EEB326 Electrical Machines I
Lectures Weeks 1-3
a) Unlike poles attract b) Like poles repel
Terms and symbolsMagnetic flux (crossing a surface of area A m2)
[Weber , abbreviated Wb]
Fm = NI (Amperes ) where N is number of turns and I is
Magnetic field strength or magnetising force
length. is current, is turns,ofnumber is Where lINl
Example 1: An
square cross section
produces a flux
density of 0.45 T
when the flux is 720
Wb. Determine the dimensions of the cross section.
l= mean length
A coil of 750 turns is uniformly wound on a circular magnetic
circuit of mean diameter 30 cm. Find the current in the coil
when the magnetising force is 8000 A/m.
Magnetic Flux Density
B=H [Tesla (T)] where is permeability of the material
=r0 where r is relative permeability of the material, and
0 = 4 x 10-7 H/m is permeability of free space.
Notes: 1T = 1 Wb/m2.
Example 3: A magnetic pole face has a rectangular section of 300 mm by 150 mm. Calculate the flux density when the total flux from the pole is 300 Wb.
B = (300*10-6 Wb)/(300*10-3*150*10-3 m2)=0.0066667 T
in vacuumdensity flux
materialin density flux r
An iron ring of mean diameter 10 cm is uniformly wound with 2000
turns of wire. When a current of 0.25A is passed through the coil a flux
density of 0.4 T is set up in the iron. Find (a) the magnetising force and
(b) the relative permeability of the iron under these conditions.
a.l = d = 10 cm = 10 10-2 m,N = 2000 turns, I = 0.25A and B = 0.4 T
b. B = 0rH, hence
F =NI= Magneto Motive Force or
MMF = # of turns * Current passing through it
F = NI = Hl (why!)
= Reluctance of magnetic path
Find (a) The reluctance of a piece of mumetal of length
150 mm and cross-sectional area 1800 mm2 when the
relative permeability is 4000. (b) The absolute permeability
of the mumetal.
= 0r = (4 10-7)(4000)
= 5.027 10-3 H/m
A mild steel ring has a radius of 50 mm and a cross-sectional
area of 400 mm2. A current of 0.5A flows in a coil wound
uniformly around the ring and the flux produced is 0.1 mWb. If
the relative permeability at this value of current is 200 find (a)
the reluctance of the mild steel and (b) the number of turns on
l = 2r = 2 50 10-3 m, A = 400 10-6 m2,I = 0.5A, = 0.1 10-3 Wb and r = 200
from whichm.m.f. = S i.e. NI = S Hence, number of turns
Composite series magnetic circuits
For a series magnetic circuit having n parts the total
reluctance is the sum of the individual reluctances.
This is similar to the series connection of resistances
A closed magnetic circuit of cast steel contains a 6 cm
long path of cross-sectional area 1 cm2 and a 2 cm path
of cross-sectional area 0.5 cm2. A coil of 200 turns is
wound around the 6 cm length of the circuit and a
current of 0.4A flows. Determine the flux density in the 2
cm path, if the relative permeability of the cast steel is
Total circuit reluctance S = S1 + S2= (6.366 + 4.244) 105
= 10.61 105/H
Flux density in the 2 cm path,
Electric and magnetic circuit similarities
Electric and magnetic quantities correspondences
Electric quantity Magnetic quantity
Current I Magnetic flux
Current density J Magnetic flux density B
Electromotive force =
resistance x I
Magnetomotive force = reluctance x
Electric field intensity E Magnetic field intensity H
Conductance = 1/resistance Permeance = 1/reluctance
Resistance =l/A Reluctance =l/A
Magnetic circuit laws
Consider a magnetic circuit with an air gap. Note that the fringing at
the air gap increases the effective cross sectional area.
lyrespect ive sources
and sreluctance of numbers are and Where
law voltageKichholfs toanalogous is Which
T herefore 0 magnet icsin law sGauss' From
energy theincrease torequired force T he
gap.air theof volumein the increase the torelated isenergy addit ional T he
energy addit ional T he
done. work the todue is
of gapair of volumeaddit ional in the storedenergy magnet ic addit ional T he
.N force aby of distance aaway surfaces theof one movingConsider
gapair in storeddensity energy Magnet ic
Tutorial: Lifting electromagnet design.
