CHAP6 Electromagnetism - taiwan921.lib.ntu.edu.t · 2 導入在探討電的現象時，必須同時探討磁的現象。因為，電與磁的關係相當密切，沒有「磁」的作用，記錄系

1

CHAP6 CHAP6 ElectromagnetismElectromagnetism

2

Hall effect sensorsHall effect sensorsmagnetic magnetic reed switchesreed switchesMRIMRI

Edwin H. Hall 1879NPmajority carrierconcentrationmobilityelectric probe

3

Basic Properties of Basic Properties of MagnetismMagnetism

4

magnetmagnet

magnetizationmagnetizationnatural magnetnatural magnetartificial magnetartificial magnet

permanent permanent magnetmagnettemporary magnettemporary magnet

5

1/21/2

magnetic fieldmagnetic field

magnetic lines of forcemagnetic lines of force

6

1/21/2

NNSSNN

lines of forceflux lines

7

1/21/2

magnetic polemagnetic poleSSNN

8

2/22/2

9

pole strengthpole strength

221

rmmkF

mm11mm22weberweberwbwb permeabilitypermeability

41k

10

rr

r2IH

Magnetic flux lines surrounding a current-carrying conductor

magnetizing forcemagnetizing forceHH

11

1/21/2

ElectromagnetElectromagnet

retentivityretentivityRetentivityRetentivity

12

2/22/2

13

Direction of Flux vs.Direction of Flux vs.

14

AmpereAmperes rights right--hand hand rulerule

15

Exercise 2

For the electromagnet in Fig. 6.58, determine the direction of IFor the electromagnet in Fig. 6.58, determine the direction of Ineeded to establish the flux pattern, and label the induced nortneeded to establish the flux pattern, and label the induced north and h and south poles.south poles.

N S

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ApplicationApplication

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Magnetic RelayMagnetic Relay

Coil Coil normally closednormally closedNCNCnormally openednormally openedNONOCCAANONONCNCCCBB

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Circuit BreakerCircuit Breaker

movable armmovable arm

19

Magnetically Controlled LockMagnetically Controlled Lock

corecorecorecore

corecore

20

Electric BellElectric Bell

CoilCoil

21

Magnetic CircuitMagnetic Circuit

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ReluctanceReluctance

reluctancereluctanceairairwoodwoodglassglass

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PermeabilityPermeability

PermeabilityPermeabilityflux lineflux line

Relative permeabilityRelative permeability100100

current

length

24

Magnetic FluxMagnetic Flux

magnetic fluxmagnetic fluxelectromagnetic coreelectromagnetic corereluctancereluctance

magnetic fluxmagnetic flux

25

Magnetomotive ForceMagnetomotive Force

Magnetomotive force Magnetomotive force mmfmmf

magnetic magnetic circuitcircuitfluxflux

fluxflux

26

vs. vs.

27

Exercise 4

a.a. If the length of a magnetic core is increased for the same If the length of a magnetic core is increased for the same magnetomotive force, what will happen to the magnitude of the magnetomotive force, what will happen to the magnitude of the resulting flux?resulting flux?

b.b. If the area of a magnetic core is doubled and the length reducedIf the area of a magnetic core is doubled and the length reduced to to oneone--third, what will be the effect on the resulting flux if the third, what will be the effect on the resulting flux if the magnetomotive force is held constant?magnetomotive force is held constant?

AA

A 1 1/36

28

Flux densityFlux densityCGSCGScentimetercentimeter--gramgram--secondsecondGuassGuassMKSMKSmetermeter--kilogramkilogram--secondsecondteslateslaweberweber

1 Tesla=10,000 Guass1 Tesla=10,000 Guass

11,000mG

Wb = 108

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EXAMPLE 6.1EXAMPLE 6.10.4T

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Gauss MeterGauss Meter

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HH

magnetizing force Hmagnetizing force Hmagnetomotive forcemagnetomotive forcemmfmmf

A

NINI

AB

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B vs.B vs.H H 1/41/4BrBrthe retentivity levelthe retentivity levelcoilcoilcorecore

