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9. Maxwell’s Equations
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9. Maxwell’s Equationsiirc.khu.ac.kr/uploads/6/3/4/3/63434825/sadiku_ch09.pdf · 2019-11-23 · 4 9.2 Faraday’s Law Faraday’s law states that the induced emf (electromotive

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Page 1: 9. Maxwell’s Equationsiirc.khu.ac.kr/uploads/6/3/4/3/63434825/sadiku_ch09.pdf · 2019-11-23 · 4 9.2 Faraday’s Law Faraday’s law states that the induced emf (electromotive

9. Maxwell’s Equations

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* Stationary Charge Electrostatic Fields

𝛻 ∙ E = ρ

𝛻 × E = −𝜕B/𝜕t

* Steady Current Magnetostatic Fields

𝛻 × H = 𝜕D/𝜕t + J

= J

H1 = 0

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* Time-Varying Current Electromagnetic Fields (or Waves)

Fig 9.1 Various type of time-varying current

(a) sinusoidal (b) rectangular

(b) triangular

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ψ=1 ψ=2 ψ=3

ψ=4

Magnetic Flux: 𝛙 = 𝐁 ∙ 𝐝 𝐒

𝐁

at 2D

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9.2 Faraday’s Law

Faraday’s law states that the induced emf (electromotive force), Vemf [volt], in

any closed circuit is equal to the time rate of change of the magnetic flux

linkage by the circuit.

Vemf = −dλ

dt= −N

dt(9.1)

“- sign” is from Lenz’s Law & diamagnetic.

𝐁

𝐈

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5

)b3.9(IRLdE -LdE V

battery)(through )a3.9(LdE 0 LdE LdE

)2.9(EEE

P

N eP

N femf

L

P

N ffL

ef

From Fig. 9.2 (Ef : emf-produced field, Ee : electrostaic field)

• Ee cannot maintain a steady current in a closed circuit• Ef is nonconservative

Fig 9.2 Circuit showing emf-producing field Ef and electrostatic field Ee.

P

N

EeEf R

Ee

I

전지

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6

ef

PP

L

LL

kLL

EEE

?0E

0V

dV-

Ld)V( LdE

0LELdE

0SdE

0t

BE

k

P

N

E’eEf R

Ee

I

전지

P

N

Ef R

Ee

I

전지

∆𝐋𝐤

𝐄𝐋𝐤실제의 경우

Electrostatic 경우

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zt

z

yt

y

xt

x

tdt

d

)(

dV

dzz

Vdy

y

Vdx

x

V

dzadyadxaz

Va

y

Va

x

VaLdV

)(

zyxzyx

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0v)ex(

)veconservatiisE(0dL

dEE)( L

** 물 풍선: 부피가 보존

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9.3 Transformer and Motional EMFs (EMF=electromotive force)

A. Stationary Loop in Time-Varying B Field (Transformer emf)

)5.9(SdB dt

d

SdBt

Sdt

BSdELdE V

)4.9(dt

d- V

S

SSSLemf

emf

)8.9(t

BE

)7.9(Sdt

B Sd)E(

)6.9(Sdt

B LdE V

SS

L Semf

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Fig 9.3 Induced emf due to a stationary

loop in a time varying B field.

applied 𝐁

𝐈

𝐄 induced 𝐁

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B. Moving Loop in Static B Field (Motional emf)

(9.14))B u(E

(9.13)uB V

(9.12)BI F

(9.11)B I F

)10.9(Ld)B u(LdEV

(9.9)B uQ

FE

(8.2)B IB uIt B uQ F

m

emf

m

m

LL memf

mm

m

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Fig 9.4 A direct-current machine.

I

N S

+ -

𝐅 𝐅

𝐁

I

I

발전기

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Fig 9.5 Induced emf due to a moving loop

in a static B field.

