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Magnetic Materials

Iron, Cobalt and Nickel and various other alloys andcompounds made using these three basic elements

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Electric Current and Magnetic Field

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A Few Definitions Related to

Electromagnetic Field* (Unit is Weber (Wb)) = Magnetic Flux Crossing a Surface of Area A in m 2.

B (Unit is Tesla (T)) = Magnetic Flux Density = * /A

H (Unit is Amp/m) = Magnetic Field Intensity =QB

Q = permeability = Qo Qr

Qo = 4 T*10 -7 H/m (H Henry) = Permeability of free space (air)

Qr = Relative Permeability

Qr >> 1 for Magnetic Material

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Ampres Law

!pp

idl.H

The line integral of the magnetic field intensity around a closed path is

equal to the sum of the currents flowing through the area enclosed bythe path.

U!pppp

cosdlHdl.H

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Example of Ampres LawFind the magnetic field along a circular path around an infinitely long

Conductor carrying I ampere of current.

90 0B,H

r dl

Since bothp

dlp

Hand are perpendicular to radius r at any point A

on the circular path, the angle Uis zero between them at all points.Also since all the points on the circular path are equidistant fromthe current carrying conductor is constant at all points on thecircle

p

H

Ir 2HdlHdl.H !T!!ppppp

or r 2I

HT

!p

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They are basically ferromagnetic structures(mostly Iron, Cobalt, Nickel alloys and compounds) with coils wound around them

Because of high permeability most of the magnetic flux is confinedwithin the magnetic circuit

Thus is always aligned with

Examples: Transformers,Actuators, Electromagnets, Electric

Machines

Magnetic Circuits

pH )0(dl

p

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P = mean length

N

I

d

w

Magnetic Circuits (1)

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F =NI= Magneto Motive Force or MMF = # of turns *Current passing through it

NIB !

Q

Nor NI

A

!

Q

* Nor

or )A/(

NIQ

!*N

!* NIor

F = NI = H P (why!)

Magnetic Circuits (2)

= Reluctance of magnetic path

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Analogy Between Magnetic and ElectricCircuits

F =MMF is analogous to Electromotive force (EMF) =E

* = Flux is analogous to I = Current

= Reluctance is analogous to R = Resistance

= ermeance ! 1 = Analogous to conductanceR 1

G !

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H

B B

H

Linear

knee saturation

Magnetization curve(linear) (Ideal)

Magnetization curve(non-linear) (Actual)

(see also Fig. 1.6 in the text)

Magnetization Curves

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One can linearize magnetic circuits by including air-gaps

However that would cause a large increase in ampere-turnrequirements.

Ex: Transformers dont have air-gaps. They have very littlemagnetizing current (5% of full load)

Induction motors have air-gaps. They have large magnetizing

current (30-50%)

Question: why induction motors have air gap andtransformers dont?

Magnetization Curves(2)

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Magnetization Circuits with Air-gap

N

i

d

wP c

P

g

cc

cc A

l Q! g g

g g A

l

Q! g c Ni

!*

g g cc l H l H Ni ! )( fr inging eglecting wd A A g c !!

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N

iwP c

Fringing

With large air-gaps, flux tends to leak outside the air gap. This iscalled fringing which increases the effective flux area. One way toapproximate this increase is:

nn gn g n g n d l d d l !!! ;;

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Example of Magnetic Circuits Onreenboard

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Magnetization Curves (for examples)

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Inductance(L)

Definition: Flux Linkage( P ) per unit of current(I) in a magnetic circuit

IIL

*!

P!

!*I

!@2

L

Thus inductance depends on the geometry of construction

*

- 2

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Example of Inductances On reenboard

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How to find exact Inductances withmagnetic circuit with finite thickness (saya torroid with finite thickness)

see problem 1.16

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Faradays law of Electromagnetic

InductionThe EMF (Electromotive Force) induced in a magnetic circuit isEqual to the rate of change of flux linked with the circuit

dtd

Ndt

) N(ddtd

e*!*!

P!

