11 Magnetic Circuit

8/13/2019 11 Magnetic Circuit

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Magnetic Flux density B

Unit for magnetic flux density is Tesla

The definition of 1 tesla is the flux density thatcan produce a force of 1 Newton per meter acting

upon a conductor carrying 1 ampere of current.

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 oneturn coil if the flux reduce to zero with uniform rate.

Magnetic field strength H

Unit for magnetic field strength is Ampere/m

A line of force that produce flux


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F = B l I newton

Where F = force ; B = magnetic flux density ; l =the length of

conductor and I = current in the conductor

ABWhere = magnetic flux ; B = magnetic flux density and A =

area of cross-section

HB Where H= magnetic field strength ; B = magnetic flux density

and = permeability of the medium

o= B/H = 4x 10-7 H/mPermeability in free space


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Relative permeability is defined as a ratio of flux densityproduced in a material to the flux density produced in a

vacuum for the same magnetic filed strength. Thus

Relative Permeability (r)

r = /o

= ro = B/H

or B = roH


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H vs r

Relative

permeability

Magnetic field strength

Cast iron

Mild steel


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Relative permeability vs magnetic flux


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B vs H


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Electromagnetic Force (mmf)

turns

H

NIH mmf

Where H= magnetic field strength ; l =the path length of ; N

number of turns and I = current in the conductor


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Example 1A 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

N = 200 turns

l = 600 x 10

-3

mA = 500 x 10-6m2

I = 4 A

(a) H = NI/l = 200 x 4 / 600 x 10-3 = 1333 A

(b) B = oH = 4x 10-7x 1333 = 0.001675 T = 1675 T

(c) Total Flux = BA = 1675 x 10-6x 500 x 10-6

= 0.8375 Wb

turns


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Ohmslaw 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=areaof cross-section [meter square]

Analogy to the ohmslaw

V=NI=H l I= and R=S

Reluctance ( S )

weberS

Hl weberampere

AS

or

/

where


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Example 2

A mild steel ring, having a cross-section area of 500 mm2and a

mean circumference of 400 mm is wound uniformly by a coilof 200 turns. Calculate(a) reluctance of the ring and (b) a

current required to produce a flux of 800 Wb in the ring.

TA

B 6.110500

10800

6

6

Dari graf r/B, pada B = 1.6;

r = 380

turns

47 105104380

4.0 AS or

]/[10667.1 6 WbA

(a)


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(b)

SH SH

NIAH

][134210667.110800

66mmf

][7.6200

13421342A

NI


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Magnetic circuit with different materials

11

1Aa

S l

22

2B

aS

l

BA SSS 22

2

11

1

aa

ll

and

For A: area of cross-section = a1mean length = l1absolute permeability = 1

ForB: area of cross-section= a2mean length= l2absolute permeability= 2


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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


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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 throughthe given circuit, and that the relative permeability r is 1300,

estimate the coil current to produce such a flux density.

WbAB CC 44 1045.0105.13.0


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mAAI 4.45104.454000

8.181 3

AtAS aor

aa 1.4410501041300

10801045.067

34

AtSSSNIcba

8.1813.1194.181.44

and

Mmf = S = H l = N I

AtA

Sbor

bb 4.18

10901041300

10601045.067

34

AtA

Scor

cc 3.119

101501041300

105.01045.067

34

Material a

Material b

airgap

Total mmf


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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

Magnetic circuit with air-gap Leakages and fringing of flux

fluxusable

fluxtotalfactorLeakage

fringing

leakage


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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 and2500 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.


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TA

Ba

a 1102500

105.26

3

]/[796000104

17 mAT

BH

o

aa

factorleakageairgapinfluxfluxTotal T

][1594002.0796000 ATHmmf

Wb003.02.10025.0

92% of the depth is laminated steel, thus the area of crosssection is

AS = 40 x 50 x 0.92 = 1840 mm2=0.00184m


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TA

BS

TS 63.1

00184.0

103 3

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

I = 3992/800 = 5 A

NI = 3992


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Mmf in loop C = NI = HLlL + HMlMoutside loop NI= HLlL+ HNlNAnd in loop D 0 = HMlM+ HNlNIn general (m.m.f) = Hl

Magnetic circuit applying voltage law Analogy to electrical circuit

D

applying voltage Kirchoffs law


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L = M+ N

Or L- M- N = 0

In general: = 0

At node P we can also applying current Kirchoffs law

P

L M N

E

Q

IL

IN

IM


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Example 5

340

mm

340

mm 150

mm

1mm

A magnetic circuit made of silicon steel is arranged as in the

Figure. The center limb has a cross-section area of 800mm

2

andeach 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.


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aBAA SSSfmm 21..

AB TA

B 25.110800

1016

3

159151050010434000

1034067

3

1

11

AS

or

Looking at graph at B=1.25T r=34000

Apply voltage law in loops A and B 340mm

340

mm 150

mm

1mm

AB

43881080010434000

10150

67

3

2

S

99471810800104

10167

3

aS


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10079998

994718438810115915105.0.. 33 fmm

Since the circuit is symmetry A=B

In the center limb , the flux is 1mWb which is equal to 2

Therefore =0.5mWb

aSSSfmm 21 2..

A


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Weight

Iron yoke

KeeperAir gap

Example 6

A U-shaped electromagnet shown in

Fig. is designed to lift a mass. The

material for the yoke has a relativepermeability of 2900. The yoke has a

uniform cross-sectional area of 4000

mm2and 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

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 Ais passed through the coil. You may

neglect the fringing effect and flux

leakage; and assume that


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aB

a

a

o

aa

aa

B

B

lBlH

2

10

104

101.02

22

3

7

3

aa

or

iaii B

BlBlH

6.11

106

1042900

10600 3

7

3

Calculation of maximum weight lifted by and electromagnet.

Let the flux density in the air gap be

For the air gap

For the iron yoke


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NIBBlHlHHl aaiiaa

8.32310

6.11

6

2

1 3

T11.18.323

360

8.323

NIBa

N3922104

10400011.1

22

7

6222

o

a

o

a ABABF

3922mgkg400

81.9

39223922

gm

Total mmf;

Since there are two air gaps;


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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 themolecules or structures are in one direction, thus the

materials become magnetized. The more H applied

the more magnetic flux (B )will be produced


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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 byincreasing positively the also

decrease and certain point it again

change the polarity to negative until

it reach C.


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Eddy current

metal insulator

When a sinusoidal current enterthe 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 eddycurrent we use laminated core

11 Magnetic Circuit

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