𝐮

𝐁

𝐱

𝐲

L

𝐏

𝐈

𝐅𝐑

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C. Moving Loop in Time-Varying Field

(9.16))Bu(t

BE

)15.9(Ld)Bu( Sdt

B LdE V

L LSemf

A. Stationary Loop in Time-Varying B Field (Transformer emf)

)8.9(t

BE

)7.9(Sdt

B Sd)E(

)6.9(Sdt

B LdE V

SS

L Semf

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참고

II-4. Faraday 방정식

Eulerian Description

날아가는 야구공이 있을 때 야구공의 속도, 가속도는 야구공에서 측정, 혹은 계산한 값이다. 우리가 사용하는 d/dt의 미분자는 관찰자의 시선이 움직이는야구공을 따라가며 측정, 혹은 계산한 시간에 대한변화율이다.

시선을 움직이는 야구공에 고정하고 초기의 위치와속도를 기억하여 지금의 속도를 계산하고 위치를 계산하는 것이 Lagrangian Description이다.

시선을 움직이지 않고 시간과 공간 좌표로 야구공이나는 것을 관찰, 묘사하는 것이 Eulerian Description

이다. 대부분의 경우 Field를 기술하는데 Eulerian

Description이 편리하게 사용된다. 하지만 움직이는야구공을 묘사하는 경우에는 Lagrangian Description

이 편리하다.

그림 2.8 Eulerian Description에서Total Derivative d/dt 설명도.

입자들이 A에서 B로 이동.

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그림 2.8에서 u, v, w는 좌표 x, y, z 방향의 속도이다.

시간이 t에서 t+dt로 지나가 입자들이 A에서 B로 이동했을 때 임의의 특성 변화를 dH라고 가정 한다.

Euler Description에 의거 Taylor Series에 의해 dH는방정식 (2.45)과 같이 나타낼 수 있다.

dH =𝜕H

𝜕tdt +

𝜕H

𝜕xdx +

𝜕H

𝜕ydy +

𝜕H

𝜕zdz (2.45)

t의 미분 의 의미는 t 이외의 독립변수는 상수처럼 여기면서 미분하라는 수학적인 도구이며 현재 위치에서의 시간에 대한 변화율이다. 과거에 현재 위치에있던 것은 그림 2.8에서 B로 갔고 지금 현재 위치에있는 것은 다른 것이다. d/dt는 흐름을 따라가며 측정한 시간에 대한 변화율이다. dx=udt, dy=vdt, dz=wdt

이므로 방정식 (2.45)는 방정식 (2.46)이 된다.

dH

dt=𝜕H

𝜕t+ u

𝜕H

𝜕x+ v

𝜕H

𝜕y+ w

𝜕H

𝜕z

=𝜕

𝜕t+ u

𝜕

𝜕x+ v

𝜕

𝜕y+ w

𝜕

𝜕zH (2.46)

𝜕/𝜕𝑡 = 0인 것을 Steady State라 하고 𝜕/𝜕𝑡 ≠ 0이면Transient State라고 한다.

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Faraday 방정식 유도

Faraday의 정성 어린 관찰의 결과 초전도 물질로 만든 폐회로를 통과하는 자장의 시간변화율에 비례하여 폐회로에 전류를 만드는 기전력이 커진다는 것이알려졌다.

그림 2.9 Faraday 방정식을 유도하기 위한 설명도.

단위를 volt로 가지는 φ는 기전력이라 하고 F는 그림2.9의 폐회로를 수직으로 관통하는 자장의 성분에면적을 곱한 값이라고 하면 방정식 (2.47)을 얻는다.

S는 Contour C로 둘러 쌓인 면적이다.

φ = −k𝜕F

𝜕t(2.47)

where φ = E′ ∙ dL

F = B ∙ ndS

E′은 실험실에서 실험자가 폐회로에 생긴 전장을 측정한 값이다. 상수 k가 1 이 되도록 단위를 조정한다.

폐회로가 움직이지 않으면서 측정한 전장을 E′′라고하면 움직이지 않는 폐회로에 대한 방정식 (2.47)은다음과 같이 정리 된다.

