Chap20_1pg - the hysteresis loop for the material....

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An^Fe-8O% Ni alloy has a maximum permeability of 300,000 when an inductance of 3500 gauss is obtained. The alloy iJphced in a 20-rurn coil that is 2 cm in length. what current must flow ttrough oe conductor coil to obain this field? Solntlon: SinceB = FH, H = B/tt = 3500 G/300,000 G/Oe : 0.0il2 Oe = 0.928 A/m Then: I = H(/n= (0.928 A/m)(0.02 m)/Znturns = 0.qOO93 A AnFe-49% Ni alloy has a maximum permeability of 64,000 when a magnetic field of 0.125 oersled is applied. TVhat inductance is obtained and what cunent is needed-' to obtain this inductance in a 200-turn, 3-cm long coil? Solutfon: B = pH: (64,000G/Oe)(0.125 Oe) = 8000c lf weconvert units,If = O.125 Oe/4rr X 10*3 OelAlm = 9.947 Alm I = H(/n = (9.947 A/m)(0.03 d/Zn turns = 0.00149 A = 1.49 mA A magnetic material has a coercive field of 167 A/m, a saturation magnetization of 0.616 Tesla, and a residual inductance
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Unformatted text preview: the hysteresis loop for the material. Sofutlon: M-t = Bo, = 0.616 T = 6160 G B, : 3000 G H, : 167 A/m x 4n x 10-3 Oe/A/m : /.1 Oe tlZO-Zg J:sing Figure ?n-L6,determine the following properties of the magnetic material' \ - - / , r \ . ! . ! - ' - - - ^ - ^ L : . r : . . , v (a) remanance (d) initial permeability.-. ib) saturation magnetization (e) mzudmum permeability ici coercive field (0 power (maximum BII product) solutlon: (a) remanance = 13'000 G (b) saturation magnetization : 14,000 G (c) coercive field : 800 Oe (d) initial permeability = 7000 G/L200oe = 5.8 G/Oe (e) mildmum penneability = 14,000 G/900 oe : 15.6 G/Oe (f) we can bry several Blproducts in the 4th quadrant: 12,000G x 450Oe:5.4 x 106G'Oe 10,000 G x 680 Oe : 6.8 x 106 G'Oe 8,000Gx 720Oe:5,76 x 106G'Oe The ma,ximum 8I{ product, or Power, is about 6.8 X 106 G ' Oe...
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This note was uploaded on 02/01/2012 for the course MSEG 302 taught by Professor Snively during the Spring '08 term at University of Delaware.

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