Download Chapter 4 Magnetic Circuits

Chapter 4:
Magnetic Circuits
Lo Yew Chiong
Course Contents
Magnetic materials:




diamagnetic material,
paramagnetic material,
ferromagnetic material.

Hysterisis loss and eddy current loss, reluctance and
permeance.

Analysis of linear magnetic circuits (with air-gap
problems).
2
Magnetic Materials
All materials are composed of atoms, each with a positively charged
nucleus and a number of orbiting negatively charged electrons.

Orbiting electron
Spinning electron
nucleus
(1)Orbiting electron
(2)Spinning electron
Magnetization describes to what extent they are affected by magnetic
fields, and also determines the magnetic field that the material itself creates.
Magnetization in a material is mostly due to the magnetic moment of
electrons, generated by two principal mechanisms:
(1) orbital motion of the electrons, mo
(2) spinning motions of the electrons, ms


3
Magnetic Moment
Magnetic moment, m = loop area A  current I.

e
eu

T
2r
m0  IA
I 
T = time for one revolution =
2r
u
u = velocity of the electron
 
 eu  2
 
 r
2

r


eur

2

e 
 Le
m0   
 2me 
Le = angular momentum = meur
me is electron mass
e
2me
e
m


Spin magnetic moment, s
2me
Orbital magnetic moment, m0  
4
h  6.626 10-34 J  s
ħ = h/2 ,
h is Planck’s constant
Le = 0, ħ ,2ħ, …
Magnetic Permeability
Magnetization vector M = vector sum of magnetic dipole moments of the
atoms contained in a unit volume of the material. (Unit= (A x m2)/m3 = A/m)
Total Magnetic flux density,
B= µ0H +
µ0M
= µ0(H+M)


External applied
magnetic field
Magnetization of
the material
M = mH where m is magnetic susceptibility constant.


m - Degree of magnetization of a material in response to an applied magnetic field
B = µ0 (H + mH) = µ0(1 + m)H
B = µH


µ= µ0 (1 + m) , µr = µ/µ0 = 1 + m
Permeability, µ : Degree of magnetization of a material that responds
linearly to applied magnetic field.


5
H
Diamagnetism
In this material, all the orbital moments pair off.

Fig: Orbital diagram for
Neon (1s22s22p6)- 10 electrons

When H is applied, electrons in this material rearrange their
orbital motion and the magnetic moment will oppose H.

Diamagnetic materials have a very weak and negative
susceptibility (m) to external magnetic fields.
6
Diamagnetism

Diamagnetic materials do not exhibit permanent magnetism,
and the induced magnetic moment disappears when the
applied field is withdrawn.

Ex.: hydrogen, copper, gold, silicon, germanium, graphite,
bismuth, helium, sulfur.
7
H
Paramagnetism
Paramagnetic materials have some unpaired electrons that produce the net
spin magnetic moments which tend to align themselves in the direction of
the external magnetic field.

Fig: Orbital diagram for
Aluminium

They are weakly attracted to magnets and have a small positive
susceptibility (m) to magnetic fields.

Paramagnetism is temperature dependent:
It becomes more magnetic when its temperature reduces.
It becomes less magnetic when its temperature increases.

Ex.: Aluminum, Platinum, air, potassium, tungsten, liquid oxygen.
8
Ferromagnetism

In these materials strong interactions between atomic magnetic
moments cause them to line up parallel to each other in
regions called magnetic domains.

When there is no externally applied field, the orientations of
the domain magnetizations are random.

But when H field is applied, they tend to orient themselves
parallel to the field. The domain boundaries also shift, the
domains magnetized in the field direction grow (A), and those
magnetized in other directions shrink (C).
Fig: Domain
theory
9
Ferromagnetism

Ferromagnetic materials have a large and positive
susceptibility to an external magnetic field.

Ferromagnetic materials lose all their magnetic
properties if they are heated to a high enough
temperature, due to the magnetized domains will
organize themselves randomly after their atoms are
being heated.

The temperature at which a ferromagnetic material
loses its magnetism is called the Curie temperature
and it is different for every metal.

