Magnetizing current for ring with 5 mm air gap

With leakage and fringing neglected, the same flux passes through the iron and the gap, and the area is the same. Iron path length is taken as $l_i = 0.8 - 0.005 = 0.795$ m.

Data: $\phi = 1$ mWb, $A = 120$ cm² $= 0.012$ m², $l_g = 5$ mm, $N = 220$, $\mu_r = 500$.

Flux density

$$B = \frac{\phi}{A} = \frac{1\times10^{-3}}{0.012} = 0.0833\ \text{T}$$

Ampere-turns for the air gap

$$\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{0.0833}{4\pi\times10^{-7}} = 66315\ \text{AT/m} \\ AT_g &= H_g l_g = 66315 \times 0.005 = 331.6\ \text{AT} \end{aligned}$$

Ampere-turns for the iron

$$\begin{aligned} H_i &= \frac{B}{\mu_0\mu_r} = \frac{66315}{500} = 132.6\ \text{AT/m} \\ AT_i &= H_i l_i = 132.6 \times 0.795 = 105.4\ \text{AT} \end{aligned}$$

Magnetising current

$$\begin{aligned} NI &= 331.6 + 105.4 = 437.0\ \text{AT} \\ I &= \frac{437.0}{220} = 1.986\ \text{A} \end{aligned}$$

Answer: magnetising current ≈ 1.99 A. (Taking the iron length as the full 0.8 m gives 1.989 A, practically the same.) Note that the 5 mm gap needs about three times the ampere-turns of the 0.795 m of iron.