E′′ ∙ dL = 𝛻 × E′′ ∙ ndS

=𝜕

𝜕t B ∙ ndS

𝛻 × E′′ +𝜕B

𝜕t= 0 (2.48)

방정식 (2.48)은 폐회로와 관계없는 일반적인 현상을묘사하는 방정식이다. 폐회로가 움직이는 경우를 고찰한다. 폐회로가 그림 2.10과 같이 등속도 를 가지고 이동하며 폐회로에 생긴 전장 를 측정할 경우를살핀다.

그림 2.10 폐회로가 속도 v로 이동.

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방정식 (2.49)에 의하여 방정식 (2.47)을 정리하면 다음과 같다.

𝛻 × E′ − v × B +𝜕B

𝜕t= 0 (2.50)

일정하게 상대적으로 움직이는 2 개의 좌표계에서각 관찰자가 관찰하는 물리적 현상은 같아야 한다는Galilean Invariance에 의해 폐회로에 생긴 전장은 같

아야 한다. 방정식 (2.48)와 (2.50)은 같다. E′′ = E라하면 방정식 (2.51)을 얻을 수 있다.

E′′ ∙ dL = 𝛻 × E′′ ∙ ndS

=𝜕

𝜕t B ∙ ndS

𝛻 × E′′ +𝜕B

𝜕t= 0 (2.51)

다음과 같이 Faraday 방정식이 유도 되었다.

𝛁 × 𝐄 +𝛛𝐁

𝛛𝐭= 𝟎 (𝟐. 𝟓𝟐)

방정식 (2.46)의 d/dt 의미를 새기면서 방정식 (2.47)

의 우변을 정리한다.

d

dt B ∙ ndS

= dB

dt∙ ndS

= 𝜕B

𝜕t+ v ∙ 𝛻B ∙ ndS

𝛻 × B × v = v ∙ 𝛻B − B ∙ 𝛻v + B𝛻 ∙ v − v𝛻 ∙

B

= 𝜕B

𝜕t+ 𝛻 × (B × v) + v(𝛻 ∙ B) ∙ ndS

= 𝜕B

𝜕t∙ ndS + 𝛻 × (B × v) ∙ ndS

= 𝜕B

𝜕t∙ ndS + B × v dL (2.49)

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참고: Vector identity

𝛻 ∙ φu = φ𝛻 ∙ u + u ∙ 𝛻φ

𝛻 × φu = φ𝛻 × u + 𝛻φ × u

𝛻 ∙ u × v = v ∙ 𝛻 × u − u ∙ 𝛻 × v

𝛻 × u × v = v ∙ 𝛻u − u ∙ 𝛻v + v ∙ 𝛻u − v ∙ 𝛻u

𝛻 u ∙ v = u ∙ 𝛻v + v ∙ 𝛻u + u × (𝛻 × v) +v × (𝛻 × u)

𝛻 × 𝛻φ = 0

𝛻 ∙ 𝛻 × u = 0

𝛻 × 𝛻 × u = 𝛻(𝛻 ∙ u) − 𝛻2u

𝛻 ∙ 𝛻φ1 × 𝛻φ2 = 0

a × b × c = (a ∙ c)b − (b ∙ a) c

a × b ∙ c × d = a ∙ c ∙ b ∙ d − a ∙ d ∙ b ∙ c

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Fig 9.6 For example 9.1.

예제 9.1 그림 9.6의 Circuit에 유도되는 전압을 계산하라.

(a) 막대가 y=8 cm에 있고 𝐁 = 𝟒𝐜𝐨𝐬(𝟏𝟎𝟔𝐭) 𝐚𝐳 𝐦𝐖𝐛/𝐦𝟐

(b) 막대가 속도 𝐮 = 𝟐𝟎 𝐚𝐲 𝐦/𝐬𝐞𝐜로 이동하고 𝐁 = 𝟒 𝐚𝐳 𝐦𝐖𝐛/𝐦𝟐

(c) 막대가 속도 𝐮 = 𝟐𝟎 𝐚𝐲 𝐦/𝐬𝐞𝐜로 이동하고 𝐁 = 𝟒𝐜𝐨𝐬(𝟏𝟎𝟔𝐭 − 𝐲) 𝐚𝐳 𝐦𝐖𝐛/𝐦𝟐

LS

S

Lemf

Ld)Bu(Sdt

B

SdE

LdE V(9.15)

