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Power Electronic Systems & Chips Lab., NCTU, Taiwan
Magnetic Circuits
電力電子系統與晶片實驗室
Power Electronic Systems & Chips Lab.
交通大學 • 電機控制工程研究所
Chapter 1 Magnetic Circuits and Magnetic Materials, Fitzgerald & Kingsley's Electric Machinery, 7th Ed, S.D.
Umans, McGraw-Hill Book Company, 2013.
1/91
台灣新竹‧交通大學‧電機控制工程研究所‧電力電子實驗室~鄒應嶼 教授
Modeling the Stator Inductance of an IPMSM
L (t0 )
R
vi
vR
vi
1
s
iL
L
vL
1
L
iL
vR
RL
L1 ( r )  ?
Assume the rotor produces a sinusoidal flux distribution across the air gap,
how to model the stator winding inductance as a function of the rotor
position of an interior permanent magnet synchronous motor (IPMSM)?
REF: Chapter 4 Inductances, Design of Rotating Electrical Machines, Juha Pyrhonen, Tapani Jokinen, Valeria Hrabovcova,
2nd Ed., October 2013, Wiley.
2/91
Modeling of Synchronous Machine in dq-Frame

q
s
is
b  axis
id Ld

q
iq Lq d
λs
ib
vb
i1q
PM
s
1q

λPM
e

a'
r
1d
i1d
fd
vF
vs
d
ia
iF
va
a  axis
vc
r

PM
id
Electric Equations:
a
ic
Rs
vd  Rs  sLd e Lq id   sPM 
v     L
i   
 vd

R
sL
q
e
d
s
q
  
 q  ePM 
c  axis
Ld
e Lqiq
iq
sPM
Torque Equations:
3P
3P
d iq  (Ld  Lq )id iq  
(id Ld  PM )iq  (Ld  Lq )id iq 
Te 
22
22
Rs
Lq e Ld id
vq
ePM
Te  Tf  TL  J
dm
 Bm
dt
Torque Characteristics of Synchronous Machines
q
q
q
q
d
d
d
d
Ld  Lq
Ld  Lq
Ter  0
Tem  Ter
Te 
Ld  Lq
Ld  Lq
m  0
Tem  0
Tem  Ter
3P
[(id Ld  PM )iq  iq id ( Ld  Lq )]
22
Tem
Ter
Inductance Plays a Key Role in Motor Characteristics
L1 ( r )  ?
The rotor structure determines the major characteristics of a synchronous machine (SM). For
SM with concentrated winding stator, the inductance of the coil of a segmented teeth can be
calculated as a function its rotor position if the rotor has an anisotropic structure.
Generated electric torque of synchronous machine:
Te 
3P
[ I q m  I q I d ( Ld  Lq )]
22
Design of Rotating Electrical Machines, 2nd Ed., [Chapter 4: Inductances]
Juha Pyrhonen, Tapani Jokinen, Valeria Hrabovcova,
October 2013, Wiley.
[1] I. A. Viorel, A. Banyai, C. S. Martis, B. Tataranu, and I. Vintiloiu, “On the segmented rotor reluctance synchronous motor saliency ratio
calculation,” Advances in Electrical and Electronic Engineering, vol. 5, vo. 1-2, June, 2011.
[2] B.J. Chalmers and A. Williamson, AC Machines Electromagnetics and Design, Research Studies Press Ltd., John Wiley and sons Inc., 1991.
[3] Jong-Bin Im, Wonho Kim, Kwangsoo Kim, Chang-Sung Jin, Jae-Hak Choi, and Ju Lee, “Inductance calculation method of synchronous
reluctance motor including iron loss and cross magnetic saturation,” IEEE Transactions on Magnetics, vol. 45, no. 6, pp. 2803-2806, 2009.
Basic Notations for Electromagnetism
電場強度
Electric field strength
E
[V/m]
磁場強度
Magnetic field strength
H
[A/m]
電力線密度
Electric flux density
D
[C/m2]
磁力線密度
Magnetic flux density
B
[Vs/m2], [T]
電流密度
Current density
J
[A/m2]
電荷密度
Electric charge density, dQ/dV
ρ
[C/m3]
D  E
 電容率 permittivity of free space (Farads/m)
B  H
 磁導率 permeability of free space (Henrys/m)
6/91
1865年～電磁理論誕生～馬克士威電磁方程式
1865年英國物理學家詹姆斯·馬克士威以先前的電磁研究成果為基礎，經
由有系統地整理和綜合提出了由20個等式和20個變量組成的電磁理論。
現今電磁理論的數學形式分別由英國的奧利弗·黑維塞(Oliver Heaviside)
、 美 國 的 約 西 亞 · 吉 布 斯 (Josiah Gibbs) 、 與 德 國 的 海 因 里 希 ‧ 赫 茲
(Heinrich Hertz)於1884年前後以向量形式重新表達成目前的四組方程式。





Source of Electric Field   E 
0
Changing Electric Field
Source of Magnetic Field   B  0
Changing Magnetic Field
B
t
E 

  B  0  J   0

t 

E  
電磁學天堂祕笈：輕鬆解析最實用的馬克士威方程式
A Student’s Guide to Maxwell’s Equations
作者：夫雷胥 (Daniel Fleisch), 譯者：鄭以禎
出版社：天下文化, 出版日期：2010年10月18日
7/91
Maxwell's Equations (1860s~1970s)
現今使用的馬克士威的方程式包含四個方程式組，它們分別描述了靜電場
、靜磁場、感應電場、與感應磁場的關係，其積分與微分形式表示如下：
 電場的高斯定律 (Gauss’s Law for Electric Field)
Q enc

ˆ
nda
E




E


0
S
0
 法拉第定律 (Faraday’s Law)

C
E  dl  
d
dt

S
ˆ
B  nda
E  
B
t
 磁場的高斯定律 (Gauss’s Law for Magnetic Field)

S
ˆ
B  nda
0
B  0
 安陪-馬克士威定律 (The Ampere-Maxwell Law)
d

B
l





d
I
0  enc
0
 C
dt


E 

ˆ
E

nda






B
J

 S
0
0


t 

Symbols and Units of Electromagnetic Quantities
Summary of Quasi-Static Electromagnetic Equations
Electromechanical Dynamics (MIT Course Notes)
s
r
is
ir
s  is Ls  Lsr ( )ir
r  Lsr ( )is  ir Lr
dL ( )
Te  is ir sr
d
Te

Electromechanical Dynamics, Discrete Systems (Part 1),
Herbert H. Woodson and James R. Melcher,
Wiley, 1st Ed., January 15, 1968.
REF: Electromechanical Dynamics - Part 1 Discrete Systems (Woodson & Melcher, MIT 1968)
Basic Relations of Electrical and Magnetic Field
v(t )
Faraday’s Law
terminal
characteristics
B (t ),  (t )
Core
characteristics
H (t ), F (t )
i (t )
Ampere’s Law
Electrical Circuits
Magnetic Circuits
12/91
Magnetic Field
Magnetic fields are produced by electric currents, which can be macroscopic currents in
wires, or microscopic currents associated with electrons in atomic orbits. The magnetic
field B is defined in terms of force on moving charge in the Lorentz force law. The
interaction of magnetic field with charge leads to many practical applications. Magnetic
field sources are essentially dipolar in nature, having a north and south magnetic pole.
The SI unit for magnetic field is the Tesla, which can be seen from the magnetic part of the
Lorentz force law Fmagnetic = qvB to be composed of (Newton x second)/(Coulomb x
meter).
A smaller magnetic field unit is the Gauss (1 Tesla = 10,000 Gauss).
13/91
Right-Handed System and Left-Handed System
z
z
x
y
y
Left-Handed System
x
Right-Handed System
14/91
Magnetic Field of Current: Right-Handed Rule
The magnetic field lines around a long wire which carries an electric current form
concentric circles around the wire. The direction of the magnetic field is
perpendicular to the wire and is in the direction the fingers of your right hand
would curl if you wrapped them around the wire with your thumb in the direction
of the current.
15/91
Ampere’s Law
(a)
H
 dl 

i
(b)
(a) General formulation of Ampere’s law.
(b) Specific example of Ampere’s law in the case of a winding on a magnetic core
with air gap.
 Direction of magnetic field due to currents
 Ampere’s Law: Magnetic field along a path
16/91
Ampere’s Law
 H  dl  I
 B  dl  I
B
dl
B  H
H = magnetic field intensity (Ampere-turns/m)
 = magnetic permeability of material
(Wb/A.m, or Henery/m)
B = magnetic flux density (Tesla, Weber/m2)
 = permeability of free space
 0  4   10 7 H / m
r
I, if contour encloses I
 H  dl  0, if contour does not enclose I


0
r = relative permeability (between 2000-80,000 for ferromagnetic materials)
17/91
Permeability: Relationship Between B and H
Ampere,s Law
 H  dl  I
permeability =  =
H = magnetic field intensity (Ampere-turns/m)
 = magnetic permeability of material (Wb/A.m, or Henry/m)
B = magnetic flux density (Tesla, Weber/m2)
B
H
r 

0
 = permeability of free space
 0  4   10  7 H / m
r = relative permeability (between 2000-6000 for general ferromagnetic materials used in
electrical machines)
In magnetics, permeability is the ability of a material to conduct flux. The magnitude of the
permeability at a given induction is a measure of the ease with which a core material can be
magnetized to that induction. It is defined as the ratio of the flux density B to the magnetizing
force H. Manufacturers specify permeability in units of Gauss per Oersted (G/Oe).
cgs:
 gauss 
tesla
 0 = 1

 10 4

 oersted  oersted
mks:
 henrry 
 meter 
 0 = 4  10 7 
18/91
磁通量單位：韋伯 (Wb), 磁通量密度單位：特斯拉(Tesla)
磁通量的國際制(SI)單位，紀念德國物理學家韋伯而命名。簡稱韋﹐用Wb表示。
韋伯定義如下﹕令通過單匝線圈的磁通量在 1秒鐘內均勻地減小到零。如果在該線
圈中激發產生的感應電動勢為 1 伏特，則原來通過該線圈的磁通量為 1 韋伯。即
1Wb=1V．s。
韋伯是國際單位制的導出單位﹐用基本單位表示的關係式為：
米2‧千克‧秒-2 ‧安培-1 (m 2 ‧kg‧s-2‧A-1) = 伏秒 (Volt‧sec)
1882 年西門子在英國科學進展協會上第一次提出以『韋伯』作為磁通量單位，
1895年得到英國科學進展協會承認，1948年得到國際計量大會的承認。
韋伯和 CGS電磁系中的磁通量單位馬克斯威之間的換算關係為﹕
1韋伯相當於108馬克斯威。[1 Wb = 108 Maxwell]
磁通量密度的單位是特斯拉(Tesla)。 [1 Tesla = 1 Wb/m2]
1 oersted = 1000/4 ampere/turn = 79.57747154594 ampere/meter  80 A/m
19/91
Properties of Ferromagnetic Materials

B, Wb/m2
B

A is the cross-sectional area
A
B  r0H
1.4
1.2
i
1.0
0.8
N
0.6
0.4
0.2
DC Excitation
0
0
200
400
600
800
H, A-turn/m
1000
A toroidal coil and the
magnetic field inside it.
 Ferromagnetic materials, composed of iron and alloys of iron with cobalt,
tungsten, nickel, aluminum, and other metals, are by far the most common
magnetic materials.
 Transformers and electric machines are generally designed so that some
saturation occurs during normal, rated operating conditions.
20/91
B-H Curve, Permeability, and Incremental Permeability
B  H   r  0 H
Relation between B- and H-fields.
Incremental Permeability
B
H
Bs
B
 
B

H
B
H
B
B

H H
H
Linear region
Hs
Magnetic intensity H, [A-turns/m]
 The B-H characteristics of a
core material is high nonlinear.
Depends on its average
current,
current
ripple,
switching
frequency,
and
operation temperature.
 When
measuring
the
inductance of a magnetic
circuit, it should first to
identify its operating point.
21/91
B-H Curve of Major Materials
The set of magnetization curves as shown in
left figure represents an example of the
relationship between B and H for soft-iron and
steel cores but every type of core material will
have its own set of magnetic hysteresis curves.
You may notice that the flux density increases
in proportion to the field strength until it
reaches a certain value were it can not
increase any more becoming almost level and
constant as the field strength continues to
increase.
This is because there is a limit to the amount of flux density that can be generated by the core as all the
domains in the iron are perfectly aligned. Any further increase will have no effect on the value of M, and
the point on the graph where the flux density reaches its limit is called Magnetic Saturation also known as
Saturation of the Core and in our simple example above the saturation point of the steel curve begins at
about 3000 ampere-turns per meter.
22/91
B-H Characteristics of a Magnetic Material

Hs
 Performance Tradeoffs: saturation Bs, permeability , resistivity  (core
loss), remanence Br, and coercivity Hc.
Flux Density or B-Field
B  H   r  0 H
Cross-sectional area A
H-field
H
i
  BA
N
i
  N
(a)
(b)
Determination of the magnetic field direction via the right-hand in (a) the general case
and (b) a specific example of a current-carrying coil wound on a toroidal core.
The total flux pass through the coil with N turns is called
flux linkage and named as .   N 
24/91
Continuity of Flux
  
   B dA
A(closed surface)
A
B dA  0
A2
A1
1
2
3
A3
1  2  3  0
B1 A1  B2 A2  B3 A3  0 or

k
0
k
25/91
Magnetic Cores
Ideal Inductor
i
 Negligible winding resistance
 Perfect coupling between windings
v

N
 An ideal core
v  N
d
dt
d 
1
vdt
N
 ( t1 )   ( t 0 ) 
1
N

t1
t0
vdt
The above equation shows that the change in flux during a time interval t0-t1 is
proportional to the integral of the voltage over the interval, or the volt-seconds applied
to the winding.
26/91
Ideal Inductor [Define its Initial Conduction]


i
v
i
N
(a) Circuit model.
(b) -i characteristic (or B-H curve).
v
v

0

t
0
(c) v is a step input; (t0) = 0.
(d) v = Vm sin t ; (t0) = 0.
v
v


0
t
(e) v is a square wave; (t0) = -m.
0
(f) v = Vm sin t ; (t0) = -m.
27/91
Magnetic Field Strength H of Some Configurations
long, straight wire
Toroidal Coil
Long solenoid
28/91
Inductance of Wound Magnetic Core

  magnetic flux per turnwebers (Wb) [1 Wb = 108 Maxwell]
i
v
B  magnetic flux density webers/meter2 (teslas)
  flux linkage webers
N
A  core cross-sectional area square meters
H  magnetic field strength ampere-turns/meter
N  number of turns
d
di
d
 N

dt
dt
dt
d
d

L  N
di
di
v  L
i  coil current ampere
l m  mean length of magnetic flux path meters
  permeability henrys/meter (4×10-7 in perfect vacuum)
  N
  BA
and
L  inductancehenrys
Ni
H 
lm
The inductance of a wound magnetic core is directly proportional
to the incremental permeability of the core material, which is the
slope of the B-H curve.
N 2 A dB
N 2 A
L 

l m dH
lm
29/91
Inductance of a Core
i
A
N
v
N2  A
L
  r  o
le

le
 r  o  C1  N 2
L  N2
 r  C  N 2
L  le1
(a)
Flux saturation

slope L
(b)
i
L A
C1 
A
l
C   o  C1   o 
A
l
The inductance L represents the capability of
magnetic flux density produced by unit current of
a circuit loop.
30/91
Magnetic Reluctance and Permeance
Cross-sectional
area A
Mean path length l
i
 H  dl 
H 
N
H 
Permeability 
 
Magnetic-motive force (mmf)
  Ni
l
A
Reluctance

Permeance
1


Hl  Ni
Ni
l
Ni

  B 
l
A
ANi
Ni


 
l
 l  


A




 Ni

 
31/91
Inductance of a Toroidal Core (Self Inductance)
Cross-sectional
area A
Mean path length l
Weber-turns (=N)
i
N
L

i
Amp (I)
Permeability 
For a magnetic circuit that has a linear relationship between  and i because of material of
constant permeability or a dominating air gap, we can define the -i relationship by the selfinductance (or inductance) L as
L

i

N
i

 
ANi

l
L

i

N ANi
A
  N 2
i l
l
where  =N, the flux linkage, is in weber-turns. Inductance is measured in henrys or weberturns per amp.
32/91
Flux Density Distribution of a Toroidal Core
Representing the magnetic vector potential (A), magnetic flux (B), and current
density (j) fields around a toroidal inductor of circular cross section. Thicker lines
indicate field lines of higher average intensity. Circles in cross section of the core
represent B flux coming out of the picture. Plus signs on the other cross section of
the core represent B flux going into the picture. Div A = 0 has been assumed. 33/91
A toroidal coil and the
magnetic field inside it.
Energy Stored in a Core
N: number of turns
Mean path length l
Cross-sectional
area Ac
The energy stored in the core:
t
E L   Pdt 
0

t
0
Li ' di ' 
1 2
LI
2
The energy density (energy/volume) is:
I
LI 2
1 1   N 2 Ac

B 

Ac l
Ac l 2 
l
1 B2
B2


2 
2r 0
1
2
Permeability 
A N2
LN  

l
2
  B 2l 2 
 2 2 
  N 
The energy stored in the core:
EL 
1 2
LI   BVcore
2
Vcore: volume of the core
Chapter 11 Inductance and Magnetic Energy of Introduction to Electricity and Magnetism, MIT 8.02 Course Notes, Sen-Ben Liao, Peter
Dourmashkin, and John Belche, Prentice Hall, 2011.
34/91
Typical Energy Density of a Ferrite Core
Ec
B2
B 

2r 0
Ve
For a typical ferrite, assuming the relative permeability is about r = 2000, and the
saturation flux density Bsat = 0.3 T (3000 G), we get (for most ungapped ferrite
cores) a typical power density of
0 .3 2
Ec
B2
B 


