Example 1.1 a). Determine the voltage that must be applied to the magnetizing coil in order to produce a flux density of 0.2T in the air gap. Flux fringing will be assumed negligible. Assume that the magnetization curve for the core material (which is homogeneous) is that given in Fig 1.5. The coil has 80 turns and a resistance of 0.05 Ω. The cross-sectional area of the core material is 0.0400 m2. ECE 441

Flux “Fringing” All lines of flux must leave and arrive ECE 441

Flux Distribution ECE 441

Procedure Determine Φgap and Fgap Determine Hbcde, Bbcde, and Φbcde Determine Φefab, Befab, Hefab, and Fefab Determine FT and the required current Using Ohm’s Law, determine the required voltage ECE 441

For the Gap Φgap = BgapAgap = (0.2)(0.04) = 0.008Wb Flux Density to establish 0.2 T in the center leg is determined from the magnetization curve in Fig. 1.5. (Next slide) From the curve, H0.30=H0.60=0.47 oersteds Multiply by 79.577 for A-t/m H30 = H60 = 37.4 A-t/m ECE 441

0.2 0.47 ECE 441

Magnetic Potential Differences F0.30 = H·l = (37.4)(0.30) = 11.22 A-t F0.69 = H·l = (37.4)(0.69) = 25.81 A-t Fgap = Hgap·lgap To get Hgap, μgap = Bgap/Hgap  4πx10-7 = 0.2/Hgap Hgap = 159,155 A-t/m Fgap = (159,155)(0.005) = 795.77 A-t Fbghe = 11.22 + 25.81 + 795.77 = 833 A-t ECE 441

For the bcde leg Fbghe = Fbcde Hbcde=Fbcde/lbcde=833/(1+1+1)=277.67 A-t/m In oersteds Hbcde = 277.67/79.577 = 3.49 oersteds Determine Bbcde from the magnetization curve in Fig. 1.5. (Next slide) Bbcde = 1.45 T Φbcde = BA = (1.45)(0.04) = 0.058 Wb ECE 441

1.45 3.49 ECE 441

For the efab leg Φefab=Φgap + Φbcde=0.08+0.058=0.066 Wb Befab = Φ/A = 0.066/0.04 = 1.65 T Determine H to establish this field intensity from the magnetization curve in Fig. 1.5. (Next slide) Hefab = 37 oersteds Hefab = (37)(79.577) = 2944.35 A-t/m Fefab= H·l = (2944.35)(1+0.8+0.8) Fefab =7655.31A-t ECE 441

1.65 37 ECE 441

Total mmf to be supplied by the coil FT = Fbghe + Fefab = 7655.31 + 833 FT = 8488.31 A-t FT = N·I = 8488.31 I = 106.1 A V = I·R =(106.1)(0.05) V = 5.30 V ECE 441

Example 1.1 Part b Using equations 1-5 and 1-7, determine the relative permeability of each of the legs of the core, and compare the calculated values with corresponding values obtained from the permeability curve in Fig. 1.5. ECE 441

Combining Eq (1-5) and (1-7) μ = B/H μr = μ/μ0 μr = (B/H)/4π·10-7 = B/(4π·10-7·H) μrleft = 1.65/(4π·10-7·2944) = 446 μrcenter = 0.20/(4π·10-7·37.4) = 4256 μrright = 1.45/(4π·10-7·277.67) = 4156.1 ECE 441

1.65 450 37 ECE 441

4000 0.2 0.47 ECE 441

1.45 4100 3.49 ECE 441