CHAPTER 5 MAGNETIC CIRCUIT

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Presentation transcript:

CHAPTER 5 MAGNETIC CIRCUIT OBJECTIVES Becomes aware of the similarities between the analysis of magnetic circuits and electric circuits. Develop a clear understanding of the important parameters of a magnetic circuit and how to find each quantity for a variety of magnetic circuit configurations. Begin to appreciate why a clear understanding of magnetic circuit parameters is an important component in the design of electrical/electronic systems.

MAGNETIC FIELD -Magnet and electromagnetic used widely in: Motors, generators, transformers, loudspeaker medical equipment, relays and etc. FLUX DENSITY A Flux density (b) – A measure of the flux per unit area perpendicular to amagnetic flux path.It is measured in tesla(T) or webers per square meter.(Wb/m2) B =  /A B = Wb/m2 =tesla (T)  = webers (Wb) A = m2 The pressure on the system to establish magnetic lines of force is determined by the applied magnetomotive force which is directly related realated to the number of turns and current of the magnetizing coil .

Magnetizing coil : F =NI F = ampere turns (At) N= turns (t) I= amperes (A) The level of magnetic flux established in a ferromagnetic core is a direction of the permeability of the material. Ferromagnetic materials have a very high level of permeability while non- magnetic materials such as air and wood have very low levels. The ration of the permeability of the material to that of air is called relative permeability and is defined: r =  /o o = 4 x 10-7Wb/A .m Change the magnetomotive force, and the relative permeability changes.

RELUCTANCE The resistance of a material to the flow of charge (current) is determined for electric circuits by the equation : R = ( l / A) (ohms,  ) The reluctance of a material to the setting up magnetic flux lines in the material is determined by the following equation: R = l / A (rels or At/Wb) R = reluctance l = length of the magnetic path A the cross-sectional area. Notes that the resistance and reluctance are inversely proportional to the area, indicating that an increase in area results in a reduction in each and an increase in the desired result, current and flux. For an increase in length , the opposite is true, and the desired effect is reduced. The larger the  , or the smaller the  , the smaller the reluctance and resistance .

OHM’S LAW FOR MAGNETIC CIRCUIT For the magnetic circuits, the effect desired is the flux .  = F / R MAGNETIZING FORCE The magnetomotive force per unit length is called the magnetizing force (H) : - H = F / l (At/m) ……(1) H = NI / l (At/m) ……(2) If ,NI = 40 At and l = 0.2 m , then H = NI / l = 40 At / l = 0.2 m = 200 At/m

FIGURE 1: Defining the magnetizing force of a magnetic circuit

In words, the result indicates that there are 200At of pressure per meter to establish flux in the core. The direction of the flux  can be determined by placing the fingers of your right hand in the direction of current around the core and noting the direction of the thumb. Realize that, the magnetizing force is dependent of the type of core material, it is determined solely by the number of turns, the current and the length of the core. The flux density and the magnetizing force are related by the following equation : B =H H = henries (not the magnetizing force (H) The greater the permeability, the greater the induced flux density.

HYSTERESIS FIGURE 2 : Normal magnetization curve for three ferromagnetic materials. Figure 2 is called the normal magnetization curve.an expanded view of one region appears in figure 3 This is for the higher .

Figure 3 Expanded view of figure 2 for the low magnetizing force region

AMPERE’S CIRCUITAL LAW A law establishing the fact that the algebraic sum of the rises and drops of the mmf around a closed loop of a magnetic circuit is equal to zero. SERIES MAGNETIC CIRCUITS : DETERMINING NI NI= Hl N = no. of turns I = Hl / N I = Current

EXAMPLE 1 Find the value of I required to develop a magnetic flux of  = 4 x 10-4 Wb Determine μ and μr for the material under these condition. Flux density B ; B =  /A = 4 x 10-4Wb / 2 x 10-3 m2 = 2 x 10-1 T = 0.2T using B-H curves : H = (cast steel ) = 170 At/m (Plot in the graph) Applying Ampere’s circuital law yields NI= Hl I = Hl / N = (170 At/m) (0.16m) 400 = 68 mA

The permeability of the material ;  = B /H =0.2T /170 (At/m) = 1.176 x 10-3 Wb/A.m Relative permeability ; r = /o = 1.176 x 10-3 4 x 10-7 = 935.83

AIR GAPS FIGURE 4 : Air gaps : ( a) with fringing; (b) ideal

The flux density of the air gap: Bg =g / Ag g =  core Ag = A core Magnetizing force of the air gap is determine ; Hg = Bg /0 Hg = Bg /0 =Bg / 4 x 10-7 Hg = (7.96 x 105 ) Bg (At /m)

EXAMPLE 2 Find the value of I required to establish a magnetic flux of  = 0.75 x 10-4 Wb in the series magnetic circuit in figure below.

Solution The flux density for each section is B =  /A = 0.75 x 10-4 Wb /1.5 x 10-4 Wb = 0.5 T From the B-H curves H (steel ) = 2801 At/m From the equation Hg = (7.96 x 105 ) Bg (At /m)( 0.5 T ) = (3.98 x 105 At/m) the mmf drops are Hcore lcore = ( 280 At/m)(100 x 10-3m) = 28 At Hglg = (3.98 x 105 At/m)(2 x10 -3 m ) = 796 At. Applying ampere’s circuital law, NI = Hcore lcore + Hglg = 28 At + 296 At (200t) I = 824 At I = 4.12 A

DETERMINING ,  Example 3 : Calculate the magnetic flux  for the magnetic circuit in fig.below.

Solution ; By ampere’s circuital law, NI = Habcda l abcda Or Habcda = NI /l abcda = (60t)(5A) / 0.3 m = 300 At / 0.3 m = 1000 At / m Babcda = 0.39T (based on figure Normal magnetization curve for three ferromagnetic material ) And since B =  /A , we have  = BA = (0.39 T) (2 x 10-4 m2) = 0.78 x10-4 Wb.

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