Electrical Power and Machines
Lecture 2
<Dr Ahmed El-Shenawy>
<Dr Hadi El Helw>
Magnetism
Magnetic fields are an essential element in the conversion of
mechanical
energy to electrical energy and vice versa
Sources of magnetic fields:
Permanent magnets
Current carrying conductors
(a) Permanent Magnets
In the region surrounding a permanent magnet there exists a magnetic
field, which can be represented by magnetic flux lines. Magnetic lines
exist in continuous loops, as shown in fig. The symbol for magnetic flux is
the Greek letter Φ (phi).
If a nonmagnetic material, such as glass or copper, is placed in the flux
paths surrounding a permanent magnet, there will be an almost
unnoticeable change in the flux distribution. However if a magnetic
material, such as soft iron, is placed in the flux path, the flux lines will pass
through the soft iron rather than the surrounding air because flux line
pass with greater ease through magnetic materials than through air. as
shown in fig.
Unlike magnet poles
Attract
Like magnet poles
Repulse
Magnetic lines of force that move from N to S
They form closed loops
(b) Current Carrying Conductors
Direction can be determined
using the right hand grasp rule
+
If the conductor is wound in a single-turn coil, the resulting flux will
flow in a common direction through the center of the coil. A coil of
more than one turn would produce a magnetic field that would exist in
a continuous path through and around the coil which is quite similar to
that of the permanent magnet.
The strength of the magnetic filed is determined by the density of the
flux line. The filed strength of the coil can be effectively increased by
placing certain materials, such as iron or steel, within the coil to
increase the flux density within the coil or by increase the current in
the conductor.
There are many application of the electro magnetic effect such as generator,
transformer, Relay……..
Magnetic Flux and Magnetic Flux Density
(a) Magnetic Flux (φ)
Total number of magnetic lines of force present in a magnetic field
Unit….. Weber (wb)
(b) Magnetic Flux Density (B)
It is a measure of the concentration of the magnetic flux per unit area
Unit…… Tesla
B = φ/A (wb/m2) or Tesla
Example
(1)
The total magnetic flux at the pole face of a bar magnet is 3×10-4 wb, the bar
Magnet is rectangular and has a cross sectional area of 2 cm2. what is the
Flux density within the magnet?
Solution
B = φ/A = 3×10-4 / 2×10-4
= 1.5 Tesla (wb/m2)
(c) Magnetomotive Force (Fm)
In a magnetic circuit, a magnetic flux is created when a magnetomotive Force
(m.m.f ) acts on the circuit
Fm = N×I …… (AT)
N: number of turns
I: Current flowing through coil (A)
(d) Reluctance (Rm)
In a magnetic circuit Rm, is the opposition offered by the magnetic circuit to
the establishment of the magnetic flux by the m.m.f
φ
Rm = Fm/φ …… (AT/wb)
Fm
Rm
(e) Permanence (Pm)
Permanence is defined as the ability of the magnetic circuit to permit the
Establishment of a magnetic field
Pm = 1/Rm …… (wb/AT)
(it is more convenient to use Pm in making calculations in parallel magnetic circuits)
(f) Permeability (μ)
Permeability is a measure of the ease with which a magnetic field may be
Established
It could be given as the permanence per unit length and cross sectional
area of a magnetic field
μ = Pm l /A = l /A Rm (wb / AT.m)
(l: length of magnetic circuit, A: cross sectional area of magnetic circuit, Rm: Reluctance)
Note the following
In electric circuits conductor materials are compared according to its
Resistivity (δ), R = δ l/A Ω
 In magnetic circuits conductor materials are compared by means of its
permeability (μ), μ = Pml/A
wb/AT.m
Permeability of free space (μ0)= 4π×10-7 wb/AT.m
 For magnetic materials, permeability is given as follows:
μ = μ0 × μr
Where μr = 1 for air and non magnetic materials
= (500-2500) for iron and steel
μr is dimensionless
(g) Magnetic field intensity (H)
It is the magnetomotive force per unit length
H = Fm/l (AT/m)
l: length of magnetic circuit
Note
As Rm = Fm /φ, then Pm = φ/Fm and since μ=Pml/A
Then μ = (φ×l) / (Fm×A) = B/H
Example
(2)
What is the magnetic field intensity and the m.m.f. needed to produce a
flux of 2×10-4 wb in a steel ring whose mean circumferential length is 100
cm and Has a cross-sectional area of 5 cm2, assume μr = 500.
Solution
Given: φ=2×10-4 wb, l =1 m, A = 5×10-4 m2 and μr =500
Required: H and Fm
Fm = NI = φRm = φl/μA = φl/μ0 μr A = 2×10-4 ×1 / 500 ×4π ×10-7 ×5 ×10-4
= 636.62 AT
H = Fm / l= 636.62/1 = 636.62 AT/m
Magnetization Curves
Magnetic flux (φ) is produced by passing an electric current through a coil
that creates a m.m.f
To have an effective comparison between materials , they are compared in
terms Of unit quantities (Flux density “B” and Magnetizing force “H”)
B-H Curves for different materials
1.Sheet steel
2.Silicon steel
3.Cast steel
4.Tungsten steel
5.Magnet steel
6.Cast iron
7.Nickel
8.Cobalt
9.Magnetite
 A magnetic material is said to be fully saturated when its permeability
becomes almost the same as that of free space (H is so high)
 If “H” is lowered by decreasing the current in the coil, B will not
decrease as rapid as it increased (the B-H curve will not retrace itself),
this irreversibility is called Hysteresis (B lags H)
 when H is reduced to zero, residual flux density Br will appear
meaning that the magnetic material has been magnetized
Analysis of magnetic circuits
Magnetic Circuits versus Electric Circuits
I
V
R1  R 2
l
c  c
Ac