 17 . 9 J/m 3
7
2  r  0 2  2000  4  10
Ve
0  4  10 7 H/m
 4  10 Newton/A
7
2
Ec
 18 J/m 3  18 μJ/cm
Ve
(  r  2000, B sat  3000G)
3
一般典型的鐵氧體鐵心 (Ferrite core)的儲能密度約為每立方公分18微
焦爾。假設開關頻率為100 kHz，最大開關責任比為50%，工作於臨界
導通模式(CRM)，則其可處理之平均功率約為3.6瓦。以36瓦的反馳式
電源供應器為例，其鐵心體積約為10立方公分。
35/91
Inductance of Air-Core Solenoid
 H  dl  N i
Dc
lc

lc
0
Hdl  
Long air-core solenoid
lc  d
lc
( 0 )dl  
Hl c  Ni
2 lc  d
lc  d
( 0 )dl  
2 lc  2 d
2 lc  d
( 0 )dl  Ni
H N

i
lc
d
d ( BA)
dH 4 107 N 2 A N 2 Dc2 2
N
 0 NA


107
LN
di
di
di
lc
lc
0  4  10 7 H/m
 4  10 7 Newton/A 2
where
L  inductance in henrys
N  Total number of turns
A  cross-sectional area inside of solenoid coil in square meters (   D c2 / 4 )
Dc  diameter of solenoid in meters
l c  length of solenoid in meters
36/91
Inductance of a Solenoid
This is a single purpose calculation which gives you the inductance value when you make any change in the parameters.
Small inductors for electronics use may be made with air cores. For larger values of inductance and for transformers, iron is
used as a core material. The relative permeability of magnetic iron is around 200.
This calculation makes use of the long solenoid approximation. It will not give good values for small air-core coils, since they
are not good approximations to a long solenoid.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html
37/91
Inductance of a Solenoid
D=10 mm
I
b
c
l=50 mm
N=30
WD=1.0 mm
a
d
Wire diameter
 I , if contour encloses I
does not enclose I
 H  d l   0, if contour
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html
38/91
HW: Inductance of an Air-Core Solenoid
D=10 mm
I
b
c
l=50 mm
N=30
a
WD=1.0 mm
d
Wire diameter
An air-core solenoid with construction parameters as shown above, solve the following
problems:
1. Calculate the ideal equivalent inductance of the air-coil solenoid?
2. Compute the equivalent inductance of the air-coil solenoid.
3. Make a Maxwell simulation of the flux distribution of the air-core solenoid and compare
the simulated inductance with the the analytical result.
39/91
Simple Magnetic Circuits

c
F
 
F
c
 g  c
Analogy between electric and magnetic circuits.
Electrical-Magnetic Analogy
Magnetic Circuit
mmf Ni
Flux 
reluctance 
permeability 
Electric Circuit
v
i
R
1/, where =resistivity


i

N

41/91
Equivalent Electrical Circuit of a Magnetic Circuit

Reluctance

i


N

Inductance
  
Magnetic
Ni


1
A

N N 2

L 

i
i
Electrical
Ohm' s law :
l
v
 R 
A/
i
  k   N mim
Kirchhoff' s voltage law : i  Rk   v m

Kirchhoff's current law :  ik  0
k
k
k
m
k
0
l
(unit : H -1 )
A
m
k
42/91
Magnetic Circuits of a Gapped Core
l1 = mean path length
i1
in
i1
Airgap: Hg
N1
g
H
Core: H1
(a)
(b)
mean flux path in the ferromagnetic material
43/91
Modeling of a Simple Magnetic Circuit
i
v
mean flux path in the air gap
lg
b
a
N
magnetic motive force (mmf)
(unit: Ampere-turns)
li
mean flux path in the
ferromagnetic material
 H  dl 

b
a
I
a
H i dl   H g dl  N i
b
Hi : Magnetic field intensity in the ferromagnetic material
Hg : Magnetic field intensity in the air gap
H i l i  H g l g  Ni
44/91
Modeling of a Simple Magnetic Circuit
Bi
B  H
i
li 
Bg
g
l g  Ni
Flux
 
B
A
 dS
The surface integral of flux density is equal to the flux.
If the flux density is uniformly distributed over the cross-sectional area, then
 g  B g Ag
 i  Bi Ai
The streamlines of the flux density are closed, therefore  i   g
lg
li

  Ni
 i Ai
 g Ag
li
 i Ai
i 
g 
 i    g    (  i   g )  Ni
lg
 g Ag
45/91
Modeling of the Air-Gap
mean flux path in the air gap

lg
i
v
Rg
N
li
b
a
mean flux path in the
ferromagnetic material
Ni
Rc
In general,
Rg  Rc
46/91
Inductance of a Slotted Ferrite Core
AC: Cross Section Area
i
v
N
lg
b
a
NBc Ac N 2  0 Ac
L

i
lg
The shearing of an idealized B-H loop due to an air gap.
台灣新竹‧交通大學‧電機控制工程研究所‧電力電子實驗室~鄒應嶼 教授
47/91
Air-gap Fringing Fields
Reluctance of the air gap:
g 
lg
 g Ag
The effect of the fringing fields is to increase the
effective cross-section area Ag of the air gap.
Fringing flux decreases the total reluctance of the
magnetic path and, therefore, increases the
inductance by a factor, F, to a value greater than
the one calculated.
[1] Colonel Wm. T. McLyman, Fringing Flux and Its Side Effects, AN-115 Kg Magnetics Inc.
[2] Colonel Wm. T. McLyman, Chapter 3 Magnetic Cores of Transformer and Inductor Design Handbook, Fourth Edition, CRC
Press, April 26, 2011.
[3] W.A. Roshen, “Fringing field formulas and winding loss due to an air gap,” IEEE Transactions on Magnetics, vol. 43, no. 8, pp.
3387-3394, 2007.
48/91
Fringing Flux at the Gap
The effect of the fringing fields is to increase the effective cross-section area Ag of the air gap. The fringing
flux effect is a function of gap dimension, the shape of the pole faces, and the shape, size and location of the
winding. Its net effect is to shorten the air gap. Fringing flux decreases the total reluctance of the magnetic
path and, therefore, increases the inductance by a factor, F, to a value greater than the one calculated. In most
practical applications, this fringing effect can be neglected.
49/91
A Simple Wound-Rotor Synchronous Machine
Calculate the air-gap flux density Bg
The magnetic structure of a synchronous machine is
shown schematically in the right figure. Assuming that
rotor and stator iron have infinite permeability ( ),
find the air-gap flux  and flux density Bg. For this
example I = 10 A, N = 1000 turns, g = 1 cm, and Ag =
2000 cm2.
2  lg
2  10 2
g 

 g Ag 4  10 7  0.2

F
1000  10

 0.13 Wb
g
g
Bg 

Ag

0.13
 0.65 T
0.2
50/91
Flux linkage, Inductance, and Energy
Faraday’s Law
 When magnetic field varies in time an electric field is produced in space as
determined by Faraday’s Law:
 E  ds  
C
d
B  da

dt S
v(t ) 
d
dt
 Line integral of the electric field intensity E around a closed contour C is equal to
the time rate of the magnetic flux linking that contour.
 Since the winding (and hence the contour C) links the core flux N times then
above equation reduces
 The induced voltage is usually referred as electromotive force to represent the
voltage due to a time-varying flux linkage.
v(t ) 
d
d
N
dt
dt
51/91
Direction of EMF
The direction of emf: If the winding terminals were short-circuited a current
would flow in such a direction as to oppose the change of flux linkage.
 (t )  max sin t  Ac Bmax sin t
e(t )   Nmax cos t  Emax cos t
e(t)
N
Emax   Nmax  2 f NAc Bmax
Erms  2 f NAc Bmax
52/91
Example: Estimate the Inductance of a Gapped Core
The magnetic circuit of Fig. (a) consists of an N-turn winding on a magnetic
core of infinite permeability with two parallel air gaps of lengths g1 and g2 and
areas A1 and A2, respectively.
Find (a) the inductance of the winding and (b) the flux density Bl in gap 1 when
the winding is carrying a current i. Neglect fringing effects at the air gap.
53/91
Example: Plot the Inductance as a Function of Relative
Permeability
The magnetic circuit as shown below has dimensions Ac = Ag = 9 cm2, g = 0.050 cm, lc = 30
cm, and N = 500 tums. With the given magnetic circuit, using MATLAB to plot the inductance
as a function of core relative permeability over the range 100 ≤ r ≤ 100,000.
(a)
(b)
54/91
Example: Plot the Inductance as a Function of the Air
Gap Length
The magnetic circuit as shown below has dimensions Ac = Ag = 9 cm2, lc = 30 cm, and N = 500
tums. With the given magnetic circuit, r = 70,000. , using MATLAB to plot the inductance as
a function of the air gap length over the range 0.01 cm ≤ g ≤ 0.1 cm.
(a)
55/91
Example: Magnetic circuit with two windings
The following figure shows a magnetic circuit with an air gap and two windings. In this case
note that the mmf acting on the magnetic circuit is given by the total ampere-turns acting on the
magnetic circuit (i.e., the net ampere turns of both windings) and that the reference directions
for the currents have been chosen to produce flux in the same direction.
 0 Ac
 g
1  N1  N12 

 0 Ac

i
N
N
1
1 2

 g

 i2

56/91
Example: Analysis of a Switching Inductor
A switching inductor can be used as a fundamental energy storage cell with a switching power
converting system. Assume components in the following circuit are all ideal, make an analysis
of the given problems. Assume the duty ratio for the MOSFET switch is 20%.
iL
L
R
Vdc
iS
S
D
iD
VDC  48 V
f s  20 kHz
L  5 mH
R  10 
 Calculate the current ripple (peak-to-peak) of the inductor current.
57/91
Recommended Books
電磁學天堂祕笈：輕鬆解析最實用的馬克士威方程式
A Student’s Guide to Maxwell’s Equations
作者：夫雷胥 (Daniel Fleisch), 譯者：鄭以禎
出版社：天下文化, 出版日期：2010年10月18日
A Student's Guide to Vectors and Tensors,
Daniel A. Fleisch,
Cambridge University Press, 1st Ed., November 14, 2011.
Introduction to Electricity and Magnetism,
MIT 8.02 Course Notes,
Sen-Ben Liao, Peter Dourmashkin, and John Belche,
Prentice Hall, 2011.
Electricity and Magnetism,
W. N. Cottingham and D. A. Greenwood,
Cambridge University Press, 1st Ed., November 29, 1991.
58/91
References: Magnetic Circuits
[1]
[2]
[3]
[4]
G. K. Dubey, Fundamentals of Electrical Drives, Alpha Science International, Ltd, March 30th 2001.
Chapter 1: Magnetic Circuits and Magnetic Materials, Fitzgerald & Kingsley's Electric Machinery, S.D. Umans, 7th Ed, McGraw-Hill
Book Company, 2013.
Chapter 11 Inductance and Magnetic Energy, Introduction to Electricity and Magnetism, MIT 8.02 Course Notes, Sen-Ben Liao, Peter
Dourmashkin, and John Belche, Prentice Hall, 2011.
Chapter 4 Inductances, Design of Rotating Electrical Machines, Juha Pyrhonen, Tapani Jokinen, Valeria Hrabovcova, 2nd Ed., October
2013, Wiley.
Introduction to Electrodynamics,
David Griffiths,
4th Ed., Addison-Wesley, October 6, 2012.
59/91
Power Electronic Systems & Chips Lab., NCTU, Taiwan
Modeling of Practical Inductors
電力電子系統與晶片實驗室
Power Electronic Systems & Chips Lab.
交通大學 • 電機控制工程研究所
台灣新竹‧交通大學‧電機控制工程研究所‧電力電子實驗室~鄒應嶼 教授
60/91
Modeling of Practical Inductors
Ideal impedance model is for a simple linear relationship between
frequency and impedance.
Z ( j )
RAC
L
RDC
RC
1
C
C
R
L
0 
(a) Examples of inductor
1
LC

(c) Frequency response
(b) Equivalent circuit
 Not true across the whole frequency range for real components!
 For practical capacitors and inductors with nonlinear characteristics, its frequency responses
are only valid for small signal perturbation around its operating point this operating point are
generally highly dependent on its dc value, frequency, and temperature.
61/91
Building a Model of a Real Inductor
 Ideal inductor
 Perfectly conducting wire
 Core of ‘ideal’ magnetic
material

 Practical inductor
 Real wires have small DC resistance
 Real wire resistance is frequency dependent
 Real inductor may saturate
 Real magnetic materials for inductors are both
frequency and temperature dependent!
 Parasitic capacitance exists between turns of
the coil, between layers if wound in layers
 Parasitic lead capacitance
iL
iL
vL
vL
diL
dt
L( f )
R( f )
C
62/91
Simple Model for Real Inductors
RESR
L
 The inductor is modeled as a constant inductance with a series
connected resistance (RESR).
 As frequency increases, the inductive impedance increases
 This model does not resonate (no capacitance)
 There is a corner frequency where the inductive impedance begins
to dominate
63/91
Simple Model for Real Inductors
Example Parameter: L=100 nH, R=2 
L
R
Z ( j )  j L  R

f 3dB
L
R
Time Constant [sec]
1/ 
R
Corner frequency [Hz]


2 2  L
f3dB 
R
2

 3.18 MHz
2  L 2 100109
64/91
Improved Model for Practical Inductors
L
R
 Parallel resonant circuit
 Resonant frequency is r  0 1   2
C
L
1

Q  C 0 
2Q
R
R = series resistance
C = parallel capacitance
1
Z ( j )  ( R  j L ) ||
jC
Z ( j ) 
R  j L
1  j RC   2 LC
1
LC
r 
1
R2

LC 4 L2
 In general, R and C are quite small, and the
resonant frequency can be approximated to the
undamped natural frequency 0:
L  CR 2
r  0 
1
LC
[1] R. L. Boylestad, Introductory Circuit Analysis, 12th Edition, Prentice Hall, 2010.
[2] Electromagnetic Compatibility Handbook, Kenneth L. Kaiser, CRC Press, 2005.
[3] Cartwright, K., E. Joseph, and E. Kaminsky, “Finding the Exact Maximum Impedance Resonant Frequency of a Practical Parallel
Resonant Circuit without Calculus,” Technology Interface Internat. J., vol.11, no. 1, Fall/Winter 2010, pp. 26-36.
Improved Model for Practical Inductors
Example Parameter: L=100 nH, R=2 C=10 pF
L
C
R
Z ( j )  ( R  j L) ||
1
jC
66/91
More Complex Models for HF Inductor
Fairly accurate model for SMT
chip inductor
67/91
Simple Electro-Magnetic Circuits
Equivalent Circuit
Toroidal Inductance
Block Diagram
68/91
Transient Response of Inductance
Vdc
u (t )
If the above PWM voltage is applied to an ideal
inductor, what will be the current waveform?
What about a practical inductor?
69/91
Inductor with Resistance
Block Diagram
Equivalent circuit of a linear
inductor with coil resistance
70/91
Magnetic Saturation

Weber-turns (=N)
L

i
Amp (I)
71/91
AC Excitation of Ferromagnetic Materials

B (or  )
i
b
c
N
a
H (or F)
i(t)
e
d
t
Hysteresis Loop
72/91
Magnetic Domains
magnetic moment (dipole)
magnetic domain
Magnetic domains oriented randomly.
Magnetic domains lined up in the presence
of an external magnetic field.
73/91
Hysteresis Curves of a Ferromagnetic Core in AC
Excitation
B
B
Magnetization or B-H Curve
Residual Flux Density
Br
Coercive Force
saturation
-Hc
H
H
area  hysteresis loss
Hysteresis Loop
台灣新竹‧交通大學‧電機控制工程研究所‧電力電子實驗室~鄒應嶼 教授
74/91
AC Excitation of a Magnetic Circuit

Ac: cross-section surface area
i
N
v(t)
mean flux path in the
lc ferromagnetic material
Assume a sinusoidal variation of the core flux (t); thus
(t)= maxsint=Ac Bmax sint
amplitude of the flux density
From Faradays law, the voltage induced in the N-turn winding is
v(t) 
where
d
 Nw  max  2  fNA c B max
dt
v rms 
1
v max  2fNAc Bmax
2
and
v max   N  max  2  fNA c B max
75/91
AC Excitation of a Magnetic Circuit
Excitation phenomena. (a) Voltage, flux, and exciting current; (b) corresponding
hysteresis loop.
76/91
AC Excitation Phenomena of a Magnetic Circuit

i
vs

vs

 
 


i
N
i
t i

t 
i
t
i
(a)
i
HC
(b)
(a) Voltage, flux, and excitation current; (b) corresponding hysteresis loop.
To produce the magnetic field in the core requires current in the excitation winding know as the
excitation current i
For the given , we can obtain the corresponding i from the B-H hysteresis loop. Because  =
BcAc , and i = HcBLc /N
The saturated hystersis loop will result peakly excitation current with sinusoidal flux variation
77/91
Core Saturation Due to Over-Excitation

i
N
vs
 ( t1 ) 
1
N

t1
t0
v s ( t ) dt   ( t0 )
 ( t0 )  0
78/91
Core Saturation Due to Over-Excitation

i
N
vs
79/91
Core Saturation Due to Residual Flux

i
vs
N
 ( t1 ) 
1
N

t1
t0
v s ( t ) dt   ( t0 )
 ( t0 )  0
Power transformer inrush current caused by residual flux at switching instant; flux (green), iron
core's magnetic characteristics (red) and magnetizing current (blue).
80/91
Inrush Electric Current in a Transformer

What really happen
i
vs
N
What you assume
81/91
Typical Waveform of Magnetizing Inrush Current
 In practical applications, the winding resistance and losses of the core will
decay residual flux and the dc offset due to initial volts-sec integration.
 A a soft start procedure can be used to reduce this effect and a balance control
loop can be used to eliminate this dc offset in application to inverters.
82/91
Saturation of an Inductor (Incremental Inductance)
Weber-turns (=N)
Cross-sectional
area Ac
Mean path length l
L(I x ) 
N
i

i
iIx
Amp (I)
Permeability 
I
x
 A practical inductor will saturate as the current is increased.
 The incremental inductance is defined as the inductance at a specified current
with small signal perturbation, this is equivalent to a linear inductance for current
around this operating point.
 Note: In the given example, the current source as a perturbation source.
83/91
Measuring the Incremental Inductance at Specific
Operating Points (1/3)
Weber-turns (=N)
S1
Vdc
B
N
S2
C
S3
L(I x ) 
S4

i
iIx
Amp (I)
A
 Practical inductors are nonlinear and its incremental inductance (smallsignal inductance) is highly dependent on its operating point, such as
its average current, the magnitude of current ripples, the switching
frequency, and the core temperature, etc.
 The winding inductance of a synchronous machine is nonlinear, especially
for an interior PMSM. This characteristics is useful for the detection of its
rotor pole position under sensorless control. Devise a scheme to measure
the incremental inductance for different operation points (A, B, and C)?
84/91
Measuring the Incremental Inductance at Specific
Operating Points (2/3)
0.1mH, IL(pp)=2A, 20 kHz
inductor
S1
S3
8 Ohm, 50W,
Cement Resistors
e 
A
v AB
Vdc
S2
S4
Ro
B
v AB
L
L
L

R R ESR  Ro  2 R DS ( ON )
Ts   e
RESR
 Design an inductor with 0.1 mH, average current from 1 A to 6A, and operating with
switching frequency of 20 kHz. An illustrated design example can be found in [1].
 Devise an incremental inductance measurement scheme for operation points of A (0A), B
(4A), and C (8A) with a current ripple of 20 kHz, 2 A (peak-to-peak).
 Make a simulation study in consideration of the RDS(ON) and the diode forward voltage drop.
 Make experimental verifications for the proposed scheme.
REF: [1] Inductor Design in Switching Regulators (Technical Bulletin SR-1A, Magnetics).pdf
85/91
Compute the Inductance of a Toroidal Ferrite Core
60-turn toroidal inductor with
the TN23/14/7 ferrite core
TN23/14/7 Ring Core
[1] Rosa Ana Salas and Jorge Pleite, “Simple procedure to compute the inductance of a toroidal ferrite core from the linear to the saturation
regions,” Materials, no. 6, pp. 2452-2463, 2013.
86/91
TN23147-3R1 - Ferroxcube
Effective Core Parameters
Permeability as a Function of Frequency of Different Materials
B and H Magnetic Fields Inside the Toroidal Core
Moduli of the B and H magnetic fields as a function of the distance from the center of the
inductor core (x = 30 mm) obtained by 2D (red dashed line) and 3D (black solid line) simulations,
for (a,b) I = 0.0057 A (linear region); (c,d) I = 0.16 A (intermediate region); (e,f) I = 3 A
(saturation region).
89/91
Winding Inductances of an IPMSM
a
La-bc
1.5Lq
1.5Ld
La-bc
b
IPMSM stator coil
N
N
S
La-bc 
3/2
2
Rotor pole position ()
S
N
S

