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barcode reader in asp.net codeproject lgap, Agap m I m0 in Software
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Bar Code Generation In None Using Barcode drawer for Software Control to generate, create barcode image in Software applications. Bar Code Drawer In None Using Barcode generation for Software Control to generate, create bar code image in Software applications. Known Quantities: Relative permeability; number of coil turns; coil current; structure
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Scanning Barcode In Visual Basic .NET Using Barcode recognizer for .NET framework Control to read, scan read, scan image in VS .NET applications. Linear Barcode Drawer In VB.NET Using Barcode encoder for .NET Control to generate, create 1D image in .NET framework applications. Find: gap ; Bgap Schematics, Diagrams, Circuits, and Given Data: r ; N = 1,000 turns; i = 10 A; lgap = 001 m; Agap = 01 m2 The magnetic circuit geometry is de ned in Figure 1621 Assumptions: All magnetic ux is linked by the coil; the ux is con ned to the magnetic core; the ux density is uniform The reluctance of the magnetic structure is negligible Analysis: Calculation of magnetomotive force From equation 1628, we calculate the magnetomotive force: F = mmf = N i = (1,000 turns)(10 A) = 10,000 A t Calculation of reluctance Knowing the magnetic path length and cross sectional area, we can calculate the equivalent reluctance of the two gaps: Rgap = lgap lgap 001 = 397 104 A t/Wb = = gap Agap 0 Agap 4 10 7 02 Req = 2Rgap = 794 104 A t/Wb 3 Calculation of magnetic ux and ux density From the results of steps 1 and 2, we calculate the ux 10,000 A t F = 0126 Wb = = Req 794 104 A t/Wb and the ux density: Bbar = 0126 Wb = = 126 Wb/m2 A 01 m2 Comments: Note that the ux and ux density in this structure are signi cantly larger
than in the preceding example because of the larger mmf and larger gap area of this magnetic structure The subject of electric motors will be formally approached in 17 EXAMPLE 165 Equivalent Circuit of Magnetic Structure with Multiple Air Gaps
Problem
Figure 1623 depicts the con guration of a magnetic structure with two air gaps Determine the equivalent circuit of the structure Solution
Known Quantities: Structure geometry Find: Equivalent circuit diagram Assumptions: All magnetic ux is linked by the coil; the ux is con ned to the magnetic core; the ux density is uniform The reluctance of the magnetic structure is negligible Analysis: Calculation of magnetomotive force F = mmf = N i
16
Principles of Electromechanics
I A Igap 1 gap 1 Igap 2 Agap 2
Calculation of reluctance Knowing the magnetic path length and cross sectional area we can calculate the equivalent reluctance of the two gaps: Rgap 1 = Rgap 1 = lgap 1 lgap 1 = gap 1 Agap 1 0 Agap 1 lgap 2 lgap 2 = gap 2 Agap 2 0 Agap 2 N turns
Calculation of magnetic ux and ux density Note that the ux must now divide between the two legs, and that a different airgap ux will exist in each leg Thus: 1 = 2 = N i 0 Agap 1 Ni = Rgap 1 lgap 1 N i 0 Agap 2 Ni = Rgap 2 lgap 2 f Ni + _ Figure 1622 Magnetic structure with two air gaps
and the total ux generated by the coil is = 1 + 2 The equivalent circuit is shown in the bottom half of Figure 1622 Comments: Note that the two legs of the structure act like resistors in a parallel circuit
EXAMPLE 166 Inductance, Stored Energy, and Induced Voltage
Problem
1 2 Determine the inductance and the magnetic stored energy for the structure of Fig 1617(a) The structure is identical to that of Example 162 except for the air gap Assume that the ux density in the air gap varies sinusoidally as B(t) = B0 sin( t) Determine the induced voltage across the coil, e Solution
Known Quantities: Relative permeability; number of coil turns; coil current; structure geometry; ux density in air gap Find: L; Wm ; e Schematics, Diagrams, Circuits, and Given Data: r ; N = 500 turns; i = 01 A The magnetic circuit geometry is de ned in Figures 1614 and 1615 The air gap has lg = 0002 m B0 = 06 Wb/m2 Assumptions: All magnetic ux is linked by the coil; the ux is con ned to the magnetic core; the ux density is uniform The reluctance of the magnetic structure is negligible Analysis: Part 1 To calculate the inductance of this magnetic structure, we use equation 1630: N2 R Thus, we need to rst calculate the reluctance Assuming that the reluctance of the structure is negligible, we have: L= Rgap = lgap lgap 0002 = 159 107 A t/Wb = = gap Agap 0 Agap 4 10 7 00001

