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Three-coil core, flux in air gap

Short answer

Net mmf is 1136 AT (assuming two coils aid, one opposes). Iron reluctance = 1.414×10⁶ AT/Wb, gap reluctance = 9.947×10⁶ AT/Wb, total = 1.136×10⁷ AT/Wb. Flux in gap = 1136 / 1.136×10⁷ = 1.0×10⁻⁴ Wb (0.1 mWb).

A rectangular iron core is shown in figure 1. It has a mean length of magnetic path of 100 cm, cross-section of (2 cm ×\times 2 cm), relative permeability of 1400 and an air-gap of 5 mm cut in the core. The three coils carried by the core have number of turns Na=335N_a = 335, Nb=600N_b = 600 and Nc=600N_c = 600; and the respective currents are 1.6 A, 4 A and 3 A. The directions of the currents are as shown in the figure. Find the flux in the air-gap. [Figure: rectangular iron core with an air gap in the top limb; coil NaN_a with current IaI_a on the left limb, coil NbN_b with current IbI_b on the right limb and coil NcN_c with current IcI_c on the bottom limb; the current directions are drawn in the figure and are not legible in the scan]

The flux in the air gap is the net mmf of the three coils divided by the total reluctance of the core and gap.

Assumption: the current directions in the figure are not legible. I assume coils aa and bb magnetise in the same sense and coil cc opposes them. If the figure shows a different arrangement, only the net mmf changes and the method stays the same.

Dimensions: lg=5l_g = 5 mm, li=100−0.5=99.5l_i = 100 - 0.5 = 99.5 cm, A=2×2=4A = 2\times2 = 4 cm2^2 =4×10−4= 4\times10^{-4} m2^2, μr=1400\mu_r = 1400.

Net mmf

Fa=335×1.6=536 ATFb=600×4=2400 ATFc=600×3=1800 ATFnet=536+2400−1800=1136 AT\begin{aligned} F_a &= 335\times1.6 = 536\ \text{AT} \\ F_b &= 600\times4 = 2400\ \text{AT} \\ F_c &= 600\times3 = 1800\ \text{AT} \\ F_{net} &= 536 + 2400 - 1800 = 1136\ \text{AT} \end{aligned}

Reluctance

Si=0.9954π×10−7×1400×4×10−4=1.414×106 AT/WbSg=0.0054π×10−7×4×10−4=9.947×106 AT/WbS=Si+Sg=1.1361×107 AT/Wb\begin{aligned} S_i &= \frac{0.995}{4\pi\times10^{-7}\times1400\times4\times10^{-4}} = 1.414\times10^{6}\ \text{AT/Wb} \\ S_g &= \frac{0.005}{4\pi\times10^{-7}\times4\times10^{-4}} = 9.947\times10^{6}\ \text{AT/Wb} \\ S &= S_i + S_g = 1.1361\times10^{7}\ \text{AT/Wb} \end{aligned}

Flux

Φ=FnetS=11361.1361×107=1.0×10−4 Wb\Phi = \frac{F_{net}}{S} = \frac{1136}{1.1361\times10^{7}} = 1.0\times10^{-4}\ \text{Wb}

The flux density in the gap is B=Φ/A=0.25B = \Phi/A = 0.25 T.

Answer: Φgap≈0.1\Phi_{gap} \approx 0.1 mWb (1.0×10−41.0\times10^{-4} Wb)

✓ Answer checked by Claude Opus 5.5 (high effort)

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