𝐮

𝐁

𝐱

𝐲

L=6 cm

𝐏

𝐐

𝐋

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예제 9.1 그림 9.6의 Circuit에 유도되는 전압을 계산하라.

volt)t10sin(2.19

)06.008.0()t10sin(4000

dxdy)t10sin(10004.0

Sdt

B

Ld)Bu(Sdt

BV)15.9(

6

6

08.0

0y

06.0

0x66

emf

(a) 막대가 y=8 cm에 있고 𝐁 = 𝟒𝐜𝐨𝐬(𝟏𝟎𝟔𝐭) 𝐚𝐳 𝐦𝐖𝐛/𝐦𝟐

𝐮

𝐁

𝐱

𝐲

L=6 cm

𝐏

𝐐

𝐋

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(b) 막대가 속도 𝐮 = 𝟐𝟎 𝐚𝐲 𝐦/𝐬𝐞𝐜 로 이동하고 𝐁 = 𝟒 𝐚𝐳 𝐦𝐖𝐛/𝐦𝟐

mV8.4

06.0004.020uB

dx)uB(

dxaa)uB(

adx)aBau(

Ld)Bu(

Ld)Bu(Sdt

BV(9.15)

0

x

0

x xx

0

x xzy

L

LSemf

예제 9.1 그림 9.6의 Circuit에 유도되는 전압을 계산하라.

𝐮

𝐁

𝐱

𝐲

L=6 cm

𝐏

𝐐

𝐋

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예제 9.1 그림 9.6의 Circuit에 유도되는 전압을 계산하라.

volt)t10cos(240)yt10cos(240

)yt10cos(0048.0)t10cos(240)yt10cos(240

)yt10cos()06.0)(08.0()yt10cos(240

dxaa)]yt10cos()004.0)(20[(

dxyd)yt10sin()10)(004.0(

dxa]a)yt10cos()004.0(a20[

dxyd)yt10sin()10)(004.0(

Ld)Bu(Sdt

BV(9.15)

66

666

6y

0

6

0

06.0 xx6

06.0

0x

y

066

0

06.0 xz6

y

06.0

0x

y

066

LSemf

(c) 막대가 속도 𝐮 = 𝟐𝟎 𝐚𝐲 𝐦/𝐬𝐞𝐜 로 이동하고 𝐁 = 𝟒𝐜𝐨𝐬(𝟏𝟎𝟔𝐭 − 𝐲) 𝐚𝐳 𝐦𝐖𝐛/𝐦𝟐

방법 1

𝐮

𝐁

𝐱

𝐲

L=6 cm

𝐏

𝐐

𝐋

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예제 9.1 그림 9.6의 Circuit에 유도되는 전압을 계산하라.

mWb)t10sin(24.0)t20t10sin(24.0

)t20uty(

mWb)t10sin(24.0)yt10sin(24.0

)yt10sin()06.0(4

ydxd)yt10cos(4

SdB

66

66

y

0y

6

y

0y

06.0

0x6

S

volt)t10cos(240)yt10cos(240

)t10cos()2010)(001.0(24.0)t20t10cos()2010)(001.0(24.0

t

)001.0(V

66

6666

emf

(c) 막대가 속도 𝐮 = 𝟐𝟎 𝐚𝐲 𝐦/𝐬𝐞𝐜로 이동하고 𝐁 = 𝟒𝐜𝐨𝐬(𝟏𝟎𝟔𝐭 − 𝐲) 𝐚𝐳 𝐦𝐖𝐛/𝐦𝟐

방법 2

𝐮

𝐁

𝐱

𝐲

L=6 cm

𝐏

𝐐

𝐋

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25

Figure 9.7 For example 9.2; polarity is for increasing emf.

