Transcript Cct component - Universiti Sains Malaysia

Magnetic Flux 
•Unit for flux is weber
•The definition of 1 weber is the amount of flux
that can produce an induced voltage of 1 V in a one
turn coil if the flux reduce to zero with uniform rate.
Magnetic Flux density B
•Unit for magnetic flux density is Tesla
•The definition of 1 tesla is the flux density that
can produce a force of 1 Newton per meter acting
upon a conductor carrying 1 ampere of current.
Magnetic field strength H
•Unit for magnetic field strength is Ampere/m
•A line of force that produce flux
F = B l I newton
Where F = force ; B = magnetic flux density ; l =the length of
conductor and I = current in the conductor
  B A
Where  = magnetic flux ; B = magnetic flux density and A =
area of cross-section
B  H
Where H = magnetic field strength ; B = magnetic flux density
and  = permeability of the medium
Permeability in free space
o = B/H = 4 x 10-7 H/m
Relative Permeability (r)

Relative permeability is defined as a ratio of flux density
produced in a material to the flux density produced in a
vacuum for the same magnetic filed strength. Thus
r = /o

 = ro = B/H
or
B = roH
H vs r
Mild steel
Cast iron
Relative
permeability
Magnetic field strength
Relative permeability vs magnetic flux
B vs H
Electromagnetic Force (mmf)
H
turns
mmf
H  NI
Where H = magnetic field strength ; l =the path length of ; N
number of turns and I = current in the conductor
Example 1
A coils of 200 turns is uniformly wound around a wooden ring
with a mean circumference of 600 mm and area of cross-section
of 500 mm2. If the current flowing into the coil is 4 A, Calculate
(a) the magnetic field strength , (b) flux density dan (c) total flux
turns
N = 200 turns
l = 600 x 10-3 m
A = 500 x 10-6 m2
I = 4A
(a)
H = NI/l = 200 x 4 / 600 x 10-3 = 1333 A
(b)
B = oH = 4 x 10-7 x 1333 = 0.001675 T = 1675 T
(c)
Total Flux  = BA = 1675 x 10-6 x 500 x 10-6
= 0.8375 Wb
Reluctance ( S )
Ohm‘s law I = V/R [A]
Where I =current; V=voltage and R=resistance
And the resistance can be relate to physical parameters as
R =  l /A ohm
Where =resistivity [ohm-meter], l= length in meter and A=area
of cross-section [meter square]
Analogy to the ohm‘s law
V=NI=H l
I= and R=S
Hl
weber 

S
where S 

 r o A
am pere/ weber
Example 2
A mild steel ring, having a cross-section area of 500 mm2 and a
mean circumference of 400 mm is wound uniformly by a coil
of 200 turns. Calculate(a) reluctance of the ring and (b) a
current required to produce a flux of 800 Wb in the ring.
(a)
 800106
B 
 1.6T
6
A 50010
Dari graf r/B, pada B = 1.6;
r = 380

0.4
S

r o A 380 4 107  5 104
 1.667106[ A / Wb ]
turns
(b)
mmf
H

S
H   S
H  800106 1.667106  1342[ A]  NI
1342 1342
I 

 6.7[ A]
N
200
Magnetic circuit with different materials
l2
SB 
 2a 2
l1
SA 
1a1
and
S  SA  SB
l1
l2


1a1  2a 2
For A:
ForB:
area of cross-section = a1
mean length = l1
absolute permeability = 1
area of cross-section = a2
mean length = l2
absolute permeability = 2
Mmf for many materials in series
total mmf = HAlA + HBlB
HA =magnetic strength in material A
lA=mean length of material A
HB =magnetic strength in material B
lB=mean length of material B
In general
(m.m.f) = Hl
Example 3
A magnetic circuit comprises three parts in series, each of
uniform cross-section area(c.s.a). They are :
(a)A length of 80mm and c.s.a 50 mm2;
(b)A length of 60mm and c.s.a 90mm2;
(c)An airgap of length 0.5mm and c.s.a 150 mm2.
A coil of 4000 turns is wound on part (b), and the flux density
in the airgap is 0.3T. Assuming that all the flux passes through
the given circuit, and that the relative permeability r is 1300,
estimate the coil current to produce such a flux density.
  BC AC  0.31.5 104  0.45104Wb
Mmf =  S = H l = N I
Material a
0.45104  80103
Sa  

 44.1At
7
6
r o Aa 1300 4 10  5010
Material b
0.45104  60103
Sb  

 18.4 At
7
6
r o Ab 1300 4 10  9010
airgap
Total mmf
and
a
b
0.45104  0.5 103
Sc  

 119.3 At
7
6
r o Ac 1300 4 10 15010
c
NI  Sa  Sb  Sc  44.1  18.4  119.3  181.8 At
181 .8
I
 45.4 10 3 A  45.4mA
4000
Leakages and fringing of flux
leakage
fringing
Magnetic circuit with air-gap
Leakages and fringing of flux
Some fluxes are leakage via paths a, b and c . Path d is
shown to be expanded due to fringing. Thus the usable
flux is less than the total flux produced, hence
total flux
Leakage factor 
usable flux
Example 4
A magnetic circuit as in Figure is made
from a laminated steel. The breadth of
the steel core is 40 mm and the depth is
50 mm, 8% of it is an insulator
between the laminatings. The length
and the area of the airgap are 2 mm and
2500 mm2 respectively.
A coil is
wound 800 turns. If the leakage factor
is 1.2, calculate the current required to
magnetize the steel core in order to
produce flux of 0.0025 Wb across the
airgap.

