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Air gap flux density in core with three windings

Short answer

Treating the core as a magnetic circuit with node potential, and assuming all MMFs aid flux in the centre limb, the centre limb flux is 4.648 mWb. With area 20 cm², the air‑gap flux density is 2.32 T.

The magnetic core with three excitation windings is shown in figure below. The mean length lA, l1 and l2 is 50 cm, 20 cm and 10 cm respectively. The length of air gap g is 10 mm. The number of turns N1 and N is 1000 and 500. If iA, iB and i1 are 20, 15 and 10 amperes, what would be the magnetic flux density at the air gap. The core area Ac = 20 cm² and relative permeability of core is μr = 4000. [Figure: rectangular core with a centre limb; the centre limb carries the N1-turn winding (current i1) and contains the air gap g; the left and right outer limbs each carry an N-turn winding with currents iA and iB respectively; lA is the outer path length, l1 and l2 are the centre-limb lengths above and below.]

Treat the core as a two-loop magnetic circuit. Assumption (polarities are not given): all three windings are connected so that their mmfs drive flux in the same sense through the centre limb, i.e. the outer coils push flux up the outer limbs and the centre coil pushes it down the centre limb. Each outer path has length lA=50l_A = 50 cm; the centre limb has iron l1+l2=30l_1 + l_2 = 30 cm plus the 10 mm gap. Fringing is neglected.

   +----------+----------+
   |  phiA ^  |  phic v  |  phiB ^
 [N iA]       [N1 i1]  [N iB]
   |          |   g      |
   +----------+----------+

MMFs

FA=500×20=10000 AT,FB=500×15=7500 AT,F1=1000×10=10000 ATF_A = 500\times20 = 10000\ \text{AT}, \quad F_B = 500\times15 = 7500\ \text{AT}, \quad F_1 = 1000\times10 = 10000\ \text{AT}

Reluctances (Ac=20×10−4A_c = 20\times10^{-4} m², μr=4000\mu_r = 4000)

SA=SB=0.54π×10−7×4000×20×10−4=4.974×104 AT/WbScore,c=0.34π×10−7×4000×20×10−4=2.984×104 AT/WbSg=0.014π×10−7×20×10−4=3.979×106 AT/WbSc=Score,c+Sg=4.009×106 AT/Wb\begin{aligned} S_A = S_B &= \frac{0.5}{4\pi\times10^{-7}\times4000\times20\times10^{-4}} = 4.974\times10^{4}\ \text{AT/Wb} \\ S_{core,c} &= \frac{0.3}{4\pi\times10^{-7}\times4000\times20\times10^{-4}} = 2.984\times10^{4}\ \text{AT/Wb} \\ S_g &= \frac{0.01}{4\pi\times10^{-7}\times20\times10^{-4}} = 3.979\times10^{6}\ \text{AT/Wb} \\ S_c &= S_{core,c} + S_g = 4.009\times10^{6}\ \text{AT/Wb} \end{aligned}

Node equation

Let UU be the magnetic potential of the top junction relative to the bottom. Then ϕA=(FA−U)/SA\phi_A = (F_A - U)/S_A, ϕB=(FB−U)/SB\phi_B = (F_B - U)/S_B, ϕc=(U+F1)/Sc\phi_c = (U + F_1)/S_c, and ϕc=ϕA+ϕB\phi_c = \phi_A + \phi_B:

U=FA/SA+FB/SB−F1/Sc1/SA+1/SB+1/Sc=8634 ATϕc=8634+100004.009×106=4.648×10−3 Wb\begin{aligned} U &= \frac{F_A/S_A + F_B/S_B - F_1/S_c}{1/S_A + 1/S_B + 1/S_c} = 8634\ \text{AT} \\ \phi_c &= \frac{8634 + 10000}{4.009\times10^{6}} = 4.648\times10^{-3}\ \text{Wb} \end{aligned}

(The outer limbs carry ϕA=27.46\phi_A = 27.46 mWb up and ϕB=−22.81\phi_B = -22.81 mWb, i.e. 22.81 mWb down; their sum is 4.65 mWb.)

Flux density in the air gap

Bg=ϕcAc=4.648×10−320×10−4=2.32 TB_g = \frac{\phi_c}{A_c} = \frac{4.648\times10^{-3}}{20\times10^{-4}} = 2.32\ \text{T}

Answer: Bg≈2.32B_g \approx 2.32 T (for the assumed winding polarities, linear core, no fringing). Because the gap reluctance is about 80 times that of the iron, Bg≈μ0(Fˉouter+F1)/gB_g \approx \mu_0(\bar F_{outer} + F_1)/g, so the result is mainly set by the gap.

✓ Answer checked by Claude Opus 5.5 (high effort)

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