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Flux density in steel ring with air gap

Short answer

Mean circumference 0.754 m, iron length 0.752 m, cross-sectional area 3.14×10⁻⁴ m². Applying NI = (B/μ₀)(lᵢ/μᵣ + l_g) gives B = (4π×10⁻⁷·550·3)/(0.752/800 + 0.002) = 0.705 T.

A steel ring of 12 cm mean radius and of circular cross section 1 cm in radius has an air gap of 2 mm length. It is wound uniformly with 550 turns of wire carrying 3 A. Neglecting magnetic leakage, calculate the magnetic flux density in the core. Given that relative permeability of the steel is 800.

Flux density follows from NI=Hili+HglgNI = H_i l_i + H_g l_g with the same B in the steel and in the gap.

Data: mean radius r=0.12r = 0.12 m, section radius 1 cm, lg=2l_g = 2 mm, N=550N = 550, I=3I = 3 A, μr=800\mu_r = 800.

Dimensions

l=2πr=2π×0.12=0.7540 mli=0.7540−0.002=0.7520 mA=π(0.01)2=3.1416×10−4 m2\begin{aligned} l &= 2\pi r = 2\pi\times0.12 = 0.7540\ \text{m} \\ l_i &= 0.7540 - 0.002 = 0.7520\ \text{m} \\ A &= \pi (0.01)^2 = 3.1416\times10^{-4}\ \text{m}^2 \end{aligned}

Flux density

NI=Bμ0(liμr+lg)B=μ0NIliμr+lg=4π×10−7×550×30.7520800+0.002=2.0735×10−39.40×10−4+2×10−3=2.0735×10−32.940×10−3=0.705 T\begin{aligned} NI &= \frac{B}{\mu_0}\left(\frac{l_i}{\mu_r} + l_g\right) \\ B &= \frac{\mu_0 NI}{\dfrac{l_i}{\mu_r} + l_g} = \frac{4\pi\times10^{-7}\times550\times3}{\dfrac{0.7520}{800} + 0.002} \\ &= \frac{2.0735\times10^{-3}}{9.40\times10^{-4} + 2\times10^{-3}} = \frac{2.0735\times10^{-3}}{2.940\times10^{-3}} \\ &= 0.705\ \text{T} \end{aligned}

Flux: ϕ=BA=0.705×3.1416×10−4=2.22×10−4\phi = BA = 0.705\times3.1416\times10^{-4} = 2.22\times10^{-4} Wb.

Answer: B ≈ 0.705 T (flux ≈ 0.222 mWb).

✓ Answer checked by Claude Opus 5.5 (high effort)

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