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Transcript
```BASIC ELECTRICAL TECHNOLOGY
DET 211/3
Chapter 4 – Magnetic
Circuits
1
Magnetic Materials and Circuits
Introduction
Magnet contains a north pole and south pole.
Magnet flux leaves the magnet as the north pole
and the place where the flux returns to the magnet
as the south pole.
Two types of magnet:
• Permanent magnet
• Electromagnet
2
Right Hand Rule and Ampere’s Law
When a conductor carries current a magnetic
field is produced around it.
Fingers– indicate current direction
Thumb – indicate the direction of magnetic
flux is wrapping around the wire
3
Right Hand Rule and Ampere’s Law
The relationship between current and magnetic field
intensity can be obtained by using Ampere’s Law.
Ampere’s Law states that the line integral of the
magnetic field intensity, H around a closed path is equal
to the total current linked by the contour.
 H .dl   i
H: the magnetic field intensity at a point on the contour
dl: the incremental length at that point
If θ: the angle between vectors H and dl then
 Hdl cos    i
4
Right Hand Rule and Ampere’s Law
Consider a rectangular core with N winding
 i  Ni
dl  lc
 Hl c  Ni
Therefore
Ni
H
lc
5
Relationship between B-H
The magnetic field intensity, H produces a magnetic
flux density, B everywhere it exists.
B  H ( weber / m2 ) or Tesla
B   r  0 H ( wb / m ) or T
2

- Permeability of the medium
0
- Permeability of free space, 4 x 10-7 wb

r 
0
A.t .m
- Relative permeability of the medium
unity
r
is
6
MAGNETIC EQUIVALENT CIRCUIT
A simple magnetic circuit having a ring shaped
magnetic core (toroid) and a coil that extends
around the entire circumference
When current i flows through the coil of N
turns, a magnetic flux is produced.
7
MAGNETIC EQUIVALENT CIRCUIT
Assumption:
•All fluxes are confined to the core
•The fluxes are uniformly distributed in the core
The flux outside the toroid (called leakage
flux), is so small (can be neglected)
Use Ampere’s Law,
 H .dl  Ni
Hl  Ni
H .2r  Ni
Hl  Ni  F
F = Magnetomotive force (mmf)
8
MAGNETIC EQUIVALENT CIRCUIT
B  H
Ni
H
( At / m )
l
Ni
B
(T )
l
Where;
N – no of turns of coil
i – current in the coil
H – magnetic field intensity
l – mean length of the core
9
Magnetic Flux Density (B) and Magnetizing
Curve (B-H Curve)
Case 1: Non magnetic material core (Cu, Al,
air, plastic, wood,..)
B  0 H
 0 Ni
B
(T )
l
10
Magnetic Flux Density (B) and Magnetizing
Curve (B-H Curve)
Case 2: Ferromagnetic material core (iron,
steel, ferrite,..)
B  H  0 r H
 0 r Ni
B
(T )
l
The magnetic flux density, B increases almost
linear in the region of low values of magnetic
intensity, H.
At higher value of H, the change of B is nonlinear.
11
MAGNETIC EQUIVALENT CIRCUIT
The flux in the coil,
  BA( weber )
Where
Ni
Ni
Ni F

A


l
l
R R
A
Ф – flux in the coil (wb)
F – magnetomotive force (mmf)
R – 1/μA = 1/P ,Reluctance
P = permeance
A – cross sectional area
12
ANALOGY BETWEEN MAGNETIC CIRCUIT
AND ELECTRIC CIRCUIT
a) Magnetic equivalent circuit
b) Electric equivalent circuit
To solve magnetic equivalent circuit – Kirchhoff Voltage and
Current Laws (KVL & KCL)
Electric circuit
Magnetic circuit
Driving force
EMF (E)
MMF (F)
Produces
Current (i)
Flux (Ф)
Limited by
Resistance (R)
Reluctance (R)
13
MAGNETIC CIRCUIT WITH AIR GAP
In electric machines, the rotor is physically isolated
from the stator by the air gap.
Practically the same flux is present in the poles
(made by magnetic core) and the air gap.
To maintain the same flux density, the air gap will
require much more mmf than the core.
14
MAGNETIC CIRCUIT WITH AIR GAP
lc
Rc 
 c Ac
Rg 
lg
 g Ag
Where
Ni

Rc  Rg
Ni  H clc  H g I g
c
Bc 
Ac
g
Bg 
Ag
lc – mean length of the core
lg – the length of the air gap
15
FRINGING EFFECT
Fringing Effect: Bulging of the flux lines in the air gap.
Effect: The effective cross section area of air gap
increase so the reluctance of the air gap decrease. The
flux density Bg < Bc, Bc is the flux density in the core.
If the air gaps is small, the fringing effect can be
neglected. So
Ag  Ac

Bg  Bc 
Ac
In practical, large air gap will be divided into several
small air gaps to reduce the fringing effect.
16
INDUCTANCE
A coil wound on a magnetic core as shown in figure above,
is frequently used in electric circuits. This coil may be
represented by an ideal circuit element, called inductance,
which is defined as the flux linkage of the coil per ampere
of its current.
Inductance L 

i
N NBA NHA NHA N 2 N 2
L





Hl
l
i
i
i
R
N
A
17
Example
15cm
30cm
10cm

15cm
i
N=200turns
l1
l2
15cm
30cm
30cm
15cm
10cm
A ferromagnetic core is shown in Figure above. Three sides of this core are of
uniform width, while the fourth side is somewhat thinner. The depth of the core
(into the page) is 10cm and the other dimensions are shown in figure. There is 200
turn coil wrapped around the left side of the core. Assuming relative permeability
r of 2500, how much flux will be produce by a 1A input current?
18
Solution Example
The mean path length of region 1 is 45cm and the cross-sectional area is 10 x 10
cm = 100cm2. Therefore, the reluctance in the first region is:
l1
l1
R1 

A1  r o A1
0.45m

(2500)( 4x10 7 )(0.01m 2 )
 14,300 A.turns / Wb
The mean path length of region 2 is 130cm and the cross-sectional area is 15 x 10
cm = 150cm2. Therefore, the reluctance in the second region is:
R2 
l2
l2

A2  r o A2
1.3m
(2500)( 4x107 )(0.015m 2 )
 27,600 A.turns / Wb

19
Solution Example
Therefore, the total reluctance in the core is:
Req  R1  R2
 14,300 A.turns / Wb  27,600 A.turns / Wb
 41,900 A.turns / Wb
The total magnetomotive force (MMF) is:
F  NI  (200turns)(1.0 A)  200 A.turns
The total flux in the core is given by:
F
200 A.turns
 
R 41,900 A.turns / Wb
 0.0048Wb
20
Assignment 3
15cm
30cm
15cm

15cm
i
30cm
N=200turns
lc
15cm
30cm
15cm
15cm
A ferromagnetic core is shown in Figure above. All side of this core are uniform
width. The depth of the core is 10cm. Assuming relative permeability r of 2500,
how much flux will be produce by a 1A input current?
21
```
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