예제 9.2 그림 9.7에서 자기장 𝐁 = 𝟓𝟎 𝐚𝒙 mWb/m2 내에 loop가 존재 한다.loop가 50 Hz로 회전하며 자속과 쇄교 하며 t=0 에서 yz 평면에 있다.

(a) t=1 msec에서의 기전력(b) t=3 msec에서 저항이 0.1 Ω일 때 유도 전류

R=0.1 Ω

𝐁

3 cm

A

B

C

D

y

x

z

φ

ω

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26

R=0.1 Ω

𝐁

3 cm

A

B

C

D

y

x

z

φ

ω

예제 9.2 그림 9.7에서 자기장 𝐁 = 𝟓𝟎 𝐚𝐱𝐦𝐖𝐛/𝐦𝟐 내에 loop가 존재 한다.

loop가 50 Hz로 회전하며 자속과 쇄교 하며 t=0 에서 yz 평면에 있다.(a) t=1 msec에서의 기전력

mVcos6

03.0cos5010004.0

)dzcosB(

adz)acosB(V

acosB

0sinBcosB

00

aaa

Bu

)asina(cosBaBB

sec]/radian[100f2

aadt

d

dt

Ldu

adzLdLd

DCLine

Ld)Bu(Ld)Bu(Sdt

BV(9.15)

03.0

0z 0

03.0

0z zz0emf

z0

00

z

0x0

zDC

LLSemf

에서

0Ld)Bu(

mV825.5

mV)10/sin(6

)10/sin()2/10/cos(

2/10/

2/t

mVcos6V

001.0t

emf

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27

𝐁

A

B

C

D

y

x

z

ω

asinacosa x

x

y

a

ya

xa

a

a

𝐮

𝐝 𝐋

𝐮 × 𝐁

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28

예제 9.2 그림 9.7에서 자기장 𝐁 = 𝟓𝟎 𝐚𝒙𝐦𝐖𝐛/𝐦𝟐 내에 loop가 존재 한다.

loop가 50 Hz로 회전하며 자속과 쇄교 하며 t=0 에서 yz 평면에 있다.(b) t=3 msec에서 저항이 0.1 Ω일 때 유도 전류

A1525.01.0

)10/3sin(006.0

R

VI

mV)10/3sin(6

)10/3sin()2/10/3cos(

1002/10/3

2/t

mVcos6V

emf

003.0t

emf

R=0.1 Ω

𝐁

3 cm

A

B

C

D

y

x

z

φ

ω

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29

volt)t100cos(6

1.02

)t100cos(300001.0)104(500200100

)t100cos(3002

SNN

dt

di

2

SNN

dt

dNV

2

SiN

S/

iN

R

7

0

211

0

2122

0

1111

예제 9.3 그림 9.8에서 단면적 10-3 m2, Turn N1=200회 인 Coil에 전류 i1=3sin(100πt) A가흐를 때 N2=100회 인 Coil에 유도되는 기전력을 구하라.매질의 투자율은 μ=500μ0 이다.

N1=200

ρ0=10 cm

N2=100

i1=3sin(100πt)

V1 V2

ψ

+

-

+

-

Fig 9.8 Magnetic circuit of example 9.3.

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30

9.4 Displacement Current

(9.23)Jt

DH

)b22.9(currentntdisplaceme:t

DJ

t

D)D(

ttJJ

JJ0)H(

JJH

0t

-J

J0)H(

(9.17)JH

d

vd

d

d

v

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31

L

2S

I

Fig 9.10 Two surfaces of integration

showing the need for J𝐝 in Ampère’s circuit law.

SL

SS

S

SdJLdH

SdJSdH

Sd)JH(

JH

ISdJ1S

0SdJ2S

1S

L

I

전자가 통과하지 않는다.

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32

Fig 9.10 Two surfaces of integration

showing the need for 𝐉𝐝 in Ampère’s circuit law.