Ex: cobalt, nickel, iron are usually used to fabricate
permanent magnets due to the ability to retain their
magnetism properties for long time.
10
Fig: Curie point
Ferromagnetism





Experiment of Curie
Temp.: Nickel-iron as
ferromagnetic.
Domains of electrons
align causing the sample
to be attracted to the
magnet.
Heating the sample
scrambles the alignment
of the electrons.
The sample is no longer
attracted to the magnet.
After the sample has
cooled down, the
electrons realign and
the ferromagnetism
returns.
11
After 25 seconds
Properties of Magnetic Materials
12
Hysteresis




Definition: A property of systems that
do not instantly react to the forces
applied to them, but react slowly, or do
not return completely to their original
state.
Measuring the magnetic flux B of a
ferromagnetic material while the
magnetizing force H is changed.
o-a – The material has never been
previously magnetized or has been
thoroughly demagnetized will follow the
dashed line as H is increased. The
greater the amount of current applied
(H+), the stronger the magnetic field in
the component (B+).
a – Almost all of the magnetic domains
are aligned and an additional increase in
the magnetizing force will produce very
little increase in magnetic flux. The
material has reached the point of
magnetic saturation.
13
H
H
0
H=0
Fig: Hysterisis Loop
Hysteresis

a-b - H is reduced back down to
zero.

b - Some magnetic flux remains in
the material even though the
magnetizing force is zero. This is
referred to as the point of
retentivity on the graph and
indicates the remanence or level
of residual magnetism in the
material. (Some of the magnetic
domains remain aligned but some
have lost their alignment.)

H
H
0
H=0
b-c - As the magnetizing force is
reversed, the curve moves to
point c.
Fig: Hysterisis Loop
14
Hysteresis



c - Where the flux has been
reduced to zero. This is called the
point of coercivity. (The reversed
magnetizing force has flipped
enough of the domains so that the
net flux within the material is
zero.) The force required to
remove the residual magnetism
from the material, is called the
coercive force or coercivity of the
material.
c-d - Magnetizing force is
increased in the negative direction.
d - The material will again become
magnetically saturated but in the
opposite direction.
0
H
15
Fig: Hysterisis Loop
Hysteresis

d-e - Reducing H to zero brings the
curve to point e. It will have a level
of residual magnetism equals to that
achieved in the other direction.

e-f - Increasing H back in the
positive direction will return B to
zero. Notice that the curve did not
return to the origin of the graph
because some force is required to
remove the residual magnetism. The
curve will take a different path from
point f back the saturation point
where it with complete the loop.

0
H=0
a-b-c-d-e-f-a - Complete close loop is
called as a hysteresis loop.
H
16
Fig: Hysterisis Loop
Hysteresis

Consider the solenoid that shown in figure, when the current I is increasing,
the electromotive force (e.m.f) induced in the winding opposes the
increase in current according to Lenz’s law, the extra power spent by the
source is
where A is the cross sectional area of the core, N is the number of turns, B is the
magnetic flux density induced in the core, and  is the magnetic flux generated.

From the relation of NI = HL,
where L = the length of the core and
V = AL is the volume.
17
Hysteresis

Energy supplied by the source in going from the point g to the point b,

This integral corresponds to the shaded area in the figure and is equal to
the energy supplied per unit volume of the magnetic core.
When the current is in the same direction but is decreasing, the polarity of
the induced e.m.f. is reversed, according to Lenz’s law, with the result that
the energy is returned to the source.

18
Hysteresis

Finally, the energy supplied by the source during one cycle is
where the integral is evaluated around the hysteresis loop (gbcdefg).

Work done per cycle = volume of material x area of loop (Joule/cycle)
 Work done per cycle per volume = area of loop (Joule/m3cycle)

Sometimes the energy dissipated per unit volume per cycle is also named
as the hysteresis loss. This energy loss is due to the friction encountered
during domain-wall motion and rotation.
19
Hysteresis
Fig: (a) Hard Material
(b) Soft Material

Hard material
High retentivity and coercivity - suitable for permanent
magnets.
Large hysteresis loss - not suitable for alternating
magnetization.

Soft material
Small hysteresis loss - suitable for armature and transformer
core.
20
Modeling Hysteresis Loss

Power of the hysteresis loss is empirically given by Steinmetz
hysteresis law,
Ph  fkh Bpx

where f is the frequency of excitation, kh is a constant
determined by the nature of the ferromagnetic material
experimentally, Bp is the peak value of the magnetic flux
density, and x is the Steinmetz coefficient ranging from 1.5 to
2.5.
Unit = W/m3
21
Eddy Current

It is caused when a moving (or changing) magnetic
field intersects a conductor, or vice-versa.

