7. Magnetostatic Field

1

Matrix, Cross Product, Determinant

4 0 10

14

132

212

102

a)BABA(

a)BABA(

a)BABA(

BBB

AAA

aaa

BA

)B,B,B(B

)A,A,A(A

zxyyx

yxzzx

xyzzy

zyx

zyx

zyx

zyx

zyx

2

22

22

2

2

2

yx3

]x)yx[(xx

f

?x

f

x)z,y,x(f

xy2x

f

zyx)z,y,x(f

)8.2(

zz

siny

cosx

)7.2(

zz

x

ytan

yx

1

22

φ

z

x

y

z

ρ

az

aρ

aφ

dρ

ρdφ

dz

편미분 (Partial Derivative)

3

)Spherical(A

sinr

1)sinA(

sinr

1)Ar(

rr

1A

)lCylindrica( z

AA1)A(

1A

(Catesian) z

A

y

A

x

AA

Divergence

)Spherical( aV

sinr

1a

V

r

1a

r

VV

)lCylindrica( az

Va

V1a

VV

)(Cartesian az

Va

y

Va

x

VV

Gradient

r2

2

z

zyx

r

z

zyx

Del Operator

4

AsinrrAAr

asinrara

sinr

1A

AAAz

aaa

1A

AAA

zyx

aaa

A

Curl

r

r

2

z

z

zyx

zyx

zxyyx

yxzzx

xyzzy

zyx

zyx

zyx

a)BABA(

a)BABA(

a)BABA(

BBB

AAA

aaa

BA

5dzddxddv

add

adzd

adzdSd

adzadadLd

coordinate ylindricalC

dzdxdyxddv

adS

adydx

adxdz

adzdy Sd

adzadyadxLd

coordinateCartesian

3

z

z

3

n

z

y

x

zyx

적분자 (Integrator)

φ

z

x

y

z

ρ

az

aρ

aφ

dρ

ρdφ

dz

dL dzaz

dxax

dyay

6

ddrdsinrxddv

addr

addrsinr

addsinrSd

adsinrardadrLd

coordinate Spherical

23

r2

r

θ는 Zenith Angle이고φ는Azimuthal Angle.

φ

z

x

y

θ

rdφ

r

dr

r sinθdφ

rdθr sinθ

7

v3

vS

S nL

xdAdvASdA

theorem)(Gauss theoremDivergence

Sda)A(LdA

theoremsStoke'

선적분, 면적분

8

densitychargeVolume:

densityCurrent:J

)(ty Permittivi : m/F10854.8m/F36/10

)(tyPermeabili:m/H104

fieldElectric:E

densityfluxElectric:ED

fluxMagnetic:B

fieldMagnetic:/BH

EqAmperet

DJH

VE0EEqFaradayt

BE

EqGauss0B

EqGaussD

v

1290

70

v

유전율

투자율

Maxwell Equation

9

Ex 3.7 G r = 10e−z(ρaρ + az)일때 ρ=1, 0 ≤ z ≤1 인원통밖으로

나오는 G의 Flux를구하라.

0

2

1102)e1(102

2

12e10

dd10dzde10dde10

)0z(dd)a()aa(e10

)1(dzda)aa(e10

)1z(dda)aa(e10

dSaG

11

1

0

2

0

1

0

2

0z1

0

2

01

b1

0

2

0 zzz

s1

0

2

0 zz

t1

0

2

0 zzz

bstS n

zadd

adzd

adzdSd

coordlCylindrica

Ψt

Ψs

Ψb

z

yx

10

2014.1 학기 중간고사:

다음그림의 Closed Path에서 (zax + yay + yaz) ∙ dL 을구하라.

xy

z1

1

②

①

③

2

1

0)adzLd,0yx(

Ld)ayayaz(

02

11

2

1dz)z1(dy)y(

)adzadyLd,0x,y1z(

Ld)ayayaz(

2/1dy)y()adyLd,0zx(

Ld)ayayaz(

Ld)ayayaz(

z

3 zyx

1

0

0

1

zy

2 zyx

1

0y

1 zyx

L zyx

zyx adzadyadxLd

coordCartesian

dL dzaz

dxax

dyay

11

2013. 1 학기 기말고사:

J = (x, 0,0)일때중심이원점에있고반지름이 a인구의표면을통과하여흐르는전류의양을구하라.

