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
1
Expert's answer
2020-07-01T17:22:19-0400

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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS