Question 5.6: Calculate the flux produced in the air gap in the magnetic c......
Calculate the flux produced in the air gap in the magnetic circuit shown in Fig. 5.26, which is excited by the MMF of two windings. The mean length of the flux path is 40 cm. The permeability of iron is 2000. The uniform core cross-sectional area is 10 cm² .
Applying Fleming’s thumb rule we can see that the ampere turns of coil 2 produce flux in the opposite direction as the flux produced by the ampere turns of coil 1.
\begin{aligned}\text{Total MMF}\quad & =\mathrm{N}_1 \mathrm{I}_1-\mathrm{N}_2 \mathrm{I}_2 \\ & =100 \times 10-80 \times 1.5 \\ & =880 \mathrm{AT} \end{aligned}Total reluctance = Reluctance of iron + Reluctance of air gap
=\frac{l_{\mathrm{i}}}{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \mathrm{A}}+\frac{l_{\mathrm{g}}}{\mu_{\mathrm{o}} \mathrm{A}}=\frac{1}{\mu_{\mathrm{o}} \mathrm{A}}\left[\frac{l_{\mathrm{i}}}{\mu_{\mathrm{r}}}+l_{\mathrm{g}}\right]Substituting values
\begin{aligned}\text{Total Reluctance }\quad & =\frac{1}{4 \pi \times 10^{-7} \times 10 \times 10^{-4}}\left[\frac{\left(40 \times 10^{-2}-1 \times 10^{-3}\right)}{2000}+1 \times 10^{-3}\right] \\ & =\frac{10^{10}}{4 \pi}\left[\frac{399 \times 10^{-3}}{2000}+1 \times 10^{-3}\right] \\ & =\frac{10^7}{4 \pi}\left[\frac{399}{2000}+1\right] \\ & =0.955 \times 10^6 \mathrm{AT} / \mathrm{Wb} . \end{aligned}Core Flux =\frac{\text { MMF }}{\text { Reluctance }}=\frac{880}{0.955 \times 10^6}=921.5 \times 10^{-6}
= 0.9215 \times 10^{-3} \mathrm{~Wb}
The air gap flux is the same as the core flux as the whole of the core flux crosses the air gap and there is no fringing.