3

3

3

3zyx

3

n

a3

4

xd

xdz

0

y

0

x

x

xdz

J

y

J

x

J

xdJ

dSaJ

a

dS

an=ar

12

2013. 1 학기 기말고사:

J = (x, 0,0)일때중심이원점에있고반지름이 a인구의표면을통과하여흐르는전류의양을구하라.

3

3

2

0

1

023

2

02

033

2

rx

n

a3

4

)(3

2a2

d2

)2cos(1dt)t1(a2

tcos

dcosdsina

)ddsinr)(cos)(sincossinr(

dSaax

dSaJ

)23.2(

asinacosa

acos

asincos

asinsina

asin

acoscos

acossina

)22.2(

cosrz

sinsinry

cossinrx

rz

ry

rx

addraddrsinraddsinrSd

coordinate Spherical

r2

13

tan 1 cot

sin cos

cscsec

2

22

secd

tand

sind

cosd

cosd

sind

tantan1

tantan)tan(

sincos)2cos(

sinsincoscos)cos(

cossin2)2sin(

sincoscossin)sin(

14ab

h1

b

1

a

1

ba

h1

zh

bab

ba

h1

dzzh

bab

1

zh

bab

dz1

dzz

S

LR

h/z)ba(b

h

0

1

h

0

2

h

0 2

h

0 22

2013. 1학기 기말고사:

전도율이 σ 인물질로만들어진다음그림의원뿔대의상하양단사이의저항R을구하라.

Δz ~ dzρ

a

b

hz

ρ

15JA/VeqPoisson

2/HBw2/EDwdensityEnergy

dt/LdIVdt/CdVI

LICVQ

SdBSdDFlux

r4

LIdA

r4

dLVPotential

)0J(VHVEPotential

HBEDField

]m/Wb[S/B]m/C[S/DdensityFlux

]m/A[L/IH]m/volt[L/VEensityintField

LIdvQdQelementSource

BvQFEQFForce

ILdHQSdD

R4

aLIdBda

r4

QQFlawBasic

2v

2

mE

L

m

22

encenc

2r0

r221

관계

자장전장

표 7.1 전기장과 자기장의 상사성

t

DJH

t

BE

0B

D v

Maxwell

16

7.2 BIOT-SAVART’S LAW

)4.7(R4

RLdI

R4

aLdIHd

)3.7(R4

sindLIdH

)2.7(R

sindLkIdH

)1.7(R

sindLIdH

32R

2

2

2

Fig 7.1 𝐈𝐝 𝐋에 의한 자기장 𝐝𝐇

𝐁~𝐇

𝐑

𝐝 𝐋 I

P

𝛂

17

Fig 7.2 Determining of dH using

(a) the right-hand rule, or

(b) the right=handed screw rule.

H (or I) is out

⊙

(a)

H (or I) is in

⊗

(b)

Fig 7.3 Conventional representation

of H (or I)

