Answer to Question #124266 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 permeability of cast steel is 5.0×10⁻³ H/m.

Calculate the magnetic reluctance of the elements of the chain. For steel:

$$R_s=\frac{L_s}{\mu_s\mu_0A}.$$

For two airgaps:

$$R_a=2\frac{L_a}{\mu_0A}.$$

The flux is the magnetomotive force divided by magnetic reluctance:

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

The relative permeability of steel is

$$\alpha=\frac{\mu_s}{\mu_0}=3979.$$