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Three-limb core, ampere-turns for flux

Short answer

Assuming 5 cm core depth, central limb reluctance = 1.061×10⁴ AT/Wb, each side path reluctance = 5.57×10⁴ AT/Wb. Flux divides equally. NI = Φ S_c + (Φ/2) S_side = 0.75×1.061e4 + 0.375×5.57e4 = 2.88×10⁴ AT for 0.75 Wb; if flux is 0.75 mWb, NI ≈ 28.8 AT.

For the magnetic circuit shown below, calculate the Amp-turn (NI) required to establish a flux of 0.75 Wb in the central limb. Given that μr=4000\mu_r = 4000 for iron core. [Figure: double-window (three-limb) iron core, overall width 44 cm and height 20 cm, with 4 cm thick top and bottom yokes; bottom dimensions from left to right 4 cm, 15 cm, 6 cm, 15 cm, 4 cm (outer limbs 4 cm, central limb 6 cm, windows 15 cm); a coil of N turns carrying current I is on the central limb; three hatched regions at the bottom of the limbs [?]]

The coil is on the central limb, so the central flux divides equally between the two outer limbs, which are identical. This gives a parallel-series circuit, solved with reluctances.

Assumptions (the figure is incomplete):

  • The core depth (thickness) is not shown, so I take d=5d = 5 cm throughout. The result is inversely proportional to dd.
  • Mean paths are taken through the centre of each member. Central limb and each outer limb: 20−4=1620 - 4 = 16 cm. Top or bottom yoke between centre limb and outer limb: 15+3+2=2015 + 3 + 2 = 20 cm.
  • The flux is taken as given, 0.75 Wb. The answer is proportional to flux, so for 0.75 mWb divide by 1000.
   +---+---------------+------+---------------+---+
   |   top yoke 4 cm                              |
   |  outer      window    centre    window outer |
   |  4 cm       15 cm     6 cm      15 cm   4 cm |
   |   bottom yoke 4 cm                           |
   +----------------------------------------------+

Areas

Ac=0.06×0.05=3×10−3 m2,Ao=Ay=0.04×0.05=2×10−3 m2A_c = 0.06\times0.05 = 3\times10^{-3}\ \text{m}^2,\qquad A_o = A_y = 0.04\times0.05 = 2\times10^{-3}\ \text{m}^2

Reluctances with μ=4π×10−7×4000=5.027×10−3\mu = 4\pi\times10^{-7}\times4000 = 5.027\times10^{-3} H/m:

Sc=0.165.027×10−3×3×10−3=1.061×104 AT/WbSo=0.165.027×10−3×2×10−3=1.592×104 AT/WbSy=0.205.027×10−3×2×10−3=1.989×104 AT/Wb\begin{aligned} S_c &= \frac{0.16}{5.027\times10^{-3}\times3\times10^{-3}} = 1.061\times10^{4}\ \text{AT/Wb} \\ S_o &= \frac{0.16}{5.027\times10^{-3}\times2\times10^{-3}} = 1.592\times10^{4}\ \text{AT/Wb} \\ S_y &= \frac{0.20}{5.027\times10^{-3}\times2\times10^{-3}} = 1.989\times10^{4}\ \text{AT/Wb} \end{aligned}

One side path (outer limb + top yoke + bottom yoke) has Sside=1.592×104+2(1.989×104)=5.570×104S_{side} = 1.592\times10^4 + 2(1.989\times10^4) = 5.570\times10^{4}. Flux in each side path is Φc/2=0.375\Phi_c/2 = 0.375 Wb.

Mmf

NI=ΦcSc+Φc2Sside=0.75×1.061×104+0.375×5.570×104=7957+20 889=28 847 AT\begin{aligned} NI &= \Phi_c S_c + \frac{\Phi_c}{2}S_{side} \\ &= 0.75\times1.061\times10^{4} + 0.375\times5.570\times10^{4} \\ &= 7957 + 20\,889 = 28\,847\ \text{AT} \end{aligned}

Answer: NI≈2.88×104NI \approx 2.88\times10^{4} AT for 0.75 Wb (about 28.8 AT if the flux is 0.75 mWb), with d=5d = 5 cm assumed.

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