Iron ring: reluctance and flux with saw cut

Given: $D = 100$ cm, $A = 10$ cm$^2 = 10^{-3}$ m$^2$, $N = 1000$, $\mu_r = 2000$, $I = 1$ A.

(i) Reluctance (no saw cut)

$$\begin{aligned} l &= \pi D = \pi\ \text{m} \\ S &= \frac{l}{\mu_0\mu_r A} = \frac{\pi}{4\pi\times10^{-7}\times2000\times10^{-3}} = 1.25\times10^{6}\ \text{AT/Wb} \end{aligned}$$

(ii) Flux with 1 A

$$\Phi = \frac{NI}{S} = \frac{1000\times1}{1.25\times10^{6}} = 8\times10^{-4}\ \text{Wb} = 0.8\ \text{mWb}$$

(iii) Flux with a 1 mm saw cut

The iron path becomes $l_i = \pi - 0.001 = 3.1406$ m, and the gap is $l_g = 1$ mm.

$$\begin{aligned} S_i &= \frac{3.1406}{4\pi\times10^{-7}\times2000\times10^{-3}} = 1.2496\times10^{6} \\ S_g &= \frac{0.001}{4\pi\times10^{-7}\times10^{-3}} = 7.958\times10^{5} \\ S_{total} &= 2.0454\times10^{6}\ \text{AT/Wb} \\ \Phi &= \frac{1000}{2.0454\times10^{6}} = 4.889\times10^{-4}\ \text{Wb} \end{aligned}$$

A gap of only 1 mm cuts the flux from 0.8 mWb to 0.489 mWb, because air has a much higher reluctance than iron.

Answer: (i) $1.25\times10^6$ AT/Wb; (ii) 0.8 mWb; (iii) 0.489 mWb