Ly Thuyet Truong Dien Tu - DHCN

8/6/2019 Ly Thuyet Truong Dien Tu - DHCN

1/108


2/108

1

LI NI U

K t khi Hertz bng thc nghim chng t nng lng in c th bc

x trong khng gian v s tn ti ca trng in t mu k nguyn ng

dng sng in t trong thng tin lin lc, truyn s liu, gii tra phng tin,

iu khin t xa ... H thng thng tin v tuyn ny ngy cng tr nn quan

trng v thit yu trong x hi hin i. Do vic hiu bit bn cht ca sng

in t, tnh cht lan truyn ca trng in t cng nh cc ng dng ca n l

rt cn thit. tch lu phn kin thc ny ngi hc cn phi c kin thc nn

tng v gii tch vector, php tnh tensor, phng trnh vi phn v o hm

ring, gii tch hm mt bin v hm nhiu bin trong Ton hc cao cp; quang

hc sng v in hc trong Vt l i cng.

Gio trnh L thuyt trng in tc bin son trong khun kh ca

chng trnh hon thin b sch gio trnh dng ging dy v hc tp ca

Khoa Cng nghin t, Trng i hc Cng nghip TP H Ch Minh, bao

gm cc ni dung c trnh by trong 5 chng nh sau:

Chng 0 Mt scng thc ton hc

Chng 1 Cc nh lut v nguyn l cbn ca trng in t

Chng 2 Tch phn cc phng trnh Maxwell

Chng 3 Sng in tphng

Chng 4Nhiu x sng in t

Do thi gian v ti liu tham kho cn nhiu hn ch, cho nn chc chn

gio trnh cn nhiu thiu st. Rt mong c sng gp, ph bnh ca bn c

gio trnh c hon thin hn.

Tc gi

V Xun n


3/108

2

MC LC

TrangLi ni u 1

Chng 0 Mt s cng thc ton hc 3

Chng 1 Cc nh lut v nguyn l cbn ca trng in t 8

Chng 2 Tch phn cc phng trnh Maxwell 32

Chng 3 Sng in t phng 60

Chng 4 Nhiu x sng in t 90

Ti liu tham kho 107


4/108


5/108

4

+

+

=

==

y

a

x

ak

x

a

z

aj

z

a

y

ai

aaazyx

kji

aarot xyzxyz

zyx

rrr

rrr

rr

S phc

Hm m

( )ysiniycoseee xiyxz +== +

Hm m l mt hm tun hon c chu k l 2i. Thc vy, ta c

1k2sinik2cose ik2 =+=

Suy razik2zik2z ee.ee == +

Cng thc Euler

eiy = cosy +isiny

Khi s phc z = r ei = r(cos +isin)

Phng trnh vi phn tuyn tnh cp hai

Phng trnh vi phn t trng cp hai l phng trnh bc nht i vi

hm cha bit v cc o hm ca n:

)x(fyayay 21 =++ (1)

Trong :

a1, a2 v f(x) l cc hm ca bin c lp x

f(x) = 0 (1) gi l phng trnh tuyn tnh thun nht

f(x) 0 (1) gi l phng trnh tuyn tnh khng thun nht

a1, a2 const (1) gi l phng trnh tuyn tnh c h s khng i

Phng trnh vi phn tuyn tnh cp hai thun nht

Phng trnh vi phn t trng cp hai thun nht c dng:

0yayay 21 =++ (2)

a1, a2 l cc hm ca bin x


6/108

5

nh l 1. Nu y1 = y1(x) v y2 = y2(x) l 2 nghim ca (2) th y = C1y1 + C2y2

(trong C1, C2 l 2 hng s tu ) cng l nghim ca phng trnh y.

Hai hm y1(x) v y2(x) l c lp tuyn tnh khi( )

( )

const

xy

xy

2

1 , ngc li l ph

thuc tuyn tnh

nh l 2. Nu y1(x) v y2(x) l 2 nghim c lp tuyn tnh ca phng trnh vi

phn t trng cp hai thun nht (2) th y = C1y1 + C2y2 (trong C1, C2 l 2

hng s tu ) l nghim tng qut ca phng trnh y.

nh l 3. Nu bit mt nghim ring y1(x) ca phng trnh vi phn t

trng cp hai thun nht (2) th c th tm c mt nghim ring y2(x) caphng trnh , c lp tuyn tnh vi y1(x) bng cch t y2(x) = y1(x).u(x)

Phng trnh vi phn tuyn tnh cp hai khng thun nht

Phng trnh vi phn t trng cp hai l phng trnh bc nht i vi

hm cha bit v cc o hm ca n:

)x(fyayay 21 =++ (3)

Trong :a1 v a2 l cc hm ca bin c lp x; f(x) 0

nh l 1. Nghim tng qut ca phng trnh khng thun nht (3) bng

nghim tng qut ca phng trnh thun nht (2) tng ng v mt nghim

ring no ca phng trnh khng thun nht (3).

nh l 2. Cho phng trnh khng thun nht

)x(f)x(fyayay 2121 +=++ (4)

Nu y1(x) l nghim ring ca phng trnh

)x(fyayay 121 =++ (5)

v y2(x) l nghim ring ca phng trnh

)x(fyayay 221 =++ (6)

th y(x) = y1(x) + y2(x) cng l nghim ring ca phng trnh (4)

Phng trnh vi phn tuyn tnh cp hai c h s khng i


7/108

6

Phng trnh vi phn t trng cp hai thun nht c dng:

0qyypy =++ (7)

p, q l cc hng s

Gi s nghim ring ca (7) c dngkxey = (8)

Trong : k l hng s sc xc nh

Suy rakxkey = , kx2eky = (9)

Thay (8) v (9) vo (7) ta c

( ) 0qpkke 2kx =++ (10)

V ekx 0 nn

0qpkk 2 =++ (11)

Nu k tho mn (11) th y = ekx l mt nghim ring ca phng trnh vi

phn (7). Phng trnh (11) gi lphng trnh c trng ca phng trnh vi

phn (7)

Nhn xt: Phng trnh c trng (7) l phng trnh bc 2 c 2 nghim k1

v k2 nh sau

- k1 v k2 l 2 s thc khc nhau, khi 2 nghim ring ca phng trnh

vi phn (7) lxk

11ey = , xk2 2ey = (12)

Hai nghim ring (12) l c lp t trng v

( ) consteyy xkk

2

1 21 = (13)

Do nghim tng qut ca phng trnh vi phn (7) lxk

2xk

12121 eCeCyyy +=+= (14)

- k1 v k2 l 2 s thc trng nhau: k1 = k2

Hai nghim ring c lp t trng: xk1 1ey = ,xk

21xey =


8/108

7

Nghim tng qut ca phng trnh vi phn (7) l

( ) xk21xk

2xk

1111 exCCxeCeCy +=+= (15)

- k1 v k2 l 2 s phc lin hp: k1 = + i v k2 = - i

Hai nghim ring ca phng trnh vi phn (7) l

( )

( ) xixxi2

xixxi1

eeey

eeey

+

==

== (16)

Theo cng thc Euler ta c

xsinixcose

xsinixcosexi

xi

=

+=

(17)

Suy ra

( )

( )xsinixcoseeey

xsinixcoseeey

xxix2

xxix1

==

+==

(18)

Nu

1y v

2y l 2 nghim ca phng trnh vi phn (7) th cc hm

xsinei2yy

y

xcose2yyy

x212

x211

=+

=

=+=

(19)

cng l nghim ca phng trnh vi phn (7) v c lp t trng v

constxtgyy

2

1 = (20)

Do nghim tng qut ca phng trnh vi phn (7) l( )xsinCxcosCexsineCxcoseCy 21

xx2

x1 +=+=

(21)


9/108

8

Chng 1

CC NH LUT

V NGUYN L CBN CA TRNG IN T

1.1. Cc i lng c trng cho trng in t

1.1.1. Vector cng in trng

in trng c c trng bi lc tc dng ln in tch t trong in

trng

EqFrr

= (1.1)

Hay:

q

FE

r

r

= (1.2)

Ct Er

ti mt im bt k trong in trng l i lng vector c tr s

bng lc tc dng ln mt n vin tch im dng t ti im

Lc tc dng gia 2 t im Q v q

20

0 rr

4QqF

r

r

= (1.3)

- m/F10.854,8 120= - hng sin

- - in thm tng i

- 0rr

- vector n v ch phng

Ht im n21 q,...,q,q

==

==n

1i2

i

i0i

0

n

1ii

rrq

41EE

r

rr

(1.4)

i0rr

- cc vector n v ch phng

Trong thc t h thng l dy mnh, mt phng hay khi hnh hc, do :

=

l2l

0

l r

rdl

4

1E

r

r

(1.5)