/2
0
c

3   Ld  Lq   Ld  Lq 

cos 2 

2
2
2


S
S
N
N
90/91
Mid-Term Report (April 15, 2016)
Modeling of the Stator Winding Inductance
L1  ?
L1 ( r )  ?
1. Define the stator structure, mechanical dimensions, winding
mechanism, and material parameters of the segmented motor.
2. Construct an equivalent circuit for a single segment of the stator
teeth and calculate its inductance. Make a Maxwell simulation to
verify the calculation.
3. Put the segmented teeth into the stator but without the rotor, make
a Maxwell simulation to calculate the inductance of a single stator
segment.
4. Define the rotor structure and material parameters and make a
Maxwell simulation to calculate the inductance of a single stator
segment as a function of the rotor pole position.
91/91
UNITS
FOR MAGNETIC
PROPERTIES
Quantity
Symbol
Gaussian & cgs emu a
Conversion
factor, C b
SI & rationalized mks C
Magnetic flux density,
magnetic induction
B
gauss (0) d
10- 4
tesla (T), Wb/m 2
maxwell (Mx), G·cm 2
Magnetic flux
weber (Wb), volt second (V·s)
Magnetic potential difference,
ma~netomotive force
U,F
gilbert (Gb)
Magnetic field strength,
,magnetizing force
H
oersted (Oe),e Gb/cm
A/m!
(Volume) magnetization g
M
emu/cm 3 h
A/m
(Volume) magnetization
477M
G
A/m
Magnetic polarization,
intensity of magnetization
1,1
emu/cm 3
477 X 10- 4
T, Wb/m 2i
(Mass) magnetization
(F,
emu/g
1
477 X 10- 7
A.m 2/kg
Wb·m/kg
Magnetic moment
m
emu, erg/G
10- 3
A.m 2, joule per tesla (l/T)
Magnetic dipole moment
j
emu, erg/G
477 X 10- 10
Wb·m'
M
10/477
dimensionless
henry per meter (H/m), Wb/(A.m)
dimensionless, emu/cm 3
(V 01 ume) susceptibility
ampere (A)
(Mass) susceptibility
Xp, K p
cm 3/g, emu/g
477 X 10- 3
(477)2 X 10- 10
m3/kg
H.m 2/kg
(Molar) susceptibility
Xmo}, K mo]
cm 3/mol, emu/mol
477 X 10- 6
(477)2 X 10- l3
m3/mol
H·m 2/mol
Permeability
1J-
dimensionless
477 X 10- 7
H/m, Wb/(A·m)
Relative permeability j
1J-r
not defined
(Volume) energy density,
energy product k
W
erg/cm 3
10- l
J/m 3
Demagnetization factor
D,N
dimensionless
1/477
dimensionless
dimensionless
a. Gaussian units and cgs emu are the same for magnetic properties. The defining relation is B =H + 477M.
b. Multiply a number in Gaussian units by C to convert it to SI (e.g., 1 G X 10- 4 T /G = 10- 4 T).
c. SI (Systeme International d'Unites) has been adopted by the National Bureau of Standards. Where two conversion factors are
given, the upper one is recognized under, or consistent with, SI and is based on the definition B = 1J-o(H + M), where
1J-o = 477 X 10- 7 H/m. The lower one is not recognized under SI and is based on the definition B =1J-oll +J, where the symbol
I is often used in place of J.
d. 1 gauss = 105 gamma (1').
e. Both oersted and gauss are expressed as em -1I2. g 1l2.S -1 in terms of base units.
/. A/m was often expressed as "ampere-turn per meter" when used for magnetic field strength.
g. Magnetic moment per unit volume.
h. The designation "emu" is not a unit.
i. Recognized under SI, even though based on the definition B = 1J-oll +J. See footnote c.
j. 1J-r = 1J-/1J-o = 1 + X, all in SI. 1J-r is equal to Gaussian 1J-.
k. B·H and 1J-oM·H have SI units J/m 3; M·H and B·H /477 have Gaussian units erg/cm 3.
R. B. Goldfarb and F. R. Fickett, U.S. De~artment of Commerce, National Bureau of Standards, Boulder, Colorado 80303, March 1985
NBS Special Publication 696 For sale by the Superintendent of Documents, U.S. Government Printing Office, Washington, DC 20402
CHAPTER
Fitzgerald & Kingsley's Electric Machinery,
1
Stephen Umans,
McGraw-Hill Education, 7th Ed., Jan. 28, 2013.
agnetic Circuits and
agnetic Materials
he objective of this book is to study the devices used in the interconversion
of electric and mechanical energy. Emphasis is placed on electromagnetic rotating machinery, by means of which the bulk of this energy conversion takes
place. However, the techniques developed are generally applicable to a wide range of
additional devices including linear machines, actuators, and sensors.
Although not an electromechanical-energy-conversion device, the transformer is
an important component of the overall energy-conversion process and is discussed
in Chapter 2. As with the majority of electromechanical-energy-conversion devices
discussed in this book, magnetically coupled windings are at the heart of transformer
performance. Hence, the techniques developed for transformer analysis form the basis
for the ensuing discussion of electric machinery.
Practically all transformers and electric machinery use ferro-magnetic material
for shaping and directing the magnetic fields which act as the medium for transferring and converting energy. Permanent-magnet materials are also widely used in
electric machinery. Without these materials, practical implementations of most familiar electromechanical-energy-conversion devices would not be possible. The ability
to analyze and describe systems containing these materials is essential for designing
and understanding these devices.
This chapter will develop some basic tools for the analysis of magnetic field
systems and will provide a brief introduction to the properties of practical magnetic
materials. In Chapter 2, these techniques will be applied to the analysis of transformers. In later chapters they will be used in the analysis of rotating machinery.
In this book it is assumed that the reader has basic knowledge of magnetic
and electric field theory such as is found in a basic physics course for engineering
students. Some readers may have had a course on electromagnetic field theory based
on Maxwell's equations, but an in-depth understanding of Maxwell's equations is not
a prerequisite for mastery of the material of this book. The techniques of magneticcircuit analysis which provide algebraic approximations to exact field-theory solutions
T
1
2
CHAPTER 1
Magnetic Circuits and Magnetic Materials
are widely used in the study of electromechanical-energy-conversion devices and form
the basis for most of the analyses presented here.
1.1
The complete, detailed solution for magnetic fields in most situations of practical
engineering interest involves the solution of Maxwell's equations and requires a set
of constitutive relationships to describe material properties. Although in practice
exact solutions are often unattainable, various simplifying assumptions permit the
attainment of useful engineering solutions. 1
We begin with the assumption that, for the systems treated in this book, the frequencies and sizes involved are such that the displacement-current term in Maxwell's
equations can be neglected. This term accounts for magnetic fields being produced
in space by time-varying electric fields and is associated with electromagnetic radiation. Neglecting this term results in the magneto-quasi-static form of the relevant
Maxwell's equations which relate magnetic fields to the currents which produce
them.
(1.1)
i
B-da=O
(1.2)
Equation 1.1, frequently referred to as Ampere's Law, states that the line integral
of the tangential component of the magnetic field intensity H around a closed contour C
is equal to the total current passing through any surface S linking that contour. From
Eq. 1.1 we see that the source of His the current density J. Eq. 1.2, frequently referred
to as Gauss' Law for magnetic fields, states that magnetic flux density B is conserved,
i.e., that no net flux enters or leaves a closed surface (this is equivalent to saying that
there exist no monopolar sources of magnetic fields). From these equations we see
that the magnetic field quantities can be determined solely from the instantaneous
values of the source currents and hence that time variations of the magnetic fields
follow directly from time variations of the sources.
A second simplifying assumption involves the concept of a magnetic circuit. It is
extremely difficult to obtain the general solution for the magnetic field intensity Hand
the magnetic flux density B in a structure of complex geometry. However, in many
practical applications, including the analysis of many types of electric machines, a
thtee-dimensional field problem can often be approximated by what is essentially
1 Computer-based numerical solutions based upon the finite-element method form the basis for a number
of commercial programs and have become indispensable tools for analysis and design. Such tools are
typically best used to refine initial analyses based upon analytical techniques such as are found in this
book. Because such techniques contribute little to a fundamental understanding of the principles and
basic performance of electric machines, they are not discussed in this book.
1.1
Introduction to Magnetic Circuits
Mean core
length lc
Cross-sectional
areaAc
Magnetic core
permeability f1,
Figure 1.1 Simple magnetic circuit. )._ is the winding flux
linkage as defined in Section 1.2.
a one-dimensional circuit equivalent, yielding solutions of acceptable engineering
accuracy.
A magnetic circuit consists of a structure composed for the most part of highpermeability magnetic material. 2 The presence of high-permeability material tends
to cause magnetic flux to be confined to the paths defined by the structure, much as
currents are confined to the conductors of an electric circuit. Use of this concept of
the magnetic circuit is illustrated in this section and will be seen to apply quite well
to many situations in this book. 3
A simple example of a magnetic circuit is shown in Fig. 1.1. The core is assumed
to be composed of magnetic material whose magnetic permeability fJ., is much greater
than that of the surrounding air (JJ., >> tJvo) where /Jvo = 4n x 1o- 7 Him is the magnetic
permeability of free space. The core is of uniform cross section and is excited by a
winding of N turns carrying a current of i amperes. This winding produces a magnetic
field in the core, as shown in the figure.
Because of the high permeability of the magnetic core, an exact solution would
show that the magnetiC flux is confined almost entirely to the core, with the field lines
following the path defined by the core, and that the flux density is essentially uniform
over a cross section because the cross-sectional area is uniform. The magnetic field
can be visualized in terms of flux lines which form closed loops interlinked with the
winding.
As applied to the magnetic circuit of Fig. 1.1, the source of the magnetic field
in the core is the ampere-turn product N i. In magnetic circuit terminology N i is the
magnetonwtive force (mmf) F acting on the magnetic circuit. Although Fig. 1.1 shows
only a single winding, transformers and most rotating machines typically have at least
two windings, and N i must be replaced by the algebraic sum of the ampere-turns of
all the windings.
2
In its simplest definition, magnetic permeability can be thought of as the ratio of the magnitude of the
magnetic flux density B to the magnetic field intensity H.
3 For a more extensive treatment of magnetic circuits see A.E. Fitzgerald, D.E. Higgenbotham, and A.
Grabel, Basic Electrical Engineering, 5th ed., McGraw-Hill, 1981, chap. 13; also E.E. Staff, M.I.T.,
Magnetic Circuits and Transformers, M.I.T. Press, 1965, chaps. 1 to 3.
3
4
CHAPTER 1
Magnetic Circuits and Magnetic Materials
The net magnetic flux ¢ crossing a surface S is the surface integral of the normal
component of B; thus
¢=
1
B·da
(1.3)
In SI units, the unit of¢ is the weber (Wb).
Equation 1.2 states that the net magnetic flux entering or leaving a closed surface
(equal to the surface integral of B over that closed surface) is zero. This is equivalent
to saying that all the flux which enters the surface enclosing a volume must leave
that volume over some other portion of that surface because magnetic flux lines form
closed loops. Because little flux "leaks" out the sides of the magnetic circuit of Fig. 1.1,
this result shows that the net flux is the same through each cross section of the core.
For a magnetic circuit of this type, it is common to assume that the magnetic
flux density (and correspondingly the magnetic field intensity) is uniform across the
cross section and throughout the core. In this case Eq. 1.3 reduces to the simple scalar
equation
(1.4)
where
c/Je =core flux
Be = core flux density
Ae = core cross-sectional area
From Eq. 1.1, the relationship between the mmf acting on a magnetic circuit and
the magnetic field intensity in that circuit is. 4
(1.5)
The core dimensions are such that the path length of any flux line is close to
the mean core length le. As a result, the line integral of Eq. 1.5 becomes simply the
scalar product Hele of the magnitude of H and the mean flux path length le. Thus,
the relationship between the mmf and the magnetic field intensity can be written in
magnetic circuit terminology as
(1.6)
where He is average magnitude of H in the core.
The direction of He in the core can be found from the right-hand rule, which can
be stated in two equivalent ways. (1) Imagine a current-carrying conductor held in the
right hand with the thumb pointing in the direction of current flow; the fingers then
point in the direction of the magnetic field created by that current. (2) Equivalently, if
the coil in Fig. 1.1 is grasped in the right hand (figuratively speaking) with the fingers
4
In general, the mmf drop across any segment of a magnetic circuit can be calculated as
portion of the magnetic circuit.
JHdl over that
1.1
Introduction to Magnetic Circuits
pointing in the direction of the current, the thumb will point in the direction of the
magnetic fields.
The relationship between the magnetic field intensity H and the magnetic flux
density B is a property of the material in which the field exists. It is common to assume
a linear relationship; thus
(1.7)
where f.1, is the material's magnetic permeability. In SI units, His measured in units of
amperes per meter, B is in webers per square meter, also known as teslas (T), and JL
is in webers per ampere-turn-meter, or equivalently henrys per 1neter. In SI units the
permeability of free space is /Lo = 4Jr X 1o- 7 henrys per meter. The permeability of
linear magnetic material can be expressed in terms of its relative permeability JLr, its
value relative to that of free space; JL = /Lr/LO· Typical values of /Lr range from 2,000 to
80,000 for materials used in transformers and rotating machines. The characteristics
of ferromagnetic materials are described in Sections 1.3 and 1.4. For the present we
assume that J.l,r is a known constant, although it actually varies appreciably with the
magnitude of the magnetic flux density.
Transformers are wound on closed cores like that of Fig. 1.1. However, energy
conversion devices which incorporate a moving element must have air gaps in their
magnetic circuits. A magnetic circuit with an air gap is shown in Fig. 1.2. When
the air-gap length g is much smaller than the dimensions of the adjacent core faces,
the core flux c/Jc will follow the path defined by the core and the air gap and the
techniques of magnetic-circuit analysis can be used. If the air-gap length becomes
excessively large, the flux will be observed to "leak out" of the sides of the air gap
and the techniques of magnetic-circuit analysis will no longer be strictly applicable.
Thus, provided the air-gap length g is sufficiently small, the configuration of
Fig. 1.2 can be analyzed as a magnetic circuit with two series components both
carrying the same flux¢: a magnetic core of permeability f.l,, cross-sectional area Ac
and mean length lc, and an air gap of permeability fLo, cross-sectional area Ag and
length g. In the core
(1.8)
+
Mean core
length lc
i
---+-
+--Air gap,
permeability fl-o,
AreaAg
Magnetic core
permeability fl-,
AreaAc
Figure 1.2 Magnetic circuit with air gap.
5
6
CHAPTER 1
Magnetic Circuits and Magnetic Materials
and in the air gap
¢
(1.9)
Ba=Ac
1:>
Application of Eq. 1.5 to this magnetic circuit yields
+ Hgg
:F = Hclc
(1.10)
and using the linear B-H relationship of Eq. 1.7 gives
B
+ ___!
g
(1.11)
JLo
Here the :F = Ni is the mmf applied to the magnetic circuit. From Eq. 1.10 we
see that a portion of the mmf, Fe = Hclc, is required to produce magnetic field in the
core while the remainder, :Fg = Hgg produces magnetic field in the air gap.
For practical magnetic materials (as is discussed in Sections 1.3 and 1.4), Be
and He are not simply related by a known constant permeability JL as described by
Eq. 1.7. In fact, Be is often a nonlinear, multi-valued function of He. Thus, although
Eq. 1.10 continues to hold, it does not lead directly to a simple expression relating
the mmf and the flux densities, such as that of Eq. 1.11. Instead the specifics of the
nonlinear Be-He relation must be used, either graphically or analytically. However, in
many cases, the concept of constant material permeability gives results of acceptable
engineering accuracy and is frequently used.
From Eqs. 1.8 and 1.9, Eq. 1.11 can be rewritten in terms of the flux ¢cas
:F = ¢ (__!_:__
J-LAc
+
_g_)
JLoAg
(1.12)
The terms that multiply the flux in this equation are known as the reluctance (R)
of the core and air gap, respectively,
(1.13)
g
Ra=-1:>
JLoAg
(1.14)
and thus
(1.15)
"Finally, Eq. 1.15 can be inverted to solve for the flux
:F
¢=---
Rc+Rg
(1.16)
or
:F
¢ = ---;---
+-g!LoAg
(1.17)
1.1
I
Introduction to Magnetic Circuits
~
---+-
nc
Rl
+
+
v
:F
ng
R2
I---v-
¢=
- (R 1 +R2)
:F
('Rc + 'Rg)
(b)
(a)
Figure 1.3 Analogy between electric and magnetic circuits.
(a) Electric circuit. (b) Magnetic circuit.
In general, for any magnetic circuit of total reluctance Rtob the flux can be found as
¢
= _!__
(1.18)
Rtot
The term which multiplies the mmf is known as the permeance P and is the inverse
of the reluctance; thus, for example, the total permeance of a magnetic circuit is
Ptot
=
1
(1.19)
rn
/'\-tot
Note that Eqs. 1.15 and 1.16 are analogous to the relationships between the current and voltage in an electric circuit. This analogy is illustrated in Fig. 1.3. Figure 1.3a
shows an electric circuit in which a voltage V drives a current I through resistors R1
and R2 • Figure 1.3b shows the schematic equivalent representation of the magnet~c
circuit of Fig. 1.2 . Here we see that the mmf :F (analogous to voltage in the electric
circuit) drives a flux ¢ (analogous to the current in the electric circuit) through the
combination of the reluctances of the core Rc and the air gap Rg. This analogy between the solution of electric and magnetic circuits can often be exploited to produce
simple solutions for the fluxes in magnetic circuits of considerable complexity.
The fraction of the mmf required to drive flux through each portion of the magnetic
circuit, commonly referred to as the mnif drop across that portion of the magnetic
circuit, varies in proportion to its reluctance (directly analogous to the voltage drop
across a resistive element in an electric circuit). Consider the magnetic circuit of
Fig. 1.2. From Eq. 1.13 we see that high material permeability can result in low core
reluctance, which can often be made much smaller than that of the air gap; i.e., for
(fl,Ac/lc) >> (f.loAg/ g), Rc << Rg and thus Rtot ~ Rg. In this case, the reluctance
of the core can be neglected and the flux can be found from Eq. 1.16 in terms of :F
and the air-gap properties alone:
¢ ~ :F
Rg
= :FfJ,oAg = Ni
g
f.loAg
g
(1. 20)
7
8
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Fringing
fields
++--+---+- Air gap
H-+++-t-1-HI-H--+-H-+H
Figure 1.4 Air-gap fringing fields.
As will be seen in Section 1.3, practical magnetic materials have permeabilities which
are not constant but vary with the flux level. From Eqs. 1.13 to 1.16 we see that as
long as this permeability remains sufficiently large, its variation will not significantly
affect the performance of a magnetic circuit in which the dominant reluctance is that
of an air gap.
In practical systems, the magnetic field lines "fringe" outward somewhat as they
