Reluctances and air gap flux of two-winding core

Both windings are on the same single-loop core, so their mmfs add (assuming they are wound to aid each other) and drive one flux through the core and gap in series. Gap area is taken equal to core area (fringing neglected).

i) Reluctances

$$\begin{aligned} R_c &= \frac{l_c}{\mu_0\mu_r A_c} = \frac{0.6}{4\pi\times10^{-7}\times2000\times1.8\times10^{-3}} = 1.326\times10^{5}\ \text{AT/Wb} \\ R_g &= \frac{g}{\mu_0 A_c} = \frac{2.3\times10^{-3}}{4\pi\times10^{-7}\times1.8\times10^{-3}} = 1.017\times10^{6}\ \text{AT/Wb} \end{aligned}$$

Total reluctance: $R = R_c + R_g = 1.149\times10^{6}$ AT/Wb.

ii) Flux in the air gap

$$\begin{aligned} F &= N_1 i_1 + N_2 i_2 = 83\times1.5 + 20\times1.25 = 124.5 + 25 = 149.5\ \text{AT} \\ \phi &= \frac{F}{R} = \frac{149.5}{1.149\times10^{6}} = 1.30\times10^{-4}\ \text{Wb} \end{aligned}$$

Gap flux density: $B_g = \phi/A_c = 1.30\times10^{-4}/1.8\times10^{-3} = 0.0723$ T.

Answer: $R_c = 1.33\times10^5$ AT/Wb, $R_g = 1.02\times10^6$ AT/Wb, $\phi = 0.130$ mWb (B = 0.072 T).

(If the two windings opposed each other, $F = 124.5 - 25 = 99.5$ AT and $\phi = 0.0866$ mWb.)