10/108

9

=

S2S

0

S r

rdS

4

1E

r

r

(1.6)

=

V

2V

0

V

r

rdV

4

1E

r

r

(1.7)

1.1.2. Vector in cm

n gin khi tnh ton i vi cc mi trng khc nhau, ngi ta s

dng vector in cm Dr

ED 0rr

= (1.8)

1.1.3. Vector tcm T trng c c trng bi tc dng lc ca t trng ln in tch chuyn

ng hay dng in theo nh lut Lorentz

BvqFr

r

r

= (1.9)

T trng do phn t dng in lIdr

to ra c xc nh bi nh lut thc

nghim BVL

( )rlIdr4

Bd2

0 rrr

= (1.10)

- m/H10.257,110.4 670 == - hng s t

- - t thm tng i

T trng ca dy dn c chiu di l

=

l

20

r

rlId

4

Br

r

r

(1.11)

1.1.4. Vector cng ttrng

n gin khi tnh ton i vi cc mi trng khc nhau, ngi ta s

dng vector cng t trng Hr


11/108

10

0

BH

=

r

r

(1.12)

1.2. nh lut Ohm v nh lut bo ton in tch

1.2.1. nh lut Ohm dng vi phn Cng dng in I chy qua mt S t vung gc vi n bng lng in

tch q chuyn qua mt S trong mt n v thi gian

dt

dqI =

(1.13)

Du tr ch dng in I c xem l dng khi q gim

m ty s chuyn ng ca cc ht mang in trong mi trng dnin, ngi ta a ra khi nim mt dng in

EvvenJ 0r

rr

r

=== (1.14)

dng vi phn ca nh lut Ohm

- n0 - mt ht in c in tch e

- - mt in khi

- vr

- vn tc dch chuyn ca cc ht in- - in dn sut

Dng in qua mt S c tnh theo

===SSS

SdESdJdIIrrrr

(1.15)

Mt vt dn dng khi lp phng cnh L, 2 mt i din ni vi ngun p

U, ta c

(lu : p dng c/t S = L2 vLS

LR

== )

R

ULU)EL)(L(ESEdSI

S

===== (1.16)

dng thng thng ca nh lut Ohm

V Er

v Sdr

cng chiu, t


12/108

11

RL

1=

(1.17)

- in dn sut c n v l 1/m

1.2.2. nh lut bo ton in tch in tch c th phn b lin tc hay gin on, khng t sinh ra v cng

khng t mt i, dch chuyn t vng ny sang vng khc v to nn dng

in.

Lng in tch i ra khi mt kn S bao quanh th tch V bng lng in

tch gim i t th tch V .

Gi s trong th tch V c bao quanh bi mt S, ta c

=V

dVQ (1.18)

sau thi gian dt lng in tch trong V gim i dQ

==V

dVdt

d

dt

dQI (1.19)

Mt khc

=S

SdJIr

r

(1.20)

Suy ra

=VS

dVt

SdJrr

(1.21)

Theo nh l OG

( )

==VVS

dVt

dVJ.SdJvrr

(1.22)

Suy ra

0t

J. =

+

v

(1.23)

y l dng vi phn ca nh lut bo ton in tch hayphng trnh lin

tc.

1.3. Cc c trng cbn ca mi trng


13/108

12

Cc c trng cbn ca mi trng: , ,

Cc phng trnh:

ED 0rr

= (1.24)

=

0

BH

r

r

(1.25)

gi l cc phng trnh vt cht

, , cng trng : mi trng tuyn tnh

, , const : mi trng ng nht v ng hng

, , theo cc hng khc nhau c gi tr khng i khc nhau: mi trng

khng ng hng. Khi , biu din bng cc tensor c dng nh bng

s. Chng hn ferrite b t ho hoc plasma b t ho l cc mi trng

khng ng hng khi truyn sng in t

, , v tr : mi trng khng ng nht

Trong t nhin a s cc cht c > 1 v l mi trng tuyn tnh.

Xecnhec c >> 1 : mi trng phi tuyn

> 1 : cht thun t : cc kim loi kim, Al, NO, Phng trnh, O, N,

khng kh, ebonic, cc nguyn tt him

< 1 : cht nghch t : cc kh him, cc ion nh Na+, Cl- c cc lp

electron ging nh kh him, v cc cht khc nh Pb, Zn, Si, Ge, S, CO2, H2O,

thu tinh, a s cc hp cht hu c

>> 1 : cht st t : mi trng phi tuyn : Fe, Ni, Co, Gd, hp kim cc

nguyn t st t hoc khng st t Fe-Ni, Fe-Ni-Al. t ho ca cht st t

ln hn t ho ca cht nghch t v thun t hng trm triu ln.

Cn c vo dn in ring : cht dn in, cht bn dn v cht cch

in hay in mi

Cht dn in: > 104 1/m, = : cht dn in l tng

Cht bn dn: 10-10 < < 104


14/108


15/108

14

(c giao tuyn l AB). Php tuyn ngoi ca S v S' s c chiu ngc nhau.

Do tch phn trn S v S' c cng gi tr nhng tri du. Khi thng

lng ca Dr

qua ton mt kn S bng 0.

Xt hin tch im q1, q2, ..., qnt trong mt kn S, ta c

=

=n

1iiDD

rr

(1.29)

Thng lng ca Dr

do h q1, q2, ..., qn gy ra qua ton mt kn S

QqSdDSdDn

1ii

n

1i Si

S

==== ==

rrrr

(1.30)

Vy: Thng lng ca vector in cm Dr

qua mt kn S bt k bng tng

i s cc in tch nm trong th tch V c bao quanh bi S

Lu : V Q l tng i s cc in tch q1, q2, ..., qn, do c th m

hoc dng

Nu trong th tch V c bao quanh bi S c mt in khi th c

tnh theo

QdVSdDVS

E === rr

(1.31)

Cc cng thc (1.30) v (1.31) l dng ton hc ca nh l Ostrogradski-

Gauss i vi in trng.

Nguyn l lin tc ca t thng

Thc nghim chng tng sc t l khp kn d ngun to ra n l

dng in hay nam chm. Tm biu thc ton hc biu din cho tnh cht ny

Dr

Sdr

A

B

q


16/108

15

Gi s c mt kn S tu nm trong t trng vi vector t cm Br

. Thng

lng ca Br

qua mt kn S bng tng s cc ng sc ti qua mt S ny.

Do ng sc t khp kn nn sng sc ti vo th tch V bng s

ng sc ti ra khi th tch V . V vy thng lng ca Br

c tnh

theo

0SdBS

M == rr

(1.32)

Cng thc (1.32) gi l nguyn l lin tc ca t thng. y l mt phng

trnh cbn ca trng in t

1.5. Lun im thnht - Phng trnh Maxwell-FaradayKhi t vng dy kn trong mt t trng bin thin th trong vng dy ny

xh dng in cm ng. Chng t trong vng dy c mt in trng Er

c chiu

l chiu ca dng in cm ng .

Th nghim vi cc vng dy lm bng cc cht khc nhau, trong iu kin

nhit khc nhau u c kt qu tng t. Chng t vng dy dn khng phi

l nguyn nhn gy ra in trng m ch l phng tin gip ch ra s c mtca in trng . in trng ny cng khng phi l in trng tnh v

ng sc ca in trng tnh l ng cong h. in trng tnh khng lm

cho ht in dch chuyn theo ng cong kn to thnh dng in c (v

ho ra trong in trng tnh khng cn tn cng m vn sinh ra nng lng

in !).

Mun cho cc ht in dch chuyn theo ng cong kn to thnh dng

in th cng phi khc 0, c ngha l

0ldEql

rr

(1.33)

v .sc ca in trng ny phi l cc .cong kn v gi l in trng xoy.

Pht biu lun im I: Bt k mt t trng no bin i theo thi gian

cng to ra mt in trng xoy.


17/108


18/108

17

=

Sl

Sdt

BldE

r

r

rr

(1.38)

y l phng trnh Maxwell-Faraday di dng tch phn, cng l mt

phng trnh cbn ca trng in t.

Vy: Lu s ca vector cng in trng xoy dc theo mt ng

cong kn bt k bng v gi tr tuyt i nhng tri du vi tc bin thin theo

thi gian ca t thng gi qua din tch gii hn bi ng cong kn .

Theo gii tch vector (cng thc Green-Stock)

( ) =Sl

SdEldErrrr

(1.39)

Theo cc phng trnh (1.38) v (1.39)

t

BE

=

r

r

(1.40)

y l phng trnh Maxwell-Faraday di dng vi phn, c th p dng

i vi tng im mt trong khng gian c t trng bin thin.