cross the air gap, as illustrated in Fig. 1.4. Provided this fringing effect is not excessive,
the magnetic-circuit concept remains applicable. The effect of these fringing fields is to
increase the effective cross-sectional area Ag of the air gap. Various empirical methods
have been developed to account for this effect. A correction for such fringing fields
in short air gaps can be made by adding the gap length to each of the two dimensions
making up its cross-sectional area. In this book the effect of fringing fields is usually
ignored. If fringing is neglected, Ag = A c.
In general, magnetic circuits can consist of multiple elements in series and
parallel. To. complete the analogy between electric and magnetic circuits, we can
generalize Eq. 1.5 as
(1.21)
where F is the mmf (total ampere-turns) acting to drive flux through a closed loop of a
magnetic circuit, and Fk = Hklk is the mmf drop across the k'th element of that loop.
This is directly analogous to Kirchoff's voltage law for electric circuits consisting of
voltage sources and resistors
(1.22)
where V is the source voltage driving current around a loop and Rkik is the voltage
drop across the k'th resistive element of that loop.
1.1
Introduction to Magnetic Circuits
Similarly, the analogy to Kirchoff's current law
(1.23)
n
which says that the net current, i.e. the sum of the currents, into a node in an electric
circuit equals zero is
(1.24)
1l
which states that the net flux into a node in a magnetic circuit is zero.
We have now described the basic principles for reducing a magneto-quasi-static
field problem with simple geometry to a magnetic circuit model. Our limited purpose
in this section is to introduce some of the concepts and terminology used by engineers
in solving practical design problems. We must emphasize that this type of thinking
depends quite heavily on engineering judgment and intuition. For example, we have
tacitly assumed that the permeability of the "iron" parts of the magnetic circuit is a
constant known quantity, although this is not true in general (see Section 1.3), and
that the magnetic field is confined solely to the core and its air gaps. Although this is a
good assumption in many situations, it is also true that the winding currents produce
magnetic fields outside the core. As we shall see, when two or more windings are
placed on a magnetic circuit, as happens in the case of both transformers and rotating
machines, these fields outside the core, referred to as leakage fields, cannot be ignored
and may significantly affect the performance of the device.
The magnetic circuit shown in Fig. 1.2 has dimensions Ac = Ag = 9 cm2, g = 0.050 em,
lc = 30 em, and N =500 turns. Assume the value Mr =70,000 for core material. (a) Find the
reluctances Rc and Rg. For the condition that the magnetic circuit is operating with Be= 1.0 T,
find (b) the flux ¢ and (c) the current i.
II Solution
a. The reluctances can be found from Eqs. 1.13 and 1.14:
lc
0.3
10-7)(9
Rc
=- = 70,000 (4n
fJ.-rfJ.-oAc
Rg
8
=- = -----'---= 4.42 X 105
fJ.,oAg
(4n X 10-7)(9 X
X
5 x 10-4
b. From Eq. 1.4,
c. From Eqs. 1.6 and 1.15,
.
l=
F
N
X
10-4)
= 3.79 X
103
A· turns
Wb
A · turns
Wb
9
10
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Find the flux cjJ and ctment for Example 1.1 if (a) the number of turns is doubled to N = 1000
turns while the circuit dimensions remain the same and (b) if the number of turns is equal to
N =500 and the gap is reduced to 0.040 em.
Solution
a. cjJ
b. dJ
= 9 x IQ-4 Wb and i = 0.40 A
= 9 X IQ-4 Wb and i = 0.64 A
The magnetic structure of a synchronous machine is shown schematically in Fig. 1.5. Assuming
that rotor and stator iron have infinite permeability (J.L -7 oo ), find the air-gap flux cjJ and flux
density Bg. For this example I= 10 A, N = 1,000 turns, g = 1 em, and Ag = 200 cm2 •
II Solution
Notice that there are two air gaps in series, of total length 2g, and that by symmetry the\ flux
density in each is equal. Since the iron permeability is assumed to be infinite, its reluctance is
negligible and Eq. 1.20 (with g replaced by the total gap length 2g) can be used to find the flux
7
- N I J.LoAg - 1000(10)(4n X 10- )(0.02) - 12 6 Wb
¢. m
2g
0.02
and
cjJ
0.0126
Ba = - =
= 0.630 T
" Ag
0.02
Figure 1.5 Simple synchronous machine.
1.2 Flux Linkage, Inductance, and Energy
For the magnetic structure of Fig. 1.5 with the dimensions as given in Example 1.2, the air-gap
flux density is observed to be Bg = 0.9 T. Find the air-gap flux¢ and, for a coil of N = 500
turns, the current required to produce this level of air-gap flux.
Solution
¢ = 0.018 Wb and i = 28.6 A.
1.2 FLUX LINKAGE, INDUCTANCE,
AND ENERGY
When a magnetic field varies with time, an electric field is produced in space as
determined by another of Maxwell's equations refened to as Faraday's law:
i
d { B · da
dt Js
E·ds=
(1.25)
Equation 1.25 states that the line integral of the electric field intensity E around a
closed contour C is equal to the time rate of change of the magnetic flux linking
(i.e., passing through) that contour. In magnetic structures with windings of high
electrical conductivity, such as in Fig. 1.2, it can be shown that the E field in the wire
is extremely small and can be neglected, so that the left-hand side ofEq. 1.25 reduces
to the negative of the induced voltage5 e at the winding terminals. In addition, the flux
on the right-hand side ofEq. 1.25 is dominated by the core flux¢. Since the winding
(and hence the contour C) links the core flux N times, Eq. 1.25 reduces to
e=Nd<p=dA
dt
dt
(1.26)
where Ais the flux linkage of the winding and is defined as
A= N<p
(1.27)
Flux linkage is measured in units of webers (or equivalently weber-turns). Note that
we have chosen the symbol <p to indicate the instantaneous value of a time-varying
flux.
In general the flux linkage of a coil is equal to the surface integral of the normal
component of the magnetic flux density integrated over any surface spanned by that
coil. Note that the direction of the induced voltage e is defined by Eq. 1.25 so that if
the winding terminals were short-circuited, a cunent would flow in such a direction
as to oppose the change of flux linkage.
For a magnetic circuit composed of magnetic material of constant magnetic
permeability or which includes a dominating air gap, the relationship between A
5
The term electromotive force (emf) is often used instead of induced voltage to represent that component
of voltage due to a time-varying flux linkage.
11
12
CHAPTER 1
Magnetic Circuits and Magnetic Materials
and i will be linear and we can define the inductance L as
A.
L=i
(1.28)
Substitution ofEqs. 1.5, 1.18 and 1.27 into Eq. 1.28 gives
L=
N2
(1.29)
Rtot
from which we see that the inductance of a winding in a magnetic circuit is proportional
to the square of the turns and inversely proportional to the reluctance of the magnetic
circuit associated with that winding.
For example, from Eq. 1.20, under the assumption that the reluctance of the core
is negligible as compared to that of the air gap, the inductance of the winding in
Fig. 1.2 is equal to
(1.30)
Inductance is measured in henrys (H) or weber-turns per ampere. Equation 1.30
shows the dimensional form of expressions for inductance; inductance is proportional
to the square of the number of turns, to a magnetic permeability and to a crosssectional area and is inversely proportional to a length. It must be emphasized that
strictly speaking, the concept of inductance requires a linear relationship between
flux and mmf. Thus, it cannot be rigorously applied in situations where the nonlinear characteristics of magnetic materials, as is discussed in Sections 1.3 and 1.4,
dominate the performance of the magnetic system. However, in many situations of
practical interest, the reluctance of the system is dominated by that of an air gap
(which is of course linear) and the non-linear effects of the magnetic material can be
ignored. In other cases it may be perfectly acceptable to assume an average value of
magnetic permeability for the core material and to calculate a corresponding average
inductance which can be used for calculations of reasonable engineering accuracy.
Example 1~3 illustrates the former situation and Example 1.4 the latter.
The magnetic circuit of Fig. 1.6a consists of an N -turn winding on a magnetic core of infinite
permeability with two parallel air gaps of lengths g 1 and g2 and areas A1 and A2 , respectively.
Find (a) the inductance of the winding and (b) the flux density B1 in gap I when the
winding is carrying a current i. Neglect fringing effects at the air gap.
Solution
a. The equivalent circuit of Fig. 1.6b shows that the total reluctance is equal to the parallel
~ combination of the two gap reluctances. Thus
<P=
Ni
1.2 Flux Linkage, Inductance, and Energy
+
Ni
(a)
(b)
Figure 1.6 (a) Magnetic circuit and (b) equivalent circuit for Example 1.3.
where
From Eq. 1.28,
).
L=
N¢
b. From the equivalent circuit, one can see that
Ni
JJ, 0 A 1Ni
¢1=-=-nl
gl
and thus
In Example 1.1, the relative permeability of the core material for the magnetic circuit of Fig. 1.2
is assumed to be Jl,r = 70, 000 at a flux density of 1.0 T.
a. In a practical device, the core would be constructed from electrical steel such as M-5
electrical steel which is discussed in Section 1.3. This material is highly nonlinear and its
relative permeability (defined for the purposes of this example as the ratio BI H) varies
from a value of approximately Jl,r = 72,300 at a flux density of B = 1.0 T to a value of on
the order of Jl,r = 2,900 as the flux density is raised to 1.8 T. Calculate the inductance
under the assumption that the relative permeability of the core steel is 72,300.
b. Calculate the inductance under the assumption that the relative permeability is equal to
2,900.
13
14
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Solution
a. From Eqs. 1.13 and 1.14 and based upon the dimensions given in Example 1.1,
Rc = _lc_·_ =
f.Lrf.LoAc
72,300 (4n
X
0.3
10-7)(9
= 3.67
X
X
103 A ·turns
Wb
while Rg remains unchanged from the value calculated in Example 1.1 as
Rg = 4.42 x 105 Aturns/Wb.
Thus the total reluctance of the core and gap is
Rtot
= Rc + Ra = 4.46 X
o
A· turns
105 -Wb
--
and hence from Eq. 1.29
N2
L = =
Rtot
5002
= 0.561 H
4.46 X 105
b. For f.Lr = 2,900, the reluctance of the core increases from a value of 3.79 x 103 A· turns I
Wb to a value of
Rc = __
Zc_ = _____0_.3_____ = 9.1 5 X 104 _A_·_tu_rn_s
f.Lrf.LoAc
2,900 (4n X 10-7)(9 X
Wb
and hence the total reluctance increases from 4.46 x 105 A· turns I Wb to 5.34 x 105 A ·
turns I Wb. Thus from Eq. 1.29 the inductance decreases from 0.561 H to
N2
L = =
Rtot
5002
= 0.468 H
5.34 X 105
This example illustrates the linearizing effect of a dominating air gap in a magnetic
circuit. In spite of a reduction in the permeablity of the iron by a factor of 72,300/2,900 =
25, the inductance decreases only by a factor of 0.46810.561 = 0.83 simply because the
reluctance of the air gap is significantly larger than that of the core. In many situations, it
is common to assume the inductance to be constant at a value corresponding to a finite,
constant value of core permeability (or in many cases it is assumed simply that f.Lr ~ oo ).
Analyses based upon such a representation for the inductor will often lead to results which
are well within the range of acceptable engineering accuracy and which avoid the
immense complication associated with modeling the non-linearity of the core material.
Repeat the inductance calculation of Example 1.4 for a relative permeability f.Lr = 30,000.
Solution
L
= 0.554H
Using MATLAB,6 plot the inductance of the magnetic circuit of Example 1.1 and Fig. 1.2 as
; function of core permeability over the range 100 :S f.Lr :S 100,000.
6
"MATLAB" ia a registered trademarks of The Math Works, Inc., 3 Apple Hill Drive, Natick, MA
01760, http://www.mathworks.com. A student edition ofMatlab is available.
1.2 Flux Linkage, Inductance, and Energy
II Solution
Here is the MATLAB script:
clc
clear
% Permeability of free space
muO = pi*4.e-7;
%All dimensions expressed in meters
Ac = 9e-4; Ag = 9e-4; g
Se-4; lc = 0.3;
N
500;
%Reluctance of air gap
Rg = g/(muO*Ag);
mur = 1:100:100000;
Rc = lc./(mur*muO*Ac);
Rtot = Rg+Rc;
L = W'2. /Rtot;
plot(mur,L)
xlabel('Core relative permeability')
ylabel('Inductance [H] ')
The resultant plot is shown in Fig. 1.7. Note that the figure clearly confirms that, for the
magnetic circuit of this example, the inductance is quite insensitive to relative permeability
0.6
0.5
53'
'";; 0.4
§
u
.§ 0.3
,s
0.2
0.1
2
3
7
4
5
6
Core relative permeability
8
9
X
10
104
Figure 1. 7 MATLAB plot of inductance vs. relative permeability for
Example 1.5.
15
16
CHAPTER 1
Magnetic Circuits and Magnetic Materials
until the relative permeability drops to on the order of 1,000. Thus, as long as the effective
relative permeability of the core is "large" (in this case greater than 1,000), any non-linearities
in the properties of the core material will have little effect on the terminal properties of the
inductor.
Write a MATLAB script to plot the inductance of the magnetic circuit of Example 1.1 with
= 70,000 as a function of air-gap length as the the air gap is varied from 0.01 em to 0.10 em.
f.Lr
Figure 1.8 shows a magnetic circuit with an air gap and two windings. In this case
note that the mmf acting on the magnetic circuit is given by the total ampere-turns
acting on the magnetic circuit (i.e., the net ampere-turns of both windings) and that
the reference directions for the currents have been chosen to produce flux in the same
direction. The total mmf is therefore
(1.31)
and from Eq. 1.20, with the reluctance of the core neglected and assuming that Ac =
Ag, the core flux¢ is
, = (N JZJ.
'f'
+ N2Z2. )JloAc
-g
(1.32)
In Eq. 1.32, ¢is the resultant core flux produced by the total mmf of the two windings.
It is this resultant ¢ which determines the operating point of the core material.
If Eq. 1.32 is broken up into terms attributable to the individual currents, the
resultant flux linkages of coil 1 can be expressed as
)q
JloAc) i! + NtN2
= NI¢ = NI 2 ( -g-
( -gJloAc) i2
(1.33)
which can be written
(1.34)
Magnetic core
permeability f.L,
mean core length lc,
cross-sectional area Ac
Figure 1.8 Magnetic circuit with two windings.
1.2 Flux Linkage, Inductance, and Energy
where
2J.LoAc
L u= N1 - g
(1.35)
is the self-inductance of coil 1 and L IIi I is the flux linkage of coil 1 due to its own
current iz. The mutual inductance between coils 1 and 2 is
J.LoAc
= N1Nz--
L12
g
(1.36)
and L12i2 is the flux linkage of coill due to curr-ent i 2 in the other coil. Similarly, the
flux linkage of coil 2 is
A2
J.LoAc) i1 + N2 2 ( -g~toAc) i2
= Nz¢ = N1N2 ( -g-
(1.37)
= L21i1 + L2ziz
(1.38)
or
Az
where L2 I
= L 12 is the mutual inductance and
L22
= N221LoAc
-g
(1.39)
is the self-inductance of coil 2.
It is important to note that the resolution of the resultant flux linkages into the
components produced by iz and i 2 is based on superposition of the individual effects
and therefore implies a linear flux-mmf relationship (characteristic of materials of
constant permeability).
Substitution of Eq. 1.28 in Eq. 1.26 yields
d
e = dt (Li)
(1.40)
for a magnetic circuit with a single winding. For a static magnetic circuit, the inductance is fixed (assuming that material nonlinearities do not cause the inductance to
vary), and this equation reduces to the familiar circuit-theory form
di
e = L(1.41)
dt
However, in electromechanical energy conversion devices, inductances are often timevarying, and Eq. 1.40 must be written as
di
dL
+
i(1.42)
dt
dt
Note that in situations with multiple windings, the total flux linkage of each
winding must be used in Eq. 1.26 to find the winding-terminal voltage.
The power at the terminals of a winding on a magnetic circuit is a measure of the
rate of energy flow into the circuit through that particular winding. The power, p, is
determined from the product of the voltage and the current
e = L-
p
dA.
= ie = idt-
(1.43)
17
18
CHAPTER 1
Magnetic Circuits and Magnetic Materials
and its unit is watts (W), or joules per second. Thus the change in magnetic stored
energy ~ W in the magnetic circuit in the time interval t 1 to t2 is
(1.44)
In SI units, the magnetic stored energy W is measured in joules (J).
For a single-winding system of constant inductance, the change in magnetic
stored energy as the flux level is changed from A1 to A2 can be written as
~w=
1
A.2
idA=
1
A.,
A.2
A.,
A
1
- dA =-(A~- AI)
L
2L
(1.45)
The total magnetic stored energy at any given value of A can be found from
_setting A1 equal to zero:
(1.46)
For the magnetic circuit of Example 1.1 (Fig. 1.2), find (a) the inductance L, (b) the magnetic
stored energy W for Be = 1.0 T, and (c) the induced voltage e for a 60-Hz time-varying core
flux of the form Be= 1.0 sinwt T where w = (2n)(60) = 377.
Ill Solution
a. From Eqs. 1.16 and 1.28 and Example 1.1,
L- ~- N¢-
N2
i - Re +Rg
- i -
5002
4.46 X 105 = 0.56 H
Note that the core reluctance is much smaller than that of the gap (Rc « Rg). Thus
to a good approximation the inductance is dominated by the gap reluctance, i.e.,
N2
L
~
Rg
= 0.57H
b. In Example 1.1 we found that when Be= 1.0 T, i =0.80 A. Thus from Eq. 1.46,
1
·2
1
W = 2Lz = 2:(0.56)(0.80)
2
= 0.18 J
c. From Eq. 1.26 and Example 1.1,
e=
d).
d<p
dBe
=N-=NAedt
dt
dt
500
X
(9
X
10-4)
X
= 170 cos (377t) v
(377
X
1.0cos(377t))
1.3
Properties of Magnetic Materials
19
Practice Problem 1.5
Repeat Example 1.6 for Be =0.8 T and assuming the core flux varies at 50 Hz instead of 60Hz.
Solution
a. The inductance L is unchanged.
b. w = 0.115 J
c. e = 113 cos (314t) V
1.3 PROPERTIES OF MAGNETIC MATERIALS
In the context of electromechanical-energy-conversion devices, the importance of
magnetic materials is twofold. Through their use it is possible to obtain large magnetic
flux densities with relatively low levels of magnetizing force. Since magnetic forces
and energy density increase with increasing flux density, this effect plays a large role
in the performance of energy-conversion devices.
In addition, magnetic materials can be used to constrain and direct magnetic
fields in well-defined paths. In a transformer they are used to maximize the coupling between the windings as well as to lower the excitation current required for
transformer operation. In electric machinery magnetic materials are used to shape the
fields to obtain desired torque-production and electrical terminal characteristics. Thus
a knowledgeable designer can use magnetic materials to achieve specific desirable
device characteristics.
Ferromagnetic materials, typically composed of iron and alloys of iron with
cobalt, tungsten, nickel, aluminum, and other metals, are by far the most common magnetic materials. Although these materials are characterized by a wide range of properties, the basic phenomena responsible for their properties are common to them all.
Ferromagnetic materials are found to be composed of a large number of domains,
i.e., regions in which the magnetic moments of all the atoms are parallel, giving rise
to a net magnetic moment for that domain. In an unmagnetized sample of material,
the domain magnetic moments are randomly oriented, and the net resulting magnetic
flux in the material is zero.
When an external magnetizing force is applied to this material, the domain magnetic moments tend to align with the applied magnetic field. As a result, the domain magnetic moments add to the applied field, producing a much larger value of
flux density than would exist due to the magnetizing force alone. Thus the effective
permeability J.L, equal to the ratio of the total magnetic flux density to the applied
magnetic-field intensity, is large compared with the permeability of free space fLo.
As the magnetizing force is increased, this behavior continues until all the magnetic
moments are aligned with the applied field; at this point they can no longer contribute