The relative motion causes a circulating flow of
electrons, or current, within the conductor.

These circulating eddies of current create
electromagnets with magnetic fields that oppose the
effect of the applied magnetic field (Lenz's law).
22
Eddy Current - (Lenz's law)

As the magnet approaches the loop, the applied B field
in the centre increases. This is a change.

An Induced Field is created which attempts to negate
the applied field - ie to keep the total field at zero - its
original value.

This Induced field must be associated with a current the INDUCED CURRENT in the loop whose
direction is determined by the first version of the RH
rules
23
Example 1: Eddy Current

A rotating disk in a magnetic field.
Fig: Eddy current on a
rotating disk





Magnetic flux is flowing through a portion of disk in the direction perpendicular to
the disk surface.
Element Ob: Moving across the magnetic field; e.m.f. will be induced along it.
Elements Oa and Oc: Not in the magnetic field, but connected to the disk and
the element Ob.
Applying Lenz’s Law, these would serve as the return path for charges in element
Ob to move back from b to O
-> EDDY CURRENT.
24
Example 1: Eddy Current
B
F
I
Fig: Eddy current on a
rotating disk



Eddy current along Ob (in magnetic field) will cause a force that goes
against the disk rotation.
The current in the return path is outside of the magnetic field, and thus,
does not produce any force.
The interaction between the eddy current and the magnetic field would act
to stop the disk rotation.
25
Example 2: Eddy Current



Consider the core of an AC transformer.
The a.c. current in the primary coil of transformer will supply
alternating flux in the core, and an e.m.f. will be induced in the
secondary coil.
Since the iron core is also a conductor and the cross section like AA
can be considered as a closed conducting circuit, with one in the
other one, thus there will be eddy current flowing. The flowing of
eddy current will dissipate energy and therefore is not needed.
26
Example 2: Eddy Current


To reduce the eddy current losses, the resistivity of the material is
increased by adding silicon in the metal or ferromagnetic materials.
Another effective way to achieve low eddy current loss is by using
laminations of electrical metal sheets. These metal sheets are coated
with electric insulation which break the eddy currents path.
Fig: Laminated core transformer
showing edge of laminations
at top of unit.
27
Eddy current for domain wall model

The power due to the eddy current loss, Pe is given as
where f is the frequency of excitation
ke is a constant determined by the nature of the magnetic material,
Bp is the peak value of the magnetic flux density,
d is the thickness of the lamination.

The formula is obtained under the assumption of global eddy current.
This is incorrect for ferromagnetic materials due to the magnetic domains.
Fig: Eddy current for
domain wall model

When the excitation field varies, the domain walls move accordingly and
local eddy currents are induced by the fluctuating of the local flux density
caused by the domain.
28
Eddy current for domain wall model

The total eddy current caused by the local eddy currents is
generally higher than that predicted by the formulation under
the global eddy current assumption. The different is known as
the excess loss (Pex).
where f is the frequency of excitation,
Cex is a constant determined by the nature of the metal or magnetic material,
Bp is the peak value of the magnetic flux density.

The total loss can be calculated as

Disadv. of Eddy Current: Reduces the efficiency of a electromagnetic device
Adv. of Eddy Current : Used as an non-destructive tool to measure material
thickness and coating thickness. It also can be used in the crack detection,
material identification, heat damage detection, and heat treatment
monitoring.

29
The Concept of Magnetic Circuit


A magnetic circuit is a closed path containing a
magnetic flux.
Consider a ferromagnetic core wound with a coil of
N turns carrying a current I, as shown in the
following figure. The magnetic flux  through the
cross section of the core, A can be determined by
Ampere’s circuital law.
30
The Concept of Magnetic Circuit

Applying Ampère’s circuital law to a circular path of radius R1
going all around inside the toroid, and let R2 to be the inner
radius of the toroid, the flux through the core is determined.
 