1.6. Lun im thhai - Phng trnh Maxwell-Ampere

Theo lun im I, t trng bin thin theo thi gian sinh ra in trng

xoy. Vy ngc li in trng bin thin c sinh ra t trng khng ?

m bo tnh i xng trong mi lin h gia in trng v t trng, Maxwell

a ra lun im II:

Bt k mt in trng no bin thin theo thi gian cng to ra mt t

trng.

( chng minh bng thc nghim)Lu : in trng ni chung c th khng p.b ng u trong khng

gian, c ngha l thay i tim ny sang im khc, nhng theo lun im II

sbin thin ca in trng theo khng gian khng to ra ttrng, chc

sbin thin ca in trng theo thi gian mi to ra ttrng.

Thit lp phng trnh Maxwell-Ampere:


19/108


20/108

19

t

PJdP

=

v

r

- mt dng in p.cc trong in mi do s x dch ca cc

in tch

tEJ 00d

=

r

r

- in trng bin thin trong chn khng v gi l mt dng

in dch

chng minh s tn ti ca dng in dch, xt th d sau: c mt mt

kn S bao quanh 1 trong 2 bn ca tin. Do c in p xoay chiu t vo t

in nn gia 2 bn t c in trng bin thin Er

v dng in bin thin chy

qua t. Dng in ny chnh l dng in dch trong chn khng v gia 2 bn

t khng tn ti in tch chuyn ng v c gi tr:

t

ESI 00d

=

r

(1.44)

Theo nh lut Gauss

SESdEq 0S

0==

rr

(1.45)

SSdS

=

r

v in trng ch tn ti gia 2 bn t

i vi mi trng chn khng, ta c: = 1

Dng in dn chy trong dy dn ni vi t c gi tr bng

S

S'+q

-q

Er

~


21/108

20

t

ESSdE

dt

d

dt

dqI 0

S0

===

r

rr

(1.46)

Suy ra

I = Id0 (1.47)Vy: dng in dch chy gia 2 bn t bng dng in dn chy mch

ngoi tin.

Bng cch b sung dng in dch vo v phi ca phng trnh (1.42), ta

c

(b sung c v v kha cnh to ra t trng dng in dch tng

ng dng in dn)

+=SSl

Sdt

DSdJldH

r

r

rrrv

(1.48)

Hay

+=

Sl

SdtD

JldHr

r

rrv

(1.49)

y l phng trnh Maxwell-Ampere di dng tch phnTheo gii tch vector (cng thc Green-Stock)

( ) =Sl

SdHldHrrrv

(1.50)

Suy ra

dJJtD

JHrr

r

rr

+=

+=

(1.51)

y l phng trnh Maxwell-Ampere di dng vi phn, cng l mtphng trnh cbn ca trng in t

Nu mi trng c in dn sut = 0 (in mi l tng v chn khng)

th do 0EJ ==rr

, ta c:

0d0 Jt

EH

r

r

r

=

=

(1.52)


22/108

21

Vy: dng in dch hay in trng bin thin theo thi gian cng to ra

t trng nh dng in dn.

1.7. Trng in tv h phng trnh Maxwell

Theo cc lun im ca Maxwell, t trng bin thin theo thi gian to ra

in trng xoy, v ngc li in trng bin thin theo thi gian to ra t

trng. Vy trong khng gian in trng v t trng c thng thi tn ti

v c lin h cht ch vi nhau

in trng v t trng ng thi tn ti trong khng gian to thnh mt

trng thng nht gi l trng in t.

Trng in t l mt dng vt cht c trng cho s tng tc gia ccht mang in.

- Phng trnh Maxwell-Faraday

Dng tch phn

=

Sl

Sdt

BldE

r

r

rr

(1.53)

Dng vi phn

t

BE

=

r

r

(1.54)

Din t lun im thnht ca Maxwell vmi lin h gia ttrng bin

thin v in trng xoy.

- Phng trnh Maxwell-Ampere

Dng tch phn

+=

Sl

Sdt

DJldH

r

r

rrv

(1.55)

Dng vi phn

t

DJH

+=

r

rr

(1.56)


23/108

22

Din t lun im thhai ca Maxwell: in trng bin thin cng sinh

ra ttrng nhdng in dn.

- nh l OG i vi in trng

Dng tch phn

qSdDS

=rr

(1.57)

Theo gii tch vector: =VS

dVD.SdDrrr

v =V

dVq , ta c

Dng vi phn

= D.r

(1.58)

Din t tnh khng khp kn ca cc ng sc in trng tnh lun t

cc in tch dng i ra v i vo cc in tch m: trng c ngun

- nh l OG i vi ttrng

Dng tch phn

0SdBS

=rr

(1.59)

Dng vi phn0B. =

r

(1.60)

Din t tnh khp kn ca cc ng sc ttrng: trng khng c ngun

Cc phng trnh (1.54), (1.56), (1.58), (1.60) gi l h phng trnh

Maxwell

t

BE

=

r

r

t

DJH

+=

r

rr

(1.61)

= D.r

0B. =r

- H phng trnh Maxwell vi ngun ngoi


24/108

23

Trong l thuyt anten bc xin t pht ra t ngun v i vo khng gian.

Dng in trong anten l ngun bc xin t. Ngun dng in ny c lp

vi mi trng v khng chu nh hng ca trng do n to ra, gi l ngun

ngoi. Cc ngun ngoi c bn cht in hoc khng in. c trng cho

ngun ngoi ca trng in t ta c khi nim mt dng in ngoi OJr

.

.lut Ohm dng vi phn:

( )OO EEJJrrrr

+=+ (1.62)

Nhn xt: h phng trnh Maxwell (1.61) ch m t trng in t ti

nhng im trong khng gian khng tn ti ngun ngoi ca trng haytrng

in ttdo. Khi c ngun ngoi h phng trnh Maxwell c vit li

t

BE

=

r

r

t

DJJH O

++=

r

rrr

(1.63)

= D.r

0B. =

r

Trong mi trng ng nht v ng hng c , v , tc l

mi trng in mi: ED 0rr

=

mi trng dn in: EJrr

=

mi trng t ho: HB 0rr

= , ta c

t

HE 0

=

r

r

t

EJEH 0O

++=

r

rrr

(1.64)

0

E.

=

r

0H. =r

- Nguyn li ln ca h phng trnh Maxwell


25/108

24

Xt trng hp mi trng ng nht v ng hng, khng dng in

dn, khng in tch t do v ngun ngoi 0JJ O ===rr

t

H

E 0

=

r

r

t

EH 0

=

r

r

(1.65)

0E. =r

0H. =r

Nhn xt: Er

v Hr

i xng v c thi ln cho nhau

h phng trnh Maxwell trong trng hp c ngun ngoi vn ixng, cn phi a thm 2 i lng hnh thc

MJr

- mt dng t ngoi

M - mt t khi

Trong mi trng ng nht v ng hng, khng dng in dn, khng

in tch t do, vi ngun in v t ngoi

t

HJE 0M

=

r

rr

t

EJH 0E

+=

r

rr

,JEJO(1.66)

0

E.

=

r

0

MH.

=

r

ng dng: nu kt qu bi ton cho mt ngun in (ngun t) bit, th

s dng nguyn l i ln xc nh kt qu bi ton cho mt ngun t (ngun

in), m khng cn phi gii c hai.

- H phng trnh Maxwell i vi trng in tiu ho


26/108


27/108

26

Khi S = 0 ta c: D1n = D2n hay1

2

n2

n1

E

E

=

- i vi thnh phn tip tuyn ca in trng

E1 = E2,1

2

2

1

DD

=

(1.71)

- i vi thnh phn php tuyn ca t trng

B1n = B2n,1

2

n2

n1

H

H

= (1.72)

- i vi thnh phn tip tuyn ca t trng

H1 - H2 = IS

ISdng in mt

Khi IS = 0 ta c: H1 = H2 hay1

2

2

1

B

B

=

(1.73)

- Trng hp c bit mi trng 1 l in mi v mi trng 2 l vt dn

l tng c 2 = . Trong vt dn l tng trng in t khng tn ti, c ngha

l 0HE 22 ==rr

.

Thc vy, nu vt dn l tng tn ti trng in t 0H;E 22 rr

th di tc

dng ca trng cc in tch t do s phn b li in tch trn b mt ca n

cho n khi trng ph do chng to ra trit tiu vi trng ban u v kt qu

trng tng hp trong vt dn l tng bng 0. Trn b mt S ca vt dn l

tng c dng in mt v in tch mt tn ti trong mt lp mng v hn.