to increasing the magnetic flux density, and the material is said to be fully saturated.
In the absence of an externally applied magnetizing force, the domain magnetic
moments naturally align along certain directions associated with the crystal structure
of the domain, known as axes of easy magnetization. Thus if the applied magnetizing
20
CHAPTER 1
Magnetic Circuits and Magnetic Materials
1.8
1.6
I
I
~
1.4
I/
1.2
/;
('IE 1.0
::0
I
~
t:4 0.8
/
------
I
r
~
I
-..,..,.....
-
~
~f-
~[.."
-~ ~
~
;
I
0.6
I
0.4
I
~
bJ}
c:,
~I
u,
0.2
~,
0
~I
I
-10
0
Cll
10
20
40
30
H, A • turns/m
50
I
70 90 110 130 150 170
1.9 B-H loops for M-5 grain-oriented electrical steel 0.012 in thick. Only the
top halves of the loops are shown here. (Armco Inc)
force is reduced, the domain magnetic moments relax to the direction of easy magnetism nearest to that of the applied field. As a result, when the applied field is reduced
to zero, although they will tend to relax towards their initial orientation, the magnetic
dipole moments will no longer be totally random in their orientation; they will retain
a net magnetization component along the applied field direction. It is this effect which
is responsible for the phenomenon known as magnetic hysteresis.
Due to this hysteresis effect, the relationship between B and H for a ferromagnetic
material is both nonlinear and multivalued. In general, the characteristics of the material cannot be described analytically. They are commonly presented in graphical form
as a set of empirically determined curves based on test samples of the material using
methods prescribed by the American Society for Testing and Materials (ASTM). 7
The most common curve used to describe a magnetic material is the B-H curve
or hysteresis loop. The first and second quadrants (corresponding to B 2: 0) of a
set of hysteresis loops are shown in Fig. 1.9 for M-5 steel, a typical grain-oriented
~umerical data on a wide variety of magnetic materials are available from material manufacturers. One
problem in using such data arises from the various systems of units employed. For example,
magnetization may be given in oersteds or in ampere-turns per meter and the magnetic flux density in
gauss, kilogauss, or teslas. A few useful conversion factors are given in Appendix D. The reader is
reminded that the equations in this book are based upon SI units.
7
1.3 Properties of Magnetic Materials
21
2.4
2.2
2.0
~
1.8
v
1.6
':§
~---
f.- -:-
~,.-I-
v'
1.4
/
J
~ 1.2
I
~ 1.0
1/
018
0.6
0.4
~
0.2
0
1
-1----"v
I
I
10
100
1000
10,000
100,000
H, A • turns/m
Figure 1.10 De magnetization curve for M-5 grain-oriented electrical steel 0.012 in. thick.
(Armco Inc.)
electrical steel used in electric equipment. These loops show the relationship between
the magnetic flux density B and the magnetizing force H. Each curve is obtained
while cyclically varying the applied magnetizing force between equal positive and
negative values of fixed magnitude. Hysteresis causes these curves to be multivalued.
After several cycles the B-H curves form closed loops as shown. The arrows show the
paths followed by B with increasing and decreasing H. Notice that with increasing
magnitude of H the curves begin to flatten out as the material tends toward saturation.
At a flux density of about 1.7 T this material can be seen to be heavily saturated.
Notice also that as H is decreased from its maximum value to zero, the flux
density decreases but not to zero. This is the result of the relaxation of the orientation
of the magnetic moments of the domains as described above. The result is that there
remains a remanent magnetization when His zero.
Fortunately, for many engineering applications, it is sufficient to describe the
material by a single-valued curve obtained by plotting the locus of the rriaximum
values of B and H at the tips of the hysteresis loops; this is known as a de or normal
magnetization curve. A de magnetization curve for M-5 grain-oriented electrical steel
is shown in Fig. 1.1 0. The de magnetization curve neglects the hysteretic nature of
the material but clearly displays its nonlinear characteristics.
Assume that the core material in Example 1.1 is M-5 electrical steel, which has the de magnetization curve of Fig. 1.1 0. Find the current i required to produce Be = 1 T.
22
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Solution
The value of He for Be = 1 T is read from Fig. 1.10 as
He= 11 A· turns/m
The mmf drop for the core path is
Fe = Hele = 11 (0.3) = 3.3 A· turns
Neglecting fringing, Bg = Be and the mmf drop across the air gap is
Bgg
1(5 x w- 4)
Fg = Hgg = =
= 396 A · turns
/Lo
4.rr x IQ-7
The required current is
i
= _Fe_+___c:. = _39_9 = 0.80 A
N
500
Repeat Example 1.7 but find the current i for Be= 1.6 T. By what factor does the current have
to be increased to result in this factor of 1.6 increase in flux density?
Solution
The current i can be shown to be 1.302 A. Thus, the current must be increased by a factor of
1.302/0.8 = 1.63. Because of the dominance of the air-gap reluctance, this is just slightly in
excess of the fractional increase in flux density in spite of the fact that the core is beginning to
significantly saturate at a flux density of 1.6 T.
1
In ac power systems, the waveforms of voltage and flux closely approximate sinusoidal
functions of time. This section describes the excitation characteristics and losses
associated with steady-state ac operation of magnetic materials under such operating
conditions. We use as our model a closed-core magnetic circuit, i.e., with no air gap,
such as that shown in Fig. 1.1. The magnetic path length is lc, and the cross-sectional
area is Ac throughout the length of the core. We further assume a sinusoidal variation
of the core flux cp (t); thus
cp(t) =¢max sin wt = AcBmax sin wt
where
¢max =amplitude of core flux cp in webers
Bmax = amplitude of flux density Be in teslas
w = angular frequency = 2nf
f = frequency in Hz
(1.47)
1.4 AC Excitation
From Eq. 1.26, the voltage induced in theN-turn winding is
e(t) = wN<Pmax cos (wt) =
Emax
coswt
(1.48)
where
Emax
= wN ¢max = 2n f N AcBmax
(1.49)
In steady-state ac operation, we are usually more interested in the root-meansquare or rms values of voltage$ and currents than in instantaneous or maximum
values. In general, the rms value of a periodic function of time, f (t), of period T is
defined as
(1.50)
From Eq. 1.50, the rms value of a sine wave can be shown to be 11J2 times its peak
value. Thus the rms value of the induced voltage is
2n
Erms
r;;
= J2fNAcBmax = v2nfNAcBmax
(1.51)
An excitation current irp, corresponding to an excitation mmf Nirp(t), is required
to produce the flux cp(t) in the core. 8 Because of the nonlinear magnetic properties
of the core, the excitation current corresponding to a sinusoidal core flux will be
non sinusoidal. A curve of the exciting current as a function of time can be found
graphically from the magnetic characteristics of the core material, as illustrated in
Fig. l.lla. Since Be and He are related to cp and irp by known geometric constants,
the ac hysteresis loop of Fig. 1.11 b has been drawn in terms of cp = BeAe and is
irp = Hele/N. Sine waves of induced voltage, e, and flux, cp, in accordance with
Eqs. 1.47 and 1.48, are shown in Fig. l.lla.
At any given time, the value of irp corresponding to the given value of flux can
be found directly from the hysteresis loop. For example, at time t' the flux is cp' and
the current is i~; at timet" the corresponding values are cp" and i~. Notice that since
the hysteresis loop is multivalued, it is necessary to be careful to pick the rising-flux
values (cp' in the figure) from the rising-flux portion of the hysteresis loop; similarly
the falling-flux portion of the hysteresis loop must be selected for the falling-flux
values (cp" in the figure).
Notice that, because the hysteresis loop "flattens out" due to saturation effects,
the waveform of the exciting current is sharply peaked. Its rms value lrp,rms is defined
by Eq . .1.50, where T is the period of a cycle. It is related to the corresponding rms
value Hrms of He by the relationship
I
_:_ leHrms
rp,rms-
N
(1.52)
The ac excitation characteristics of core materials are often described in terms
of rms voltamperes rather than a magnetization curve relating B and H. The theory
8 More generally, for a system with multiple windings, the exciting mmf is the net ampere-turns acting to
. produce flux in the magnetic circuit.
23
24
CHAPTER 1
Magnetic Circuits and Magnetic Materials
(b)
(a)
Figure 1.11 Excitation phenomena. (a) Voltage, flux, and exciting current;
(b) corresponding hysteresis loop.
behind this representation can be explained by combining Eqs. 1.51 and 1.52. From
Eqs. 1.51 and 1.52, the rms voltamperes required to excite the core of Fig. 1.1 to a
specified flux density is equal to
~
Ermsl<p,rms = v L nf N AcBmax
lcHrms
N
= .Ji nf BmaxHrms(Aclc)
(1.53)
In Eq. 1.53, the product Aclc can be seen to be equal to the volume of the core and
hence the rms exciting voltamperes required to excite the core with sinusoidal can be
seen to be proportional to the frequency of excitation, the core volume and the product
of the peak flux density, and the rms magnetic field intensity. For a magnetic material
of mass density Pc, the mass of the core is AclcPc and the exciting rms voltamperes
per unit mass, Sa, can be expressed as
Sa= ErmJ<p,rms =.JinJ(BmaxHrms)
mass
Pc
(1.54)
Note that, normalized in this fashion, the rms exciting voltamperes depends only
on the frequency and Bmax because Hrms is a unique function of Bmax as determined by
the shape of the material hysteresis loop at any given frequency f. As a result, the ac
excitation requirements for a magnetic material are often supplied by manufacturers
in terms of rms voltamperes per unit mass as determined by laboratory tests on
closed-core samples of the material. These results are illustrated in Fig. 1.12 for M-5
grain-oriented electrical steel.
The exciting current supplies the mmf required to produce the core flux and the
power input associated with the energy in the magnetic field in the core. Part of this
1.4 AC Excitation
25
2.2
- --
2.0
I--'
1.8
~
/
1.6
('I
_§
..0
1.4
I
/
1.2
~
~ 1.0
J
.... ~---
/
/
0.8
I
0.6
0.4
/
/
/
....
0.2
0
0.001
--
~
~
0.01
0.1
10
100
Sa, rms VA/kg
Figure 1.12 Exciting rms voltamperes per kilogram at 60 Hz for M-5 grain-oriented electrical
steel 0.012 in. thick. (Armco Inc.)
energy is dissipated as losses and results in heating of the core. The rest appears as
. reactive power associated with energy storage in the magnetic field. This reactive
power is not dissipated in the core; it is cyclically supplied and absorbed by the
excitation source.
Two loss mechanisms are associated with time-varying fluxes in magnetic materials. The first is due to the hysteretic nature of magnetic material. As has been
discussed, in a magnetic circuit like that of Fig. 1.1, a time-varying excitation will
cause the magnetic material to undergo a cyclic variation described by a hysteresis
loop such as that shown in Fig. 1.13.
Equation 1.44 can be used to calculate the energy input W to the magnetic core
of Fig. 1.1 as the material undergoes a single cycle
W
=
f
i• d/..
=
f (i:')
(A,N dB,)= A,l,
f
H, dB,
(1.55)
Recognizing that Aclc is the volume of the core and that the integral is the area of the
ac hysteresis loop, we see that each time the magnetic material undergoes a cycle,
there is a net energy input into the material. This energy is required to move around
the magnetic dipoles in the material and is dissipated as heat in the material. Thus
for a given flux level, the corresponding hysteresis losses are proportional to the area
of the hysteresis loop and to the total volume of material. Since there is an energy
loss per cycle, hysteresis power loss is proportional to the frequency of the applied
excitation.
The second loss mechanism is ohmic heating, associated with induced currents in
the core material. From Faraday's law (Eq. 1.25) we see that time-varying magnetic
26
CHAPTER 1
Magnetic Circuits and Magnetic Materials
B
H
Figure 1.13 Hysteresis loop;
hysteresis loss is proportional to the
loop area (shaded).
fields give rise to electric fields. In magnetic materials these electric fields result
in induced currents, commonly referred to as eddy currents, which circulate in the
core material and oppose changes in flux density in the material. To counteract the
corresponding demagnetizing effect, the current in the exciting winding must increase.
Thus the resultant "dynamic" B-H loop under ac operation is somewhat "fatter"
than the hysteresis loop for slowly varying conditions and this effect increases as
the excitation frequency is increased. It is for this reason that the charactersitics
of electrical steels vary with frequency and hence manufacturers typically supply
characteristics over the expected operating frequency range of a particular electrical
steel. Note for example that the exciting rms voltamperes of Fig. 1.12 are specified at
a frequency of 60 Hz.
To reduce the effects of eddy currents, magnetic structures are usually built
with thin sheets or laminations of magnetic material. These laminations, which are
aligned in the direction of the field lines, are insulated from each other by an oxide
layer on their surfaces or by a thin coat of insulating enamel or varnish. This greatly
reduces the magnitude of the eddy currents since the layers of insulation interrupt the
current paths; the thinner the laminations, the lower the losses. In general, as a first
approximation, eddy-current loss can be considered to increase as the square of the
excitation frequency and also as the square of the peak flux density.
In general, core losses depend on the metallurgy of the material as well as the flux
density and frequency. Information on core loss is typically presented in graphical
form. It is plotted in terms of watts per unit mass as a function of flux density; often
a family of curves for different frequencies is given. Figure 1.14 shows the core loss
density Pc for M-5 grain-oriented electrical steel at 60Hz.
1.4 AC Excitation
27
2.2
2.0
I
1.8
I
1.6
1/
1.4
N
.§
..0
1.2
~
~ 1.0
E
>l::l
0.8
v
0.6
0.4
v
)
II
~..--~.--~--
0.2
1-
1---
0
0.0001
0.001
0.01
0.1
Figure 1.14 Core loss density at 60 Hz in watts per kilogram for M-5 grain-oriented
electrical steel 0.012 in. thick. (Armco Inc.)
Nearly all transformers and certain components of electric machines use sheetsteel material that has highly favorable directions of magnetization along which the
core loss is low and the permeability is high. This material is termed grain-oriented
steel. The reason for this property lies in the atomic structure of a crystal of the siliconiron alloy, which is a body-centered cube; each cube has an atom at each corner as
well as one in the center of the cube. In the cube, the easiest axis of magnetization
is the cube edge; the diagonal across the cube face is more difficult, and the diagonal
through the cube is the most difficult. By suitable manufacturing techniques most of
the crystalline cube edges are aligned in the rolling direction to make it the favorable
direction of magnetization. The behavior in this direction is superior in core loss
and permeability to nonoriented steels in which the crystals are randomly oriented to
produce a material with characteristics which are uniform in all directions. As a result,
oriented steels can be operated at higher flux densities than the nonoriented grades.
Nonoriented electrical steels are used in applications where the flux does not
follow a path which can be oriented with the rolling direction or where low cost is
of importance. In these steels the losses are somewhat higher and the permeability is
very much lower than in grain-oriented steels.
The magnetic core in Fig. 1.15 is made from laminations of M-5 grain-oriented electrical
steel. The winding is excited with a 60-Hz voltage to produce a flux density in the steel of
B = 1.5 sin wt T, where w = 2n60 ~ 377 rad/sec. The steel occupies 0.94 of the core crosssectional area. The mass-density of the steel is 7.65 g/cm3 • Find (a) the applied voltage, (b) the
peak current, (c) the rms exciting current, and (d) the core loss.
10
28
CHAPTER 1
Magnetic Circuits and Magnetic Materials
em---+~
25 em
e
Figure '1. '15 Laminated steel core with winding for
Example 1.8.
Ill Solution
a. From Eq. 1.26 the voltage is
dcp
dB
e=N-=NAc-
dt
= 200
dt
X
25 cm2
X
0.94
X
1.5
X
377
COS
(377t)
= 266 cos (377t) v
b. The magnetic field intensity corresponding to Bmax = 1.5 Tis given in Fig. 1.10 as
Hmax = 36 A turns/m. Notice that, as expected, the relative permeability
fJ-r = Bmax/(tJ-oHmax) = 33,000 at the flux level of 1.5 Tis lower than the value of /J-r =
72,300 found in Example 1.4 corresponding to a flux level of 1.0 T, yet significantly larger
than the value of 2,900 corresponding to a flux level of 1.8 T.
lc = (15
+ 15 + 20 + 20) em= 0.70 m
The peak current is
X 0.70 _ O
- .13A
200
c. The rms current is obtained from the value of Sa of Fig. 1.12 for Bmax = 1.5 T.
_ Hmaxlc _ 36
1---N
Sa= 1.5 VA/kg
The core volume and mass are
Vole= 25 cm2 x 0.94 x 70 em= 1645 cm3
7.65g) = 12.6kg
Me= 1645 cm 3 x ( l.Ocm3
1.5 Permanent Magnets
The total rms voltamperes and current are
S = 1.5 VA/kg x 12.6 kg= 18.9 VA
s
18.9
fcp,rms = Erms = 266j,J2 = 0.10 A
d. The core-loss density is obtained from Fig. 1.14 as
Pcore
= 1.2 W/kg
X
Pc
= 1.2 W/kg. The total core loss is
12.6 kg= 15.1 W
Repeat Example 1.8 for a 60-Hz voltage of B = 1.0 sin wt T.
Solution
a. V = 177 cos 377 t V
b. I= 0.042 A
c. lcp = 0.041 A
d. p = 6.5 w
1 .. 5 PERMANENT MAGNETS
Figure 1.16a shows the second quadrant of a hysteresis loop for Alnico 5, a typical
permanent-magnet material, while Fig. 1.16b shows the second quadrant of a hysteresis loop for M-5 steel. 9 Notice that the curves are similar in nature. However, the
hysteresis loop of Alnico 5 is characterized by a large value of residual or remanent
magnetization, Br, (approximately 1.22 T) as well as a large value of coercivity, He,
(approximately -49 kA/m).
The residual magnetization, Br, corresponds to the flux density which would
remain in a section of the material if the applied mmf (and hence the magnetic field
intensity H) were reduced to zero. However, although the M-5 electrical steel also has
a large value of residual magnetization (approximately 1.4 T), it has a much smaller
value of coercivity (approximately -6 Aim, smaller by a factor of over 7500). The
coercivity He is the value of magnetic field intensity (which is proportional to the
mmf) required to reduce the material flux density to zero. As we will see, the lower
the coercivity of a given magnetic material, the easier it is to demagnetize it.
The significance of residual magnetization is that it can produce magnetic flux
in a magnetic circuit in the absence of external excitation such as is produced by
winding currents. This is a familiar phenomenon to anyone who has afixed notes to
a refrigerator with small magnets and is widely used in devices such as loudspeakers
and permanent-magnet motors.
9
To obtain the largest value of residual magnetization, the hysteresis loops of Fig. 1.16 are those which
would be obtained if the materials were excited by sufficient mmf to ensure that they were driven heavily
into saturation. This is discussed further in Section 1.6.
29
30
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Energy product, kJ/m 3
B,T
Point of
1.0
0.5
-40
H,kA/m
-30
-20
-10
0
(a)
B,T
1.5
B,T
1.0
'
3.8 x
'''
w-5
'
~'
w-
Slope=
-6.28 x
Wb/A • m
H,Aim
0
(b)
w-5
2x
w-5
Load line for
',Example 1.9
''
0.5
4x
6
'
',
'
''
H,A!m -6
'0
(c)
Figure 1.16 (a) Second quadrant of hysteresis loop for Alnico 5; (b) second
quadrant of hysteresis loop for M-5 electrical steel; (c) hysteresis loop for M-5
electrical steel expanded for small B. (Armco Inc.)
1.5 Permanent Magnets