S dL

S dS

S d

d

Sd)JJ(LdH

Sd)JJ(SdH

Sd)JJH(

JJH

0SdJ

ISdJ

1S d

S 1

ISdJ

0SdJ

2S d

2S

1S

L

I

L

2S

I

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33

예제 9.4 극판 면적이 5 cm2, 사이 간격이 3 mm 인 평행 평판 Capacitor에전압 V=50sin(1000t) 를 인가하였다. ε=2ε0 라고 할 때변위전류 (Displacement Current)를 계산하라.

nA)t1000cos(4.147

)t1000cos(50000003.0

0005.0

36

102

)t1000cos(50000d

S

t

VC

t

V

d

SJSI

t

V

dt

DJ

d

VED

9

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34

9.5 Maxwell’s Equation in Final Forms

Table 9.1 Generalized Forms of Maxwell’s Equations.

densityeargchVolume:

densityCurrent:J

m/F10854.8m/F36/10

)(tyPermeabili:m/H104

fieldElectric:E

densityfluxElectric:ED

fluxMagnetic:B

fieldMagnetic:/BH

EqAmpereSdt

DJLdH

t

DJH

EqFaradaySdBt

LdEt

BE

EqGauss0SdB0B

EqGaussdxdydzSdDD

formIntegralformalDifferenti

v

1290

70

SL

SL

S

vSv

투자율

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35

Continuity 방정식 유도

vfdnwhere

0)vn(t

n

dAfm

Knxd)fv(vfd

t

xdfm

Kxd)fv(vfd

t

xdfm

Kxd)fv(vd

t

f

vd0fm

Kfv

t

f

3

r

3r

3

3v

3r

3

3v

3r

3

3vr

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36

Ohm 방정식

VIR

L

VAIJA

EJ

Em

qvqn

Emn

qv

vmn)Bv(qEq)nkT(dt

vdmn

2

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37

)c30.9(uEJ

(9.30b))M H( H B

(9.30a)PEED

)29.9(ut

J

)28.9()BuE(QF

v

0

0

v

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38

Control Volume을 이용하여 유도한다. 그림 2.7과 같이 dy와 dx 길이를 가지고 두께가 1 인 상상의 2차원 Control Volume을 설정한다. 2차원으로 Control Volume을 설정하는 이유는 도표에 나타내기가 쉽고 덜 복잡하기 때문이다. 이 Control Volume에 들어 있는 물질의 질량은 물질이 새롭게 생성되거나 없어지지 않는다면 안으로 흘러 들어가는 물리양의 Flux와 흘러 나가는 Flux의 차이 만큼 증가하게 된다.

𝜕ρ

𝜕t= dxdy ∙ 1

= ρudy ∙ 1 − ρu +𝜕 ρu

𝜕xdx dy ∙ 1

+ρvdx ∙ 1 − ρu +𝜕 ρv

𝜕ydy dx ∙ 1 (2.38)

z 방향의 성분을 고려하여 이를 3차원으로다시 정리하면 방정식 (2.39)와 같다.

𝜕ρ

𝜕t+𝜕 ρu

𝜕x+𝜕 ρv

𝜕y+𝜕 ρw

𝜕z= 0 2.39

𝜕ρ

𝜕t+ 𝛻 ∙ ρv = 0 (2.40)

그림 2.7 연속 방정식을 유도하기위한 Control Volume.

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39

0B0E

0J0H0E

conductorperfectFor

(9.31d)0a)BB(or 0BB

(9.31c)a)DD(or DD

(9.31b)Ka)HH(or KHH

)a(9.310a)EE(or EE

nt

12n21n2n1

s12n21sn2n1

12n21t2t1

12n21t2t1

tE

na

nE

E

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40

(a) compatibility equations

(b) constitutive equations

(c) equilibrium equations

)35.9(Jt

B-E

(9.34)0 B

m

m

(9.37)ED

(9.36)H B

ρm: Free magnetic density자하밀도

Jm: magnetic current density자기전류밀도

)39.9(t

DJH

)38.9(D v

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41

Figure 9.11 electromagnetic flow diagrams showing the relationship between the

potentials and vector fields: (a) electrostatic system, (b) magnetostatic system, (c)

electromagnetic system. [adapted with permission from the publishing department of the

institution of electrical engineers.]