Khi ta c

E1n =1

S

E1 = 0

H1n = 0

H1 = IS

(1.74)


28/108

27

Vy: trng in t trong in mi st mt vt dn l tng ch c thnh

phn php tuyn ca Er

v thnh phn tip tuyn ca Hr

1.9. Nng lng trng in t- nh l Umov Poynting

- Nng lng ca trng in t

W = WE + WM = ( ) +V

ME dV =

+

V

20

20 dV

2

H

2

E

- nh l Umov Poynting

chng minh c

Ot

S

PP

dt

dWSd =rr

(1.75)

Trong

HErrr

= (W/m2) vector Poynting

Phng trnh = =V

2

V

dVEdVEJrrr

cng sut tiu hao nhit do dng in dn

Jr

gy ra trong V

PO = V

E dVEJrr

cng sut ca ngun ngoi trong th tch V

(1.75) gi l nh l Umov Poynting m t s cn bng ca trng in t

trong th tch V

Pht biu: Tng cc bin i nng lng trng in t, cng sut tn

hao nhit v cng sut ngun ngoi trong th tch V bng thng lng ca

vector Poynting qua mt kn S bao th tch V .

Vector Poynting r

biu th s dch chuyn nng lng ca trng in t.1.10. nh l nghim duy nht

H phng trnh Maxwell c nghim duy nht khi trng in t tho mn

cc iu kin sau


29/108

28

1. Bit cc vector cin trng v t trng ti thi im t0 = 0 ti bt

k im no trong vng khng gian kho st hay cn gi l iu kin ban u,

tc l

( )0,z,y,xEE 0rr

= khi t = 0

( )0,z,y,xHH0rr

= (1.76)

2. Bit thnh phn tip tuyn ca Er

v thnh phn tip tuyn ca Hr

ti mt

gii hn S bao min khng gian kho st trong khong thi gian 0 < t < hay

cn gi l iu kin bin

E = E|S hoc H = H|S vi 0 < t < (1.77)

Nhn xt: nh l nghim duy nht c ngha quan trng v bng cch no

ta nhn c nghim ca h phng trnh Maxwell v nu n tho mn cc

iu kin trn th nghim nhn c l duy nht.

1.11. Nguyn l tng h

Nguyn l tng h phn nh mi quan h tng h gia trng in t v

cc ngun to ra n ti hai im khc nhau trong khng gian.

1. BLorentz

Dng vi phn

=

m1m2Mm2m1M

m1m2Em2m1Em1m2m2m1

HJHJ

EJEJHE.HE.

rrrr

rrrrrrrr

(1.78)

Dng tch phn

=

=

V

m1m2Mm2m1Mm1m2Em2m1E

S

m1m2m2m1

dVHJHJEJEJ

dSHEHE

rrrrrrrr

rrrr

(1.79)

V , ta c


30/108

29

0dVHJHJEJEJV

m1m2Mm2m1Mm1m2Em2m1E =

rrrrrrrr

(1.80)

2. Nguyn l tng h

Gi s trong mi trng ng nht v ng hng, ngun in v t 1 phnb trong V1, ngun in v t 2 phn b trong V2 v 2 th tch ny khng c

min chung. Do v tri ca phng trnh (1.80) tch phn trong min V

chia thnh 3 min V1, V2 v min cn li. Tuy nhin tch phn trong min cn

li bng 0 v min ny khng tn ti ngun cho nn phng trnh (1.80) c

vit li

=

2V

m1m2Mm1m2E

1V

m2m1Mm2m1E dVHJEJdVHJEJrrrrrrrr

(1.81)

gi l nguyn l tng h ca trng in t v ngun ca chng 2 min khc

nhau.

1.12. Nguyn lng dng in ng

Nguyn lng dng in ng hay cn gi l nguyn l mu ho xc nh

mi quan h gia trng in t. Cc tham sin v hnh hc ca hin t vmi trng i vi 2 hin tng dng in ng vi nhau.

Tham s ho cc i lng ca trng in t

665544M33E2211 at;al;aJ;aJ;aE;aH ======r

r

r

r

r

r

r

r

(1.82)

4321 a;a;a;arrrr

l cc vector n v khng c th nguyn ch s ph thuc ca

cng trng v ngun vo cc to khng gian v thi gian

65 a;a l cc n v v hng xc nh to khng gian v thi gian

Cc h s t li c th nguyn tng ng l

1 [A/m], 2 [V/m], 3 [A/m2], 4 [V/m

2], 5 [m], 6 [s]

Thay cc i lng trong (1.82) vo cc phng trnh Maxwell sau y

t

EJEH 0E

++=

r

rrr

,JEJO(1.83)


31/108

30

t

HJE 0M

=

r

rr

Ta c

33

6

2211 aca

accar

r

r

++= (1.84)

6

15442 a

acaca

=

r

rr

Cc h s t l ci khng c th nguyn tng ng vi cc biu thc sau

1

521c

= ;

6

522c

= ;

1

533c

= ;

2

544c

= ;

62

515c

=

H phng trnh (1.84) l dng khng c th nguyn, m t cc hin t

khc nhau qua h s ci. Hai hin t c cc h s ci tng ng bng nhau gi

l 2 hng dng in ng vi nhau.

1.13. Trng tnh in

Trng tnh in c to ra bi cc in tch ng yn v khng bin i

theo thi gian, ta c h phng trnh Maxwell nh sau

0E =r

= D.r

(1.85)

ED 0rr

=

1.14. Ttrng ca dng in khng i

0E =r

= D.

r

(1.86)ED 0rr

=

JHrr

=

0B. =r

(1.87)

HB 0rr

=


32/108

31

Nhn xt: in trng ca dng in khng i cng tng t nhin

trng tnh v l mt trng th, ch khc nhau l in trng ca dng in

khng i tn ti ngay c trong vt dn EJrr

= , cn in trng tnh th khng

tn ti bn trong vt dn.


33/108


34/108

33

MM

0M

0

E02

2

002 J

t

J1J

t

H

t

HH

r

r

r

rr

r

+

+

+=

(1)

t

J1J

t

E

t

EE E0

0

M02

2

002

+

+=

r

r

rr

r

(2)

Nhn xt: V tri ca cc phng trnh (1) v (2) trong (2.3) ch cn Er

hoc Hr

. y l cc phng trnh vi phn cp 2 c v phi. Rt kh gii v v

phi l cc hm rt phc tp. Thng ch gii trong trng hp khng c ngun

v in mi l tng = 0, ta c

0t

HH

2

2

002 =

r

r

(1)

0tE

E2

2

002 =

r

r

(2)(2.4)

2.2. Phng trnh cho cc thin ng

Nhn xt: h phng trnh Maxwell (2.1) l tuyn tnh, cc ngun in v

t thng c kch thch ring r v c lp vi nhau.

2.2.1. i vi ngun in

n gin xt trng trong in mi l tng = 0 h phng trnh

Maxwell (2.1) c vit li

tE

JH 0E

+=

r

rr

(1)

tH

E 0

=

r

r

(2)(2.5)

0

E.

=

r

(3)

0H. =r

(4)

t:

( )E0

A1

Hrr

= (2.6)


35/108

34

EAr

gi l th vector in

D thy rng: ( ) 0A.1H. E0

=

=rr

a (2.6) vo (2) ca h phng trnh (2.5) ta c

0t

AE E =

+

r

r

(2.7)

Suy ra

EE

tA

E

=

r

r

(2.8)

Lu 0E = (2.9)

E l th v hng in

EAr

v Ec gi chung l cc thin ng ca ngun in

Nh vy: Hr

v Er

c biu din qua EAr

v E theo cc cng thc (2.6) v

(2.8) tng ng.

Tm EAr

v E ?

T cc cng thc (2.6) v (2.8) thay Hr

v Er

vo (1) ca (2.5) ta c

E0E

00E2E

2

00E2 J

tA.

tA

Arr

r

r

=

+

(2.10)

EAr

v Ec chn tu . V vy n gin ta c th chn iu kin ph

0tA.E

00E =

+

r

(2.11)

(2.11) cn gi l h thc chun

Phng trnh sng (2.10) c vit li

E02E

2

00E2 J

tA

Ar

r

r

=

(2.12)

T cng thc (2.8) thay Er

vo (3) ca (2.5) v p dng (2.11) ta c


36/108

35

02

E2

00E2

t

=

(2.13)

Cc phng trnh (2.12) v (2.13) gi l cc phng trnh sng khng

thun nht hay cc phng trnh dAlambert cho cc thin ng ca trng

in ti vi ngun in. EAr

v E

2.2.2. i vi ngun t

H phng trnh Maxwell (2.1) i vi ngun t trong in mi l tng

= 0 c dng

t

EH

0

=

r

r

(1)

tH

JE 0M

=

r

rr

(2)(2.14)

0E. =r

(3)

0

MH.