From Fig. 1.16, it would appear that both Alnico 5 and M-5 electrical steel would
be useful in producing flux in unexcited magnetic circuits since they both have large
values of residual magnetization. That this is not the case can be best illustrated by
an example.
As shown in Fig. 1.17, a magnetic circuit consists of a core of high permeability (p, --+ oo),
an air gap of lengt~ g = 0.2 em, and a section of magnetic material of length lm = 1.0 em.
The cross-sectional area of the core and gap is equal to Am = Ag = 4 cm2 • Calculate the flux
density Bg in the air gap if the magnetic material is (a) Alnico 5 and (b) M-5 electrical steel.
• Solution
a. Since the core permeability is assumed infinite, H in the core is negligible (otherwise a
finite H would produce an infinite B). Recognizing that there is zero mmf acting on the
magnetic circuit of Fig. 1.17, we can write
or
where Hg and Hm are the magnetic field intensities in the air gap and the magnetic
material, respectively.
Since the flux must be continuous through the magnetic circuit,
¢ = AgBg = AmBm
or
Bg = ( Am)
Ag Bm
where Bg and Bm are the magnetic flux densities in the air gap and the magnetic material,
respectively.
Figure 1.17 Magnetic circuit
for Example 1.9.
31
32
CHAPTER 1
Magnetic Circuits and Magnetic Materials
These equations can be solved to yield a linear relationship for Bm in terms of Hm
B,
= -JJ.o
U:) (~)
Hm
-5
f1n
Hm
= -6.28 X
w-'Hm
To solve for Bm we recognize that for Alnico 5, Bm and Hm are also related by the
curve of Fig. 1.16a. Thus this linear relationship, commonly referred to as a load line, can
be plotted on Fig. 1.16a and the solution obtained graphically, resulting in
Bg = Bm = 0.30 T = 3,000 gauss
b. The solution for M-5 electrical steel proceeds exactly as in part (a). The load line is the
same as that of part (a) because it is determined only by the permeability of the air gap and
the geometries of the magnet and the air gap. Hence from Fig. 1.16c
Bg
= 3.8 X w-s T = 0.38 gauss
which is much less than the value obtained with Alnico 5 and is essentially negligible.
Example 1.9 shows that there is an immense difference between permanentmagnet materials (often referred to as hard magnetic materials) such as Alnico 5 and
soft magnetic materials such as M-5 electrical steel. This difference is characterized
in large part by the immense difference in their coercivities He. The coercivity is a
measure of the magnitude of the mmf required to reduce the material flux density to
zero. As seen from Example 1.9, it is also a measure of the capability of the material
to produce flux in a magnetic circuit which includes an air gap. Thus we see that
materials which make good permanent magnets are characterized by large values of
coercivity He (considerably in excess of 1 kA/m).
A useful measure of the capability of permanent-magnet material is known as
its maximum energy product. This corresponds to the largest magnitude of the B-H
product (B · H) max found in the second quadrant of that material's hysteresis loop.
As can be seen from Eq. 1.55, the product of Band H has the dimensions of energy
density (joules per cubic meter). We now show that operation of a given permanentmagnet material in a magnetic circuit at this point will result in the smallest volume
of that material required to produce a given flux density in an air gap.
In Example 1.9, we found an expression for the flux density in the air gap of the
magnetic circuit of Fig. 1.17:
(1.56)
We also found that
(1.57)
1.5 Permanent Magnets
Equation 1.57 can be multiplied by
Eq. 1.56 yields
~-to
to obtain Bg =
~-toHg.
Multiplying by
(1.58)
where Volmag is the volume of the magnet, Volair gap is the air-gap volume, and the
minus sign arises because, at the operating point of the magnetic circuit, H in the
magnet (Hm) is negative.
Solving Eq. 1.58 gives
(1.59)
·which is the desired result. It indicates that to achieve a desired flux density in the air
gap, the required volume of the magnet can be minimized by operating the magnet
at the point of the largest possible value of the B-H product HmBm, i.e., at the point
of maximum energy product. Furthermore, the larger the value of this product, the
smaller the size of the magnet required to produce the desired flux density. Hence the
maximum energy product is a useful performance measure for a magnetic material and
it is often found as a tabulated "figure of merit" on data sheets for permanent-magnet
· materials. As a practical matter, this result applies to many practical engineering
applications where the use of a permanent-magnet material with the largest available
maximum energy product will result in the smallest required magnet volume.
Equation 1.58 appears to indicate that one can achieve an arbitrarily large airgap flux density simply by reducing the air-gap volume. This is not true in practice
because a reduction in air-gap length will increase the flux density in the magnetic
circuit and as the flux density in the magnetic circuit increases, a point will be reached
at which the magnetic core material will begin to saturate and the assumption of
infinite permeability will no longer be valid, thus invalidating the derivation leading
to Eq. 1.58.
The magnetic circuit of Fig. 1.17 is modified so that the air-gap area is reduced to Ag = 2.0 cm2,
as shown in Fig. 1.18. Find the minimum magnet volume required to achieve an air-gap flux
density of 0.8 T.
II Solution
Note that a curve of constant B-H product is a hyperbola. A set of such hyperbolas for different
values of the B-H product is plotted in Fig. 1.16a. From these curves, we see that the maximum
energy product for Alnico 5 is 40 kJ/m 3 and that this occurs at the point B = 1.0 T and
H = -40 kA/m. The smallest magnet volume will be achieved with the magnet operating at
this point.
33
34
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Figure 1.18 Magnetic circuit for Example 1.1 0.
Thus from Eq. 1.56,
= 2cm2 (0.8)
= 1.6cm2
1.0
and from Eq. 1.57
= -0.2cm (
(4n
0.8
X
lQ- 7)(-40
)
X
103)
= 3.18 em
Thus the minimum magnet volume is equal to 1.6 cm2 x 3.18 em= 5.09 cm3 •
Repeat Example 1.10 assuming the air-gap area is further reduced to Ag = 1.8 cm2 and that the
desired air-gap flux density is 0.6 T.
Solution
Minimum magnet volume= 2.58 cm3 •
1
Examples 1. 9 and 1.10 consider the operation of permanent-magnetic materials under
the assumption that the operating point can be determined simply from a knowledge
of the geometry of the magnetic circuit and the properties of the various magnetic
1.6 Application of Permanent-Magnet Materials
1.5
-
-...........
-- - ----
1.4
neodymium-iron-boron
Alnico 5
samarium-cobalt
Alnico 8
Ceramic 7
1.3
..-6
/:''
/
/
/
/
/
/
/
/
-1000 -900
/
/
/
/
/
/
-600
1.0
iL 0.9
v-1
/
0.8
'
:I
/
B,T
!t
0.7
~
/•
0.6
I
II
0.5
I
L
I
I
l ....v'
0.4
y'"
0.3
0.2
~,/J
L,
/
-700
1.1
'
''
'
/
//
-800
.
/
/
/
1.2
-500
-400
-300
-200
0.1
I
I
-100
0
H,kA/m
Figure 1.19 Magnetization curves for common permanent-magnet materials.
materials involved. In fact, in practical engineering devices, the situation is often more
complex. 10 This section will expand upon these issues.
Figure 1.19 shows the magnetization characteristics for a few common permanentmagnet materials. These curves are simply the second-quadrant characteristics of the
hysteresis loops for each material obtained when the material is driven heavily into
saturation. Alnico 5 is a widely-used alloy of iron, nickel, aluminum, and cobalt originally discovered in 1931. It has a relatively large residual flux density. Alnico 8 has
a lower residual flux density and a higher coercivity than Alnico 5. It is hence less
subject to demagnetization than Alnico 5. Disadvantages of the Alnico materials are
their relatively low coercivity and their mechanical brittleness.
Ceramic permanent-magnet materials (also known as ferrite magnets) are made
from iron-oxide and barium- or strontium-carbonate powders and have lower residual
flux densities than Alnico materials but significantly higher coercivities. As a result,
they are much less prone to demagnetization. One such material, Ceramic 7, is shown
in Fig. 1.19, where its magnetization characteristic is almost a straight line. Ceramic
magnets have good mechanical characteristics and are inexpensive to manufacture.
10por a further discussion of permanent magnets and their application, see P. Campbell, Permanent
Magnet Materials and Their Application, Cambridge University Press, 1994; R.J. Parker, Advances in
Permanent Magnetism, John Wiley & Sons, 1990; R.C. O'Handley, Modern Magnetic Materials:
Principles and Applications, John Wiley & Sons, 2000; and E.P. Ferlani, Permanent Magnet and
Electromechanical Devices, Academic Press, 2001.
35
36
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Samarium-cobalt represents a significant advance in permanent-magnet technology
which began in the 1960s with the discovery of rare-earth permanent-magnet materials. From Fig. 1.19 it can be seen to have a high residual flux density such as is found
with the Alnico materials, while at the same time having a much higher coercivity
and maximum energy product.
The newest of the rare-earth magnetic materials is the family of neodymium-ironboron materials. They feature even larger residual flux density, coercivity, and maximum energy product than does samarium-cobalt. The development of neodymiumiron-boron magnets has had a tremendous impact in the area of rotating machines
and as a result permanent-magnet motors with increasingly large ratings are being
developed by manufacturers around the world.
Note that in Fig. 1.19 the hysteretic nature of the magnetization characteristics
of Alnico 5 and Alnico 8 is readily apparent while the magnetization characteristics
of the remaining materials appear to be essentially straight lines. This straight-line
characteristic is deceiving; in each case the material characteristic bends sharply
downward just as does that of the Alnico materials. However, unlike the Alnico
materials, this bend, commonly referred to as the knee of the magnetization curve,
occurs in the third quadrant and hence does not appear in Fig. 1.19.
Consider the magnetic circuit of Fig. 1.20. This includes a section of hard magnetic material in a core of highly permeable soft magnetic material as well as an
N -turn excitation winding. With reference to Fig. 1.21, we assume that the hard magnetic material is initially unmagnetized (corresponding to point (a) of the figure) and
consider what happens as current is applied to the excitation winding. Because the
core is assumed to be of infinite permeability, the horizontal axis of Fig. 1~21 can
be considered to be both a measure of the applied current i = H lm/ N as well as a
measure of H in the magnetic material.
As the current i is increased to its maximum value, the B-H trajectory rises from
point (a) in Fig. 1.21 toward its maximum value at point (b). To fully magnetize the
material, we assume that the current has been increased to a value irnax sufficiently large
Figure 1.20 Magnetic circuit
including both a permanent magnet
and an excitation winding.
1.6 Application of Permanent-Magnet Materials
B,T
(d)
I
I
(e) 1
fl
/I
I
:
I
I
I
I
I
0
Hmax
H, kA/m
i,A
Figure 1.21 Portion of a B-H characteristic showing a minor loop and a
recoil line.
that the material has been driven well into saturation at point (b). When the current
is then decreased to zero, the B-H characteristic will begin to form a hysteresis loop,
arriving at point (c) at zero current. At point (c), notice that H in the material is zero
but B is at its remanent value Br.
As the current then goes negative, the B-H characteristic continues to trace out a
hysteresis loop. In Fig. 1.21, this is seen as the trajectory between points (c) and (d). If
the current is then maintained at the value i (d), the operating point of the magnet will
be that of point (d). Note that, as in Example 1.9, this same operating point would be
reached if the material were to start at point (c) and, with the excitation held at zero,
an air gap of length g = lm (Ag/ Am)( -JhoH(d) / B(d)) were then inserted in the core.
Should the current then be made more negative, the trajectory would continue
tracing out the hysteresis loop toward point (e). However, if instead the current is
returned to zero, the trajectory does not in general retrace the hysteresis loop toward
point (c). Rather it begins to trace out a minor hysteresis loop, reaching point (f) when
the current reaches zero. If the current is then varied between zero and i(d), the B-H
characteristic will trace out the minor loop as shown.
As can be seen from Fig. 1.21, theB-Htrajectorybetween points (d) and (f) can be
represented by a straight line, known as the recoil line. The slope of this line is called
the recoil permeability fhR· We see that once this material has been demagnetized to
point (d), the effective residual magnetization of the magnetic material is that of point
(f) which is less than the residual magnetization Br which would be expected based
on the hysteresis loop. Note that should the demagnetization be decreased past point
(d), for example, to point (e) of Fig. 1.21, a new minor loop will be created, with a
new recoil line and recoil permeability.
The demagnetization effects of negative excitation which have just been discussed
are equivalent to those of an air gap in the magnetic circuit. For example, clearly the
37
38
CHAPTER 1
Magnetic Circuits and Magnetic Materials
magnetic circuit of Fig. 1.20 could be used as a system to magnetize hard magnetic
materials. The process would simply require that a large excitation be applied to the
winding and then reduced to zero, leaving the material at a residual magnetization Br
(point (c) in Fig. 1.21).
Following this magnetization process, if the material were removed from the
core, this would be equivalent to opening a large air gap in the magnetic circuit,
demagnetizing the material in a fashion similar to that seen in Example 1.9. At this
point, the magnet has been effectively weakened, since if it were again inserted in
the magnetic core, it would follow a recoil line and return to a residual magnetization
somewhat less than Br. As a result, hard magnetic materials, such as the Alnico
materials of Fig. 1.19, often do not operate stably in situations with varying mmf
and geometry, and there is often the risk that improper operation can significantly
demagnetize them.
At the expense of a reduction in value of the residual magnetization, hard magnetic
materials such as Alnico 5 can be stabilized to operate over a specified region. This
procedure, based on the recoil trajectory shown in Fig. 1.21, can best be illustrated
by an example.
Figure 1.22 shows a magnetic circuit containing hard magnetic material, a core and plunger of
high (assumed infinite) permeability, and a 100-turn winding which will be used to magnetize
the hard magnetic material. The winding will be removed after the system is magnetized. The
plunger moves in the x direction as indicated, with the result that the air-gap area varies over
the range 2 cm2 .::::: Ag .::::: 4 cm2 • Assuming that the hard magnetic material is Alnico 5 and
that the system is initially magnetized with Ag = 2 em, (a) find the magnet length lm such that
the system will operate on a recoil line which intersects the maximum B-H product point on
the magnetization curve for Alnico 5, (b) devise a procedure for magnetizing the magnet, and
(c) calculate the flux density Bg in the air gap as the plunger moves back and forth and the air
gap varies between these two limits.
Hard magnetic
material, area
Am= 2cm2
magnetizing
coil
Figure 1.22 Magnetic circuit for Example 1.11.
1.6 Application of Permanent-Magnet Materials
II Solution
a. Figure 1.23a shows the magnetization curve for Alnico 5 and two load lines
corresponding to the two extremes of the air gap, Ag = 2 cm2 and Ag = 4 cm2 • We see
that the system will operate on the desired recoil line if the load line for Ag = 2 cm2
intersects the B-H characteristic at the maximum energy product point (labeled point (a)
in Fig. 1.23a), B~> = 1.0 T and H~a> = -40 kA/m.
From Eqs. 1.56 and 1.57, we see that the slope of the required load line is given by
B~~>
Bg Ag lm
-H~a>
= Hg Am
g
and thus
= 0.2cm
2) (
( -2
4n
1.0
X
IQ-? X
4 X 104
) = 3.98cm
b. Figure 1.23b shows a series of load lines for the system with Ag = 2 cm2 and with current
i applied to the excitation winding. The general equation for these load lines can be
readily derived since from Eq. 1.5
and from Eqs. 1.3 and 1.7
Thus
=JL+G) e~ nHm+2:~~-3 G}l
9
= -2.50 X w-s Hm + 6.28 X w-1 i
From this equation and Fig. 1.23b, we see that to drive the magnetic material into
saturation to the point (Hmax• Bmax), the current in the magnetizing winding must be
increased to the value imax where
imax
=
Bmax
+ 2.50 X 1o-s Hmax
6.28
A
X IQ- 2
In this case, we do not have a complete hysteresis loop for Alnico 5, and hence we
will have to estimate Bmax and Hmax. Linearly extrapolating the B-H curve at H = 0 back
to 4 times the coercivity, that is, Hmax = 4 x 50 = 200 kA/m, yields Bmax = 2.1 T. This
value is undoubtedly extreme and will overestimate the required current somewhat.
However, using Bmax = 2.1 T and Hmax = 200 kA/m yields imax = 113 A.
Thus with the air-gap area set to 2 cm2, increasing the current to 113 A and then
reducing it to zero will achieve the desired magnetization.
39
40
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Br= 1.24
1.08
1.0
Point of maximum
energy product
0.5
Hm,kA!m
-so
-40
-30
-20
-10
0
(a)
Load
line for
i=O
Material initially
not magnetized
(b)
Figure 1.23 (a) Magnetization curve for Alnico 5 for Example 1.11; (b) series of load
lines for Ag = 2 cm 2 and varying values of i showing the magnetization procedure for
Example 1.11.
c. Because we do not have specific information about the slope of the recoil line, we shall
assume that its slope is the same as that of the B- H characteristic at the point H = 0,
B = Br. From Fig. 1.23a, with the recoil line drawn with this slope, we see that as the
air-gap area varies between 2 and 4 cm2 , the magnet flux density Bm varies between 1.00
1.6 Application of Permanent-Magnet Materials
and 1.08 T. Since the air-gap flux density equals Am/ Ag times this value, the air-gap flux
density will equal (2/2)1.00 = 1.0 T when Ag = 2.0cm2 and (2/4)1.08 = 0.54 T when
Ag = 4.0 cm2 • Note from Fig. 1.23a that, when operated with these air-gap variations, the
magnet appears to have an effective residual flux density of 1.17 T instead of the initial
value of 1.24 T. As long as the air-gap variations limited to the range considered here, the
system will continue to operate on the line labeled "Recoil line" in Fig. 1.23a and the
magnet can be said to be stabilized.
As has been discussed, hard magnetic materials such as Alnico 5 can be subject
to demagnetization, should their operating point be varied excessively. As shown in
Example 1.11, these materials can be stabilized with some loss in effective residual
magnetization. However, this procedure does not guarantee absolute stability of operation. For example, if the material in Example 1.11 were subjected to an air-gap
area smaller than 2 cm2 or to excessive demagnetizing current, the effect of the stabilization would be erased and the material would be found to operate on a new recoil
line with further reduced magnetization.
However, many materials, such as samarium-cobalt, Ceramic 7, and neodymiumiron-boron (see Fig. 1.19), which have large values of coercivity, tend to have very
low values of recoil permeability, and the recoil line is essentially tangent to the B-H
characteristic for a large portion of the useful operating region. For example, this can
be seen in Fig. 1.19, which shows the de magnetization curve for neodymium-iron. boron, from which we see that this material has a residual magnetization of 1.25 T
