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However, the air gap re- mains unsaturated, since the B-H curve for the air medium is linear (J.t is constant). A magnetic circuit having two or more media-such as the magnetic core and air gap in Fig. 1.8-is known as a composite structure. For the purpose of analysis, a magnetic equivalent circuit can be derived for the composite structure. Let us consider the simple composite structure of Fig. 1.9a. The driving force in this magnetic circuit is the mmf, F = Ni, and the core medium and the air gap medium can be represented by their corresponding reluctances. The equivalent magnetic circuit is shown in Fig. 1.9b. ffi =l g f.toAg <I>= Ni ffic + ffig Ni = Hclc + Hglg where lc is the mean length of the core lg is the length of the air gap The flux densities are B = <l>c c Ac ( 1.15) (1.16) (1.17) (1.18) ( 1.19) (1.20) In the air gap the magnetic flux lines bulge outward somewhat, as shown in Fig. 1.1 0; this is known as fringing of the flux. The effect of the fringing <I> ll + ffic N Tg Ni 9kg (a) (b) FIGURE 1.9 Composite structure. (a) Magnetic core with air gap. (b) Magnetic equivalent circuit. 8 chapter 1 Magnetic Circuits FIG{Tl?,E 1.10 Fringing flux. is to increase the cross- fringing effect can be n sectional areas of the ' EXAMPLE 1.1 :tional area of the air gap. For small air 'rected. If the fringing effect is neglected, ·e and the air gap are the same and the Ag =Ac <I> B =B =- g c Ac ps the cross- .re Figure E 1.1 represents the magnetic circuit of a primitive relay. The coil has 500 turns and the mean core path is lc = 360 mm. When the air gap lengths are 1.5 mm each, a flux density of 0.8 tesla is required to actuate the relay. The cc is cast steel. (a) Find the ct ;ent in the coil. (b) Compute + e values of permeability and relative permeability of the core. (c) If the · -sap is zero, find the current in the coil for the same flux dens; 8 T) in the core. Solutior (a) The den the, N p is small and so fringing can be neglected. Hence the flux s the same in both air ga·"1 2nd core. From the B-H curve o~ c steel core (Fig. 1. 7). FIGURE E1.1 N = 500 turns, lc = 36 em. - I I Magnetic Circuits 9 For Be= 0.8 T, He= 510 At/m mmf Fe= Hele = 510 X 0.36 = 184 At For the air gap, mmf Fg = Hg2lg = :: 2lg = 4:i~- 7 X 2 X 1.5 X 10-3 = 1910At Total mmf required: F =Fe+ Fg = 184 + 1910 = 2094At Current required: . F 2094 z = N = SOO = 4.19 amps Note that although the air gap is very small compared to the length of the core (lg = 1.5 mm, lc = 360 mm), most of the mmf is used at the air gap. (b) Permeability of core: (c) - Be - 0.8 - 1 57 X o-3 J.-te-He-510-. 1 Relative permeability of core: = f.-te = 1.57 X 10-3 = 1250 f.-tr }-to 47T10-7 F = Hcle = 510 X 0.36 = 184 At i = ~~~ = 0.368 A Note that if the air gap is not present, a much smaller current is required to establish the same flux density in the magnetic circuit. \u2022 EXAMPLE 1.2 Consider the magnetic system of Example 1.1. If the coil current is 4 amps when each air gap length is 1 mm, find the flux density in the air gap. Solution In Example 1.1, the flux density was given and so it was easy to find the magnetic intensity and finally the mmf. In this example, current (or mmf) is given and we have to find the flux density. The B-H characteristic for the 10 chapter 1 Magnetic Circuits air gap is linear, whereas that of the core is nonlinear. We need nonlinear magnetic circuit analysis to find out the flux density. Two methods will be discussed. 