=

r

(4)

Cch lm tng t nhi vi ngun in ta c

( )M0

A1

Err

=

MM

tA

H

=

r

r

(2.15)

M02M

2

00M2 J

t

AA

r

r

r

=

0

M2M

2

00M2

t =

(2.16)

0t

A. M00M =

+

r

(2.17)

MAr

v M l cc thin ng i vi ngun t


37/108

36

Nu trong mi trng in mi l tng tn ti ng thi c ngun in v

ngun t th trng in t tng hp bng chng cht trng ca ngun in v

ngun t, c ngha l

( ) EM0

E A1

tA

E

=

r

r

r

( ) MME0 t

AA

1H

=

r

rr

(2.18)

Nhn xt: Er

v Hr

c biu din qua EAr

v E hoc MAr

v M lm cho h

phng trnh Maxwell n gin hn. y chnh l u im ca phng php

dng cc thin ng.2.2.3. i vi trng iu ho

Nu cc ngun ca trng bin thin iu ho theo thi gian vi tn s gc

th cc phng trnh sng dAlambert (2.12), (2.13) v (2.16) vit di dng

bin phc nh sau

Em02

Em2

2

Em

2

Jt

A

kA

=

r

r

r

0

m2Em

22

Em2

tk

=

Mm02

Mm2

2Mm

2 Jt

AkA

=

r

r

r

(2.19)

0

Mm2Mm

22

Mm

2

tk

=

Trong : 00k = l s sng trong mi trng

(2.19) l cc phng trnh khng thun nht, cn gi l phng trnh

Hemholtz

Biu thc ca Er

v Hr

c dng


38/108

37

EmMm

0

Em A1

AiE

=

rrr

(2.20)

Mm

Mm

Em0 t

A

A

1

H

=

r

rr

Gia th vector v th v hng c mi quan h sau

Em

00

Em A.1

=r

(2.21)

Mm

00

Mm A.1

=r

Nhn xt: Theo (2.20) v (2.21) cho thy rng i vi trng in tiuho ch cn tm nghim ca hai phng trnh Hemholtz i vi cc th vector

EmAr

v MmAr

2.3. Phng trnh sng cho cc vector Hertz

2.3.1 Vector Hertz in

t

tA E00E

=

r

r

(2.22)

Trong : Er

gi l vector Hertz in

Thay (2.22) vo (2.6) ta c

( ) ( )E0E0 t

A1

H

=

=

rrr

(2.23)

Thay (2.22) vo h thc chun (2.11) ta c

( ) 0.t EE

=+

r

(2.24)

Suy ra

EE .=r

(2.25)

Thay (2.22) v (2.25) vo (2.8) ta c


39/108

38

( )2E

2

00EEE

t.

tA

E

=

=

r

r

r

r

(2.26)

Nhn xt: Er

v Hr

c biu din qua vector Hertz in Er

Tm Er

?

Thay (2.22) vo (2.12) ta c

E02E

2

00E2

002E

2

00E2 J

tttA

Ar

r

r

r

r

=

=

(2.27)

Hay

E

02

E2

00E2 J

1

tt

r

r

r

=

(2.28)

Ly tch phn 2 v ca (2.28) t 0 n t ta c

=

t

0E

02E

2

00E2 dtJ

1t

r

r

r

(2.29)

t

=t

0

EE dtJPrr

(2.30)

EPr

gi l vector phn cc ca ngun in

Phng trnh (2.29) c vit li

0

E2E

2

00E2 P

t =

rr

r

(2.31)

Nh vy: vector phn cc EPr

l ngun to ra vector Hertz in Er

. Do

Er

cn gi l th vector phn cc in.

2.3.2 Vector Hertz t

Tng t cch lm ca vector Hertz in hoc p dng nguyn li ln

ca h phng trnh Maxwell ta c

tA M00M

=

r

r

(2.32)


40/108


41/108

40

Trng hp cc vector Hertz in Er

v vector Hertz t Mr

ch c mt

thnh phn. Trong h to Decac cc vector Hertz in Er

v vector Hertz t

Mr

theo phng z l

EE k=rr

(2.40)

MM k=rr

(2.41)

- Trng ca ngun in (ng vi vector Hertz in Er

mt thnh phn) s

c Hr

theo phng z bng 0 (Hz = 0), cn cc thnh phn khc ca Hr

ni chung

khc 0. Trng in t loi ny gi l trng loi in dc E hay t ngang TM

- Trng ca ngun t (ng vi vector Hertz t Mr

mt thnh phn) s c

Er

theo phng z bng 0 (Ez = 0), cn cc thnh phn khc ca Er

ni chung khc

0. Trng in t loi ny gi l trng loi t dc H hay in ngang TE

Nh vy: trong trng hp tng qut v iu kin bin nht nh, trng

in t c th xem nh tng hp ca 2 loi trng: loi in v loi t

2.4. Tm nghim ca phng trnh sng

Nhn xt: p dng nguyn li ln, vic tm nghim ca cc phng trnh

d Alambert ch cn xc nh Er

hoc Hr

. Do c th s dng mt hm v

hng i din cho E v M hoc bt c thnh phn no trong h to

Decac ca Er

, Mr

, EAr

v MAr

, phng trnh d Alambert c vit li

gt 2

2

002 =

(2.42)

g - hm ngun ca trng phn b trong th tch V

Nghim ca (2.42) bng tng nghim ca phng trnh sng thun nht

khng v phi v nghim ring ca phng trnh sng thun nht c v phi, tc

l tm nghim ca phng trnh sau

0t 22

002 =

(2.43)


42/108

41

i vi trng hp ngun im t gc to. V ngun im c tnh

i xng cu nn hm ch ph thuc r v t. Trong h to cu ta c

( )

=

+

= r

rrr

1

rr

2

r 2

2

2

22

(2.44)

t = rta c

0tr 22

002

2

=

(2.45)

Nghim ca phng trnh vi phn (2.45) l

++

=

vr

tfvr

tf 21 (2.46)

Suy ra

rvr

tf

rvr

tf 21

+

+

=

(2.47)

Trong :00

1v

= l vn tc truyn sng trong mi trng; f1 v f2 l

cc hm tu

rvr

tf1

m t sng cu phn k truyn t ngun v cng

rv

rtf2

+

m t sng cu hi t truyn t v cng ngun

iu kin bc x ti v cng:

0EiktE

rlimr

=

+

r

r

0Hikt

Hrlim

r=

+

r

r

(2.48)

Trong : 00k = l s sng


43/108

42

Nhn xt: v l ngun im t ti gc to v khng gian l v hn nn

theo iu kin bc x ti v cng ta chn nghim ca phng trnh sng (2.43)

cho ngun im l hm f1 v loi b hm f2

Vy

rvr

tf1

=

(2.49)

Nu r 0 (ti gc to) th nghim (2.49) khng tho mn phng trnh

sng thun nht m phi tho mn phng trnh sng d Alambert v th ta phi

chn dng ca f1 sao cho l nghim ca phng trnh sng d Alambert v

phi tho mn trng trng thi dng.

trng thi dng, phng trnh sng d Alambert c vit li

g2 = (2.50)

gi l phng trnh sng Poisson v c nghim l

=

V

dVr

g

4

1 (2.51)

Lu :

r l khong cch t v tr quan st trng n yu t vi phn gdV. Theo

(2.49) v (2.51) ta chn dng hm ca f1 nh sau

=

vr

tg41

vr

tf1 (2.52)

Nh vy, nghim ca phng trnh sng d Alambert l

( )

=

V

dVr

vrt,rg

41

t,r

(2.53)

Nhn xt: trng thi im t ti v tr quan st bng gi tr ca ngun

thi im t sm hn t mt khong thi gian l

vr

t = (2.54)


44/108

43

Nh vy, trng ti v tr quan st chm pha so vi ngun mt khong thi

gian t nn (2.53) gi l th chm ca trng in t.

Tng t nh nghim (2.53) ta c

( )

=

V

E0

E dVrvrt,rJ

4t,rA

r

r

(2.55)

( )

=

V

M0

M dVrvr

t,rJ

4t,rA

r

r

(2.56)

i vi trng iu ho ta c

ikrtiikrm

v

rti

m egeegegvr

tg

===

(2.57)

( ) ikrEvr

ti

EmE etAeAvr

tA

==

rrr

(2.58)

( ) ikrMvr

ti

MmM etAeAvr

tA

==

rrr

(2.59)

Cc th chm ME A,A,

rr

c tnh l

( )( )

=

V

ikr

dVr

et,rg41

t,r (2.60)

( )( )

=

V

ikrE0

E dVr

et,rJ4

t,rA

r

r

(2.61)


45/108


46/108

45

Lu : Sd tnh c tch phn (2.64) l do gi thit bin v pha ca

dng in cung cp l khng i trn ton lng cc in v do r >> l nn

khong cch t bt cim no trn lng cc in n v tr xc nh trng

u bng r.

