Question 5.8: An iron ring is made up of two different materials A and B h......
An iron ring is made up of two different materials A and B having a relative permeability of 1000 and 1500, respectively. The length L_A\ and\ L_B of the two materials used are 75 cm and 25 cm, respectively. The air gap length is 2 mm. The cross-sectional area of the core is 10 cm² . The magnetizing coil has 1000 turns and a current of 5 A is allowed to flow through it. Calculate the flux produced in the air gap.
We know, \phi=\frac{\text { MMF }}{\text { Total reluctance }}
Total reluctance, S = Reluctance of part A + Reluctance of part B + Reluctance of air gap
\begin{aligned} \mathrm{S} & =\frac{\mathrm{L}_{\mathrm{A}}}{\mu_{\mathrm{o}} \mu_{\mathrm{r}_1} \mathrm{~A}}+\frac{\mathrm{L}_{\mathrm{B}}}{\mu_{\mathrm{o}} \mu_{\mathrm{r}_1} \mathrm{~A}}+\frac{1_{\mathrm{g}}}{\mu_{\mathrm{o}} \mathrm{A}} \\ & =\frac{0.75}{4 \pi \times 10^{-7} \times 1000 \times 10 \times 10^{-4}}+\frac{0.25}{4 \pi \times 10^{-7} \times 1500 \times 10 \times 10^{-4}} \\ & +\frac{2 \times 10^{-3}}{4 \pi \times 10^{-7} \times 10 \times 10^{-4}} \\ & =10^4[59.7133+13.2696+1592.3566] \\ & =10^4 \times 1665.3395 \\ & =16.653395 \times 10^6 \mathrm{AT} / \mathrm{Wb} \end{aligned}.
\begin{aligned}\text{Flux, }\quad \phi & =\frac{\mathrm{mmf}}{\text { Reluctance }}=\frac{\mathrm{NI}}{\mathrm{S}}=\frac{1000 \times 5}{16.653395 \times 10^6} \\ & =0.3 \times 10^{-3} \mathrm{~Wb} \\ & =0.3 \mathrm{mWb} \end{aligned}