Answer to Question #124257 in Electricity and Magnetism for victoria

Question

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

Answer 1

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:

$$R_s=\frac{L_s}{\mu_s\mu_0A},\\\space\\ R_a=\frac{2L_a}{\mu_a\mu_0A}.$$

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

$$F_M=\Phi R=\Phi(R_s+R_a)=\Phi\bigg(\frac{L_s}{\mu_s\mu_0A}+\frac{2L_a}{\mu_0A}\bigg),\\\space\\ F_M=\frac{\Phi}{\mu_0A}\bigg(\frac{L_s}{\mu_s}+2L_a\bigg),\\\space\\ F_M=\frac{0.0005}{4\pi\cdot10^{-7}\cdot200\cdot10^{-6}}\bigg(\frac{0.12}{0.005}+2\cdot0.0002\bigg)=\\=47.7\cdot10^6\text{ A}.$$

Find the relative permeability of steel:

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