Question #124257

A magnetic circuit consists of a cast steel yoke which has a cross-sectional area of 200 mm2 and a mean

length of 120 mm. There are two air gaps, each 0.2 mm long. Calculate the mmf required to produce a

flux of 0.5 mWb in the air gaps and the value of the relative permeability of cast steel at this flux

density. The magnetization curve for cast steel is given by the following:

B (T) 0.1 0.2 0.3 0.4

H (A/m) 170 300 380 460

Expert's answer

The absolute permeability of cast steel is 5.0×10−3 H/m.

Find the magnetic reluctance of the steel and two air gaps respectively:


Rs=Lsμsμ0A, Ra=2Laμaμ0A.R_s=\frac{L_s}{\mu_s\mu_0A},\\\space\\ R_a=\frac{2L_a}{\mu_a\mu_0A}.

The magnetomotive force (MMF) is the flux times total reluctance:

FM=ΦR=Φ(Rs+Ra)=Φ(Lsμsμ0A+2Laμ0A), FM=Φμ0A(Lsμs+2La), FM=0.00054π107200106(0.120.005+20.0002)==47.7106 A.F_M=\Phi R=\Phi(R_s+R_a)=\Phi\bigg(\frac{L_s}{\mu_s\mu_0A}+\frac{2L_a}{\mu_0A}\bigg),\\\space\\ F_M=\frac{\Phi}{\mu_0A}\bigg(\frac{L_s}{\mu_s}+2L_a\bigg),\\\space\\ F_M=\frac{0.0005}{4\pi\cdot10^{-7}\cdot200\cdot10^{-6}}\bigg(\frac{0.12}{0.005}+2\cdot0.0002\bigg)=\\=47.7\cdot10^6\text{ A}.

Find the relative permeability of steel:


μr=μsμ04000.\mu_\text{r}=\frac{\mu_s}{\mu_0}\approx4000.
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