Chapter 1
Introduction to Machinery
Principles
Magnetic materials form a major part in the
construction of electrical machines.
In this chapter
• Production of a Magnetic Field
• Magnetic Circuits
• Magnetic Behaviour of Ferromagnetic Materials
• Energy Losses in a Ferromagnetic Core
Production of Magnetic Field
f
i
Ni
N
Cross-sectional area
Faraday’s Law
Whenever a conductor is moved through a magnetic field, or
whenever the magnetic field near a conductor is changed, currents
flow in the conductor. This effect is called electromagnetic
induction. Faraday's law gives the magnitude and direction of the
emf produced in a conducting loop whenever the loop is moved, or
whenever the magnetic field near the loop is changed:
e
dl
dt
where l=Nf
f   B  dA  BA
and
A
B is the flux density and A is the area
f is the magnetic flux in webers (Wb)
l is the flux linkage in weber-turns
N is the number of turns
Magnetic Circuits
Transformers: the magnetic
circuits may be formed by
ferromagnetic materials only.
Rotating Machines: the
magnetic circuits may be
formed by ferromagnetic
materials in conjunction with
an air medium.
Magnetic Circuits
The magnetomotive force (mmf) of an N-turn current
carrying coil is given by
=Ni
The reluctance of a path for magnetic flux, such as the bar or
iron shown in the following figure, is given by:
lc

mA
where,
lc = mean length of the path
m = permeability of the material
A = cross-sectional area
l
f
A
Reluctance is analogous to resistance in an electrical circuit.
When the bar is not straight, the length of the path is taken to
be the length of the centerline. l is therefore considered as
the mean length of the path.
Magnetic flux, f, in a magnetic circuit is equivalent of
current in an electrical circuit. Magnetic flux, reluctance and
mmf are related by:
  f
which is equivalent to Ohm’s Law (V=iR).
The current carrying coil is the magnetic circuit equivalent to
a voltage source in an electrical circuit. Therefore, mmf can
be considered to be analogous to a source voltage.
Example 1
Using the magnetic circuit concepts, analyze the toroid coil shown in
Fig. 1. The magnetic circuit is analogous to the electrical circuit shown
in Fig. 2, with a resistance connected across a voltage source.
H
Core centreline
f
R

+
R
N turn coil
i
Fig. 1
Fig. 2
Fig. 1. Coil on a toroid iron core
Fig. 2. The magnetic circuit for the toroid coil
Find the expressions for the reluctance and magnetic flux.
Solution to Example 1
The mean length of the magnetic path is: l=2pR
The cross-section of the core is circular with a radius, r.
Thus, the area of the cross section is: A=pr2.
The expression for the reluctance (from Eq. 1-2):

l
2pR 2 R

 2
2
mA mpr
mr
The mmf is: =Ni
H
Core centreline
R
 mNr I
And the magnetic flux: f    2 R
2
N turn coil
i
Example 2
Consider the magnetic core with an air gap as shown in Fig. 3.
The core material has a relative permeability of 6000 and a
rectangular cross-section of 2cm by 3cm. The coil has 500 turns.
Determine the current required to establish a flux density of
Bgap=0.25T in the air gap.
8 cm
Core thickness=3 cm
f
2 cm
i

N=500
+
Rgap
Rcore
8 cm
0.5 cm
Fig. 3. Iron core with an air gap
Fig. 4. Magnetic circuit
Solution to Example 2
The mean length of the iron core and the cross-sectional area:
lc=4*6-0.5=23.5cm, Ac=(2*10-2)*(3*10-2)=6*10-4 m2
The permeability of the core is: mcore=mrm0=6000*4p*10-7=7.54*10-3
lcore
The reluctance of the core:  c 
m core Acore
Air gap cross-sectional area:
Agap=2*3 cm2=6*10-4 m2
Solution to Example 2 (cont’d)
The permeability of air is approximately equal to that of free space:
mgap=m0=4p*10-7
The reluctance of the air gap = lgap/(mgap*Agap)= 6.63*106
The total reluctance = 5.195*104 + 6.63*106 = 6.68*106
Even though the gap is much shorter than the iron core, the reluctance of
the gap is higher than that of core because of the much higher permeability
of the iron. Most of the mmf is dropped across the air gap.
The flux is given by: f=Bgap Agap=0.25*8.75*10-4 =2.188*10-4 Wb
The flux in the core is the same as that in the air gap. However, the flux
density is higher in the core, because the area is smaller.
The mmf is given by: =f =2.188*10-4*6.68*106 = 1,461
And the required current: i = /N = 1461/500 = 2.922 A
Example 3
Consider the magnetic core with an air gap as shown in Fig. 5.
The core material has a relative permeability of 6000 and a
rectangular cross-section of 2cm by 3cm. The coil has 500 turns.
Determine the current required to establish a flux density of
Bgap=0.25T in the air gap considering fringing at the air gap.
8 cm
Core thickness=3 cm
f
2 cm
i

