Chapter 2 Electromagnetism Dr. Mohd Junaidi Abdul Aziz

Chapter 2 Electromagnetism

Dr. Mohd Junaidi Abdul Aziz

Aim: impart an understanding of 　electromagnetic principles

Important as electromagnetism underpins the operation of many electrical machines

Linkage between electrical and mechanical worlds


Electromagnetism

Describes the relationship between electricity and magnetism Is essentially the foundation for all of electrical engineering Use electromagnets to generate electricity, store memory on our computers, generate pictures on a television screen, diagnose illnesses,

Electromagnetism


Electromagnetism works on the principle that an electric current through a wire generates a magnetic field In a bar magnet, the magnetic field runs from the

north to the south pole. In a wire, the magnetic field forms around the wire. If we wrap that wire around a metal object, we can

often magnetize that object. In this way, we can create an electromagnet.

Electromagnetism


Magnetism is a force field that acts on magnetic materials but not on other materials.

Magnetic field around a bar magnet Two “poles” dictated by direction of

the field Opposite poles attract (aligned

magnetic field) Same poles repel (opposing

magnetic field)

Electromagnetism


Electromagnetism


Field Detector Can use a compass to

map out magnetic field Field forms closed “flux

lines” around the magnet

Magnetic flux measured in Webers (Wb)

Symbol

Electromagnetism


Magnetic Flux Magnetic flux lines are assumed to have the following

properties: Leave the north pole (N) and enter the south pole (S)

of a magnet. Like magnetic poles repel each other. Unlike magnetic poles create a force of attraction. Magnetic lines of force (flux) are assumed to be

continuous loops.

Electromagnetism


Magnetic Field conductor A magnetic field also forms

round a conductor along which a current is flowing

Field can be described using “right hand screw rule”

Electromagnetism


Right Hand Rule Thumb indicates

direction of current flow

Finger curl indicates the direction of field

Electromagnetism


Wire Coil Notice that a coil

of wire will produce a perpendicular field

Electromagnetism


Magnetic Field: Coil A series of coils produces a field

similar to a bar magnet – but weaker!

Electromagnetism


Magnetic Field : Coil Placing a ferrous material

inside the coil increases the magnetic field

Acts to concentrate the field also notice field lines are parallel inside ferrous element

‘flux density’ has increased

Electromagnetism


Flux Density

Electromagnetism


Permeability μ is a measure of the ease by which a magnetic flux can pass through a material (Wb/Am)

Permeability of free space μo = 4π x 10-7 (Wb/Am)

Relative permeability:

Electromagnetism- Permeability


Reluctance: “resistance” to flow of magnetic flux

@

Associated with “magnetic circuit” – flux equivalent to current

What’s equivalent of voltage?

Electromagnetism- Reluctance


A

lS

r0

Magnetomotive Force Coil generates magnetic

field in ferrous toroidal Driving force F needed to

overcome toroidal reluctance

Magnetic equivalent of ohms law

Electromagnetism


Circuit Analogy

Electromagnetism


Magnetomotive Force (MMF) The MMF is generated by the coil Strength related to number of turns and

current, measured in Ampere turns (At)

Electromagnetism- Magnetomotive Force


• The longer the magnetic path the greater the MMF required to drive the flux

• Magnetomotive force per unit length is known as the “magnetizing force” H

• Magnetizing force and flux density related by:

Electromagnetism- Field Intensity


B(T)

H(A/m)

Magnetization curve (B-H characteristic)

Saturation

HB r0

Free space, electrical conductors (aluminium or copper), insulators:

= unity.

Ferromagnetic materials (iron, cobalt and nickel):

= several hundred - several thousand

A large value of : a small current can produce a large flux density

r

rr

Electromagnetism


Magnetic Field Intensity and Ampère’s LawHB

AmWb104 70

0 r

Ampère’s Law:

idlH

Electromagnetism


Flux Linkages and Faraday’s Law

AB dA

N

Faraday’s law of magnetic induction:

dt

de

Electromagnetism


Ampere’s Law

Electromagnetism


Magnetic Field Around a Long Straight Wire

r

IHB

2

Electromagnetism


• Ampere’s Law

Electromagnetism


Lenz’s Law states that the polarity of the induced voltage is such that the voltage would produce a current (through an external resistance) that opposes the original change in flux linkages

Electromagnetism


Lenz’s Law Voltages Induced in Field-Cutting

Conductors

Blue

Electromagnetism


In many engineering applications, we need to compute the magnetic fields for structures that lack sufficient symmetry for straight-forward application of Ampère’s law. Then, we use an approximate method known as magnetic-circuit analysis.

Electromagnetism- magnetic circuit


• Advantage of the Magnetic-Circuit Approach is that it can be applied to unsymmetrical magnetic cores with multiple coils.

