Flux and flux density in iron ring with 2 mm gap

Flux is found from $\phi = NI/(S_i + S_g)$. Iron length is taken as $l_i = 80 - 0.2 = 79.8$ cm; gap area = core area (no fringing).

Data: $A = 16$ cm² $= 16\times10^{-4}$ m², $l_g = 2$ mm, $N = 1000$, $I = 6$ A, $\mu_r = 2000$.

MMF

$$F = NI = 1000 \times 6 = 6000\ \text{AT}$$

Reluctances

$$\begin{aligned} S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{0.798}{4\pi\times10^{-7}\times2000\times16\times10^{-4}} = 1.984\times10^{5}\ \text{AT/Wb} \\ S_g &= \frac{l_g}{\mu_0 A} = \frac{0.002}{4\pi\times10^{-7}\times16\times10^{-4}} = 9.947\times10^{5}\ \text{AT/Wb} \\ S &= S_i + S_g = 1.193\times10^{6}\ \text{AT/Wb} \end{aligned}$$

Flux and flux density

$$\begin{aligned} \phi &= \frac{F}{S} = \frac{6000}{1.193\times10^{6}} = 5.03\times10^{-3}\ \text{Wb} \\ B &= \frac{\phi}{A} = \frac{5.03\times10^{-3}}{16\times10^{-4}} = 3.14\ \text{T} \end{aligned}$$

Answer: φ ≈ 5.03 mWb, B ≈ 3.14 T in the core (and in the gap). The 2 mm gap has five times the reluctance of the whole iron path. (If the full 80 cm is used as the iron length, φ = 5.027 mWb, which still rounds to 5.03 mWb; the difference is negligible.)