Number of turns for iron ring with air gap

Required mmf = flux × total reluctance; then $N = F/I$. Iron length is taken as $l_i = 1.25 - 0.005 = 1.245$ m; gap area = core area.

Data: $A = 0.02$ m², $l_g = 5$ mm, $\phi = 0.5$ Wb, $I = 2$ A, $\mu_r = 1000$.

Reluctances

$$\begin{aligned} S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{1.245}{4\pi\times10^{-7}\times1000\times0.02} = 4.954\times10^{4}\ \text{AT/Wb} \\ S_g &= \frac{l_g}{\mu_0 A} = \frac{0.005}{4\pi\times10^{-7}\times0.02} = 1.989\times10^{5}\ \text{AT/Wb} \\ S &= S_i + S_g = 2.485\times10^{5}\ \text{AT/Wb} \end{aligned}$$

MMF and turns

$$\begin{aligned} F &= \phi S = 0.5\times2.485\times10^{5} = 1.2424\times10^{5}\ \text{AT} \\ N &= \frac{F}{I} = \frac{1.2424\times10^{5}}{2} = 62120\ \text{turns} \end{aligned}$$

Answer: N ≈ 62,120 turns. (Using the full 1.25 m as iron length gives N ≈ 62,170.) Note: B = 0.5/0.02 = 25 T, which no iron can reach; the figure follows only from the constant μr given.