The yoke of a lifting electromagnet is made up of 20
laminations each 1.5 mm thick. Each lamination is cut
from a 80 mm x 35 mm rectangular plate. A portion 40
mm x 15 mm is cut out of the plate so that each pole face
is 20 mm wide. The coil of N turns is wound on the yoke.
The supply to the coil will be 0.5 Amps DC. The
electromagnet is to lift a mild steel block of mass 0.5 kg
and dimensions of 100 mm length, 20 mm width and 30
mm depth. The air gap can have a maximum value of 1
Determine the number of turns when the air gap is 0.1
mm. Use the BH curves provided.
Lifting electromagnet design
.element of area sect ional cross is Where
element each in density flux Calculate 5
circuitin flux magnet ic Calculate 4
area gapair is Where2
densityflux desired Calculate 3
. force lift ing desired thehence and lifted be tomass Determine 2
net .electromag lift ing of dimensions Determine 1
proceduredesign net electromag Lift ing
turnsofnumber determinecurrent available From 9
force ivemagnetomot Calculate 8
elementcircuit each for reluctance Calculate 7
curves BH smaterial' from and Find 6
Consider a magnetic circuit with a T joint
analysiscircuit magnet icin applicable are
analysis nodal andmesh assuch s, techniqueanalysiscircuit DC
junct ion aat branches ofnumber theis Where
joint T the tolaw Gauss Applying
magnets are in
magnet theacross drop" voltage" magnet ic T he
.nearly ismagnet permanent theofty permeabili the,
formula thehas curveat ion demagnet iz T he
st ic.characteri linear a and length
, area sect ional cross ofmagnet permanent aConsider
Model of a Permanent Magnet of cross sectional area Am, length lm and a linear B-H characteristic . Note that Hm is in opposite direction to Bm.
.ofdirection opposite in the being negative is Where
isty permeabili magnetic thecurve,ation demagnetiznonlinear aFor
Where r typically is 1000-100000in magnetic cores
= r 0
0ab: magnetising curve, from
by: Saturation flux density
0c: remanent flux density or
remanence, H reduced to zero.
0d: Coercive force required to reduce flux to zero.
bcde: curve formed by reducing H.
ex: Saturation flux density.
efgb: Curve formed by increasing H.
Hysteresis: Changes in B lag behind
changes in H.
bcdefgb: Hysteresis loop
Hysteresis lossEnergy is expended in ferromagnetic
materials to change the alignment of
domains during cycles of magnetisation.
The energy appears as heat and is
called hysteresis loss.
Hysteresis loss is proportional to the
area of the hysteresis loop.
The area of the hysteresis loop varies
with the hardness/softness of materials.
Figure a: hard material, high remanence,
Figure b: soft material large remanence,
Figure c: Ferrite small hysteresis
a) Unmagnetised domains
b) Magnetised domains
Material Relative Permeability
Mild steel 200 800
Silicon iron 1000 - 5000
Cast steel 300 - 900
Numetal 200 000
Stalloy 500 - 6000
Magnetic field due to an electric current
Direction of magnetic field inside coil given by either
screw rule or grip rule
Electric bell -single stroke.
When the push button is pressed a current flows and the armature is attracted to the electromagnet. The striker on the armature hits the gong. When the circuit is broken the electromagnet demagnetises and the armature is pulled back by the spring.
striking a bell.
The contacts are
Coil C is wound around
cast iron core P. A
covering R is over the
face of the
magnetic paths M pass
through the magnetic
material to be lifted.
signals to sound
according to the
transmitted to the
Force on a current carrying conductor
Where B is magnetic flux density, I is current, l is length
of conductor and is angle between conductor and magnetic field.
When an electric current flows in the coil it produces a force that moves the cone backwards and forwards according to the direction of the current. The cone transfers the force to the air, producing sound waves.
A current carrying
conductor in a
experiences a force
due to the
interaction of the
fluxes; the one for
the magnetic field
and that due to the
Flemings left hand rule
Let the thumb, first finger
and second finger of the
left hand be extended
such that they are all at
right-angles to each other,
If the first finger points in
the direction of the
magnetic field, the second
finger points in the
direction of the current,
then the thumb will point
in the direction of the
motion of the conductor
Simple dc motor
Magnetic field supplied by permanent magnet.
Current supplied to coil through brushes and commutator bar.