18

19

J

Ld

dvJ

LdAJ

LdAJLId

J

na

A

K

a

1

IdL = K ∙ adL

= Ka ∙ dL

= KdS

dL

an

20

𝐁

𝐑~𝐚𝐫

𝐝 𝐋 I

P

(7.8)current) (volume R4

advJH

(7.7)current) (surface R4

adSKH

(7.6)current) (line R4

aLdIH

)5.7( dvJdSKdLI

R4

RLdI

R4

aLdIHd)4.7(

V 2R

S 2R

L 2R

32R

21

Line current

z

xy

Ld

R a

z

x y

aza

a

z

B

O

α2

α1

RI

H

ρ

aρ

d L

aφP

)11.7( a]z[4

dzIH

)10.7(adz)aza(adzRLd

)9.7( R4

RLIdHd

2/322

zz

3

22

z

B

O

α2

α1

RI

H

ρ

aρd L = aLdL

aφP

Fig 7.5 직선선도체에의한 P 점의자기장

)12.7(a)cos(cos4

IH

a)cos(cos4

I

dsin4

aI

accs

dccs

4

I

dccsdz

cotz

)11.7(a]z[4

dzIH

12

12

33

22

2

2/322

2

1

2

1

dccsdz

sin

sin

coscossinsin

d

dz

sin

cos

tan

1z

cotz

2

2

2

23

dccs

tan

dsec

tan

tan

dseczdz

0d

dsecztandz)tanz(d

tanz

dccsdz

cotz

2

2

2

2

2tan 1 cot

sin cos

cscsec

2

22

secd

tand

sind

cosd

cosd

sind

tantan1

tantan)tan(

sincos)2cos(

sinsincoscos)cos(

cossin2)2sin(

sincoscossin)sin(

24

z

B

O

α2

α1

RI

H

ρ

aρd L = aLdL

aφP

Fig 7.5 직선 선도체에의한 P 점의 자기장

)15.7( aaa

0,

current line infinitefor )14.7(a

2

IH

0,2/

current line infinite-semifor )13.7(a

4

IH

a)cos(cos4

IH)12.7(

L

21

21

12

25

예제 7.1 그림 7.6 과 같이 도체 Loop에 10 A 의 전류가 흐를 때 (0,0,5)에서

(1) 번 도선에 의한 𝐇 를 구하라.

x

z5

2

2

1)5,0,0(

1

그림 7.6 삼각형도체 Loop

②

①

③

1

1 2x

z

y

10 A

(0,0,5)

(a)

m/mAa1.59

)a)(029/2(54

10H

)cos(cos4

IH

a)cos(cos4

IH)12.7(

29

2

25

2cos

0cos

2/

5

y

y

12

12

222

1

1

26

m/mAa65.28a2.38

)a5

3a

5

4)(01(

54

3H

)cos(cos4

IH)12.7(

yx

yx

12

yx a5

3a

5

4

예제 7.2 그림 7.7 에서 보여준 3 A 가 흐르는 도선전류에 의한

(3,4,0)에서의 𝐇 를 구하라.

1cos

0cos

5

2

1

(𝟏) 𝐇𝟏

27

m/mAa88.23

a)5

31(

44

3H

)cos(cos4

IH)12.7(

z

z

12

za

m/mA)88.23,65.28,2.38(

HHH 21

1cos

5/3cos

4

2

1

(𝟐) 𝐇𝟐

28

29

예제 7.3 그림 7.8 (a)에서 𝐚φ 방향으로 10 A 의 전류가 흐른다. (0,0,4)와

(0,0,-4)에서 𝐇 를 구하라. Coil의 반경은 3 이다.

Fig 7.8 (a) circular current loop,

(b) flux lines due to the current loop.

x

y

z

dHz

dHρ

dL aρ

ρ𝜑

R

P(0,0,h)

I

(a) circular current loop. (b) flux lines due to the current loop.

𝐇

𝐈

30

m/A)36.0,0,0()4,0,0(H

m/A)36.0,0,0()4,0,0(H

)h(2

aI

)h(4

a2I

)h(4

a2I

)h(4

d)asina(coshIH

)adahd()h(4

IHd

adahd

h0

0d0

aaa

RLd

aha

)0,y,x()h,0,0(R

adLd

R4

RLIdHd)9.7(

2/322z

2

2/322z

2

2/322z

22

0 2/322

yx

z2

2/322

z2

z

z

3

x

y

z

dHz

dHρ

dL aρ

ρ𝜑

R

P(0,0,h)

I

yx

z

asinacosa)10.2(

adzadadLd)5.3(

31

)10.2(

aa

acosasina

asinacosa

)9.2(

aa

acosasina

asinacosa

zz

yx

yx

zz

y

x

32

예제 7.4 길이 L, 반경 a, 권선수 N 인 solenoid에전류 I가흐른다.

다음을유도하라.

(a) 축상의임의의 P 점에서

H =nI

2(cosθ2 − cosθ1)az

where n=N/L

Fig 7.9 Cross section of solenoid.