PermeabilityPermeability

33

B vs. H B vs. H 2/42/4

NN HHOBOB

BBBBrHcBBrHcH=0H=0residual magnetismresidual magnetismBrBr

34

B vs. H B vs. H 3/43/4

35

B vs. H B vs. H 4/44/4

magnetization curvemagnetization curveHysteresis loopHysteresis loopBBHHBB--HH

36

Normal Magnetization CurveNormal Magnetization CurveBBHHB=B=HH

Cast steelCast steel

BBHHRRBBHH

PermeabilityPermeabilityFig. 6.15

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BB--H MeterH Meter

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EXAMPLE 6.2EXAMPLE 6.2

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1.46T

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Descriptive Example Descriptive Example 1/31/3

331010--44 WbWb

AmpereAmperes Circuital Laws Circuital Lawmagnetic circuitmagnetic circuitelectric circuitelectric circuitKVL KVL AmpereAmperes s Circuital LawCircuital Lawclosed closed pathpathmmfmmf

0HNI HNI

41


BH

Fig.6.15Fig.6.15BBHHH = 770 At/mH = 770 At/mII

42


rr

43

Exercise 5

a.a. For the system in Fig. 6.17, determine the current I if the areaFor the system in Fig. 6.17, determine the current I if the area is is doubled.doubled.

b.b. Is the resulting current in part (a) half of that obtained in thIs the resulting current in part (a) half of that obtained in the e descriptive example? Why not ?descriptive example? Why not ?

44

Exercise 5

AA661010--44 mm22

Fig.6.15Fig.6.15BBHHH = 275 At/mH = 275 At/m

T5.0m106Wb103

AB 24

4

mA110t200

)m08.0)(m/At275(NHIHNI

A35.7%

45

Exercise 6

a. For the system in Fig. 6.17, determine the resulting flux if thecurrent is reduced to 200 mA.

b. Find the relative permeability of the core.

46

II200mA200mA

m/At500NIHHNI

Fig.6.15Fig.6.15HHBBB = 0.77TB = 0.77T

Wb1031.2AB 4

5.1225

mAWb1054.1

HB

0r

3

Exercise 6

47


corecoreHNI AmpereAmperes Circuital Laws Circuital Lawclosed pathclosed pathmmfmmf

0HNI

Find Find fluxflux

48


Fig.6.15Fig.6.15HHBB

HB

49

Exercise 7

For the magnetic system in Fig. 6.59, determine:a. The magnetomotive force.b. The magnetizing force applied to the core.c. The flux density.d. The flux in the core.

Wb104BA.d

T1)m/At400)(Am/Wb104)(2000(HHB.c

m/At400m2.0

At80H.b

At80)mA400)(t200(NI.a

4

70r

50

Exercise 8Exercise 8

Determine the current I necessary to establish the flux indicated in Fig. 6.60.

T7.0m102Wb104.1

AB 24

4

Fig.6.15Fig.6.15BBHHH =400 At/mH =400 At/m

A6.1NHI

HNI

51

Exercise 9

Determine the current I1 necessary to establish a net flux = 5 10-4 Wb in the transformer in Fig. 6.61.

T25.1m104Wb105

AB 24

4


A375.0I)m15.0)(m/At1500(A2)t75(I)t200(

HININ

1

1

2211

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Air Gap EncounteredAir Gap EncounteredAir gapBH

GapGapBBCoreCoreBBHH

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AmpereAmperes Circuital Laws Circuital Lawclosed pathclosed pathmmfmmf

KVLKVL

0HHNI gapgapcorecore

54

EXAMPLE 6.4EXAMPLE 6.4 CoreCoreAir gapAir gapBBB B corecore=B =B air gapair gap

Fig.6.15BH

55


mmf 6:1

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Exercise 10

Repeat Problem 7 if an air gap of 0.01 in. is cut through the coRepeat Problem 7 if an air gap of 0.01 in. is cut through the core.re.