and a coercivity of -940 kA/m. The portion of the curve between these points is a
straight line with a slope equal to 1.06tt0 , which is the same as the slope of its recoil
line. As long as these materials are operated on this low-incremental-permeability
portion of their B-H characteristic, they do not require stabilization, provided they are
not excessively demagnetized.
For these materials, it is often convenient to assume that their de magnetization
curve is linear over their useful operating range with a slope equal to the recoil permeability ILR· Under this assumption, the de magnetization curve for these materials
can be written in the form
(1.60)
Here, H~ is the apparent coercivity associated with this linear representation. As
can be seen from Fig. 1.19, the apparent coercivity is typically somewhat larger in
magnitude (i.e. a larger negative value) than the material coercivity He because the de
magnetization characteristic tends to bend downward for low values of flux density.
A significant (and somewhat unfortunate) characteristic of permanent-magnet
materials is that their properties are temperature dependent. For example, the residual
magnetization and coercivity of neodymium-iron-boron and samarium-cobalt magnets decrease as the temperature increases, although samarium-cobalt is much less
temperature sensitive than neodymium-iron-boron.
Figure 1.24 shows magnetization curves for a high-temperature grade of
neodymium-boron-iron at various temperatures. We see that the residual magnetism
41
42
CHAPTER 1
Magnetic Circuits and Magnetic Materials
B[T]
o~~~~~~--~~--~----~----J-----~--~---=~
-900
-800
-700
-600
-500
-400
-300
-200
-100
0
H[kAim]
1.24 Second-quadrant magnetization curves for neodymium-ironboron material showing their temperature dependence.
drops from around 1.14 Tat a temperature of20C to around 0.85 Tat a temperature of
180 C. Table 1.1 gives a more complete listing of the residual flux density as a function
of temperature for this material, which has a recoil permeability f.LR = 1.04J.Lo.
Interestingly, unlike rare-earth magnets, although ceramic magnets exhibit a decrease in residual magnetism with temperature, they exhibit a corresponding increase
in coercivity. Figure 1.25 shows the general nature of the temperature dependence of
the magnetization characteristic of a typical ceramic magnet material.
Although these permanent magnet materials display a reduction in magnetization with increasing temperature, this decrease in magnetization is often reversible.
Provided the operating point of the magnet material, which will vary as the magnet
temperature changes, remains in the linear portion of the magnetization characteristic,
Table 1.1 Residual flux density as a function of
temperature for the magnetization curves of Fig. 1.24.
20C
soc
120C
1.15
1.08
1.03
150 c
180 c
210 c
0.99
0.94
0.89
1.6 Application of Permanent-Magnet Materials
B
Figure 1.25 General form of the temperature
dependence of the magnetization characteristics
of a typical ceramic magnetic material.
it will fully recover its magnetization with a decrease in temperature. However, if the
temperature is increased to a value known as the Curie Temperature, the material
. will become fully demagnetized and magnetism will not be restored by a reduction
in temperature. II
Consider a magnetic circuit containing a permanent magnet and a winding such
as that which is shown in Fig. 1.26. Fig. 1.24 includes a zero-excitation load line,
corresponding to zero-winding-current operation of this magnetic circuit. As the
temperature varies between 20 C and 120 C, the operating point varies between
points (a) and (c). Each operating point over this temperature range lies on a portion
on the material hysteresis loop which is linear in the second quadrant. As we have
seen, operation on this linear portion of the magnetization characteristic is stabilized
and as the winding current is varied, the magnetic will continue to operate on the
linear portion of its magnetization characteristic as long as the operation remains
in the second quadrant.I 2 The material will not be permanently demagnetized and
it will recover any temperature-induced loss of magnetization as the temperature
is reduced.
As the temperature is further increased, a temperature will be reached for which
the downward bend appears in the second quadrant of the magnetization characteristic.
11 The Curie temperature of neodymium-boron-iron is on the order of 350 C and that of samarium-cobalt
and Alnico is on the order of 700 C.
12Note that the magnet will be permanently demagnetized if its operating point is driven into the third
quadrant past the point at which the magnetization characteristic ceases to be linear and begins to bend
downward.
43
44
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Nturns
Figure 1.26 Magnetic circuit
with a permanent magnet, an air
gap and an excitation winding.
This can be seen in the curves for 180 C and 210 C in Fig. 1.24. In the case of the
180 C characteristic, the winding current can be varied without demagnetizing the
magnet as long as the magnet flux density does not drop below the point where
the magnetization characteristic becomes non linear. Operation below this point is
analogous to the operation of Alnico 5 as discussed with reference to the minor
loop and recoil line of Fig. 1.21. Thus, if sufficient winding current is applied to
drive the magnet below this point and the current is then reduced, a minor loop will
be created and the magnet will be somewhat demagnetized. If the magnet temperature
is reduced, the magnet will be found to be partially demagnetized. In the case of the
210 C characteristic, we see that the zero-excitation operating point falls in the nonlinear portion of the magnetization current. As a result, any winding current which
causes an increase in magnet flux density will demagnetize the magnet.
A magnetic circuit similar to that of Fig. 1.26 has a 200-turn winding (N = 200) and
includes a neodymium-boron-iron magnet of length lm = 3 em and cross-sectional area
Am = 2.5 cm2 • The airgap has an effective area of Ag = 0.259 cm2 and an effective length of
g = 0.9 em.
a. Derive an expression for the load-line of this magnetic circuit as a function of winding
current i and show that it coincides with the zero-excitation load line of Fig. 1.24 when
the winding current is equal to zero.
b. The magnetic circuit is excited by a sinusoidal winding current of peak amplitude /peak· In
order to avoid the possibility of demagnetizing the magnet, it is desirable to limit /peak to a
value such that the magnet flux density Bm remains positive. Calculate the maximum
amplitude of /peak for magnet operating temperatures of 20 C and 120 C.
1.7 Summary
II Solution
a. This magnetic circuit is essentially identical to that of Example 1.11 and Fig. 1.22. Hence
the equation for the load line is identical to that derived in that example. Specifically,
g
Ag ) (lm)
Bm = -f.Lo ( Am
Hm
= fLo [- (0.259)
2.5
!LoN
+g
(2_)
0.9
H
m
( Ag ) .
Am
l
+ 9 X20010-3 (0.259)
2.5
il
= -4.34 X w-? Hm + 2.89 X 10-3i
With i = 0, when Hm = -600 k.A/m, this equation gives Bm = 0.26 T which closely
coincides with the zero-excitation load line of Fig 1.24.
b. From Eq.l.60, over the linear operating region the relationship between BM and Hm in the
magnet is given by
Combining this expression with the equation for the load-line of part (a) gives
Bm =
f.LRNi + lmBr
-3.
( ) ( ) = 2.17 x 10 z
z + g .!:B. Am
m
flO
+ 0.249Br
Ag
For a sinusoidal current of peak amplitude lpeab Bm will remain positive as long as
/peak
lmBr
= - - = 114.8Br
f.LRN
For a temperature of 80 C, from Table 1.1 Br = 1.15 T and thus the maximum value of
/peak is 132 A. Similarly, for a temperature of 120 C, Br =1.03 and the maximum value of /peak
is 118 A.
1.7 SUMMARY
Electromechanical devices which employ magnetic fields often use fen·omagnetic
materials for guiding and concentrating these fields. Because the magnetic permeability of ferromagnetic materials can be large (up to tens of thousands times that of
the surrounding space), most of the magnetic flux is confined to fairly well-defined
paths determined by the geometry of the magnetic material. In addition, often the
frequencies of interest are low enough to permit the magnetic fields to be considered
quasi-static, and hence they can be determined simply from a knowledge of the net
mmf acting on the magnetic structure.
As a result, the solution for the magnetic fields in these structures can be obtained
in a straightforward fashion by using the techniques of magnetic-circuit analysis.
These techniques can be used to reduce a complex three-dimensional magnetic field
solution to what is essentially a one-dimensional problem. As in all engineering
solutions, a certain amount of experience and judgment is required, but the technique
gives useful results in many situations of practical engineering interest.
45
46
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Ferromagnetic materials are available with a wide variety of characteristics. In
general, their behavior is nonlinear, and their B-H characteristics are often represented
in the form of a family of hysteresis (B-H) loops. Losses, both hysteretic and eddycurrent, are functions of the flux level and frequency of operation as well as the
material composition and the manufacturing process used. A basic understanding of
the nature of these phenomena is extremely useful in the application of these materials
in practical devices. Typically, important properties are available in the form of curves
supplied by the material manufacturers.
Certain magnetic materials, commonly known as hard or permanent-magnet materials, are characterized by large values of residual magnetization and coercivity.
These materials produce significant magnetic flux even in magnetic circuits with air
gaps. With proper design they can be made to operate stably in situations which subject them to a wide range of mmfs and temperature variations. Permanent magnets
find application in many small devices, including loudspeakers, ac and de motors,
microphones, and analog electric meters.
1
1
Jl
f-Lo
f.lr
f-LR
¢, <p, </Jmax
(J)
p
A
B,B
Br
e
e, E
E
f
F
g
H, H, Hrms
He
i, I
irp, lcf>,rms
J
l
L
N
p
Pcore
Pa
Magnetic permeability [H/m]
Permeability of free space= 4n x 10-7 [H/m]
Relative permeability
Recoil permeability [H/m]
Magnetic flux [Wb]
Angular frequency [rad/sec]
Mass density [kg/m3]
Cross-sectional area [m2 ]
Magnetic flux density [T]
Residual/remanent magnetization [T]
Electromotive force [V]
Voltage [V]
Electric field intensity [V/m]
Frequency [Hz]
Magnetomotive force [A]
Gap length [m]
Magnetic field intensity [A/m]
Coercivity [A/m]
Current [A]
Exciting current [A]
Current density [A/m2]
Linear dimension [m]
Inductance [H]
Number of turns
Power [W]
Core loss [W]
Exciting rms voltamperes per unit mass [W/kg]
1.9 Problems
Pc
P
R
R
S
Sa
t
T
T
V
Vol
W
Subscripts:
c
g
m,mag
max
rms
tot
Core loss density [W/kg]
Permeance [H]
Resistance [Q]
Reluctance [H- 1]
Exciting rms voltamperes [VA]
Exciting rms voltamperes per mass [VA/kg]
Time [sec]
Period [sec]
Temperature [C]
Voltage [V]
Volume [m3]
Energy [J]
Core
Gap
Magnet
Maximum
Root mean square
Total
1 . 9 PROBLEMS
1.1 A magnetic circuit with a single air gap is shown in Fig. 1.27. The core
dimensions are
Cross-sectional Area Ac = 3.5 cm2
Mean core length lc = 25 em
Gap length g = 2.4 mm
N = 95 turns
Assume that the core is of infinite permeability (JL -+ oo) and neglect the
effects of fringing fields at the air gap and leakage flux. (a) Calculate the
reluctance of the core Rc and that of the gap Rg. For a culTent of i = 1.4 A,
calculate (b) the total ft ux ¢, (c) the ft ux linkages A. of the coil, and (d) the coil
inductance L.
Core:
mean length lc,
areaAc,
permeability f.L
Figure 1.27 Magnetic circuit for Problem 1.1.
47
48
CHAPTER 1
Magnetic Circuits and Magnetic Materials
1.2 Repeat Problem 1.1 for a finite core permeability of Jh = 2350 JLo.
1.3 Consider the magnetic circuit of Fig. 1.27 with the dimensions of
Problem 1.1. Assuming infinite core permeability, calculate (a) the number of
turns required to achieve an inductance of 15 mH and (b) the inductor current
which will result in a core flux density of 1.15 T.
1.4 Repeat Problem 1.3 for a core permeability of JL = 1700 J.Lo.
1.5 The magnetic circuit of Problem 1.1 has a non-linear core material whose
permeability as a function of Bm is given by
JL
= Jho
( 1+
2153
Vl + 0.43(Bm)l2.1
)
where Bm is the material flux density.
a. Using MATLAB, plot a de magnetization curve for this material (Bm vs.
Hm) over the range 0 ~ Bm ~ 2.1 T.
b. Find the current required to achieve a flux density of 2.1 T in the air gap.
c. Again using MATLAB, plot the coil flux linkages as a function of coil
current as the current is varied from 0 to the value found in part (b).
1.6 The magnetic circuit of Fig. 1.28 consists of a core and a moveable plunger of
width lp, each of permeability JL. The core has cross sectional area Ac and
mean length lc. The overlap area of the two air gaps Ag is a function of the
plunger position x and can be assumed to vary as
Ag =A,
(1- ;J
You may neglect any fringing fields at the air gap and use approximations
consistent with magnetic-circuit analysis.
a. Assuming that fh -+ oo, derive an expression for the magnetic flux
density in the air gap Bg as a function of the winding current i and the
plunger position x (assume xis limited to the range 0 ~ x ~ 0.5 X0).
Write an expression for the corresponding flux density in the core?
b. Repeat part (a) for a finite permeability JL.
mean length lc,
areaAc,
Plunger
Figure 1.28 Magnetic circuit for Problem 1.6.
1.9 Problems
1.7 The magnetic circuit of Fig. 1.28 has 125 turns and the following dimension:
lc = 50 em
g = 0.25 em
lp = 4 em
Ac = 100 cm2
Xo = 10cm
With x = 0.5 Xo, the measured inductance is 52 mH. Using reasonable
approximations, calculate the relative permeability J.Lr of the core and plunger
material.
1.8 Figure 1.29 shows an inductor made up of two C-cores. Each core as area Ac
and mean length lc. There are two air gaps, each of length g and effective area
A g. Finally, there are two N-turn coils, one on each of the C-cores. Assuming
infinite core permeability and for cores of dimensions
Cross-sectional area: Ac = Ag = 38.7 cm2
Core-length: lc = 45 em
Gap length: g = 0.12 em
a. Calculate the number of turns required to achieve an inductance of
12.2 mH, assuming infinite core permeability and that the coils are
connected in series. Since the number of turns must be an integer, your
answer must be rounded to the nearest integer. Calculated the actual
inductance value based upon the resultant number of turns.
b. The inductance can be fine-tuned by adjusting the air-gap length to
achieve the desired inductance. Based upon the number of turns found in
part (a), calculate the air-gap length required to achieve the desired
inductance of 12.2 mH.
c. Based upon this final inductor design, calculate the inductor current which
will produce a core flux density of 1.5 T.
C-core:
AreaAc
mean length lc
permeability fL
Air gap: ..--::L:.c:_:_j'-------"'L-:....:.:J
AreaAg
lengthg
Figure 1.29 C-core inductor for problem 1.8.
49
50
CHAPTER 1
Magnetic Circuits and Magnetic Materials
1.9 Assuming that the coils are connected in parallel, repeat Problem 1.8.
1.10 Repeat Problem 1.8 assuming that the core has a permeability of 1800 fLo.
1.11 The magnetic circuit of Fig. 1.28 and Problem 1.6 has the following
dimensions:
Ac = 9.3 cm2
lp = 2.7 em
Xo = 2.3 em
lc
g
N
= 27 em
= 0.6mm
= 480 turns
a. Assuming a constant permeability of JL = 3150 fLo, calculate the current
required to achieve a flux density of 1.25 T in the air gap when the
plunger is fully retracted (x = 0).
b. Repeat the calculation of part (a) for the case in which the core and
plunger are composed of a a non-linear material whose permeability is
given by
Jl =flo (
1
1065
+ )1 + 0.0381Bml 9
)
where Bm is the magnetic flux density in the material.
c. For the non-linear material of part (b), use MATLAB to plot the air-gap
flux density as a function of winding current for x = 0 and x = 0.5X 0 •
1.12 An inductor of the form of Fig. 1.27 has dimensions
Cross-sectional area Ac = 3.8 cm2
Mean core length lc = 19 crri
N = 122 turns
Assuming a core permeability of JL = 3240 fLo arid neglecting the effects of
leakage flux and fringing fields, calculate the air-gap length required to
achieve an inductance of 6.0 mH.
1.13 The magnetic circuit of Fig. 1.30 consists of rings of magnetic material in a
stack of height h. The rings have inner radius Ri and outer radius R0 • Assume
that the iron is of infinite permeability (M -7 oo) and neglect the effects of
Figure 1.30 Magnetic circuit for
Problem 1.13.
1.9 Problems
magnetic leakage and fringing. For
Ri = 3.2cm
R0 = 4.1 em
h = 1.8 em
g = 0.15 em
1.14
1.15
1.16
1.17
calculate:
a. The mean core length lc and the core cross-sectional area Ac
b. The reluctance of the core Rc and that of the gap Rg
For N = 72 turns, calculate
c. The inductance L
d. Current i required to operate at an air-gap flux density of Bg = 1.25T
e. The corresponding flux linkages A. of the coil
Repeat Problem 1.13 for a core permeability of fL = 750 fLO·
Using MATLAB, plot the inductance of the inductor of Problem 1.13 as a
function of relative core permeability as the core permeability varies from
/Lr = 100 to /Lr = 10,000. (Hint: Plot the inductance versus the log of the
relative permeability.) What is the minimum relative core permeability
required to insure that the inductance is within 5 percent of the value
calculated assuming that the core pe1meability is infinite?
The inductor of Fig. 1.31 has a core of uniform circular cross-section of area
Ac, mean length lc and relative permeability fLr, and anN -turn winding. Write
an expression for the inductance L.
The inductor of Fig. 1.31 has the following dimensions.
Ac = 1.1cm2
lc = 12 em
g =0.9mm
N = 520 turns
a. Neglecting leakage and fringing and assuming /Lr = 1,000, calculate the
inductance.
b. Calculate the core flux density and the inductor flux linkages for a
winding current of 1.2 A.
.---,..-..,.------,/ Core:
mean length lc,
areaAc,
relative permeability fkr
Figure 1.31 Inductor for Problem 1.16.
51
52
CHAPTER 1
Magnetic Circuits and Magnetic Materials
1.18 The inductor of Problem 1.17 is to be operated from a 60-Hz voltage source.
(a) Assuming negligible coil resistance, calculate the rms inductor voltage
corresponding to a peak core flux density of 1.5 T. (b) Under this operating
condition, calculate the rms current and the peak stored energy.
1.19 Assume the core material of the inductor of Problem 1.17 has the permeability
given in Problem 1.5. Write a MATLAB script to calculate the core flux
density and the inductor flux linkages at a current of 1.2 A.
1.20 Consider the cylindrical magnetic circuit of Fig. 1.32. This structure, known
as a pot-core, is typically made in two halves. The N -tum coil is wound on a
cylindrical bobbin and can be easily inserted over the central post of the core
as the two halves are assembled. Because the air gap is internal to the core,
provided the core is not driven excessively into saturation, relatively little
magnetic flux will "leak" from the core, making this a particularly attractive
configuration for a wide variety of applications, both for inductors such as that
of Fig. 1.31 and transformers.
Assume the core permeability to be JL = 2, 300 J-Lo and N = 180 turns. The
following dimensions are specified:
R1 = 1.6cm
h
R2
= 4.2cm
= 0.78 em
l = 2.8cm
g = 0.45 mm
a. Although the flux density in the radial sections of the core (the sections of
thickness h) actually decreases with radius, assume that the flux density
remains uniform. Find the value of the radius R3 such that the average
flux density in the outer wall of the core is equal to that within the central
cylinder.