θ2

θ

P

dz I

L

z

a θ1

33

z12

12

z

2/322

2

2/322

2

2/322

2

z

2/322

2

z

a)cos(cos2

nIH

)cos(cos2

nI

dsin2

nIH

dsin2

nI

)za(2

dznIa

)za(2

a)Indz(

)za(2

dIadH

L/Nn

3.7from)h(2

IH

2

1

예제

dsina

)za(dz

2

2/322

θ2

θ

P

dz I

L

z

a θ1

34

3.7from)h(2

IH

2/322

2

z 예제

x

y

z

dHz

dHρ

dL aρ

ρ𝜑

R

P(0,0,h)

I

35

dsina

)za(

dsina

)za(a

dsincsca

dcscadz

sin

1a

sin

coscossinsina

d

dz

sin

cosaz

z/atan

2

2/322

32/122

3

2

22tan 1 cot

sin cos

cscsec

2

22

secd

tand

sind

cosd

cosd

sind

tantan1

tantan)tan(

sincos)2cos(

sinsincoscos)cos(

cossin2)2sin(

sincoscossin)sin(

θ2

θ

P

dz I

L

z

a θ1

36

zz

zz

z22

z12

122

2

z

aL

NIanIH

aL

NIanI

a4/La2

InL

a)cos(cos2

nIH

cos)2/L(a

2/Lcos

P

aLforcenterataL

NIH)b(

중심에서가

θ2

θ

P

dz I

L

z

a θ1

37

7.3 AMPERE’S CIRCUIT LAW – MAXWELL’S EQUATION

)19.7(JH

)18.7(SdJI

)17.7(Sd)H(

LdHI

)16.7()lawcircuits'Ampere(ILdH

Senc

S

Lenc

enc

HI

Ld

38

7.4 APPLICATION OF AMPERE’S LAW

A. Infinite Line Current

Fig 7.10 Ampere’s law applied to

an infinite filamentary line current.

a2

IH

2

IH

I2H

ILdH)16.7(

a2

IH)14.7(

a)cos(cos4

IH)12.7(

)20.7(a2

IH

enc

12

39

B. Infinite Sheet of Current

)c21.7(bH2

]b[]H[

]a[]0[

]b[]H[

]a[]0[

LdHLdH

)b21.7(0zaH

0zaHH

)a21.7(bKILdH

o

0

0

1

4

4

3

3

2

2

1

x0

x0

yenc

K

H

H0ax H0ax

-H0ax -H0ax

zx

40

𝐝𝐇는 -𝐚𝐱 방향

𝐝 𝐋은 -𝐚𝐳 방향

𝐝𝐇는 𝐚𝐱 방향

𝐝 𝐋은 -𝐚𝐳 방향

41

)23.7(aK2

1H

)22.7(

0zaK2

1

0zaK2

1

H

)c21.7(bH2LdH

)b21.7(0zaH

0zaHH

)a21.7(bKILdH

n

xy

xy

o

x0

x0

yenc

42

C. Infinitely Long Coaxial Transmission Line

)26.7(ba2

IH

)25.7(a0a2

IH

)24.7(SdJILdH

2

L enc1

43

)25.7(a2

IH

H2LdH

aISdJ

a0

)24.7(SdJILdH

2

L

2

2

enc

L enc

1

1

)26.7(2

IH

H2LdH

a

aISdJ

ba

1L

2

2

enc

44

)27.7(bt2t

b1

2

IH

H2LdH

b)tb(

bIISdJ

tbb

2

22

L

22

22

encenc

1

)28.7(0H

tb

45

)29.7(

tb0

tbbabt2t

b1

2

I

baa2

I

a0aa2

I

H

2

22

2

Fig 7.13 Plot of Hφ against ρ.

Hφ

1

2πa

1

2πb

ρ0 a b b+t

46

예제 7.5 z=0, z=4 인 평면에 𝐊 = −𝟏𝟎𝐚𝐱 𝐀/𝐦, 𝐊 = 𝟏𝟎𝐚𝐱 𝐀/𝐦 인 전류가 흐른다.

