The ring consists of two parallel magnetic circuits of semiconductor cross sections having reluctances \(R_C\) and \(R_S\) corresponding to cost iron and cost steel respectively. The reluctance of the joint,

\(R_j=\frac{l}{\mu_0A}=\frac{0.2\times{10}^{-3}}{4\pi\times{10}^{-7}\times0.001}=0.016\times{10}^7AT/Wb\)

\(R_C=\frac{l_1}{\mu_{rc}\mu_0A}=\frac{\pi D/2}{\mu_{rc}\mu_0A}=\frac{\pi\times0.21}{2\times4\pi\times{10}^{-7}\times166\times0.001}=0.633\times{10}^7AT/Wb\)

The net reluctance of semicircular section of cost iron including that of joint is

\(R_1=R_C+R_j=\left(0.633\times{10}^7+0.016\times{10}^7\right)=0.649\times{10}^7AT/Wb\)

Similarly the net reluctance of semicircular section of cost steel including that of joint is

\(R_2=R_S+R_j=\frac{l_2}{\mu_{rc}\mu_0A}+R_j=\frac{\pi\times0.21}{\left(2\times4\pi\times{10}^{-7}\times800\times0.01\right)}+R_j\)

\(R_2=0.131\times{10}^7+0.016\times{10}^7=0.147\times{10}^7AT/Wb\)

Since \(R_1\) and \(R_2\) are in parallel, equivalent reluctance is

\(R_{eq}=\frac{R_1R_2}{R_1+R_2}=\frac{(0.649\times0.147)\times{10}^{14}}{(0.649+0.147)\times{10}^7}=0.120\times{10}^7AT/Wb\)

We know that the number of ampere turns = \(flux\times reluctance=0.80\times{10}^{-3}\times0.12\times{10}^7=960\)

Fig.11A semi circular ring with cost iron and cost steel