Electromagnetism- magnetic circuit


Magnetic leakage and Fringing• Magnetic leakage/ leakage flux

• Flux not passing through in the magnetic material or in air gap

» In air gap – useful fluxs

• Occurs at the magnetic source– As shown in Figure 2.a

air gap, (useful fluxs)

magnetic Source, NI

useful fluxs, a

leakage flux, l

Total flux, T

33

Magnetic leakage and Fringing• Fringing

• Occurs at the air gap• Flux intends to bulge outwards

» Increasing the effective area

» Reduce the flux density

As shown in Figure 2.a

(still useful flux)

Contoh 1.2 page 1.11, Contoh 1.3 page 1.12, Contoh 1.4 page 1.14 and Contoh 1.5 page 1.15

Magnetic Circuit

lci

N+F-

S

Equivalent circuit

Analogy between magnetic circuit and electric circuit

E R

i

Electromagnetism


Magnetic circuit Electric circuit

Term Symbol Term　 Symbol

Magnetic　flux Electric　current I

Flux　density B Current　density J

Magnetic　field　strength H Electric　field　strength E

Magnetomotive　force F Electromotive　force E

Permeability Permittivity

Reluctance S Resistance R

Electromagnetism


Series Magnetic Circuit

with air gap lc

i

N lg

+F-

Sc

Sg

g

g

g

c

cc

ggcc

gC

g0

g

g

cc

cc

AB;

AB

densityFlux

lHlHNiSS

Ni

A

lS;

A

lS

Electromagnetism


Series composite magnetic circuit with different material

i

N

iron steel

cobalt

+F-

bS

aS

cS

ccbbaacba

cc

cc

bb

bb

aa

aa

lHlHlHNiSSS

Ni

A

lS

A

lS

A

lS

;;

Electromagnetism


Electromagnetism


Electromagnetism


Electromagnetism


Electromagnetism


Circuit Analogy

Electromagnetism


Electromagnetism


Electromagnetism


Electromagnetism


Example 3 A coil of 200 turns is wound uniformly over a wooden ring

having a mean circumference of 600mm and a uniform cross-sectional area of 500mm2. if the current through the coil is 4A, calculate

(a) the magnetic field strength (b) the flux density (c) the total flux ( 1330A/m, 1680µT,0.838µWb)

Electromagnetism


Example 4 Calculate the magnetomotive force required to produce a

flux of 0.015Wb across an air gap 2.5mm long, having effective area of 200cm2

(1492At)

Electromagnetism


Example 5 A mild-steel ring having a cross- sectional area of 500

mm2 and a mean circumference 0f 400mm has a coil 0f 200 turns wound uniformly around it. The relative permeability of the mild steel for the respective flux density is about 380. Calculate

(a) the reluctance of the ring (b) the current required to produce a flux of 800µWb in

the ring (1.68 x 106 At/Wb, 6.7A)

Electromagnetism


Example 6 The Figure represents the magnetic

circuit of a relay. The coil has 500 turns and the mean core path is lc = 360 mm. When the air-gap lengths are 1.5 mm each, a flux density of 0.8 Tesla is required to actuate the relay. The core is cast steel with the field intensity 510 At/m. Find the current in the coil.(4.19 A)Compute the values of permeability and relative permeability of the core.

(1.57 x 10-3 Wb/Am, 1250 Wb/Am)If the air-gap is zero, find the current in the coil for the same flux density (0.8 T) in the core. (0.368 A)

i

N

Movablepart

lg

Electromagnetism


Example 7

A magnetic circuit comprises three parts in series each of uniform cross-sectional area (A). They are:

(a) a length of 80 mm and A= 50 mm2 (b) a length of 60 mm and A = 90 mm2 (c) an air gap of length 0.5 mm and A = 150 mm2

A coil of 4000 turns is wound on part (b) and the flux density in the air gap is 0.3 T. Assuming that all the flux passes through the given circuit, and the relative permeability is 1300, estimate the coil current to produce such a flux density

(45.43mA)

Electromagnetism


Series Parallel Magnetic Circuit

i

N

2

+F-

1 2

1

3S 2S

2233

3311

321

lHlH2loop

lHlHNI1loop

LawsKirchoff

:

:

:

Electromagnetism


Series Parallel Magnetic Circuit i

N

+F-

1 2

1S

3S

2S

2233

133

213

lHlHNI2loop

lHlHNI1loop

LawsKirchoff

:

:

:

`

Electromagnetism


Series Parallel Magnetic Circuit with Air Gap

iN

+F-

1 2

1S 3S2S

g

22ss33

11ss33

213

lHlHlHNI2loop

lHlHlHNI1loop

LawsKirchoff

:

:

:

Electromagnetism


The relationship between B and H is not linear for the types of iron used in motors and transformers.

