CHAPTER 5 MAGNETIC CIRCUIT

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Transcript CHAPTER 5 MAGNETIC CIRCUIT

CHAPTER 5 MAGNETIC
CIRCUIT
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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
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-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 .
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Magnetizing coil :
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F =NI F = ampere turns (At)
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N= turns (t)
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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
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Change the magnetomotive force, and the relative permeability
changes.
RELUCTANCE
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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
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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) : -
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H=F/l
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(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.
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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
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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
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NI= Hl
I = Hl / N
N = no. of turns
I = Current
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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.
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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 :
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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
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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
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FIGURE 4 : Air gaps
: ( a) with fringing;
(b) ideal
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The flux density of the air gap:
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Bg =g / Ag
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g =  core
Ag = A core
Magnetizing force of the air gap is determine ;
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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 104 Wb in the series magnetic circuit in figure below.
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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 , 
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Example 3 :
Calculate the
magnetic flux  for
the magnetic circuit in
fig.below.
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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.
APPLICATION
speaker
 microphones
 Hall Effect Sensor
 Magnetic Reed Switch
 Magnetic Resonance Imaging
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