57

Exercise 11

Repeat Problem 8 if an air gap of 250Repeat Problem 8 if an air gap of 250m is cut through the core.m is cut through the core.

58

gc24

4

BBT7.0m102Wb104.1

AB


A386.4t50At3.219I

At3.219)B1096.7()m2.0)(m/At400(NI

m2.0m750,199m250m2.0HHNI

gg5

c

ggcc

Exercise 11

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Exercise 12

Determine the current required to establish a flux of 5 Determine the current required to establish a flux of 5 1010--44 Wb in Wb in the core of the transformer in Fig. 6.18.the core of the transformer in Fig. 6.18.

T0.1m105Wb105

AB 24

4


A56.1NHI

HNI

60

Exercise 13

If the air gap in Fig. 6.19 is doubled (1/16 in.), will the current required to establish the same flux increase by a factor of 2 also? Determine the resulting current and comment on the results.

Air gapAir gapfluxfluxB=1.33TB=1.33T

61

Exercise 13

m109.15in161

T33.1BBB

4g

gc

Fig.6.15Fig.6.15BBHHHHcc =1750 At/m=1750 At/m

A44.22A56.4I)m109.15)(m/At33.11096.7()m08.0)(m/At1750(I)t400(

HHNIm/At1059.10B1096.7H

45

ggcc

5g

5g

62


Find the magnetic flux established in the series magnetic circuit in Fig. 6.62.

m5027.0r2

Fig.6.15Fig.6.15HHBBB =0.7TB =0.7T

m/At88.397NIH

HNI

Wb103.6BA 3

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D'arsonval movementD'arsonval movement 1/2

D'arsonval movementD'arsonval movementdcdcacac

movablemovablestationary partsstationary partsair gapair gap

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D'arsonval movementD'arsonval movement 2/2

movable coremovable core1k1k1mA1mAmovable coremovable coremovable coremovable corenameplatenameplatefull scalefull scalemovable coremovable core

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AmmeterAmmeter

1A1AmovementmovementRRshuntshuntRRshuntshunt

movementmovement1mA1mA

1A1ARsRs

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VoltmeterVoltmeter

movementmovement1V1V100V100VmovementmovementRRseriesseriesmovementmovement1V1V

RRseriesseries

movementmovement1V1V1k1k

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Using a 50Using a 50--A A 20,00020,000 momovement, design: vement, design: a. A 10a. A 10--A ammeter. A ammeter.

Rshunt ~0.1

b. A 10b. A 10--V voltmeter.V voltmeter.

Rseries~180k

1.0A10

)k20)(A50(IVRshunt

k180

A50)k20)(A50(V10

IVR

RS

RSseries

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TransformerTransformer

69

TransformerTransformer 1/21/2

TransformerTransformer

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Transformer Transformer 2/22/2

primary windingprimary windingsecondary windingsecondary winding

corecore--type transformertype transformershellshell--type transformertype transformer

72

Application of TransformerApplication of Transformer

impedance matchingimpedance matching

blockingblocking

73

Factors makeFactors make

stray capacitancestray capacitance

reversing reversing magnetic fieldmagnetic fieldhysteresis losshysteresis loss

74

Transformer Ratio Transformer Ratio 1/21/2

NNPPNNSS

)tcos(E)t(E PP

dtdN)tcos(E)t(E PPP )tsin(

NE)t(

P

P

75

Transformer Ratio Transformer Ratio 2/22/2

)tcos(ENN)tsin(

dtd

NEN

dtdN)t(E P

P

S

P

PSSS

S

P

S

P

EE

NNa a: Transformer ratioa: Transformer ratio

a1stepstep--down transformerdown transformer

76

Current RatioCurrent Ratio

77

Voltage RatioVoltage Ratio

78

Example 6.5

79

Example 6.5

80

Impedance RatioImpedance Ratio

81

Impedance RatioImpedance Ratio

ZZPP is the impedance seen at the is the impedance seen at the primary of a transformer.primary of a transformer.primary primary primary primary

