Skip to main content

Fluxes in three-limb core with two air gaps

Short answer

Solve the magnetic circuit with 3000 AT MMF. Reluctances of left and right limbs (iron plus gap) are in parallel, then series with centre limb. Flux in middle = 36.1 mWb, left = 20.1 mWb, right = 16.0 mWb; gap flux densities 3.90 T (left) and 3.12 T (right).

A ferromagnetic core with a relative permeability of 4000 is shown below. The dimensional area as shown and depth of core is 7 cm. The air gap is 0.05 cm on left and 0.07 cm on right side. The effective area of gap is 5% larger due to fringing effect. If current in coil is 10 A, calculate flux in left, right and middle part along with flux density in each air gap. [Figure: three-limb core; all limbs and yokes are 7 cm wide; the two windows are each 30 cm wide and 30 cm high; the centre limb carries a 300-turn coil with current i; the left outer limb has a 0.05 cm air gap and the right outer limb has a 0.07 cm air gap.]

The centre limb (with the coil) carries the total flux, which divides between the left and right outer paths. Each outer path contains an air gap, so the two outer paths are in parallel and both are in series with the centre limb.

Dimensions: limbs 7 cm wide, depth 7 cm → A=49A = 49 cm² =4.9×10−3= 4.9\times10^{-3} m². Windows 30 × 30 cm, so the centre-line spacing is 30 + 7 = 37 cm.

  • Centre limb: lc=37l_c = 37 cm.
  • Each outer path: top yoke 37 + outer limb 37 + bottom yoke 37 = 111 cm (minus its gap).
  • Gap area with 5% fringing: Ag=1.05×49=51.45A_g = 1.05 \times 49 = 51.45 cm².
  • MMF: F=300×10=3000F = 300 \times 10 = 3000 AT.

Reluctances (μ0μrA=4π×10−7×4000×4.9×10−3=2.463×10−5\mu_0\mu_r A = 4\pi\times10^{-7}\times4000\times4.9\times10^{-3} = 2.463\times10^{-5})

Sc=0.372.463×10−5=15022 AT/WbSL,iron=1.10952.463×10−5=45047,SL,gap=0.00054π×10−7×5.145×10−3=77335SL=45047+77335=122381 AT/WbSR,iron=1.10932.463×10−5=45038,SR,gap=0.00074π×10−7×5.145×10−3=108269SR=45038+108269=153307 AT/Wb\begin{aligned} S_c &= \frac{0.37}{2.463\times10^{-5}} = 15022\ \text{AT/Wb} \\ S_{L,iron} &= \frac{1.1095}{2.463\times10^{-5}} = 45047, \quad S_{L,gap} = \frac{0.0005}{4\pi\times10^{-7}\times5.145\times10^{-3}} = 77335 \\ S_L &= 45047 + 77335 = 122381\ \text{AT/Wb} \\ S_{R,iron} &= \frac{1.1093}{2.463\times10^{-5}} = 45038, \quad S_{R,gap} = \frac{0.0007}{4\pi\times10^{-7}\times5.145\times10^{-3}} = 108269 \\ S_R &= 45038 + 108269 = 153307\ \text{AT/Wb} \end{aligned}

Total reluctance and fluxes

SL∥SR=122381×153307122381+153307=68055 AT/WbStot=15022+68055=83077 AT/Wbϕc=300083077=0.03611 WbFLR=ϕc×68055=2457.6 ATϕL=2457.6122381=0.02008 Wb,ϕR=2457.6153307=0.01603 Wb\begin{aligned} S_L \parallel S_R &= \frac{122381\times153307}{122381+153307} = 68055\ \text{AT/Wb} \\ S_{tot} &= 15022 + 68055 = 83077\ \text{AT/Wb} \\ \phi_c &= \frac{3000}{83077} = 0.03611\ \text{Wb} \\ F_{LR} &= \phi_c \times 68055 = 2457.6\ \text{AT} \\ \phi_L &= \frac{2457.6}{122381} = 0.02008\ \text{Wb}, \quad \phi_R = \frac{2457.6}{153307} = 0.01603\ \text{Wb} \end{aligned}

Flux density in the gaps

BgL=0.020085.145×10−3=3.90 T,BgR=0.016035.145×10−3=3.12 TB_{gL} = \frac{0.02008}{5.145\times10^{-3}} = 3.90\ \text{T}, \quad B_{gR} = \frac{0.01603}{5.145\times10^{-3}} = 3.12\ \text{T}

Answer: ϕmiddle=36.1\phi_{middle} = 36.1 mWb, ϕleft=20.1\phi_{left} = 20.1 mWb, ϕright=16.0\phi_{right} = 16.0 mWb; Bgap,left=3.90B_{gap,left} = 3.90 T, Bgap,right=3.12B_{gap,right} = 3.12 T. (These densities are far above saturation of real iron; the result holds only for the constant μr = 4000 given.)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machines I chapter 1 questions →