Skip to main content

Current for flux in cast steel series circuit with air gap

Short answer

For B=0.5 T, steel mmf = 27.9 AT, air-gap mmf = 795.8 AT; total mmf = 823.7 AT from 200 turns. Required current = 4.12 A.

Find the value of I required to establish a magnetic flux of Φ = 0.75×10⁻⁴ Wb in the series magnetic circuit as shown in figure below. Given, that the relative permeability for the steel is μr = 1424. [Figure: all cast steel rectangular core, area (throughout) = 1.5×10⁻⁴ m²; coil of N = 200 turns carrying current I on the left limb; an air gap in the right limb; mean steel path length l(cdefab) = 100×10⁻³ m and air gap length l(bc) = 2×10⁻³ m.]

The steel path and the air gap are in series and carry the same flux; the area is the same throughout (fringing neglected).

Data: ϕ=0.75×10−4\phi = 0.75\times10^{-4} Wb, A=1.5×10−4A = 1.5\times10^{-4} m², ls=100×10−3l_s = 100\times10^{-3} m, lg=2×10−3l_g = 2\times10^{-3} m, μr=1424\mu_r = 1424, N=200N = 200.

Flux density

B=ϕA=0.75×10−41.5×10−4=0.5 TB = \frac{\phi}{A} = \frac{0.75\times10^{-4}}{1.5\times10^{-4}} = 0.5\ \text{T}

Steel portion

Hs=Bμ0μr=0.54π×10−7×1424=279.4 AT/mHsls=279.4×0.1=27.9 AT\begin{aligned} H_s &= \frac{B}{\mu_0\mu_r} = \frac{0.5}{4\pi\times10^{-7}\times1424} = 279.4\ \text{AT/m} \\ H_s l_s &= 279.4\times0.1 = 27.9\ \text{AT} \end{aligned}

Air gap

Hg=Bμ0=0.54π×10−7=3.979×105 AT/mHglg=3.979×105×2×10−3=795.8 AT\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{0.5}{4\pi\times10^{-7}} = 3.979\times10^{5}\ \text{AT/m} \\ H_g l_g &= 3.979\times10^{5}\times2\times10^{-3} = 795.8\ \text{AT} \end{aligned}

Current

NI=27.9+795.8=823.7 ATI=823.7200=4.12 A\begin{aligned} NI &= 27.9 + 795.8 = 823.7\ \text{AT} \\ I &= \frac{823.7}{200} = 4.12\ \text{A} \end{aligned}

Answer: I ≈ 4.12 A. About 97% of the mmf is used by the 2 mm air gap.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Machines I chapter 1 questions →