Trong h to cu ta c cng thc

= sincosrk 00r

r

r

(2.65)

0rr

v 0r

l cc vector n v trong h to cu

Khi (2.64) c vit li

( )=

sincosrr4 leIA 00ikr

m0Em

r

r

r

(2.66)

Cng t trng ca lng cc in l

( )

=

=

sincosrr

e4

lIA

1H 00

ikrm

Em

0

m

r

r

rr

(2.67)

Suy ra

resinikr14 lIH

ikr

m0m

+

=r

r

(2.68)

0r

l vector n v trong h to cu

T h phng trnh Maxwell khng ngun in tch ta c

m0m EiH

=rr

(2.69)

Khi cng in trng ca lng cc in c tnh l

++

+

=

=

sinrik

kr1

cosrik

r1

r2.

.r

ei4

lIH

i1

E

2

2020

ikr

0

mm

0

m

r

r

rr

(2.70)


47/108

46

Nhn xt: Cc biu thc tnh

Er

v

Hr

trong (2.68) v (2.70) ca bc x

lng cc in u c tha sr

e ikrv bin t l nghch vi r, c mt ng

pha l mt cu bn knh r.

Nh vy trng bc x lng cc in c tnh cht ca sng cu. Vn tc

dch chuyn ca mt ng pha gi l vn tc pha vph

Ta c phng trnh ca mt ng pha l

= t kr = const

d = dt kdr = 0

(2.72)

V

kdt

drvph

== (2.73)

Nu nhn cc biu thc ca (2.68) v (2.70) vi eit v ly phn thc ca

Er

v

Hr

ta c gi tr tc thi ca chng l

( ) ( )

( ) ( )

( ) ( )

0HHE

krtcoskr1

krtsin1rk

1sin

r4lkI

E

krtcoskr

1krtsin

rk

1cos

r2

lkIE

krtsinkrtcoskr

1sinr4

lkIH

r

220

2m

220

2m

r

m

===

=

=

=

(2.74)

2.5.2. Trng vng gn

Khi r > l th gi l trng vng gn

Do r


48/108

47

tsinsinr4

lIE

tsincosr2

lIE

tcossinr4lI

H

30

m

30

mr

2m

=

=

=

(2.75)

Nhn xt: H lch pha so vi Er v E mt gc2

nn vector Poynting

trung bnh tbr

= re

r

= 0, c ngha l nng lng trng in t ca lng cc

in vng gn ch yu l ca dao ng xung quanh ngun, khng mang tnh

cht sng, gi l vng cm ng . Hnh 2.1 trnh by cu trc ng sc ca Er

v

Hr

2.5.3. Trng vng xa

Khi r >> th th gi l trng vng xa

Do r >> nn kr = r2

>> 1 v trong (2.74) nu b qua cc v cng b

bc cao so vikr1 ta c

( ) ( )

( ) ( )krtsinsinr2

lIkrtsinsin

r4

lkIE

krtsinsinr2lI

krtsinsinr4

lkIH

0

0m

0

2m

mm

=

=

=

=

(2.76)

Nhn xt:

IEr

Er

Hr

Er

Er

Er


49/108

48

- Trng vng xa ca lng cc in ch gm 2 thnh phn H v E

ng pha, vung gc vi nhau v vung gc vi phng truyn sng r, vector

Poynting phc ch c phn thc tbr

= re

r

0, nng lng trng in t bc

x vo trong khng gian. V vy vng xa gi l vng bc x

- Bin ca H v E t l vi , t l nghch vi . Nu c cng gi tr

dng in Im, cng khong cch v tn s cng cao th H v E cng ln

- Bin ca H v E t l vi sin nn trng bc x ca lng cc in

c tnh nh hng trong khng gian. Chng t cc i ti mt phng2

v

bng 0 theo phng ca lng cc in = 0.

- Trng bc x c tnh nh hng, thng c m t bng gin

hng. Gin hng ca lng cc in, k hiu F(, ), l hm c xc

nh bi biu thc:

( ) == sinE

E,F

max

(2.77)

2.5.4. Cng sut bc x, trbc x

Cng sut bc x ca lng cc in c tnh theo cng thc

SdPS

tbbx

rr

= (2.78)

=

00

= 900

E=

0

E = Emax

Mt phng kinh tuyn

Mt phng vtuyn

Z


50/108

49

Trong

= 2

032

322m

tb sinr32klI

rr

r

(2.79)

Vi phn mt cu

dS = r2sindd

Suy ra

bx

2m

0

0222

m

0

32

0032

322m

bx R2

I

12

klI

dsindr32

klI

P =

==

(2.80)

Trong 2

0

0

0

02

bx

1

3

2

6

lkR

=

=

(2.81)

Rbx - trbc x ca lng cc in

t

0

0cz

= [] (2.82)

zc - trsng ca mi trng

Trong chn khng hoc khng kh, ta c = = 1, do

==

= 377120z

0

00c

d

d

Hr

Er

Sdr

I

r


51/108

50

=

=

22

20bx

1790

180R

W1

I395P2

2m0bx

=

2.6. Trng in tca lng cc t

Lng cc t l yu t bc x sng in t, l thnh phn cbn ca anten

Th d vlng cc t, mt on dy dn ngn mnh bn trong c dng t

bin i do ngun cung cp bn ngoi. Cch lm tng t nhi vi lng cc

in hoc p dng nguyn li ln v trong cc cng thc (2.68) v (2.70) thay

Hr

bng Er

, thay Er

bng Hr

, thay bng - v thay mI

bng MmI

re

sinikr1

4lI

Eikr

Mm

0m

+

=r

r

(2.83)

++

+

=

sinr

ikk

r

1cos

r

ik

r

1r2

r

e

i4

lIH 2

2020

ikr

0

Mmm

r

r

r

(2.84)

Theo (2.83) v (2.84) cho thy trng bc x ca lng cc t cng l

sng cu,

Er

, Hr

~ r,

Er

, Hr

c tnh nh hng trong khng gian

IErEr Er

Er

Er

Hr


52/108

51

Vai tr ca in trng v t trng lng cc t so vi ca lng cc

in thay th cho nhau. V vy cu trc ng sc ca chng l ging nhau vi

Er

v Hr

i ch cho nhau

2.6.1 Trng in tca vng dy

Nhn xt: trong thc t, ngi ta c th to ra trng in t xung quanh 1

vng dy nh mnh c dng in bin i Im chy qua tng t nh lng cc

t. Vng dy dn ny gi l anten khung nguyn t.

Gi s:

- mt phng vng dy nm trng vi mt phng vtuyn ca h to cu

- kch thc vng dy rt nh so vi bc sng ca trng in t do npht ra

- dng in bin i iu ho theo thi gian vi tn s gc : tim eII

= vi

bin v pha dc theo ng dy c gi tr nh nhau

Theo (2.61) th chm ti im Q thuc trng in t do vng dy pht ra

=

V

ikrm

0Em dVerJ

4A

r

r

(2.85)

Trong : r l khong cch tim Q n yu t vi phn ldr

Ta c:

lSddVr

= , ldIlSdJdVJ mmmrrrr

== (2.86)

Suy ra

=

l

ikrm0Em ld

re

4IA

rr

(2.87)

V dng in chy trong dy dn ch theo phng vtuyn nn th chm

EmAr

ca n cng ch c 1 thnh phn hng theo phng vtuyn

Th d:


53/108

52

Xt 2 yu t vi phn ldr

ca vng dy t i xng vi nhau qua mt phng

P i qua im tnh trng Q v vung gc vi mt phng vng dy (mt phng

P gi l mt phng kinh tuyn). Mi mt yu t vi phn ldr

li phn tch thnh 2

yu t vi phn: ld r

// (P) v ld r

(P).

Nhn xt:

- th vector do cc yu t vi phn ld r

to ra ti Q c cng gi tr nhng

hng ngc nhau nn b trit tiu

- th vector do cc yu t vi phn ld r

to ra ti Q c cng gi tr v cng

hng vi nhau nn tng gp i.

