(Chetan Upadhyay)

Chetan Upadhyay

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 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 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 = magnetic field strength ; B = magnetic flux density and = permeability of the medium

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

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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

Relative Permeability (r)

r = /o

= ro = B/H

or B = roH

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B vs H

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Electromagnetic Force (mmf)

turns

H

NIH mmf

Where = 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 turnsl = 600 x 10-3 mA = 500 x 10-6 m2

I = 4 A

(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

turns

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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

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 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.

TA

B 6.110500

108006

6

r = 380

turns

47 105104380

4.0

A

Sor

]/[10667.1 6 WbA

(a)

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(b)

S

H SH

NIAH ][134210667.110800 66mmf

][7.6200

13421342A

NI

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Magnetic circuit with different materials

11

1A a

S

l

22

2B a

S

l

BA SSS 22

2

11

1

aa

ll

and

For A: area of cross-section = a1

mean length = l1

absolute permeability = 1

ForB: area of cross-section = a2

mean length = l2

absolute 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 3A 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.

WbAB CC44 1045.0105.13.0

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mAAI 4.45104.454000

8.181 3

AtA

Saor

aa 1.44

10501041300

10801045.067

34

AtSSSNI cba 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 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.

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TA

Ba

a 1102500

105.26

3

]/[796000104

17

mATB

Ho

aa

factorleakageairgapinfluxfluxTotal T

][1594002.0796000 ATHmmf

Wb003.02.10025.0

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

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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 + HMlM

outside loop NI= HLlL + HNlN

And in loop D 0 = HMlM + HNlN

In general (m.m.f) = Hl

Magnetic circuit applying voltage law Analogy to electrical circuit

D

applying voltage Kirchoff’s law

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L = M + N

Or L - M - N = 0

In general: = 0

At node P we can also applying current Kirchoff’s law

P

L M N

E

Q

IL INIM

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Example 5

340mm

340mm 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 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.

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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

340mm 150

mm

1mm

A B

43881080010434000

1015067

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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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

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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 by increasing 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 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