Iron ring with saw cut, 4 mWb flux

The exciting current is found from the total mmf, which is the sum of the mmf for the iron path and for the air gap, using the reluctance of each part.

Given: mean diameter $D = 30$ cm, rod diameter $d = 2.5$ cm, gap $l_g = 1$ mm, $N = 500$, $\Phi = 4$ mWb, $\mu_r = 800$.

Dimensions

$$\begin{aligned} l_{total} &= \pi D = \pi \times 0.30 = 0.9425\ \text{m} \\ l_i &= 0.9425 - 0.001 = 0.9415\ \text{m} \\ A &= \frac{\pi}{4}(0.025)^2 = 4.909\times10^{-4}\ \text{m}^2 \end{aligned}$$

Reluctances

$$\begin{aligned} S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{0.9415}{4\pi\times10^{-7}\times800\times4.909\times10^{-4}} = 1.908\times10^{6}\ \text{AT/Wb} \\ S_g &= \frac{l_g}{\mu_0 A} = \frac{0.001}{4\pi\times10^{-7}\times4.909\times10^{-4}} = 1.621\times10^{6}\ \text{AT/Wb} \end{aligned}$$

The two reluctances are in series (the same flux passes through both), so $S = S_i + S_g = 3.529\times10^{6}$ AT/Wb.

Mmf and current

$$\begin{aligned} NI &= \Phi S = 4\times10^{-3}\times3.529\times10^{6} = 14\,116\ \text{AT} \\ I &= \frac{14\,116}{500} = 28.23\ \text{A} \end{aligned}$$

The flux density is $B = \Phi/A = 8.15$ T, which is far above the saturation of real iron, so the given $\mu_r = 800$ is only an assumption of the problem.

Answer: $I \approx 28.2$ A