2.5 103
Ba 

 1T
6
Aa 250010
Ba
1
Ha 

 796000[ AT / m]
7
o 4 10
mmf  H  796000 0.002  1594[ AT ]
Total flux  T  flux in airgap leakage factor
 0.0025 1.2  0.003Wb
92% of the depth is laminated steel, thus the area of cross
section is
 AS = 40 x 50 x 0.92 = 1840 mm2=0.00184m
T 3 103
BS 

 1.63T
AS 0.00184
From the B-H graph, at B=1.63T, H=4000AT/m
 mmf in the steel core = Hl = 4000 x 0.6 = 2400 AT
Total mmf. = 1592 + 2400 = 3992 AT
 NI = 3992
I = 3992/800 = 5 A
D
Magnetic circuit applying voltage law
Analogy to electrical circuit
applying voltage Kirchoff’s law
Mmf in loop C = NI = HLlL + HMlM
outside loop
NI= HLlL + HNlN
And in loop D
0 = HMlM + HNlN
In general (m.m.f) = Hl
P
IL
IM
IN
L
M
N
E
Q
At node P we can also applying current Kirchoff’s law
L = M + N
Or
L - M - N = 0
In general:
 = 0
Example 5
A magnetic circuit made of silicon steel is arranged as in the
Figure. The center limb has a cross-section area of 800mm2 and
each of the side limbs has a cross-sectional area of 500mm2.
Calculate the m.m.f required to produce a flux of 1mWb in the
center limb, assuming the magnetic leakage to be negligible.
340
mm
1mm
150
mm
340
mm
  B A

1103
B 
 1.25T
6
A 80010
Looking at graph at B=1.25T r =34000
Apply voltage law in loops A and B
340
mm
m.m. f   AS1   A   B S2  Sa 
340103
S1 

 15915
7
6
r o A1 34000 4 10  50010
1
150103
S2 
 4388
7
6
34000 4 10  80010
1103
Sa 
 994718
7
6
4 10  80010
1mm
A
150
mm
B
340
mm
Since the circuit is symmetry A =B
m.m. f  S1  2S2  Sa 
In the center limb , the flux is 1mWb which is equal to 2 
Therefore =0.5mWb


m.m. f  0.5 103 15915  1103 4388 994718
 8  999  1007
A
Example 6
A U-shaped electromagnet shown in
Fig. is designed to lift a mass. The
material for the yoke has a relative
permeability of 2900. The yoke has a
uniform cross-sectional area of 4000
mm2 and a mean length of 600 mm.
Each of the air gaps is 0.1 mm long.
The number of turns of the coil (N) is Air gap
240. Assuming that the reluctance of
the keeper is negligible, calculate the
maximum mass in kg, which can be
lifted by the system if a current of 1.5 A
is passed through the coil. You may
neglect the fringing effect and flux
leakage; and assume that
I
N
Iron yoke
Keeper
Weight
Calculation of maximum weight lifted by and electromagnet.
Let the flux density in the air gap be Ba
For the air gap
2H a la  2
Ba l a
o
Ba  0.1 103
2
4  107
103

Ba
2
For the iron yoke
Ba  600103
6 103
H i li 


Ba
7
r o 2900 4 10
11.6
Ba li
Total mmf;
6  103
1
Hl  H a la  H i li   
Ba  323.8Ba  NI

 2 11.6  
NI
360
Ba 

 1.11 T
323 .8 323 .8
Since there are two air gaps;
 Ba 2 A  Ba 2 A 1.112  4000106

F  2

 3922N
7

o
4 10
 2 o 
mg  3922
3922 3922
m

 400 kg
g
9.81
Hysteresis loss
Materials before applying m.m.f (H), the polarity of
the molecules or structures are in random.
After applying m.m.f (H) , the polarity of the
molecules or structures are in one direction, thus the
materials become magnetized. The more H applied
the more magnetic flux (B )will be produced
When we plot the mmf (H) versus the magnetic flux will produce a
curve so called Hysteresis loop
1. OAC – when more H applied, B
increased until saturated. At this
point no increment of B when we
increase the H.
2. CD- when we reduce the H the B
also reduce but will not go to zero.
3. DE- a negative value of H has to
applied in order to reduce B to zero.
4. EF – when applying more H in the
negative direction will increase B in
the reverse direction.
5. FGC- when reduce H will reduce B
but it will not go to zero. Then by
increasing positively the also
decrease and certain point it again
change the polarity to negative until
it reach C.
Eddy current
When a sinusoidal current enter
the coil, the flux  also varies
sinusoidally according to I. The
induced current will flow in the
magnetic core. This current is
called eddy current. This
current introduce the eddy
current loss. The losses due to
hysteresis and eddy-core totally
called core loss. To reduce eddy
current we use laminated core
metal
insulator