82

Example 6.6

impedance impedance matchingmatching

83

Example 6.6

88

84

Example 6.6

loadloadpowerpowerZZLLprimaryprimaryZZPPRRThThZZPP=800=800

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Example 6.6

86


A transformer with a turns ratio a = NA transformer with a turns ratio a = Npp/N/Nss = 12 has a load of 2.2 = 12 has a load of 2.2 kk applied to the secondary. If 120 V is applied to the primary, applied to the secondary. If 120 V is applied to the primary, determine:determine:

a.a. The reflected impedance at the primary. The reflected impedance at the primary. ZP = a2 ZL = 316.84kb.b. The primary and secondary currents. The primary and secondary currents. IP = EP/ZP = 0.379 mA IS =

a IP = 4.55 mAc.c. The load voltage. The load voltage. VS = ISRL = 10.01 Vd.d. The power to the load. The power to the load. P = IS2RL = 45.55 mWe.e. TheThe powerpower supplied by source. supplied by source. P = 45.55 mW

real

87


For a power transformer, Ep = 120 V, ES = 6000 V, and Ip = 20A:a. Determine the secondary current IS. IP/IS = ES/EP = 50 IS =0.4Ab. Calculate the turns ratio a. a= NP/NS = 1/50c. Is it a step-down or a step-up transformer? Step-up transformerd. If NS = 100 turns, Np = ? NP=NS a = 2 turns

88


An inductive load ZAn inductive load ZLL = 4= 4 + j4 + j4 is applied to a 120 V /6 V is applied to a 120 V /6 V filament transformer.filament transformer.

a.a. What is the magnitude of the secondary current?What is the magnitude of the secondary current?b.b. What is the power delivered to the load?What is the power delivered to the load?c.c. What is the primary current?What is the primary current?

89

Exercise 18Secondaryinductive load ZL = 4 + j 4= 5.66 45EPP = 120 V120 VVV SS= E= ESS = 6 V= 6 V

A06.1ZVI

L

SS

b.b.real powerreal power

a. Secondary current

c. c. primary currentprimary current

W494.44IP 2S

mA53NNII

P

SSP

90


A purely capacitive load ZA purely capacitive load ZLL = = --j2j2 is applied to a 120 V /6 V is applied to a 120 V /6 V filament transformer with a 12 VA apparent power rating.filament transformer with a 12 VA apparent power rating.

a.a. Determine the magnitude of the secondary current.Determine the magnitude of the secondary current.b.b. Calculate the power delivered to the primary and secondary.Calculate the power delivered to the primary and secondary.c.c. Do you expect the transformer to heat up if operating under thesDo you expect the transformer to heat up if operating under these e

conditions?conditions?

91

Exercise 19Secondarycapacitive load ZL = -j 2 EPP = 120 V120 VVV SS= E= ESS = 6 V= 6 V

A0.3ZVI

L

SS

b.b. delivered to the primary and secondarydelivered to the primary and secondary

a.

c. Apparent powerc. Apparent power12 VA12 VAsecondarysecondary2A2A3A3Aheat upheat up

92


If E = 120 V, RIf E = 120 V, RTH TH = 0.5 k= 0.5 k, and a =5 in the network in Fig. 6.25, and a =5 in the network in Fig. 6.25, , determine the load value for max power to the load.determine the load value for max power to the load.

Determine the power to the load under these conditions. Determine the power to the load under these conditions.

93

Exercise 20a.max. power transferprimaryimpedance0.5 kZP = 0.5 kZP REFLECTED IMPEDANCE

L2

P ZaZ 20ZR LL

b.primary

W2.7RIP

mA600IaImA120ZR

V120I

L2

SL

PSPTH

P

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CHAP6 Electromagnetism - taiwan921.lib.ntu.edu.t · 2 導入 在探討電的現象時，必須同時探討磁的現象。因為， 電與磁的關係相當密切，沒有「磁」的作用，記錄系

CHAP6 Electromagnetism - taiwan921.lib.ntu.edu.t · 2 導入在探討電的現象時，必須同時探討磁的現象。因為，電與磁的關係相當密切，沒有「磁」的作用，記錄系