Do tch phn trong (2.87) ch cn ly theo yu t vi phn ld r

. Hn na

do tnh i xng ca ld r

i vi mt phng P nn tch phn trn ch cn ly theo

na vng dy v nhn i

Ta c:

dl = dl cos = Rcos d (2.88)

Trong : R l bn knh ca vng dy

Suy ra:

=

V

ikrm0

0Em dr

cose

2

RIA

r

r

(2.89)

P

rr

O aa

bRI

Q

O a

R

I

dl

dl

dl

dl

dl

dl


54/108

53

Trong : 0r

l vector n v hng theo phng v tuyn, theo hnh v

trn ta c cc h thc sau222 abaQr += , += cosROa2ROaab 222 (2.90)

Hay+=++= cossinRr2RrcosROa2ROaaQr 222222 (2.91)

Trong : r l khong cch t O n Q

Theo gi thit r >> R nn cho R2 = 0 v t (2.91) ta c

== cossinRrcossinrR2

1rcossinRr2rr 2

Suy ra

+=

+

=

=

cossinrR

r1

cossinrR

1r1

cossinrR

1

1r1

cossinRr1

r1

2

V( )

( ) ( )( )+===

cossinkRsinicossinkRcoseeeee ikr

cossinikRikrcossinRrikrik

Khi >> R th kR


55/108

54

2ikr

m00Em Rikr

1sin

r4eI

A

+

=

r

r

(2.93)

++

+=

sinr

ikkr

1cosr

ik

r

1r2r

e

4

RIH

2

2020

ikr2m

m

r

r

r

(2.94)

+

=

=

ikr

1sin

ri4

lekRIH

i

1E

0

ikr22m

0m

0

mr

rr

(2.95)

D thy rng trng bc x ca vng dy dn c tnh cht tng t nh

trng bc x ca lng cc t v s hon ton ging nhau nu tho mn iu

kin sau

2m0

MmRI

i

lI=

(2.96)

t

==

i

lIlqP

Mm

MmM

r

rr

(2.97)

M

P

r

gi l moment lng cc tt

2m00m00Mv RISSISP ==

rrr

(2.98)

MvPr

gi l moment t ca vng dy dn c dng in mI

v din tch S

Khi trng bc x ca lng cc t v vng dy dn l tng ng

nhau

MvM PP

=rr

(2.99)

T cc biu thc (2.94) v (2.95) ta tnh c thnh phn trng bc x

ca vng dy vng xa l


56/108

55

( )

( )krtcossinr4kRI

E

krtcossinr4kRI

H

0

022

m

22m

==

=

(2.100)

Cng sut bc x v trbc x ca vng dy c tnh l

bxv

2m

bxv R2

IP =

(2.101)

c

2

3bx z

S38

R

=

(2.102)

2.7. Trng in tca yu t din tch mt

Xt trng bc x ca yu t vi phn din tch m trn c dng in v

t mt chy vung gc vi nhau.

Gi s yu t vi phn din tch nm trong mt phng xOy c dng hnh ch

nht kch thc a, b

Dng in mt hng theo trc x: IESx bthin iu ho theo thi gian

Dng t mt hng theo trc y: IMSy bthin iu ho theo thi gian

S


57/108

56

=

S

ikrESxm0

Exm dSreI

4A

(2.103)

=

S

ikrMSym0

Mym dSr

eI

4A

(2.104)

V dng in mt IESx hng theo trc x nn ExmA

cng ch c thnh phn

ny, tng t dng t mt IMSy hng theo trc y nn MymA

cng ch c thnh

phn ny

Theo gi thit, bin v pha ca dng in v t mt l khng i trn

ton yu t vi phn din tch, khong cch tim quan st trng n yu tdin tch ln hn rt nhiu so vi kch thc ca yu t din tch, do c th

a cc biu thc trong du tch phn ca (2.103) v (2.104) ra ngoi

r4

eISA

ikrESxm0

Exm

=

(2.105)

r4

eISA

ikrMSym0

Mym

=

(2.106)

Trong :

r l khong cch tim quan st trng n gc to

S = ab l din tch ca yu t mt

Cc thnh phn ca th vector trong h to cu v h to Decac lin

h vi nhau nh sau

++= cosAsinsinAcossinAA zyxr

++= sinAsincosAcoscosAA zyx (2.107)

+= cosAsinAA yx

Do ch c ExmA

v MymA

khc 0, ta c

=

cossinAA ExmErm


58/108

57

=

coscosAA ExmmE (2.108)

=

sinAA ExmmE

=

sinsinAA MymMrm

=

sincosAA MymmM (2.109)

=

cosAA MymmM

p dng cc cng thc (2.6) v cng thc 1 ca (2.15) cho (2.108) v

(2.109), ta c

=

Em

0

A1

Hrr

=

Mm

0

A1

Err

Kho st trng bc x ca yu tdin tch vng xa

Khi tnh trng ta ch quan tm n s hng suy gim r1

, b qua cc s

hng bc cao hnn

r1

. Do khi tnh rot trong h to cu ca (2.108) v

(2.109) ta ch gi li cc thnh phn vi o hmr

A m0

r

vr

A m0

r

c gi

li, cn cc s hng bc cao hn c b qua v ta c

ikrESxmmE e

r4

coscosIikSH

=

ikrESxmmE e

r4

sinIikSH

=

(2.110)

ikrMSymmM e

r4

sincosIikSE

=


59/108

58

ikrMSymmM e

r4

cosIikSE

=

S dng cc phng trnh Maxwell th nht v th hai

=

Em

0

Em Hi

1E

rr

=

Mm

0

Mm Ei

1H

rr

cho cc biu thc (2.110) ta c

ikrESxm00mE e

r4

sinISikE

=

ikrESxm00mM e

r4

coscosISikE

=

(2.111)

ikr

00

MSymmM e

r4

cosIikSH

=

ikr

00

MSym

mM er4

sincosIikS

H

=

Ly tng cc biu thc ca (2.110) v (2.111) theo cc thnh phn ca E

v E ta c

( )+

=+=

cos1er4

sinIikSEEE ikr

ESxm00mMmEm

(2.112)

Trong :00ESxm

MSym

I

I

=

Tng t, theo cc thnh phn ca H v H ta c

+

=+=

cos1

1er4

cosIikSHHH ikr

00

MSymmMmEm


60/108

59

( )+

=+=

cos1er4

sinIikSHHH ikr

ESxmmMmEm

(2.113)

Nhn xt:

- Cc cng thc (2.112) v (2.113) cho thy rng trng bc xvng xa

ca yu t vi phn din tch trong mt phng kinh tuyn c c trng hng

dng ng cong cardioid

- Trng bc x ca nguyn t Huyghens cng tng t nh trng bc x

ca lng cc in v lng cc tt vung gc v cng chung im gia

mt

C(1+cos)

z


61/108

60

Chng 3

SNG IN TPHNG

Sng phng: mt ng pha l mt phng

Sng tr: mt ng pha l mt tr

Sng cu: mt ng pha l mt cu

Trong thc t, sng in tc to ra t cc ngun nhn to u l sng

tr v sng cu. Sng phng ch l mu l tng ca sng in t.

Mc tiu: kho st cc tnh cht ca sng in t phng lan truyn trong

mi trng ng nht ng hng v khng ng hng, s phn x v

khc x ti cc mt phn cch, s phn cc v cc hiu ng khc. Ngun

sng in t l iu ho vi v rt xa vi im kho st.

3.1. Nghim phng trnh sng i vi sng phng

3.1.1. Sng phng ng nht TEM (transverse electromagnetic wave)

- Nu trong mt ng pha ca sng in t c bin ca Er

v Hr

bng

nhau tng ng ti mi im th sng phng c gi l ng nht

- Phng trnh Maxwell ca sng phng iu ho trong mi trng ng

nht v ng hng vi cc bin phc ca Er

v Hr

trong h to Decac c

dng

xmP

ymzmEi

zH

yH

=

(1)

ymP

zmxmEi

x

H

z

H

=

(2)

zmP

xmymEi

yH

xH

=

(3)

xm0

ymzmHi

zE

yE

=

(4)


62/108

61

ym0

zmxmHi

xE

zE

=

(5)

zm0

xmym

Hiy

E

x

E

=

(6)

Trong :

Oz phng truyn sng

mt phng ng pha v ng bin ca sng phng chnh l mt phng P //

mt phng xOy v c phng trnh z = l

=

0

0P i1

Er

v Hr

c gi tr nh nhau trn ton mt phng P v x, y; ch z, t. Khi

:

0yH

xH

yE

xE

=

=

=

=

(3.1)

0HE zmzm ==

(3.2)

Vy: sng phng ng nht lan truyn trong mi trng ng nht v ng

hng khng c cc thnh phn dc theo phng truyn sng z ca Er

v Hr

.

Cc Er

v Hr

nm trong mt phng vung gc vi phng truyn sng. Sng

phng ng nht c tnh cht nh vy gi l sng in t ngang, k hiu l sng

TEM.

