chapter 3_magnetic cct (1)
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MAGNETIC CIRCUITS
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INTRODUCTION Magnetic Circuit
A closed path containing magnetic flux. It also contain air gaps and other materials.
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INTRODUCTION (Cont…)
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PROPERTIES OF MAGNETIC CIRCUIT Flux Density, B
Occur when magnetomotive force (mmf) is applied to magnetic material.
It will induced flux, Unit in Webers(wb) The number of flux line
per area is called flux density, B.
Unit in Tesla
AB
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PROPERTIES OF MAGNETIC CIRCUIT (cont..) Permeability,µ
Different magnetic material has it own strength of magnet.
This strength is vary depends: Type of core being used Number of flux line through the core.
Permeability of free space (vacuum) is denoted by µo.
0
70 /104
r
Amwb
Relative permeability, µr
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AMPERE’S CIRCUITAL LAW Use the same concept as electric circuit.
Flux = current. It produced in a core Magnetomotive force (mmf)= electromotive
force (emf).
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AMPERE’S CIRCUITAL LAW (cont…)
REI
ICurrentEemf
,
RNI
FluxNIFmmf
,
R = Reluctance
AlR
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AMPERE’S CIRCUITAL LAW (cont…) Electric Circuit
Apply KVL ΣV = 0
Magnetic Circuit Apply Ampere’S
circuital Law ΣF = 0 The mmf drop is
HlF
H : magnetizing forcel:length of the section
(At)
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Determine
BAA
B
LNIH
H can be obtain from B-H curve
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cba
HlNIHlHlHlNI
F
ironcobaltsteel
)()()(
0
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B-H CURVE (MAGNETIZING CURVE)
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SERIES MAGNETIC CIRCUITS Reluctance is seldom determined. Flux can be determined using ohm’s law
analogy. The value of B and H is determined using
B-H curve and µ is calculated if asked for.
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Example 2: For magnetic circuit below, find the value of I to develop a magnetic flux of 4 x 10-4wb and µ and µr for the material under these conditions.
Cast steelL = 0.16mA = 2 x 10-3 m2 NI
I
RN=400
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TBA
B
2.0102104
3
4
Flux Density, B From B-H curve : H = 170 At/m
mAI
I
NHlI
HlNIF
68400
16.0170
83.9351041018.1
/1018.11702.0
7
3
3
or
AmwbHB
Permeability, µ
Solution:
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Example 3:The electromagnetic has picked up a section of cast iron. Determine the current I required to establish the indicated flux of the core.
Sheet steel
Cast steel
GivenLab = Lcd = Lef = Lfa = 10cm
Lbc = Lde = 1cm
A = 4 x 10-4 m2
Flux = 3.5 x 10-4 wb
e
f
b
a
e c
N=50t
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AIR GAPS Exist when there is an
opening at the electromagnetic device.
Air gaps will cause the flux to bends outwards. The flux density is not
uniformly equal to the flux density in the surrounding magnetic material (fringing).
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AIR GAPS (cont…) Flux density, Bg Magnetizing force, Hg
g
gg AB
ggg
og
g
gg
lHF
BH
The permeability of air is equal to permeability of free space
The mmf, Fg across the air gapcoreg
coreg
AA
where
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Example 4: Find the value of I required to establish a magnetic flux of 0.75 x 10-4 wb in a series magnetic circuit.
Given:Air gap = 2 x 10-3 mArea = 1.5 x 10-4 m2
Lcdfeab = 100 x 10-3 m
Lbc = 2 x 10-3 m
ba
e
c
f
d
I
200t
Cast steel
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Solution
TA
B 5.0105.11075.0
4
4
Flux Density, B
From B-H curve : HCS = 280 At/m
mAtB
HO
gg /1098.3
1045.0 5
7
For air gap, Hg
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AI
I
NHl
I
AtHl
Hl
HlHlHl
HlNIF
gCS
12.4200824
824
1021098.310100280 353
Ampere’s Circuital Law
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SERIES PARALLEL MAGNETIC CIRCUITS The close analogies between electric and magnetic circuits will lead to series-parallel
magnetic circuits. The analogy will be helpful in solving the problem. Example: Find the value of I required to establish a magnetic flux of 1.5 x 10-4wb in cd. Given
that, lbcde = lefab = 0.2m, lbe = 0.05m, cross sectional area, A = 6 x 10-4 m2
Φ1
Φ2bI
50t
f
c
de
aSheet steel
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Exercise 1
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Exercise 2
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Exercise 3
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Exercise 4
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Exercise 5
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