b. Write an expression for the coil inductance and evaluate it for the given
dimensions.
~+t-.,.--'-1-
Figure 1.32 Pot-core inductor for
Problem 1.20.
N-turn
winding
1.9 Problems
c. The core is to be operated at a peak flux density of 0.6 Tat a frequency of
60 Hz. Find (i) the corresponding rms value of the voltage induced in the
winding, (ii) the rms coil current, and (iii) the peak stored energy.
d. Repeat part (c) for a frequency of 50 Hz.
1.21 A square voltage wave having a fundamental frequency of 60 Hz and equal
positive and negative half cycles of amplitude E is applied to a 575-turn
winding surrounding a closed iron core of cross sectional area Ac = 9 cm2
and of length lc = 35 em. Neglect both the winding resistance and any effects
of leakage flux.
a. Sketch the voltage, the winding flux linkage, and the core flux as a
function of time.
b. Find the maximum permissible value of E if the maximum flux density is
not to exceed 0.95 T.
c. Calculate the peak winding current if the core has a magnetic permeability
of 1,000 J-to.
1.22 Assume that iron core of Problem 1.21 can be described by a magnetic
permeability given by
where B is the core flux density.
a.. Plot the core-material B-H curve for flux densities in the range
-1.8 T S B S 1. 8 T.
b. A 110 V rms, 60-Hz sinusoidal voltage is applied to the winding. Using
MATLAB, plot one cycle of the resultant winding current as a function of
time. What is the peak current?
c. The voltage of part (b) is doubled to 220 V rms. Add a plot of the resultant
current as a function of time to the plot of part (b). What is the peak
current for this case?
1.23 Repeat parts (b) and (c) of Problem 1.22 if a 10 mm air gap is inserted in the
magnetic core.
1.24 An inductor is to be designed using a magnetic core of the form of that of
Fig. 1.31. The core is of uniform cross-sectional area Ac = 6.0 cm2 and of
mean length lc = 28 em.
a. Calculate the air-gap length g and the number of turns N such that the
inductance is 23 mH and so that the inductor can operate at peak currents
of 10 A without saturating. Assume that saturation occurs when the peak
flux density in the core exceeds 1.7 T and that, below saturation, the core
has permeability f.L = 2700 J-to.
b. For an inductor cunent of 10 A, use Eq. 3.21 to calculate (i) the magnetic
stored energy in the air gap and (ii) the magnetic stored energy in the core.
Show that the total magnetic stored energy is given by Eq. 1.46.
53
54
CHAPTER 1
Magnetic Circuits and Magnetic Materials
1.25 Write a MATLAB script to design inductors based upon the magnetic core of
1.26
1.27
1.28
1.29
Fig. 1.31. Assume the core has a cross-sectional area of 10.0 cm2 , a length of
35 em and a relative magnetic permeability of 1,700. The inductor is to be
operated with a sinusoidal current at 50-Hz and it must be designed such that
the peak core flux density will be equal to 1.4 T when the peak inductor
current is equal to 7.5 A.
Write a simple design program in the form of a MATLAB script to
calculate the number of turns and air-gap length as a function of the desired
inductance. The script should be written to request a value of inductance (in
mH) from the user, with the output being the air-gap length in mm and the
number of turns. Write your script to reject as unacceptable any designs for
which the gap length is out of the range of 0.05 mm to 6.0 mm or for which
the number of turns drops below 10.
Using your program, find (a) the minimum and (b) the maximum
inductances (to the nearest mH) which will satisfy the the given constraints.
For each of these values, find the required air-gap length and the number of
turns as well as the rms voltage corresponding to the peak core flux.
Consider an inductor composed of two C-cores as shown in Fig. 1.29. Each
C-core has cross-sectional area Ac = 105 cm2 and a mean length lc = 48 em.
a. Assuming the coils are connected in parallel, calculate the number of turns
N per coil and the air-gap length g such that the inductance is 350 mH and
such that the inductor current can be increased to 6.0 A without exceeding
a core flux density of 1.2 T, thus avoiding saturation of the core. You may
neglect the reluctance of the core and the effects of fringing at the air gap.
b. Repeat part a) assuming the coils are connected in series.
Assuming the C-cores of Problem 1.26 have a magnetic permeability of
fL = 3,500 fLo, repeat Problem 1.26.
Write a MATLAB script to automate the calculations of Problem 1.26 and
1.27. The inputs to your script should be the core area, mean core length, the
core permeability and the winding connection (parallel or series) as well as
the desired inductance and maximum core flux density and current. Exercise
your script to design an inductor of 220 mH with cores of cross-sectional area
of 40 cm2 and mean length 35 em. The inductor should be able to carry a
current of up to 9.0 A at a flux density not to exceed 1.1 T.
A proposed energy storage mechanism consists of an N -tum coil wound
around a large non-magnetic (/L = /Lo) toroidal form as shown in Fig. 1.33.
As can be seen from the figure, the toroidal form has a circular cross section
of radius a and toroidal radius r, measured to the center of the cross section.
The geometry of this device is such that the magnetic field can be considered
to be zero everywhere outside the toroid. Under the assumption that a << r,
the H field inside the toroid can be considered to be directed around the toroid
and of uniform magnitude
H=
Ni
2nr
1.9 Problems
Figure 1.33 Toroidal winding for Problem 1.29.
For a coil with N = 12,000 turns, r = 9 m and a= 0.55 m:
a. Calculate the coil inductance L.
b. The coil is to be charged to a magnetic flux density of 1.80 T. Calculate
the total stored magnetic energy in the torus when this flux density is
achieved.
c. If the coil is to be charged at a uniform rate (i.e. di / dt =constant),
calculate the terminal voltage required to achieve the required flux density
in 40 s. Assume the coil resistance to be negligible.
1.30 Figure 1.34 shows an inductor wound on a laminated iron core of rectangular
cross section. Assume that the permeability of the iron is infinite. Neglect
magnetic leakage and fringing in the two air gaps (total gap length= g). The
N-turn winding is insulated copper wire whose resistivity is p&1·m. Assume
that the fraction f w of the winding space is available for copper; the rest of the
space being used for insulation.
a. Calculate the cross-sectional area and volume of the copper in the
winding space.
b. Write an expression for the flux density B in the inductor in terms of the
current density leu in the copper winding.
c. Write an expression for the copper cun·ent density leu in terms of the coil
current I, the number of turns N and the coil geometry.
d. Derive an expression for the electric power dissipation in the coil in terms
of the current density leu·
Core:
depth h into
the page
Figure 1.34 Iron-core inductor for
Problem 1.30.
55
56
CHAPTER 1
Magnetic Circuits and Magnetic Materials
e. Derive an expression for the magnetic stored energy in the inductor in
terms of the applied current density leu.
f. From parts (d) and (e) derive an expression for the L j R time constant of
the inductor. Note that this expression is independent of the number of
turns in the coil and hence does not change as the inductance and coil
resistance are changed by varying the number of turns.
1.31 The inductor of Fig. 1.34 has the following dimensions:
a = h = w = 1.8 em b = 2.2 em g
= 0.18 em
The winding factor (i.e. the fraction of the total winding area occupied by
conductor) is fw = 0.55. The resistivity of copper is 1.73 x 10- 8 Q·m. When
the coil is operated with a constant de applied voltage of 40 V, the air-gap flux
density is measured to be 1.3 T. Find the power dissipated in the coil, coil
current, number of turns, coil resistance, inductance, time constant, and wire
size to the nearest standard size. (Hint: Wire size can be found from the
expression
AWG = 36- 4.312ln (
Awire
)
X 10- 8
1.267
where AWG is the wire size, expressed in terms of the American Wire Gage,
and Awire is the conductor cross-sectional area measured in m2 .)
1.32 The magnetic circuit of Fig. 1.35 has two windings and two air gaps. The core
can be assumed to be of infinite permeability. The core dimensions are
indicated in the figure.
a. Assuming coil 1 to be carrying a current I 1 and the current in coil 2 to be
zero, calculate (i) the magnetic flux density in each of the air gaps, (ii) the
flux linkage of winding 1, and (iii) the flux linkage of winding 2.
b. Repeat part (a), assuming zero current in winding 1 and a current / 2 in
winding 2.
c. Repeat part (a), assuming the current in winding 1 to be 11 and the current
in winding 2 to be h.
Figure 1.35 Magnetic circuit for Problem 1.32.
1.9 Problems
Figure 1.36 Symmetric magnetic circuit for Problem 1.33.
d. Find the self-inductances of windings 1 and 2 and the mutual inductance
between the windings.
1.33 The symmetric magnetic circuit of Fig. 1.36 has three windings. Windings A
and B each have N turns and are wound on the two bottom legs of the core.
The core dimensions are indicated in the figure.
a. Find the self-inductances of each of the windings.
b. Find the mutual inductances between the three pairs of windings.
c. Find the voltage induced in winding 1 by time-varying currents iA(t) and
iB (t) in windings A and B. Show that this voltage can be used to measure
the imbalance between two sinusoidal currents of the same frequency.
1.34 The reciprocating generator of Fig. 1.37 has a movable plunger (position x)
which is supported so that it can slide in and out of the magnetic yoke while
maintaining a constant air gap of length g on each side adjacent to the yoke.
Both the yoke and the plunger can be considered to be of infinite permeability.
The motion of the plunger is constrained such that its position is limited to
0 ::::;x::::; w.
There are two windings on this magnetic circuit. The first has N 1 turns and
carries a constant de current / 0 . The second, which has N2 turns, is
open-circuited and can be connected to a load.
a. Neglecting any fringing effects, find the mutual inductance between
windings 1 and 2 as a function of the plunger position x.
b. The plunger is driven by an external source so that its motion is given by
x(t) =
w(l
+ E sinwt)
2
where E ::::; 1. Find an expression for the sinusoidal voltage which is
generated as a result of this motion.
57
58
CHAPTER 1
Magnetic Circuits and Magnetic Materials
00
,----J..L_-+__
Yoke
w Depth D
_,
g
I ~·I
g
}'"_./IIh
»
g
Plunger
x(t) = '!:!!. (1
2
+ E sin wt)
Figure 1.37 Reciprocating generator for
Problem 1.34.
1.35 Figure 1.38 shows a configuration that can be used to measure the magnetic
characteristics of electrical steel. The material to be tested is cut or punched
into circular laminations which are then stacked (with interspersed insulation
to avoid eddy-current formation). Two windings are wound over this stack of
laminations: the first, with N1 turns, is used to excite a magnetic field in the
lamination stack; the second, with N2 turns, is used to sense the resultant
magnetic flux.
The accuracy of the results requires that the magnetic flux density be
uniform within the laminations. This can be accomplished if the lamination
width w = R0 - Ri is much smaller than the lamination radius and if the
excitation winding is wound uniformly around the lamination stack. For the
n laminations,
each of thickness !:l
Figure 1.38 Configuration for measurement of magnetic properties of
electrical steel.
1.9 Problems
purposes of this analysis, assume there are n laminations, each of thickness b..
Also assume that winding 1 is excited by a current i 1 = I 0 sin wt.
a. Find the relationship between the magnetic field intensity H in the
laminations and current i 1 in winding 1.
b. Find the relationship between the voltage v2 and the time rate of change of
the flux density B in the laminations.
c. Find the relationship between the voltage vo = G v2dt and the flux
density.
In this problem, we have shown that the magnetic field intensity H and the
magnetic flux density B in the laminations are proportional to the current i 1
and the voltage vo by known constants. Thus, B and H in the magnetic steel
can be measured directly, and the B- H characteristics as discussed in
Arts. 1.3 and 1.4 can be determined.
1.36 From the de magnetization curve of Fig. 1.10 it is possible to calculate the
relative permeability /Lr = Bc/(tLoHc) for M-5 electrical steel as a function of
the flux level Be. Assuming the core of Fig. 1.2 to be made of M-5 electrical
steel with the dimensions given in Example 1.1, calculate the range of flux
densities for which the reluctance of the core never exceeds 5 percent of the
reluctance of the total magnetic circuit.
1.37 In order to test the properties of a sample of electrical steel, a set of
laminations of the form of Fig. 1.38 have been stamped out of a sheet of the
electrical steel of thickness 3.0 mm. The radii of the laminations are Ri = 80
mm and R0 = 90 mm. They have been assembled in a stack of 15 laminations
(separated by appropriate insulation to eliminate eddy currents) for the
purposes of testing the magnetic properties at a frequency of 50 Hz.
a. The flux in the lamination stack will be excited from a variable-amplitude,
50-Hz voltage source whose peak amplitude is 20 V. Ignoring any voltage
drop across the winding resistance, calculate the number of turns N 1 for
the excitation winding required to insure that the lamination stack can be
excited up to a peak flux density of 1.8 T.
b. With a secondary winding of N2 = 10 turns and an integrator gain
G = 1, 000, the output of the integrator is observed to be 7.5 V peak.
Calculate (i) the corresponding peak flux in the lamination stack and (ii)
the corresponding amplitude of the voltage applied to the excitation
winding.
1.38 The coils of the magnetic circuit shown in Fig. 1.39 are connected in series so
that the mmfs of paths A and B both tend to set up flux in the center leg C in
the same direction. The coils are wound with equal turns, N 1 = Nz = 120.
The dimensions are:
J
Cross-section area of A and B legs = 8 cm2
Cross-section area of C leg = 16 cm2
Length of A path = 17 em
Length of B path = 17 em
59
60
CHAPTER 1
Magnetic Circuits and Magnetic Materials
Figure 1.39 Magnetic circuit for
Problem 1.38.
Length of C path = 5.5 em
Air gap length = 0.35 em
The material is M-5 grade, 0.012-in steel. Neglect fringing and leakage.
a. How many amperes are required to produce a flux density of 1.3 T in the
air gap?
b. Under the condition of part (a), how many joules of energy are stored in
the magnetic field in the air gap and in the core? Based upon this stored
energy, calculate the inductance of this series-connected winding.
c. Calculate the inductance of this system assuming the core to be of infinite
permeability. Compare your inductance with the value calculated in
part (b).
1.39 The following table includes data for the top half of a symmetric 60-Hz
hysteresis loop for a specimen of magnetic steel:
,B;,
:
•.···
·'··.. ::::.
. 0 0.2 0.4 0.6 0.7 0.8 0.9 1.0 0.95 0.9 0.8 0.7
H,:A-tufil~/IJi 48 52 58 73 85 103 135 193 80
42
0.6
0.4
0.2
0
2 -18 -29 -40 -45 -48
Using MATLAB, (a) plot this data, (b) calculate the area of the hysterises loop
in joules, and (c) calculate the corresponding 60-Hz core loss density in
Watts/kg. Assume the density of the steel is 7.65 g/cm3 .
1.40 A magnetic circuit of the form of Fig. 1.27 has dimension
Cross-sectional Area Ac = 27 cm2
Mean core length lc = 70 em
Gap length g = 2.4 mm
N = 95 turns
and is made up of M-5 electrical steel with the properties described in
Figs. 1.10, 1.12, and 1.14. Assume the core to be operating with a 60-Hz
sinusoidal flux density of the rms flux density of 1.1 T. Neglect the winding
resistance and leakage inductance. Find the winding voltage, rms winding
current, and core loss for this operating condition. The density of M-5 steel is
7.65 g/cm3 .
1.41 Repeat Example 1.8 under the assumption that all the core dimensions are
doubled.
1.9 Problems
C/L
Figure 1.40 Magnetic circuit for the loudspeaker
of Problem 1.44 (voice coil not shown).
1.42 Using the magnetization characteristics for samarium cobalt given in
Fig. 1.19, find the point of maximum-energy product and the corresponding
flux density and magnetic field intensity. Using these values, repeat
Example 1.10 with the Alnico 5 magnet replaced by a samarium-cobalt
magnet. By what factor does this reduce the magnet volume required to
achieve the desired air-gap flux density?
1.43 Using the magnetization characteristics for neodymium-iron-boron given in
Fig. 1.19, find the point of maximum energy product and the corresponding
flux density and magnetic field intensity. Using these values, repeat
Example 1.10 with the Alnico 5 niagnet replaced by a neodymium-iron-boron
magnet. By what factor does this reduce the magnet volume required to
achieve the desired air-gap flux density?
1.44 Figure 1.40 shows the magnetic circuit for a permanent-magnet loudspeaker.
The voice coil (not shown) is in the form of a circular cylindrical coil which
fits in the air gap. A samarium-cobalt magnet is used to create the air-gap de
magnetic field which interacts with the voice coil currents to produce the
motion of the voice coil. The designer has determined that the air gap must
have radius R = 2.2 em, length g = 0.1 em, and height h = 1.1 em.
Assuming that the yoke and pole piece are of infinite magnetic
permeability (J.L -+ oo ), find the magnet height hm and the magnet radius Rm
that will result in an air-gap magnetic flux density of 1.3 T and require the
smallest magnet volume.
(Hint: Refer to Example 1.10 and to Fig. 1.19 to find the point of
maximum energy product for samarium cobalt.)
1.45 Repeat Problem 1.44 replacing the samarium-cobalt magnet with a
neodymium-iron-boron magnet assuming the magnetization characteristics of
Fig. 1.19.
1.46 Based upon the characteristics of the neodymium-iron-boron material of
Fig. 1.24 and Table 1.1, calculate the maximum-energy product of this grade
of neodymium-iron-boron magnet material at each of the temperatures of
61
62
CHAPTER 1
Magnetic Circuits and Magnetic Materials
VAirgap:
area Ag
Figure 1.41 Magnetic circuit for
Prob 1.47.
Table 1.1 and the corresponding values of H and B. (Hint: Write an analytic
expression for the maximum energy product in terms of H using the fact that
the recoil permeability is 1.04 flo.)
1.47 It is desired to achieve a time-varying magnetic flux density in the air gap of
the magnetic circuit of Fig. 1.41 of the form
Bg
= Bo + B1 sin wt
where B0 = 0.6 T and B1 = 0.20 T. The de field Bois to be created by a
neodymium-iron-boron magnet with magnetization characteristic of Fig. 1.19,
whereas the time-varying field is to be created by a time-varying current.
For Ag = 7 cm2 , g = 0.35 em, and N = 175 turns, and based upon the
neodymium-iron-boron characteristics of Fig. 1.19, find:
a. the magnet length d and the magnet area Am that will achieve the desired
de air-gap flux density and minimize the magnet volume and
b. the amplitude of the time-varying current required to achieve the desired
time-variation of the air-gap flux density.
1.48 A magnetic circuit of the form of Fig. 1.41 is to be designed using
neodymium-iron-boron material with the characteristics of Fig. 1.24 and
Table 1.1.
The magnetic circuit core will have cross-sectional area Ag = 9 cm2 and
the air-gap length will beg = 0.32 em. The circuit is designed to be operated
at temperatures up to 180 C.
a. Find the magnet length d and the magnet area Am corresponding to the
minimum magnet volume that will produce a magnetic flux density of
0.8 T with the system operating at a temperature of 180 C.
b. For the magnet of part (a), find the flux density in the air-gap when the
operating temperature is 60 C.