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Coil current for flux in three-limb cast steel core

Short answer

Using magnetic circuit analysis, the parallel limbs have equal MMF drop. With flux 2 mWb in the right limb, total flux is 10.667 mWb, total MMF is 27299 AT, so the required current is 54.6 A.

A cast steel magnetic structure made of a bar of section 2 cm × 2 cm is shown in figure. Determine the current that the 500 turns magnetizing coil on the left limb should carry so that a flux of 2 mWb is produced in the right limb. Take μr = 600 and neglect leakage. [Figure: three-limb rectangular core; the 500-turn coil carrying current I is on the left limb A; the centre limb B carries flux Φ1 and the right limb C carries flux Φ2; the left and right windows are each 25 cm wide (left limb to centre limb 25 cm, centre limb to right limb 25 cm) and the limbs are 15 cm high; flux Φ in the left limb divides into Φ1 and Φ2.]

The centre limb and the right limb are two parallel paths between the same two junctions, so they have the same mmf drop. The left limb (with the coil) carries the total flux.

Assumed mean path lengths (from the figure): left path A = 25 + 15 + 25 = 65 cm, centre limb B = 15 cm, right path C = 25 + 15 + 25 = 65 cm. Area a=2×2=4a = 2 \times 2 = 4 cm² =4×10−4= 4\times10^{-4} m² throughout.

   +-----25-----+-----25-----+
   |            |            |
 [500T]  phi ^  | phi1  v    | phi2 v
   | A          | B (15)     | C
   +------------+------------+

Reluctances

S=lμ0μra,μ0μra=4π×10−7×600×4×10−4SA=SC=0.653.016×10−7=2.155×106 AT/WbSB=0.153.016×10−7=4.974×105 AT/Wb\begin{aligned} S &= \frac{l}{\mu_0\mu_r a}, \quad \mu_0\mu_r a = 4\pi\times10^{-7}\times600\times4\times10^{-4} \\ S_A = S_C &= \frac{0.65}{3.016\times10^{-7}} = 2.155\times10^{6}\ \text{AT/Wb} \\ S_B &= \frac{0.15}{3.016\times10^{-7}} = 4.974\times10^{5}\ \text{AT/Wb} \end{aligned}

Fluxes

MMF across the parallel limbs (B and C):

FBC=ϕ2SC=2×10−3×2.155×106=4310.4 ATϕ1=FBCSB=4310.44.974×105=8.667 mWbϕ=ϕ1+ϕ2=8.667+2=10.667 mWb\begin{aligned} F_{BC} &= \phi_2 S_C = 2\times10^{-3}\times2.155\times10^6 = 4310.4\ \text{AT} \\ \phi_1 &= \frac{F_{BC}}{S_B} = \frac{4310.4}{4.974\times10^5} = 8.667\ \text{mWb} \\ \phi &= \phi_1 + \phi_2 = 8.667 + 2 = 10.667\ \text{mWb} \end{aligned}

(Equivalently ϕ1=ϕ2×65/15\phi_1 = \phi_2 \times 65/15 because the limbs differ only in length.)

Total mmf and current

NI=ϕSA+FBC=10.667×10−3×2.155×106+4310.4=22989+4310=27299 ATI=27299500=54.6 A\begin{aligned} NI &= \phi S_A + F_{BC} \\ &= 10.667\times10^{-3}\times2.155\times10^{6} + 4310.4 \\ &= 22989 + 4310 = 27299\ \text{AT} \\ I &= \frac{27299}{500} = 54.6\ \text{A} \end{aligned}

Answer: I ≈ 54.6 A.

(Note: with μr = 600 held constant the flux densities come out very high, e.g. 5 T in the right limb; a real cast-steel core would saturate, so the result is valid only for the linear data given.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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