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42

9.6 TIME-VARYING POTENTIALS

)52.9(Jt

EBJ

t

AA

)51.9(Jt

EB

t

VV

)50.9(t

EB

t

VA

)45.9(t

BE

t

AVE

)42.9(0BAB

2

22

v2

22

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43

참고: Vector identity

𝛻 ∙ φu = φ𝛻 ∙ u + u ∙ 𝛻φ

𝛻 × φu = φ𝛻 × u + 𝛻φ × u

𝛻 ∙ u × v = v ∙ 𝛻 × u − u ∙ 𝛻 × v

𝛻 × u × v = v ∙ 𝛻u − u ∙ 𝛻v + v ∙ 𝛻u − v ∙ 𝛻u

𝛻 u ∙ v = u ∙ 𝛻v + v ∙ 𝛻u + u × (𝛻 × v) +v × (𝛻 × u)

𝛻 × 𝛻φ = 0

𝛻 ∙ 𝛻 × u = 0

𝛻 × 𝛻 × u = 𝛻(𝛻 ∙ u) − 𝛻2u

𝛻 ∙ 𝛻φ1 × 𝛻φ2 = 0

a × b × c = (a ∙ c)a − (b ∙ a) c

a × b × c × d = a ∙ c ∙ b ∙ d − a ∙ d ∙ b ∙ c

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44

9.7 Time-Harmonic Fields

FIG 9.12 Representation of a phasor

z = x + jy = r∠φ.

φ

Im

Re

ω rad/sec

y

x

r

)60.9(x

ytan

)59.9(yxzr

)58.9()sinj(cosr e rz

)57.9(rjyxz

1

22

j

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45

sinjcose

e

CCe1Cez

Cjzln

jdz

dz

jz

)sinj(cosj

cosjsind

dz

sinjcosz

j

j

0jj

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46

)f91.9(rerjyxz

)e61.9(2/r

errez

)d61.9()(r

r

er

r

er

er

z

z

)c61.9()(rr

errererzz

)b61.9()yy(j )xx(zz

)a61.9()yy(j )xx(zz

j

1

2/jj

212

1

)(j

2

1j

2

j1

2

1

2121

)(j21

j2

j121

212121

212121

21

2

1

2121

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47

)66.9(]eIRe[)t(I

)t(Iinefactortimethe

droppingbyobtained:Phasor

)65.9(IeII

currentPhasor

)b64.9()tsin(r]reIm[

)a64.9()tcos(r]reRe[

)63.9(erere

)62.9(t

tjs

tj

0j

0s

j

j

tjjj

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48

)72.6(j

AdtA

)71.6(Ajt

A

)70.6(]eARe[j

]eARe[tt

A

)69.9(aeAA

)68.9()eaeRe(AAthen

a)xtcos(AAIf

tjs

tjs

yxj

os

tjy

xjo

y0

)67.9()eARe( A tjs

(𝐀𝐬 : Phasor form of 𝐀)

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Table 9.2 Time-Harmonic Maxwell’s Equations Assuming Time Factor eiωt

S ssL ssss

S ssLss

S ss

vsS svss

SdDjJLdHDjJH

SdBjLdEBjE

0SdB0B

dxdydzSdDD

formIntegralformintPo

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50

예제 9.5 다음 복소수를 계산하라.

a z1 =j(3 − 4j)∗

(−1 + 6j)(2 + j)2b z2 =

1 + j

4 − 8j

1/2

22

2

*

1

1427

)j25150

)j1427)(j1427(

)j1427)(j34(

j1427

j34

24j143

4j3

)j43)(j61(

)j43(j

)1j44)(j61(

)j43(j

)j2)(j61(

)j43(jz)a(

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51

예제 9.5 다음 복소수를 계산하라.

b z2 =1 + j

4 − 8j

1/2

)2.54(j

9477.0j

2/)11.14/(j

11.1j

4/j

2/1

2

o

e3976.0

e3976.0

e54

2

e54

e2

j84

j1z)b(

Re

Im

1

1

4/

4/je2j1

11.1j

o

e54j84

11.1

4.63

2

4/8tan

Re

Im

4

8

80

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52

예제 9.6 A = 10 cos 108t − 10x +π

3az 의 Phasor는?