Electromagnetism- magnetic core loss


dBHAl

WW

B

v 0

Electromagnetism


The relationship between B and H is complicated by non-linearity and “hysteresis” Can be used to calculate µ

Electromagnetism- Hysteresis


Hysterisis

Electromagnetism- Hysteresis


Hysteresis loopUniform distribution

From Faraday's law

Where A is the cross section area

Electromagnetism- Hysteresis Loss


Field energyInput power :

Input energy from t1 to t2

where Vcore is the volume of the core



• One cycle energy loss

where is the closed area of B-H hysteresis loop

• Hysteresis power loss

where f is the operating frequency and T is the period



Empirical equation

Summary : Hysteresis loss is proportional to f and ABH



Eddy currentAlong the closed path, apply Faraday's law

where A is the closed areaChanges in B → = BA changes

→induce e.m.f along the closed path→produce circulating circuit (eddy current) in the core

Eddy current loss where R is the equivalent resistance along

the closed path

Electromagnetism- Eddy Current Loss


How to reduce Eddy current loss– Use high resistively core materiale.g. silicon steel, ferrite core (semiconductor)– Use laminated coreTo decrease the area closedby closed path

Lamination thickness0.5~5mm for machines, transformers at line frequency0.01~0.5mm for high frequency devices



Calculation of eddy current loss– Finite element analysis

Use software: Ansys®, Maxwell®, Femlab®, etc

– Empirical equation



Core Loss Hysterisis loss

• the loss of power in the core due to the hysterisis effect

• Proportional to frequency

Eddy current loss• power loss occurs when the flux density changes rapidly in

the core

• Proportional to the square of the frequency

losscurrenteddyP

losshysteresisPwhere

PPP

e

h

ehc

Electromagnetism- Core Loss


Electromagnetism- Core Loss


Electromagnetic Induction Faraday has made the great

discovery of electromagnet induction, namely a method of obtaining an electric current with the aid of magnetic flux.

When a conductor cuts or is cut by a magnetic flux, an e.m.f is generated in the conductor.

S

A B G

GS N

C

Electromagnetism


Direction of e.m.f Fleming’s right-hand rule

Lenz’s law• The direction of an induced

e.m.f is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f

ThumbMotion of conductor

First fingerFlux

Second fingere.m.f

Electromagnetism


If a conductor cuts or is cut by a flux of dΦ webers in dt seconds, e.m.f generated in conductor

The average e.m.f induced in one turn is

e.m.f induced in a coil:

S N

C

X

Motion

Electromagnetism


The emf induced in electric circuit

Equating expressions of e.m.f induced in magnetic circuit and electric circuit:

L is the self-inductance in Henry, or simply the inductance.

For and

dt

dN

dt

diL

dt

diLe

currentofchange

linkagesfluxofchange

di

dNL

A

lS

r0

S

F

Electromagnetism


Mutual Inductances

S

A B G

Self-inductances of A and B are

S

N

NI

N

I

NL A

AA

AA

A

AAA

22

S

N

I

NL B

B

BBB

2

Electromagnetism- Mutual Inductances


B

BB

A

AA NINIS

S

NNM

NI

NN

I

NM

BA

AA

ABA

A

AB

22

22

MS

NNLL BA

BA

Mutual Inductance:

BA LLM



Mutual Inductance:

BA LLM When there is flux leakage occurs

where k = is coupling coefficient = 0 – 1

k = 1 when the magnetic leakage is zero

BALLkM



Example 8 A ferromagnetic ring of cross-sectional 800mm2 and of

mean radius 170mm has two windings connected in series, one of 500 turns and one of 700 turns. If the relative permeability is 1200, calculate the self-inductance of each coil and the mutual inductance of each assuming that there is no flux leakage.

( 0.283H, 0.552H, 0.395H)



Energy Stored in the Magnetic Field Consider a current increasing at

uniform rate in a coil having a constant inductance L henrys.

li

N

A

Cross-sectionalarea

Electromagnetism


Energy Stored in the Magnetic Field If the current increases by di

amperes in dt seconds, the induced e.m.f

And if i is the value of the current at that instant, energy absorbed by the magnetic field during time dt seconds

dt

diLe

joulesdiLidtdt

diiL ...

Electromagnetism


Energy Stored in the Magnetic Field Hence total energy absorbed by the

magnetic field when the current increases from 0 to I amperes is

jouleLIE

iLdiiLEI

I

221

02

0 2

1.

Electromagnetism


Energy Stored in the Magnetic Field Since inductance

Hence

Henryl

NAL

2

lAH

Il

NAE

221

22

21

?

Electromagnetism


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Chapter 2 Electromagnetism Dr. Mohd Junaidi Abdul Aziz