1. Load line method. For a magnetic circuit with core length le and air gap length lg, Rearranging, ( 1.21) This is in the form y = mx + c, which represents a straight line. This straight line (also called the load line) can be plotted on the B-H curve of the core. The slope is l 360 m = -1-Lof = -47rl0-7 l = -2.26 X 10-4 g The intersection on the B axis is _ Nif.Lo _ 500 X 4 X 47710-7 _ 1 256 l c--1-- 2xto-3 -. tesa g The load line intersects the B-H curve (Fig. El.2) at B = 1.08 tesla. Another method of constructing the load line is as follows: If all mmf acts on the air gap (i.e., He = 0) the air gap flux density is Ni Bg = T f.Lo = 1.256 T g This value of Bg is the intersection of the load line on the B axis. If all mmf acts on the core (i.e., Bg = 0), Ni 500 X 4 He= T = 36 X 10_2 = 5556 At/m e This value of He is the intersection of the load line on the H axis. 1.256 1.08 B 0~--~------------------------~~~--H Ni = 5555.6 lc FIGURE E1.2 Magnetic Circuits 11 2. Trial-and-error method. The procedure in this method is as follows. (a) Assume a flux density. (b) Calculate He (from the B-H curve) and Hg (= Bglp_0). (c) Calculate Fe ( = Hclc), Fg ( = Hglg), and F ( = Fe + Fg). (d) Calculate i = FIN. (e) If i is different from the given current, assume another judicious value of the flux density. Continue this trial-and-error method until the calculated value of i is close to 4 amps. If all mmf acts on the air gap, the flux density is Ni B = T p_0 = 1.256 T g Obviously, the flux density will be less than this value. The procedure is illustrated in the following table. B 1.1 1.08 800 785 EXAMPLE 1.3 8.7535 X 105 8.59435 X 105 288 282 1750.7 1718.87 F 2038.7 2000.87 i 4.08 4.0 \u2022 In the magnetic circuit of Fig. El.3a, the relative permeability of the ferro- magnetic material is 1200. Neglect magnetic leakage and fringing. All dimen- sions are in centimeters, and the magnetic material has a square cross- sectional area. Determine the air gap flux, the air gap flux density, and the magnetic field intensity in the air gap. Solution The mean magnetic paths of the fluxes are shown by dashed lines in Fig. El.3a. The equivalent magnetic circuit is shown in Fig. El.3b. F1 = N 111 = 500 X 10 = 5000 At F2 = N 212 = 500 X 10 = 5000 At P-e = 1200p_0 = 1200 X 47710-7 (fJt _ f bafe bafe-A P-c c 3 X 52 X 10-2 1200 X 47710- 7 X 4 X 10-4 = 2.58 X 106 At/Wb l 12 chapter 1 Magnetic Circuits 10 A NJ 500 turns \Jtbafe (a) \Jtbcde + N2 500 turns + F1 = 5000 At F2 = 5000 At (b) FIGURE E1.3 From symmetry 5 x w-3 4rr1o-7 x 2 x 2 x w-4 = 9.94 X 106 At/Wb CZft _ Z be( core) be(core) - A f-tc c 51.5 x 10-2 1200 X 4rr10- 7 X 4 X 10-4 = 0.82 X 106 At/Wb The loop equations are <P1(ezftbafe + ezftbe + ezftg) + <P2(ezftbe + ezftg) = F1 <P1(ezftbe + ezftg) + <Piezftbcde + ezftbe + ezftg) = F2 Magnetic Circuits 13 or or <.1>1(13.34 X 106) + <l>il0.76 X 106) = 5000 <.1>1(10.76 X 106) + <l>i13.34 X 106) = 5000 <l>1 = <l>z = 2.067 X 10-4 Wb The air gap flux is <l>g = <l>1 + <l>z = 4.134 X 10-4 Wb The air gap flux density is · B = <I>g = 4.134 X w-4 = 1.034 T g Ag 4 X 10-4 The magnetic intensity in the air gap is H = Bg = l.034 = 0.822 X 106 At/m \u2022 g /-to 47T10-7 1.1.6 INDUCTANCE A coil wound on a magnetic core, such as that shown in Fig. 1.11a, is frequently used in electric circuits. This coil may be represented by an ideal circuit element, called inductance, which is defined as the flux linkage of the coil per ampere of its current. Flux linkage Inductance A= N<l> L=~ l (1.22) (1.23) From Eqs. 1.22, 1.23, 1.11, 1.3, and 1.14, I A / Val f i ~ N4 ~~ ~--~ (a) L _ N<I> _ NBA _ N ~-tHA --------- i i i N~-tHA Hl!N (b) L ( 1.24) FIGURE 1.11 Inductance of a coil-core assembly. (a) Coil-core assembly. (b) Equivalent induc- tance. :I ! 14 chapter 1 Magnetic Circuits (1.25) Equation 1.24 defines inductance in terms of physical dimensions, such as cross-sectional area and length of core, whereas Eq. 1.25 defines inductance in terms of the reluctance of the magnetic path. Note that inductance varies as the square of the number of turns. The coil-core assembly of Fig. 1.11a is represented in an electric circuit by an ideal inductance as shown in Fig. l.llb. EXAMPLE 1.4

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