Bs = 20/j ax + 10ej2πx/3ay 의 순시형 (instantaneous form)은?

z)3/x10(j

tjz

)3/x10t10(js

z)3/x10t(j

z)3/x10t10(j

z8

ae10

eae10A

ae10Re

ae10Re

a)3/x10t10cos(10A

8

8

yx

yx

y

tj3

x2j

x

tj2

j

tjy

3

x2j

x2

j

tjs

y3

x2j

x2

j

y3

x2j

x

y3

x2j

xs

at3

x2cos10atsin20

at3

x2cos10at

2cos20

ae10ae20Re

eae10ae20Re

]eBRe[B

ae10ae20

ae10aj20

ae10aj

20B

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53

예제 9.7 𝐄 =𝟓𝟎

𝛒𝐜𝐨𝐬 𝟏𝟎𝟔𝐭 + 𝛃𝐳 𝐚𝛗 𝐯𝐨𝐥𝐭/𝐦 𝐇 =

𝐇𝟎

𝛒𝐜𝐨𝐬 𝟏𝟎𝟔𝐭 + 𝛃𝐳 𝐚𝛒 𝐀/𝐦 일 때

Phasor Form으로 변형하고 Maxwell Eq을 만족 시키기 위한상수 H0, 𝛃를 구하라. Space Charge는 없다.

50H

ae50

jaejH

aejH

aeH

H

HjE

EjH

0H1

0H

0E1

0E

t

BE

t

EJH

0B

0E

aeH

HeHH

ae50

EeEE

0

zjzj0

zj0zj0s

ss

ss

ss

ss

zj0s

tjs

zjs

tjs

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54

ss

ss

ss

ss

zj0s

tjs

zjs

tjs

HjE

EjH

0H1

0H

0E1

0E

t

BE

t

EJH

0B

0E

aeH

HeHH

ae50

EeEE

50H

aeH

jae50

j

ae50

jae50

E

0

zj0zj

zjzjs

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55

1326.0H

1033.3

50H

aeH

jae50

j

ae50

jae50

E

0

3

0

zj0zj

zjzjs

50H

ae50

jaejH

aejH

aeH

H

0

zjzj0

zj0zj0s

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56

A

sinr

1)sinA(

sinr

1)Ar(

rr

1A)41.3(

z

AA1)A(

1A)40.3(

z

A

y

A

x

AA)39.3(

aV

sinr

1a

V

r

1a

r

VV)30.3(

az

Va

V1a

VV)29.3(

az

Va

y

Va

x

VV)28.3(

r2

2

z

zyx

r

z

zyx

AsinrrAAr

asinrara

sinr

1A)56.3(

AAAz

aaa

1A)55.3(

AAA

zyx

aaa

A)53.3(

r

r

2

z

z

zyx

zyx

참고

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57

예제 9.8 𝐄 = 𝟐𝟎𝐬𝐢𝐧 𝟏𝟎𝟖𝐭 − 𝛃𝐳 𝐚𝐲 𝐯𝐨𝐥𝐭/𝐦 이다.

𝛔 = 𝟎, 𝛍 = 𝛍𝟎, 𝛆 = 𝟒𝛆𝟎 일 때 𝛃,𝐇 를 계산하라.

m/Aa)zt10sin(

3

1

eHReH

aee3

1aee

20H

3/220

20aee

20E

aee20EeEE

ae20

az

H

j

1

j

HEEjH

aee20

az

E

j

1

j

EHHjE

aee20eEE

)ztsin(

)ztsin()2/sin()ztcos()2/cos()zt2/cos(

x8

tjs

xzj2/j

xzj2/j

s

2

2

yzj2/j

2

2

s

yzj2/j

stj

s

yzj

2

2

yxss

sss

xzj2/j

xyss

sss

yzj2/jtj

s