Question 5.3: A rectangular shape iron core has an air gap of 0.01 cm. The......
A rectangular shape iron core has an air gap of 0.01 cm. The mean length of the flux path through iron is 39.99 cm. The relative permeability of iron is 2000. The coil has 1000 turns. The cross-sectional area of the core is 9 cm² . Calculate the current required to produce a flux of 1 mWb in the core.
Total reluctance of the flux path = Reluctance of iron path + Reluctance of air gap,
i.e., \mathrm{S}=\mathrm{S}_{\mathrm{i}}+\mathrm{S}_{\mathrm{g}}
S = \frac{1_i}{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \mathrm{A}}+\frac{1_{\mathrm{g}}}{\mu_{\mathrm{o}} \mathrm{A}}
Note: the iron path permeability is μ which is equal to μ_0 μ_r whereas for the air gap the permeability is μ_0 only.
Substituting the given values,
\begin{gathered} S=\frac{39.09 \times 10^{-2}}{4 \pi \times 10^{-7} \times 2000 \times 9 \times 10^{-4}}+\frac{0.01 \times 10^{-2}}{4 \pi \times 10^{-7} \times 9 \times 10^{-4}} \\ =\frac{10^6}{4 \pi}\left[\frac{39.09}{18}+\frac{100}{10}\right]=\frac{295.45 \times 10^5}{36 \pi} \end{gathered}Flux, \phi=\frac{\mathrm{NI}}{\mathrm{S}}=\frac{1000 \mathrm{I}}{\mathrm{S}}
∴ \mathrm{I}=\frac{\phi \times \mathrm{S}}{1000}=\frac{1 \times 10^{-3} \times 295.45 \times 10^5}{36 \pi \times 1000}=\frac{29.545}{36 \pi}=0.26 \mathrm{~A}