다음 point에서 𝐇를구하라.(a) (1,1,1) (b) (0,-3,0)

an

z=4

(1,1,1)an

x y

z

z=0

Fig 7.14 Parallel infinite current sheets.

K

m/Aa10HHH

m/Aa5)a()a10(2

1aK

2

1H

m/Aa5a)a10(2

1aK

2

1H

aK2

1H)3.7(

)1,1,1()a(

y40

yzxn4

yzxn0

n

47

anz=4

(0,-3,10)

an

x y

z

z=0

K

m/Aa0HHH

m/Aa5a)a10(2

1aK

2

1H

m/Aa5a)a10(2

1aK

2

1H

aK2

1H)3.7(

)10,3,0()b(

y40

yzxn4

yzxn0

n

48

예제 7.6 toroid에서 권선수 N 이고 전류 I 가 흐른다.

toroid 안과 밖에서 𝐇를 구하라.

0H0SdJ

L

NI

)toroid(2

NI

2

NIH

NI2H

I2HLdH)a27.7(

S

00

enc

밖에서는

반지름중심

𝐋

𝐒

𝐈

49

7.5 MAGNETIC FLUX DENSTY – MAXWELL’S EQUATION

E )(

)Law'Gaussor Eq sMaxwell'4th (the

)34.7(0B

)33.7(0SdB

]Gauss000,10Tesla1

Teslam/Wb[B

)32.7(]Wb[SdB

space) free ofity (Permeabil:

)31.7(m/H104

)30.7(]Tesla[HB

2

S

70

0

50

N

S

I

MagneticFluxLine

Fig 7.16 magnetic flux line due to a

straight wire with current

coming out of the page

51

52

7.6 MAXWELL’S EQUATIONS FOR STATIC EM FIELD

Law sAmpere' SdJLdHJH

field ticelectrosta

of venessConservati 0LdE0E

monopole magnetic

of ceNonexisten 0SdB0B

Law sGauss' dxdydzSdDD

Remark formIntegralformPoint

SL

L

S

vSv

53

dSaJLdHdSa)H(

JH

EqAmperet

DJ

B

LdEdSa)E(

0E

EqFaradayt

BE

0dSaB

dV)0B(

EqGauss0B

dVdSaD

dV)E(

EqGaussE

nn

n

n

vn

v

v

54

7.7 MAGNETIC SCALAR AND VECTOR POTENTIALS

)a35.7(VE

0E0)V(

)b35.7(AB

0B0)A(

)36.7(VH

0J if0H0)V(

mm

)39.7(AB

EqAmperet

DJ

B

EqFaradayt

BE

EqGauss0B

EqGaussE v

55

)36.7(0J if0HVH m

)38.7(0V

0VH

m2

m2

EqAmperet

DJ

B

EqFaradayt

BE

EqGauss0B

EqGaussE v

56

(7.43)current for volumeR4

dvJA

(7.42)current surfacefor R4

dSKA

(7.41)current linefor R4

LdIA

)40.7(r4

dQV

v0

S0

L0

0

57

K

a

1

IdL = K ∙ adL

= Ka ∙ dL

= KdS

dL

an

J

Ld

dvJ

LdAJ

LdAJLId

J

na

A

58

유도

L0

R4

LdIA)41.7(

L0

L0

L0

L0

L 30

L 30

3

RR

LId

4A

ARR

LId

4

RR

LId

4

RR

1LId

4

RR

)RR(LId

4B

R

RLId

4B

R

RLdkIBd

RR

1f

LIdu

ufuf)uf(

)i(장뒷

)ii(장뒷

R

R

R R

R

R

vd 𝐁

𝐑~𝐚𝐫

𝐝 𝐋 I

P

59

3

2/3222

zyx

z2/32222/1222

z

y2/32222/1222

y

x2/3222

2/3222x

2/1222x

2/1222zyx

R

R

R

1

)zyx(

azayax

R

1

)az()zyx()zyx(z

a

)ay()zyx()zyx(y

a

)ax()zyx(

)x2()zyx(2

1a)zyx(

xa

)zyx(z

ay

ax

aR

1

)dzzd,dyyd,dxxd

)zz,yy,xx()z,y,x(

RRR

3

L0

L 30

RR

RR

RR

1

RR

1LId

4

RR

)RR(LId

4B

(i)