 H .dl  NI
2R1
B
R2
 NI

NI
B
2R1
R1

R2
 B  ,   NI 
A
2R1
2
Flux = Magnetomotive force × Permeance


or Flux = m.m.f. / (reluctance) where permeance = 1/(reluctance)
The eqn. is analogous to ohm’s law,
Current = e.m.f. / (resistance) where conductivity = 1/(resistance)
31
Comparison between Magnetic
Circuit and Electric Circuit

Differences between electric and magnetic circuits :



32
The path of the magnetic flux flow B is perpendicular to the current flow
J in the circuit.
For a given temperature, electric resistance is constant and does not
depend on current density. However, the magnetic reluctance depends on
magnetic field and flux intensity since the permeability is not constant.
Current flowing in a electric circuit involves dissipation of energy, but for
magnetic circuit, energy is needed to generate magnetic flux.
Magnetic Circuit with an Air Gap

A simple magnetic circuit with an air gap of length lg cut in the
middle of a leg. The winding provides NI ampere-turn. The spreading
of the magnetic flux lines outside the common area of the core for
the air gap is known as fringing field.
Fig: A simple magnetic circuit with an air gap.

Fig: (a)Air gap with fringing (b)without fringing
For simplicity, this effect is negligible and the flux distribution is
assumed to be the ideal case.
33
Magnetic Circuit with an Air Gap

Applying the Ampère’s circuital law,
NI  H c Lc  H g Lg
where c = core, g = air gap.
=µH, H  Bc   c
c
 c  c Ac
Bg
g
Hg 

 g  0 Ag

B

According to Gauss’s law of magnetism, the closed surface integral
of B must be equal to zero. Hence, the flux of B must be the same
over any cross section of the magnetic circuit and the air gap.
 B  ds  0
S
c   g  
Bc Ac  Bg Ag
34
Magnetic Circuit with an Air Gap

Solving those equations,
 Lc
Lg 
Bg Ag 

  NI

A

A
 c c
0 g 


Magnetic flux is
  Bg Ag 
35
NI
 Lc
Lg 




A

A
 c c
0 g 

Reluctances of the core
and air gap in series
Magnetic circuit equivalent and electric
circuit analogy for electromagnet circuit

The shown magnetic circuit with air gap is analogous to a series
electric circuit
  Bg Ag 

NI  Rc  Rg 
NI
F  Rc  Rg 
 Lc
Lg 




A

A

0 g 
 c c
It is analogue to the Kirchhoff ’s voltage law (KVL) in
electric circuit theory,
F  R  0
V 0

k
k

k
k

k
F  R 
k
k

k
k
k
k
k
It states that the algebraic sum of the rises and drops of the
m.m.f around a closed loop of a magnetic circuit is equal to
zero. In other words, the sum of the m.m.f. rises equals the
sum of the m.m.f. drops around a closed loop.
36
Magnetic circuit equivalent and electric
circuit analogy for electromagnet circuit

The Gauss’s law of magnetism can be interpreted as an
analogue to the Kirchhoff ’s current law (KCL) in electric
circuit theory,  c   g    I entering   I leaving

entering
   leaving
where c and g are regarded as the “current entering/leaving” a junction in the magnetic
circuit.

It states that the algebraic sum of the fluxes entering or leaving
a junction of a magnetic circuit is equal to zero. In other
words, the sum of the fluxes entering a junction is equal to the
sum of the fluxes leaving a junction.
37
Magnetic Circuit with a Permanent
Magnet

The directions of B and H vectors
in different parts of the magnetic
circuit as shown in left figure:



Permanent Magnet
Soft Iron
A magnetic circuit with
a pair of permanent magnets
38
In the soft iron core, B and H are in the same
direction and the operating point on the
hysteresis curve is situated somewhere in the
first quadrant.
In the air gap, B and H are also oriented in
the same direction.
However, B and H are oriented in opposite
direction for permanent magnet since the
operating point for the permanent magnet is
in the second quadrant on the hysteresis
curve.
Magnetic Circuit with a Permanent
Magnet


There are neither free currents nor displacement currents, the
line integral of H.dl around the magnetic circuit must be zero.
H i Li  H g Lg  H m Lm  0
H i Li  H g Lg  H m Lm
where i, g, and m refer to the soft iron core, the air gap, and the
permanent magnet.
Hi 
Hg 
Hm 
39
Bi
i

Bg
g
Bm
m
i
i Ai


g
 0 Ag
m
 m Am
Magnetic Circuit with a Permanent
Magnet

 is the same over any cross section,
i   g   m  
Ai Bi  Ag Bg  Am Bm

Solving those equations,

The magnetic flux is
  Bg Ag 

 L
Lg 
Bg Ag  i 
  H m Lm
 i Ai 0 Ag 
H m Lm
 Li
Lg 




A

A

0 g 
 i i
This equation is similar to
  Bg Ag 
NI
 Lc
Lg 




A

A
 c c
0 g 

, except that NI is replaced by HmLm. This shows that permanent
magnet could be used to replace the current source to establish the
magnetic flux in the circuit.
40
Magnetic Circuit with a Permanent
Magnet

