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Iron ring with 2 mm saw cut, exciting current

Short answer

Flux Φ = B_g A = 1 mWb. Iron path l_i = 0.4692 m, iron mmf = 466.8 AT, gap mmf = 1591.5 AT, total mmf = 2058.3 AT. Exciting current I = 2058.3/300 = 6.86 A. Inductance L = NΦ/I = 300×10⁻³/6.86 = 0.0437 H = 43.7 mH.

An iron ring of 0.15 meter diameter and 0.001 m2^2 in cross section with a saw cut 2 mm wide is wound with 300 turns of wire. The gap flux density is 1 Tesla. The relative permeability of the iron is 800. Determine the exciting current and inductance.

Given: diameter D=0.15D = 0.15 m (taken as mean diameter), A=10−3A = 10^{-3} m2^2, lg=2l_g = 2 mm, N=300N = 300, Bg=1B_g = 1 T, μr=800\mu_r = 800.

Flux and lengths

Φ=BgA=1×10−3=1 mWbli=π×0.15−0.002=0.4692 m\begin{aligned} \Phi &= B_gA = 1\times10^{-3} = 1\ \text{mWb} \\ l_i &= \pi\times0.15 - 0.002 = 0.4692\ \text{m} \end{aligned}

Since the flux and area are the same in iron and gap (no fringing), Bi=1B_i = 1 T.

Mmf

Fi=Bliμ0μr=1×0.46924π×10−7×800=466.8 ATFg=Blgμ0=0.0024π×10−7=1591.5 ATF=466.8+1591.5=2058.3 AT\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{1\times0.4692}{4\pi\times10^{-7}\times800} = 466.8\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\ F &= 466.8 + 1591.5 = 2058.3\ \text{AT} \end{aligned}

Exciting current

I=FN=2058.3300=6.861 AI = \frac{F}{N} = \frac{2058.3}{300} = 6.861\ \text{A}

Inductance

L=NΦI=300×10−36.861=0.0437 HL = \frac{N\Phi}{I} = \frac{300\times10^{-3}}{6.861} = 0.0437\ \text{H}

Answer: I≈6.86I \approx 6.86 A; L≈43.7L \approx 43.7 mH

✓ Answer checked by Claude Opus 5.5 (high effort)

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