60

RR

LId

RR

1LId

RR

LId

4

RR

1LId

4B

L0

L0

RR

1)LId(

RR

LId

RR

1)LId(

)LId(RR

1

)LId(RR

1

)LId(RR

1LId

RR

1

RR

1f

LIdu

ufuf)uf(

(ii)

61

)51.7(LdA

SdA

SdB

L

S

S

62

Wb75.3

dzd2

1

SdB)32.7(

a2

AB

1Method

5

0z

2

1

S

뒷장

뒷장

a2

AAA

z

aaa

1A

z

z

예제 7.7 A = −(ρ2/4)az 일 때 φ=π/2, 1≤ρ≤2 m, 0≤z≤5 m 인면을 통과하는 자속 ψ를 구하라.

Fig 7.20 For example 7.7

3

2 4

11

2

3

4

5

ρ

z

2

L

B A

3

⊗

S

⊗

1

63

a2

a4

1

4/00

00

aaa

1

AAA

z

aaa

1A

2

2

z

z

z

64Wb75.3

dz4

0

dz4

0

LdA

SdA

SdB

2Method

0

5

2

2

5

0

1

2

4321

L

S

S

예제 7.7 A = −(ρ2/4)az 일 때 φ=π/2, 1≤ρ≤2 m, 0≤z≤5 m 인면을 통과하는 자속 ψ를 구하라.

3

2 4

11

2

3

4

5

ρ

z

2

L

A

Fig 7.20 For example 7.7

3

1

65

0zaK2

1

0zaK2

1

H)22.7(

xy

xy

예제 7.8 z=0 인면에 K = Kyay 의 면전류가 흐를 때 다음을 유도하라.

66

x0

2

0 2/322

y0

x2/3222

y0

2/3222

xy0

xy

222

yy0

0

S0

a)z(4

)dd(zK

a)zyx(4

)ydxd(zKB

)zyx(4

a)ydxd(zK

)adA(z

AdAdBd

zyx4

)ydxd(aK

R4

dSKAd

R4

dSKA)42.7(

xy

zyxy

y

zyx

a)dA(z

a)dA(x

a)dA(z

0dA0

zyx

aaa

Ad

67

x0

2

0 2/322

y0

x2/3222

y0

a)z(4

)dd(zK

a)zyx(4

)ydxd(zKB

68

0zfor2

aK

z

1

2

azK

0zfor2

aK

z

1

2

azK

z

10

2

azK

)zt(24

azK

dt)zt(4

azK

dtd2

ta

)zt(8

zdtK2

a)z(4

dzK2

a)z(4

)dd(zKB

xy0xy0

xy0xy0

2

xy0

02/12xy0

0t2/32xy0

2

x0t 2/32

y0

x0 2/322

y0

x0

2

0 2/322

y0

69

7.8 Biot-Savart 법칙과 Ampere 법칙유도

전선에 전류 I가 흐르면 Point P에 전선에 수직인 방향으로 자장이 발생한다.

30

3

x

xLId

4

x

xLdkIBd

Biot와 Savart가 정교하고 철저한 실험에 의해 전류가 흐르는 전선에 의해point P에서 발생하는 자장은 전선과point P의 거리의 제곱에 역 비례하고전선과 𝐱 에 수직인 방향으로 발생한다는 것을 알았다.

