Question 5.10: An electromagnetic-type relay shown in Fig. 5.31 has the fol......
An electromagnetic-type relay shown in Fig. 5.31 has the following particulars. The mean length of the flux path through the iron is 20 cm and the length of air gaps is 1 mm each. The exciting coil has 8000 turns and carries a current of 50 mA when excited. The cross-sectional area of the core is 0.5 cm² . The permeability of iron is 500. Calculate the flux density and the magnetic pull produced in the armature (i.e., on the moving part).
Total reluctance of the flux path = Reluctance of the path through the iron + Reluctance of the two air gaps. The length of each of the air gaps, l_g = 1 mm. We have to take into account 2l_g as the total air gap to the flux path.
\begin{aligned}\text{Total reluctance }\quad & =\frac{l_{\mathrm{i}}}{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \mathrm{A}}+\frac{2 l_{\mathrm{g}}}{\mu_{\mathrm{o}} \mathrm{A}} \\ & =\frac{20 \times 10^{-2}}{4 \pi \times 10^{-7} \times 500 \times 0.5 \times 10^{-4}}+\frac{2 \times 1 \times 10^{-3}}{4 \pi \times 10^{-7} \times 0.5 \times 10^{-4}} \\ & =3.82 \times 10^7 \mathrm{AT} / \mathrm{Wb} \end{aligned}Flux, \phi=\frac{\mathrm{MMF}}{\text { Reluctance }}=\frac{\mathrm{NI}}{\text { Reluctance }}=\frac{8000 \times 50 \times 10^{-3}}{3.82 \times 10^7}=104.7 \mathrm{mWb}
Flux, density, \mathrm{B}=\frac{\phi}{\mathrm{A}}=\frac{104.7 \times 10^{-3}}{0.5 \times 10^{-4}}=0.2094 \mathrm{~Wb} / \mathrm{m}^2
From equation 5.17, the force or the pull on the armature by each pole,
\mathrm{F}=\frac{\mathrm{B}^2 \mathrm{~A}}{2 \mu_{\mathrm{o}}}=\frac{(0.2094)^2 \times 0.5 \times 10^{-4}}{2 \times 4 \pi \times 10^{-7}}=4.16 \mathrm{~N}