Question 5.5: An iron ring of mean length of an iron path of 100 cm and ha......
An iron ring of mean length of an iron path of 100 cm and having a uniform crosssectional area of 10 cm² is wound with two magnetizing coils as shown. The direction of current flowing through the two coils are such that they produce flux in the opposite directions. The permeability of iron is 2000. There is a cut in the ring creating an air gap of 1 mm. Calculate the flux available in the air gap.
The net MMF will be the resultant effect of MMF of the two coils in producing the flux in the core. As in Fig. 5.25, the flux produced by the MMF of the two coils are in opposite directions. Thus, the resultant MMF will be the difference of these two MMFs.
\begin{aligned}\text{Resultant MMF }\quad & =\mathrm{N}_1 \mathrm{I}_1-\mathrm{N}_2 \mathrm{I}_2 \\ & =100 \times 3-50 \times 2 \\ & =200 \end{aligned}Total reluctance, \mathrm{S}=\mathrm{S}_{\mathrm{i}}+\mathrm{S}_{\mathrm{g}}
= \frac{l_{\mathrm{i}}}{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \mathrm{A}_{\mathrm{i}}}+\frac{l_{\mathrm{g}}}{\mu_{\mathrm{o}} \mathrm{A}_{\mathrm{g}}} \text { [we have considered the reluctance of } \mu_{\mathrm{o}} \mu_{\mathrm{r}} \text { for iron and } \mu_{\mathrm{o}} \text { for air] }
Since \mathrm{A}_{\mathrm{i}}=\mathrm{A}_{\mathrm{g}}, i.e., the cross-sectional area of the iron path is the same as that of the air gap,
\mathrm{S}=\frac{1}{\mu_{\mathrm{o}} \mathrm{A}}\left[\frac{l_{\mathrm{i}}}{\mu_{\mathrm{r}}}+l_{\mathrm{g}}\right].
\begin{aligned}\text{Substituting values }\quad S & =\frac{1}{4 \pi \times 10^{-7} \times 10 \times 10^{-4}}\left[\frac{50 \times 10^{-2}}{2000}+1 \times 10^{-3}\right] \\ S & =\frac{10^{10}}{4 \pi}\left[25 \times 10^{-5}+1 \times 10^{-3}\right] \\ & =\frac{10^{10}}{4 \pi} 10^{-3}\left[25 \times 10^{-2}+1\right] \\ & =\frac{10^7}{4 \pi}[1.25]=10^6 \end{aligned}Flux, \phi=\frac{\mathrm{MMF}}{\mathrm{S}}=\frac{200}{10^6}=2 \times 10^{-4}=0.2 \mathrm{mWb}