𝐁

𝐱

𝐝 𝐋 I

P

70

xdxx

)x(J

4

)x(JA

xx

1f

AfAf)Af(

xdxx

)x(J

4

xd)x(Jxx

1

4

xdxx

1)x(J

4

xd)x(J

LAd)x(JLIdxd

xx

)xx()x(J

4)x(B

x

xLId

4Bd

30

30

30

30

3

3

30

30

장뒷

x

x

x

x

71

3

2/3222

zyx

z2/32222/1222

z

y2/32222/1222

y

x2/3222

2/3222x

2/1222x

2/1222zyx

R

R

R

1

)zyx(

azayax

R

1

)az()zyx()zyx(z

a

)ay()zyx()zyx(y

a

)ax()zyx(

)x2()zyx(2

1a)zyx(

xa

)zyx(z

ay

ax

aR

1

)dzzd,dyyd,dxxd

)zz,yy,xx()z,y,x(

RRR

3

L0

L 30

RR

RR

RR

1

RR

1LId

4

RR

)RR(LId

4B

72

RR

LId

RR

1LId

RR

LId

4

RR

1LId

4B

L0

L0

RR

1)LId(

RR

LId

RR

1)LId(

)LId(RR

1

)LId(RR

1

)LId(RR

1LId

RR

1

RR

1f

LIdu

ufuf)uf(

73xd)xx()x(J

)xx(4xx

1xd)]xx(4[)x(J

4

xdxx

1)x(J

4

xdxx

)x(J

4Sd

xx

)x(J

4

xdxx

)x(J

4xd

xx

)x(J

4

xdxx

)x(J

4xd

xx

)x(J

4

A)A()A(xdxx

)x(J])([

4

xdxx

)x(J

4)x(B

xdxx

)x(J

4)x(Axd

xx

)x(J

4)x(B

30

230

320

3200

32030

32030

2320

30

3030

74

tJ

0Jt

E

Jt

EB

Jt

EB

?0J

JB

)(

)x(f)xx()x(f)x(J)x(B

xd)xx()x(J)x(B

)xx(4xx

1xd)]xx(4[)x(J

4)x(B

v

0

00

00

0

000

30

230

75

0r

1

0r)1(

r

1

rr

rr

1

r

1

r

1

xx

1

)xx(4xx

1:

2

2

2

2

22

2

경우인

유도

76

0 2/522

22

0a

230

2

2/522

2

0a

2

2

23

2/522

2

0a

3

22

2

0a

3

22

2

0a

32

)ar(

drra12

drr4xd)drr4()ar(

1a3

rr

rr

1xd

)ar(

1a3

xdar

1

xdar

1

xdr

1

0r)2(

lim

lim

lim

lim

lim

경우인

77

4

dcosdt

sintdtt12

dcossin12

sec

dtan12

dsecadr

tanar

)1(tan

dsectan12

)ar(

drra12

xdr

1

1

02

2/

02

2/

0 3

2

2

2/

0 2/52

22

0 2/522

22

0a

32

lim

tan 1 cot

sin cos

cscsec

2

22

secd

tand

sind

cosd

cosd

sind

tantan1

tantan)tan(

sincos)2cos(

sinsincoscos)cos(

cossin2)2sin(

sincoscossin)sin(

78

δ(x-x0)

x0 x0+Δx

x

1/Δx

Δx

)x(fdx)xx()x(f 00

0r

1

0r)1(

2

경우인

4xdr

1

0r)2(

32

경우인

)xx(4xx

12

79

)x(fdx)xx()x(f

)x(f

xx

)x(f

dxx

1)x(flim

dx0)x(fdxx

1)x(fdx0)x(flim

dx)xx()x(fdx)xx()x(fdx)xx()x(f

dx)xx()x(f

)x(fdx)xx()x(f

00

0

0

xx

x0x

xx 0xx

x

x

0x

xx 0xx

x 0x

0

0

00

0

0

0

0

0

0

0

0

0

0

유도

δ(x-x0)

x0 x0+Δx

x

1/Δx

Δx

80

0t/where

)60.7(JA

Jt

EA

1

A

A)A(

AB

AB)39.7(0B

Jt

EB

02

02

0

2

2

00

)59.7(0)x(A

xx

Sd)x(J

4

xdxx

)x(J

4

xdxx

)x(J)(

4

xdxx

)x(J

4

xdxx

)x(J

4)x(A

xdxx

)x(J

4)x(A

0

30

30

30

30

30

(7.59) & (7.60) 유도