As Rg 
Lg
 0 Ag
& Ri 
Li
,   Bg Ag  
 i Ai
H m Lm
Lg 
Li



 i Ai  0 Ag 
can be
rewritten as H m Lm  Ri  Rg 

The design of a magnetic circuit energized by permanent magnet
is complicated due to the position of the operating point (the
relation between B and H) on the hysteresis curve. Thus, an
operating point is usually used to make the energy product HmBm
maximum.
2
Bg
20
energy density stored in the
air gap between the two
surfaces of the iron yokes
41
Ag Lg 
1

H g Lg ( Ag Bg )
2
volume of the air gap
Magnetic Circuit with a Permanent
Magnet


The m.m.f. - HiLi can be neglected if the reluctance of the iron core is
much smaller than the air gap
( HiLi + HgLg = HmLm , AgBg=AmBm).
Bg
2
Thus, 2
0
Ag Lg 
1
H m Lm ( Am Bm )
2

1
H m Bm ( Am Lm )
2

The magnetic energy in the air gap is therefore equal to one-half the
energy product multiplied by the volume of the permanent magnet.

As a rule, one requires that the volume AmLm be as small as possible,
for reasons of economy, size, and weight. Then the operating point on
the hysteresis curve is chosen so that the energy product HmBm is max.
42
Attractive force between two iron
surfaces separated by an air gap

If one of the iron yokes is made to be movable and
pulled away from the other iron yoke by a force of
P Newton for a distance d meters.

The additional magnetic energy stored in the
additional volume of the air gap (Agd) is caused by
the work done by the force as

This indicates that the magnetic energy stored in
the air gap can be increased by increasing the
volume of the air gap(Agd) .

The force required to increase the additional
magnetic energy is then given as
I
d
Am2
43
Force P
netwon
Example 3

Find the value of I required to establish a magnetic flux of  =
0.75  10-4 Wb in the series magnetic circuit as shown in Figure
Ex3. Calculate the force exerted on the armature (moving part)
when the flux is established. The relative permeability for the steel
is r = 1424.
Fig Ex3
44
Solution (Example 3)

The above device can be analysed by its magnetic circuit
equivalent and its electric circuit analogy as shown in figure below.
(a) Magnetic circuit equivalent and
45
(b) electric circuit analogy
Solution (Example 3)
From the Gauss law (analogy to KCL in electric circuit), the flux density for each section is
 0.75  10 -4 Wb
B 
 0.5 T .
A 1.5  10 -4 m 2
The magnetic field intensity of steel is
Hc 
B
B
0.5 T


 279 At/m.
-7
-1
 c  r  0 1424  4  10 Hm
The magnetic field intensity of air gap is
Hg 
B
B
0.5 T
5



3.98

10
At/m.
-7
-1
 g  0 1 4  10 Hm
The magnetomotive force drops are


H c l c  279 At/m  100  10 -3 m  28 At.
46



H g l g  3.98  10 5 At/m 2  10 -3 m  796 At.
Applying Ampère circuital law (analogy to KVL in electric circuit),
NI  H c l c  H g l g
 28 At  796 At
200 t I  824 At
I  4.12 A.
Energy stored in air gap, W = Volume of air gap, Vg × magnetic energy density, W0
 Ag l g 
Bg2
.
20
W = mechanical energy to close the air gap
= Force, P × length of the air gap, lg
Ag l g 
B g2
2 0
 Pl g
P
47
Ag B g2
2 0
1.5  10  4 m 2  0.5 T 

2  4  10 -7 Hm -1
 14.92 N
2
Example 4

Determine the value of I required to establish a magnetic flux of
2 = 1.510-4 Wb in the section of the core indicated in Figure
Ex4. The relative permeability for the steel at region bcde, be, and
efab are 2 = 4972, 1 = 4821, and T = 2426, respectively.
Fig Ex4
48
Solution (Example 4)
The above device can be analysed by its magnetic circuit
equivalent and its electric circuit analogy as shown in figure below.
(a) Magnetic circuit equivalent and (b) electric circuit analogy
49
Solution (Example 4)
Rbcde 
lbcde
0.2
3