g 
mmf
 c  g
lg
Ag
Magnetic Circuits versus Electric Circuits
Electric circuit
Magnetic circuit
Driving force (cause)
emf
mmf
Response (effect)
current
flux
opposition
resistance
reluctance
Equivalent circuit
I
emf
R
Example 1: A ferromagnetic core is shown in the figure below. The depth of
the core is 5 cm. The other dimensions of the core are as shown in the figure.
Find the value of the current that will produce a flux of 0.005 Wb. With this
current, what is the flux density at the top of the core? what is the flux
density at the right side of the core? Assume that the relative permeability of
the core is 1000.
Solution
There are three regions in this core. The top and bottom form one region, the
left side forms a second region, and the right side forms the third region. If
we assume that the mean path length of the flux is in the centre of each leg
of the core, then the path lengths are l1=2(27.5 cm)=55cm,l2=30cm, and
l3=30cm.
1 
l
l
0.55
3



58
.
36

10
AT / Wb
7
A  r  o A 1000  4  10  (0.05  0.15)
2 
l
l
0.3
3



47
.
75

10
AT / Wb
7
A  r  o A 1000  4  10  (0.05  0.1)
3 
l
l
0.3
3



95
.
49

10
AT / Wb
7
A  r  o A 1000  4  10  (0.05  0.05)
The total reluctance is thus:
total  1   2  3  58.36  10 3  47.75  10 3  95.49  10 3  201.6  103 AT / Wb
and the magneto-motive force required to produce a flux of 0.005 Wb is
mmf  NI    0.005  201.6  10 3  1008 AT
and the required current is
I
mmf 1008

 2.016 A
N
500
The flux density on the top of the core is

0.005
B 
 0.67T
A 0.15  0.05
The flux density on the right side of the core is
B

0.005

 2T
A 0.05  0.05
Air Gaps and their effects
In many applications, magnetic flux must cross one or more air gaps
As the magnetic lines of force cross the air gap, they spread out because the
Individual lines repel each other. This spreading out is called Fringing
Example 2: Determine the magnetic flux through the air gap in the geometry
shown below. The structure is assumed to have a square cross section of area
10-6 m2, a core with μr = 1000, and dimensions l1=1 cm, l3=3cm, and l4=2cm.
Solution
1 
2 
3 
g 
2l1  l 4
 31.83  10 6 AT / Wb
1000 o A
l 4  l g  0.001
1000 o A
2  15.44  10 6 AT / Wb
2l3  l 4
 63.66  10 6 AT / Wb
1000 o A
lg
o A
 79.58  10 6 AT / Wb
The magnetic flux through the source coil is the mmf divided by the total
reluctance seen by the source:
1 
mmf
NI

 0.286  10 6 Wb
 total 1  ( 2   g ) //  3
Using “flux division” (analogous to current division), then:
2 
3
1  0.115  10 6 Wb
 2  3   g