N=500
+
Rgap
Rcore
8 cm
0.5 cm
Fig. 5. Iron core with an air gap
Fig. 6. Magnetic circuit
Solution to Example 3
The mean length of the iron core and the cross-sectional area:
lc=4*6-0.5=23.5cm, Ac=(2*10-2)*(3*10-2)=6*10-4 m2
The permeability of the core is: mcore=mrm0=6000*4p*10-7=7.54*10-3
lcore
The reluctance of the core:  c 
m core Acore
The flux lines tend to bow out in the air gap as shown in Fig. 5. This is
called fringing. Thus, the effective area of the air gap is larger than that of
the iron core. Customarily, we take this into account by adding the length
of the air gap to each of the dimensions of the air gap cross-section giving:
Agap=2.5*3.5 cm2=8.75*10-4 m2
Solution to Example 3 (cont’d)
The permeability of air is approximately equal to that of free space:
mgap=m0=4p*10-7
The reluctance of the air gap = lgap/(mgap*Agap)= 4.547*106
The total reluctance = 5.195*104 + 4.547*106 = 4.552*106
Even though the gap is much shorter than the iron core, the reluctance of
the gap is higher than that of core because of the much higher permeability
of the iron. Most of the mmf is dropped across the air gap.
The flux is given by: f=Bgap Agap=0.25*8.75*10-4 =2.188*10-4 Wb
The flux in the core is the same as that in the air gap. However, the flux
density is higher in the core, because the area is smaller.
The mmf is given by: =f =2.188*10-4*4.552*106 = 996
And the required current: i = /N = 996/500 = 1.992 A
Example 4:
The following figure shows a electromagnet with two parallel air
gaps. Neglecting leakage, iron reluctance, and fringing at the air
gaps, calculate the flux and flux density in each leg of the magnetic
circuit. All dimensions are in inch.
I=0.251 A
4
0.02
0.04
N=1,000
1
1
2
1
6
1
The calculated flux by the magnetic circuit method is within 5%
of the actual value. These inaccuracies are due to:
It is assumed that the flux is contained in the core. A small amount of the
flux escapes the core into the surrounding low permeability air - this is
referred to as leakage flux.
The calculations we have used for reluctance rely on a measurement of
the mean path. We have assumed that the mean path lies in the center of
the core. This assumption is not necessarily true, and in particular is a
poor assumption at the corner;
Where there are air gaps in the core fringing takes place. We have
assumed that the effective area caused by fringing increases is related to
the length of the gap, this is obviously an approximation;
In ferromagnetic materials, the permeability varies according to the
amount of flux already in the core. This non-linear effect (saturation) adds
inaccuracies to the results as the reluctance calculations depend upon the
permeability.
Magnetic Behaviour of Ferromagnetic Materials
B-H Characteristic
The B-H characteristic demonstrates the magnetic behaviour of a
ferromagnetic material. Permeability m is defined by the following
equation:
B=mH
m is constant in free space but not true for iron or other ferromagnetic
materials as magnetomotive force (mmf) is applied.
i
Ni
N
f=BA
Fig. 7. Coil wounded on a ferromagnetic core.
B-H Characteristic (cont’d)
Start with 0 DC current and slowly raise current to a possible maximum
value. When flux produced in the core is plotted versus the mmf
producing it, the resulting plot looks the one shown in Fig. 8.
f (Wb)
Transition: knee of the curve
 (AT)
Unsaturated
region
Saturated
region
Fig. 8. Saturation or magnetization curve
B-H Characteristic (cont’d)
A B-H Characteristic is a representation (plot) of the magnetic flux
density as a function of the applied magnetic field. A magnetization
curve is typically nonlinear. Fig. 9 shows a plot of B versus H.
H =  /l = Ni/l
F = BA or B = f/A
H is directly proportional to 
f is directly proportional to B
Therefore B-H characteristic has the same shape as the relationship
between f - .
B (Tesla)
H (AT/m)
Fig. 9. B-H Characteristic
Energy Losses in a Ferromagnetic Core
• Hysteresis loss
• Eddy current loss
Hysteresis Loss
• Apply a magnetic field intensity, H, to a sample iron.
• Magnetic fields of the atoms in small domains are aligned with the
applied field.
• If the applied field is reduced to zero, the domains return to their
original orientations.
• If H is increased in the reverse direction, the domains align in the
reverse orientations.
H
+
e
i
Fig. 10a. Iron sample and coil for applying H
Fig. 10b. Magnetic domains
Hysteresis Loss (cont’d)
• For an ac system, if the applied field is varied through a cycle,
during some interval of time energy flows from the source to
the coil-core assembly and during some other interval of time
energy returns to the source. However, the energy flowing in
is greater than the energy returned. Therefore, during a cycle
of variation of the applied field, there is a net energy flow
from the source to the coil-core assembly. This energy loss
goes to heat the core. The loss of power in the core due to the
hysteresis effect is called hysteresis loss.
Eddy Current Loss
• This current leads to core loss for ac operation. The solid iron core
itself is an electrical conductor, acting much like shorted turns. As
the magnetic field changes, voltages are induced in the core causing
current, known as eddy currents to circulate in the core material. As
a result, power is dissipated in the core according to P = v2/R.
• The eddy current loss can be reduced in two ways:
– A high-resistivity core material can be used.
– A laminated core can be used. The core can be laminated with
thin sheets of iron that are electrical insulated from one another.
The orientation of the sheets is selected to interrupt the flow of
current. Thus, the resistance is higher for eddy currents, and the
loss is greatly reduced. Another approach is to make the core
with powdered iron held together by an insulated binder.
Questions
1. Explain the terms magnetomotive force, reluctance and flux and hence
derive the expression that relates these quantities.
2. What is a ferromagnetic material? Why is the permeability of
ferromagnetic materials so high?
3. What are the relationships between the followings:
- flux density and magnetic field strength
- flux and flux density
- mmf and flux
4. What are hysteresis and eddy current losses? What can be done to
minimize them?
5. An air gap has a length of 0.1cm. What length of iron has the same
reluctance as the air gap? The relative permeability of the iron is 5000.
The cross sectional areas of the air gap and the core are the same. [0.5m]