53
.
35

10
At/Wb
-4
2
 2 0 A 4972  0  6  10 m
lbe
0.05
3


13
.
76

10
At/Wb
-4
2
10 A 4821 0  6  10 m
lefab
0.2
3
Refab 


109
.
34

10
At/Wb
-4
2
T 0 A 2426  0  6  10 m
Rbe 
Since 1Rbe   2 Rbcde
 R
1  2 bcde  5.816 10  4 Wb
Rbe
From Gauss Law,
T  1   2  (1.5  5.816) 104 Wb  7.316 104 Wb
50
Solution (Example 4)
Applying Ampère circuital law for loop 1,
NI   T Refab  1 Rbe  0
I
 T Refab  1 Rbe
51
N

80  8
 1.76 A
50
Example 5

The core of Figure Ex5 is made of cast steel. Calculate the current
I that is needed to establish a flux of g = 6  10-3 Wb at the air
gap if fringing field is neglected.
[Hint: Additional information can be obtained from the B-H curve on Slide 55]
Fig Ex5
52
Solution (Example 5)
(a) Magnetic circuit equivalent and (b) electric circuit analogy
53
Consider each section in turn.
For air gap,
Bg = g/Ag = 6  10-3 Wb/2  10-2 m2 = 0.3 T.
Hg = Bg/0 = 0.3 T/(4  10-7 H/m) = 2.388  105 At/m.
For section ab and cd,
Bab = Bcd = Bg = 0.3 T.
Hab = Hcd = 250 At/m.
(refer to B-H curve in figure 4.26 in lec note).
Apply Ampère circuital law (KVL) at loop 2,
Since the flux in leg da flows in the opposite direction of the flux in leg ab, leg cd,
and air gap, the corresponding term of Hdalda will be subtractive. Also, NI = 0 for
loop 2. Thus,

loop2
NI  loop2 Hl
0  H ab l ab  H g l g  H cd l cd  H da l da



 2500.25  2.388  10 5 0.25  10 3  2500.25  0.2 H da
 62.5  59.7  62.5  0.2 H da
54
 184.7  0.2 H da
B-H curve (Notes)
55
Solution (Example 5)
Thus,
Hda = 923.5 At/m.
Bda = 1.12 T
(from B-H curve).
2 = BdaA = 1.12  0.02 = 2.24  10-2 Wb.
1 = 2 + 3 = 2.84  10-2 Wb.
Bdea = 1/A = (2.84  10-2)/0.02 = 1.42 T.
Hdea = 2100 At/m
(from B-H curve).
Apply Ampère circuital law at loop 1,
NI = Hdealdea + Hdalda = (2100)(0.35) + 184.7 = 919.7 At
I = 919.7/200 ≈ 4.6 A
56
Example 6

The core of the magnetic device as shown in Figure Ex6 is made of cast-iron and
it is symmetrical both left and right arms. Find the current I that needed to
establish flux density of 30 Wb at the right arm of the core.
[Hint: Additional information can be obtained from the B-H curve on Slide 55]
Fig Ex6 Series-parallel magnetic
circuit made of cast iron core
57
Solution (Example 6)
(a) Magnetic circuit equivalent and (b) electric circuit analogy
58
Solution (Example 6)
Since the magnetic device is symmetry and made by the same material at
both left and right arms, we can concentrate the analysis of the circuit on
either loop 1 or loop 2. From Gauss law, one gets
1 = 2.
T = 1 + 2.
= 22.
Apply Ampère circuital law at left hand arm,

loop1
NI  loop1 Hl
NI  H ek l ek  H g l g  H mb l mb  H bafelbafe
NI  H ek l ek  H g l g  H mb l mb  H bafe lba  l af  l fe 
(1)
 T 2 2 2  30  10 6 Wb
Bek 


 0.6 T
A
A
0.01 0.01 m 2
H ek  H mb  2650 At/m
59
(from B-H curve)
(2)
Solution (Example 6)
2 2
T
2  30  10 6 Wb
5
At/m.
10

77
.
4




Hg 
 0  0 A  0 A 4  10 7 H/m 0.01  0.01 m 2
Bg
1
30  10 6 Wb
 0.3 T

Bbafe 
2
A 0.01  0.01 m
H bafe  750 At/m
(B-H curve)
Substitute (2), (3), and (4) in (1) yields

2650  0.035  4.77  105  0.5 10-2   750  3  0.04
I
N
2568
400
 6.42 A.

60
(3)
(4)