A ring having a mean diameter of 21 cm and a cross – section of 10 cm2 is made of two semicircular section of cost iron and cost steel respectively with each joint having reluctance equal to air gap of 0.2 mm as shown in figure. Determine the ampere turns required to produce a flux of 0.8 mWb. The relative permeability of cost iron and cost steel are 166 and 800 respectively. Neglect fringing and leakage effects.
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
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Fig. 18(a)