Adding them, $8.84({\Phi }_1+{\Phi }_3)=800$, so ${\Phi }_1+{\Phi }_3=9.05\times 10^{-4}\mathrm{Wb}$. Hence

Loop 1: $5.31\times 0.83+1.77\times 0.905=4.40+1.60=6.00$, i.e. $600\mathrm{At}$ (units $10^5{\mathrm{H}}^{-1}\times \mathrm{mWb}$). Loop 2: $5.31\times 0.075+1.60=2.00$, i.e. $200\mathrm{At}$. The right coil barely drives any flux through its own limb: its $200\mathrm{At}$ are almost cancelled by the potential difference $U_{AB}=Z_2{\Phi }_2=160\mathrm{At}$ that coil 1 imposes across the centre limb. Had $I_2$ been $0.75\mathrm{A}$ ($F_2=150\mathrm{At}$), ${\Phi }_3$ would have been zero, the magnetic version of a bridge balance. The result $B_2=3\mathrm{T}$ is unphysical for real iron ($B_s\approx 2\mathrm{T}$): the linear model happily gives it, which is the warning of Section 4.3.6. With a real curve the centre limb would saturate and the fluxes would be redistributed.