1. 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 The coil has 80 turns and a resistance of 0.05 Ω. The cross-sectional area of the core material is m2.
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2. Flux “Fringing”
All lines of flux must
leave and arrive
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3. Flux Distribution
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4. 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
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5. 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 (Next slide)
From the curve, H0.30=H0.60=0.47 oersteds
Multiply by for A-t/m
H30 = H60 = 37.4 A-t/m
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6. 0.2
0.47
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7. Magnetic Potential Differences
F0.30 = H·l = (37.4)(0.30) = A-t
F0.69 = H·l = (37.4)(0.69) = 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) = A-t
Fbghe = = 833 A-t
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8. For the bcde leg Fbghe = Fbcde
Hbcde=Fbcde/lbcde=833/(1+1+1)= A-t/m
In oersteds
Hbcde = / = 3.49 oersteds
Determine Bbcde from the magnetization curve in Fig (Next slide)
Bbcde = 1.45 T
Φbcde = BA = (1.45)(0.04) = Wb
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9. 1.45
3.49
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10. 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 (Next slide)
Hefab = 37 oersteds
Hefab = (37)(79.577) = A-t/m
Fefab= H·l = ( )( )
Fefab = A-t
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11. 1.65 37
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12. Total mmf to be supplied by the coil
FT = Fbghe + Fefab =
FT = A-t
FT = N·I =
I = A
V = I·R =(106.1)(0.05)
V = 5.30 V
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13. 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.
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14. 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) =
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15. 1.65
450 37
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16. 4000
0.2
0.47
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17. 1.45
4100
3.49
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