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.
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Flux “Fringing”
• All lines of flux must
• leave and arrive
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Flux Distribution
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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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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
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0.2
0.47
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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
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For the bcde leg
• Fbghe = Fbcde
• Hbcde=Fbcde/lbcde=833/(1+1+1)=277.67 At/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
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1.45
3.49
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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
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1.65
37
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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
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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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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
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1.65
450
37
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4000
0.2
0.47
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1.45
4100
3.49
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