*! NLi3

dtdi

dtdL i

e !!@

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Lenzs Law

The polarity of the induced voltage will be such as to oppose the verycause to which it is due

dtdi

Ldt

dLie !!@

The polarity of the induced voltage is given by Lenzs law

Thus som

etim

e s

we write

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V = V m Cos( [ t)* =* m Sin( [ t)

)t(CosV)t(Cos)t(Cosdt

d Ne mmm [![![[*!*!

Ideally

A precursor to Transformer

)t(SinIL

)t(Sin NL

Ni m

m [![*

!*

!

*

i 2e

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A Precursor to Transformer(2)

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 -100

-50

0

50

100

V

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 -100

-50

0

50

10 0

e

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 -60

-40

-20

0

20

40

60

p h i , i

time

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Example on excitation of magneticcircuit with sinusoidal flux Ongreenboard

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Example on excitation of magneticcircuit with square flux on greenboard

(Important for Switched Mode Power Supplies)

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What will non-linearity in magneticcircuit lead to?

It would cause distortion in current waveforms since by Faradaysand Lenzs law the induced voltage always has to balance out the

applied voltage that happens to be sinusoidal

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Sinusoidal voltage non-sinusoidalcurrent

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Iron Losses in Magnetic Circuit

There are two types of iron losses

a) Hysteresis losses

b) Eddy Current Losses

Total iron loss is the sum of these two losses

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Hysteresis losses

Br

B

H

Br = Retentive flux density (due to property of retentivity)Hc= Coercive field intensity (due to property of coercivity)

B-H or Hysteresis loop

1 21 2

i

t3

3

0

4

4

5

5

i

saturationknee point

Hc

T1

f !

f =frequencyof sine source

0

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The lagging phenomenon of B behind H is called hysteresis

The tip of hysteresis loops can be joined to obtain themagnetization characteristics

In each of the current cycle the energy lost in the core is proportional to the area of the B-H loop

Energy lost/cycle = V core

Hysteresis losses (2)

H dB

Ph = Hysteresis loss = f V core Hd = k hBnmaxf

k h = Constant, n = 1.5-2.5, B max= Peak flux density

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Eddy current loss

flux Currentflux

Because of time variation of flux flowing through the magneticmaterial as shown, current is induced in the magnetic material,following Faradays law. This current is called eddy current.The direction of the current is determined by Lenzs law. This currentcan be reduced by using laminated (thin sheet) iron structure, with

Insulation between the laminations.

Laminations

Pe = Eddy current loss = k eB2maxf

Bmax= Peak flux densityk e = Constant ,

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Permanent Magnets

Alloys of Iron, Cobalt and Nickle

Have large B-H loops, with large B r and H c

Due to heat treatment becomes mechanically hard and are thuscalled HARD IRON

Field intensity is determined by the coercive field required todemagnetize it

Operating points defined by B m,Hm in the second quadrant of the B-H loop

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PM

SOFT IRON

SOFT IRON

lm lg

Using Permanent Magnets for providingmagnetic field

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Designing Permanent MagnetsThe key issue here is to minimize the volume V m of material

required for setting up a required Bg

in a given air gap

It can be shown that V m =B g2Vg/ oBmHm (see derivation in text)where V g= A glg Volume of air-gap,l g = length of air-gap, A g =areaof air-gap

Thus by maximizing B m, H m product V m can be minimized

Once Bm, H m at the maximum B m, H m product point are known,lm =length of permanent magnet, A m =area of permanent magnetcan be found as

lm=-l gHg/Hm (applying ampres law),Am=B gAg/Bm (same flux flows through PM as well as air-gap)

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Finding the maximum product point

-1 0 0 0 -8 0 0 -6 0 0 -4 0 0 -2 0 0 0 0

0 .2

0 .4

0 .6

0 .8

1

1 .2

1 .4

H(kA/m)

B ( T e s l a

)

D e m a g n e t iz a t io n curve f o r Neody m i um -i r o n-b o r o n magnet

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B= mH+c, m and c are constants.

To find maximum BH product, we need to differentiateBH=mH 2+cH ;

and set it equal to 0. Thus we get

Hm=-c/2m. and B m =c/2

Finding the maximum product point (2)

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Finding the maximum product point (3)

Answer:

Bm=0.64 T, H m = -475 kA/m