Commutator allows for change of direction of current.
Coil rotates due to the forces acting on it.
Brushes provide contact to a moving commutator.
A practical motor has more than one winding
A coil is wound on a rectangular former of width 24 mm and
length 30 mm. The former is pivoted about an axis passing
through the middle of the two shorter sides and is placed in a
uniform magnetic field of flux density 0.8 T, the axis being
perpendicular to the field. If the coil carries a current of 50
Determine the force on each coil side
(a) for a single-turn coil,
(b) for a coil wound with 300 turns.
a Flux density B = 0.8 T, length of conductor lying at right-angles
to field l = 30 mm = 30 10-3 m and current I = 50 mA = 50
For a single-turn coil, force on each coil side
F = BIl = 0.8 50 10-3 30 10-3
= 1.2 10-3 N or 0.0012 N
b When there are 300 turns on the coil there are effectively 300
parallel conductors each carrying a current of 50 mA. Thus the
total force produced by the current is 300 times that for a
single-turn coil. Hence force on coil side, F = 300 BIl = 300
0.0012 = 0.36 N
Elementary dc motor summary
Back emf: a voltage is induced in a motor coil as it revolves
Note that a dc motor can also operate as a dc generator
DC motor equivalent circuit: Note the two flux systems
E0 = K where =2n/60, n is speed in rpm (revolutions per minute)
The back emf Eb = E0 = Km (1) Where m is mechanical speed of rotation =2n/60, n=speed in rpm (revolutions per minute), K is a factor that depends on coil geometry.
The armature current Ia =(VT - E0)/Ra (2) Note that the starting current is very high since E0 is zero at start.
The torque developed T = KIa (3) Substituting for Iafrom (2 ) T = K(VT-E0)/Ra = (K/Ra)(VT - Km)
The power developed by the motor Pdev=E0Ia = KmIa(=Mechanical power =mT= KmIa)
Torque speed characteristic
Operating point depends on load characteristic
Speed regulation: One of the measures of performance
load fullat speed is
load noat speed is Where
motor. the toapplied is load full as speed
in change theas defined is regulat ion Speed
full loadno load
- NNlationSpeed regu
DC machine classifications Depends on arrangement
of supplies to the field and armature windings.
a. Separately excited machine when the field winding has a
b. Self excited machine when field winding is connected to
the main motor supply. Three types of connections: -
i. Shunt connected machine when the field is across the main supply
ii. Series connected machine when the field and armature windings
are in series.
iii. Compound connected machine when both shunt and series
connections are used. The connections can be cumulative
compounding (fluxes additive) or differential compounding (fluxes
Shunt excited motor, has
same torque speed
characteristic as a
separately excited motor
Series excited motor
Derivation of torque speed characteristic for series motor
(3)in for ngSubst itut i
circuit armature thefrom found is
and (1) ...
A 230 V 10 hp DC shunt motor delivers power to a load at
1200 rpm. The armature current drawn by the motor is
200 A. The armature circuit resistance of the motor is 0.2
and the field resistance is 115 .
a) If the rotational losses are 500 W, what is the value of the load
b) What is the starting current of the motor
Solution part (a)
VT = 230 V
Ra = 0.2 Ia = 200 A
Prot = 500 W
N = 1200 rpm
Back emf induced in the armature
Eb = VT - IaRa = 190 V
Power developed in the rotor
Pdev = EbIa = 38000 W
Power delivered to the load
Pload =Pdev - Prot = Pout = 37500 W
Load torque = Pload/m m = 2N/60 = 125.6637 rads/sTload = 298.4155 N m
Solution part (b)
Starting current Ia start = VT / Ra = 230/0.2 A = 1150 A
Faradays law of electromagnetic induction
If the flux linking a loop (or turn) varies as a function of
time, a voltage is induced between its terminals.
The value of the induced voltage is proportional to the rate
of change of flux.
s changesflux which theduring interval t imeist
Wb coil theinsideflux of change is
coil, of turnsofnumber is N
,V voltageinduced is E
When the magnetic field in
a coil changes a voltage is
induced in the coil. This
voltage will drive a current
in a circuit to which the
coil is connected.
When a conductor is moved across a magnetic field or
vice versa so as to cut through the lines of force (or flux),
an electromotive force (e.m.f.) is prod...