3.1.2. Nghim phng trnh sng

T cc phng trnh (1), (2), (4) v (5) ta c:

P

O

l

yz


63/108

62

0Ekz

Exm

2P2

xm2

=+

(7)

0Ekz

Eym

2

P2

ym2

=+

(8)

0Hkz

Hxm

2P2

xm2

=+

(9)

0Hkz

Hym

2P2

ym2

=+

(10)

Trong :

0

0

00PP i1k

== - s sng phc

Nhn xt:

- v cc phng trnh sng (7), (8), (9) v (10) ging nhau nn ch cn tm

nghim ca mt trong s cc phng trnh sng ny.

- y l cc phng trnh vi phn cp 2 tuyn tnh thun nht c h s

khng i, do nghim ca phng trnh sng (7), chng hn, c dng l

zikxmpx

zikxmtxm

PP eEeEE

+= (3.3)

Trong :

- zikxmt PeE

biu th sng phng truyn theo trc z > 0: sng ti ti mt

phng P

P

O

l

yz


64/108

63

- zikxmpx PeE

biu th sng phng truyn theo trc z < 0: sng phn x ti mt

phng P

- xmtE

, xmpxE

l cc bin phc ca sng ti v sng phn x tng ngTng t ta c nghim ca cc phng trnh sng (8), (9) v (10) l

zikympx

zikymtym

zikxmpx

zikxmtxm

zikympx

zikymtym

PP

PP

PP

eHeHH

eHeHH

eEeEE

+=

+=

+=

(3.4)

Suy ra

++

+=+=

++

+=+=

zikympx

zikymt

zikxmpx

zikxmtymxmm

zikympx

zikymt

zikxmpx

zikxmtymxmm

PPPP

PPPP

eHeHjeHeHiHjHiH

eEeEjeEeEiEjEiE

rrrrr

rrrrr

(3.5)

tm mi lin h gia mEr

v mHr

cho sng ti v sng phn x, bng cch

quay h to Decac sao cho trc x //Er

, do trc y // Hr

, ta c

mxmymxmm EiEiEjEiE

==+=rrrrr

v 0Eym =

mymymxmm HjHjHjHiH

==+=rrrrr

v 0Hxm =

(3.6)

T phng trnh Maxwell (1), iu kin (3.6) v cc nghim (3.3), (3.4) ta

c mi lin h gia mEr

v mHr

cho sng ti v sng phn x nh sau

x

y

mHr

mEr

ymH

xmE

O


65/108

64

mpxP

ympx

P

0ympx

P

xmpxmpx

mtPymt

P

0ymt

P

xmtmt

HZHz

H

i

1

EE

HZHz

H

i

1EE

=

=

==

=

=

==

(3.7)

Trong :

( )EE0

0

P

0P

itg1

1Z

itg1Z

=

=

= (3.8)

T (3.7) dng ca mEr

v mHr

cho sng phng TEM c vit li

zikmpx

zikmtm

zikmpx

zikmtPm

PP

PP

eHeHH

ekHekHZE

+=

=

rrr

rrrrr

(3.9)

Hoc

( ) ( )

( ) ( )zktimpx

zktimt

tim

zktimpx

zktimtP

tim

PP

PP

eHeHeHH

ekHekHZeEE

+

+

+==

==

rrrr

rrrrrr

(3.10)

n gin trong nhng phn sau ta ch xt i vi sng ti lan truyn

trong mi trng rng v hn.

O

x

y

l


66/108

65

Dng ca mEr

v mHr

ca sng phng TEM lan truyn dc theo phng z

c biu din trong (3.9) hoc (3.10). Tng t theo phng l bt k hp vi

Ox, Oy v Oz to thnh cc gc , v . Ta c:

( )lktimtt

PeHH

=rr

(3.11)

mtHr

nm trong mt phng vung gc vi phng l.

V

( )lktimtPt

PelHZE

=

rrr

(3.12)

lr

l vector n v ca phng truyn sng l.

S sng phc kP v trsng phc ZP c th vit li

=

=i

PP

P

eZZ

ik (3.13)

Trong

, v l cc s thc

l h s tn hao ca mi trng

l h s pha ca sng

argument ca trsng phc

Khi , , PZ v biu din qua , , v thi gianE nh sau

E2

00 tg12

1

2

1++= (3.14)

E2

00 tg121

21

++= (3.15)

4E

2P tg1

ZZ

+= (3.16)


67/108

66

E2

E2

tg11

tg11arctgarctg

++

++=

= (3.17)

Vn tc pha vph ca sng phng chnh l vn tc dch chuyn mt ng pha

ca n. Khi theo (3.10) v (3.13), gi s mi trng khng tn hao = 0,

mt ng pha ca sng ti c dng

constzt == (3.18)

Suy ra

0dzdtd == (3.19)

Cho nn vn tc pha vphc xc nh bi

E2

E200

ph

tg121

21

v

tg121

21

1.

1dtdz

v

++

=

++

=

==

(3.20)

Trong

v l vn tc truyn sng phng trong mi trng rng v hn

Vector Poynting trung bnh ca sng ti hng theo phng truyn z ctnh l

P

2

mt2

mtPmt*

mttb Z

E

21

kHZ21

kHEre21

rerrrrrr

==

==

(3.21)

Lu : V

Er

v

Hr

ng pha nn = 0 1e i =

3.2 Sng phng ng nht trong cc mi trng ng nht v ng hng

3.2.1. Sng phng ng nht trong in mi l tng Xt sng in t phng ng nht truyn dc theo trc z > 0 (sng ti)

trong in mi l tng ng nht, ng hng v rng v hn.


68/108

67

V mi trng truyn sng in t l in mi l tng nn = 0,

0

0

0P i1 =

= , kP = k v ZP = Z. T cc biu thc (3.14)

(3.21) ta c

Z

E

2

1HZ

2

1

v1

v

ZZ

k

0,0

2mt2

mttb

00

ph

0

0P

00

==

=

=

==

==

==

r

(3.22)

mEr

v mHr

c dng l

zimtm

zimtm

ekHZE

eHH

=

=

rrr

rr

(3.23)

Hoc

( )

( )ztimt

tim

ztimt

tim

ekHZeEE

eHeHH

==

==

rrrr

rrr

(3.24)

Nhn xt:

Er

v Hr

vung gc vi nhau v cng vung gc vi phng truyn sng

Er

v Hr

lun ng pha v c bin khng i dc theo phng truyn

sng

Vn tc pha vph l hng s bng vn tc truyn sng trong mi trng

Mi trng khng tn hao nng lng, khng tn sc sng in t, tr

sng Z l mt s thc


69/108

68

3.2.2. Sng phng ng nht trong mi trng dn in

Trong mi trng dn in 0, s sng v tr sng l cc i lng

phc,

=

== ii1k 0000PP

=

=

= iP

0

0

0

P

0P eZ

i1

Z

Nh ni trn ch xt i vi sng ti, do theo (3.10) v (3.13)

Er

v

Hr

c dng

( ) ( ) ( ) zztimt

ziztimt

zktimt eeHeHeHH P

+

===rrrr

.......

( ) ( )

( ) zztimtP

ziztimt

iP

zktimtP

eekHZ

ekHeZekHZE P

+

+

=

=

=

=

rr

rrrrr

(3.25)

Hr

Er


70/108

69

Nu mi trng c in dn sut rt ln, chng hn nh kim loi, mt

cch gn ng xem , do thi gian E >> 1 nn theo cc biu thc(3.14) (3.21) ta c

0

EE2 tgtg1

=+

2tg1

2

1

2

1 0E

200

++=

2tg121

21 0

E

2

00

++=

= 0P ZZ

0E

200

ph

2

tg121

21

v

++

=

=

( )4

1arctgtg11tg11arctgarctg

E2

E

2

=++++=

=

(3.26)

gc tn hao 0 nn sng in t b tn hao nng lng, bin ca

Er

v

Hr

suy gim theo quy lut hm m e-z dc theo phng truyn sng z.

Er

v

Hr

lch pha nhau mt gc = argZP

0mE z

0mm eEE=

z

x

y


71/108

70

vph l hm s ph thuc tn s, c ngha l thay i trong qu trnh

lan truyn sng in t sng phng trong mi trng dn in b tn

sc. Do mi trng dn in l mi trng tn sc.

3.3. Hiu ng bmt trong vt dn

Nhn xt:

Theo cng thc2

0 nhn thy rng

Trong vt dn in tt rt ln v nu tn s sng in t cng cao th

cng ln. Do bin ca Er

v Hr

suy gim rt nhanh khi truyn vo

bn trong vt dn, c ngha l sng in t ch tn ti mt lp rt mngst b mt ca vt dn in tt.

Dng in cao tn chy trong vt dn cng ch chy lp mt ngoi.

Chng hn f = 1 kHz th d = 2 mm v f = 100 kHz th d = 0,2mm.

d: lng kim thp Cu lm dy dn dng in cao tn

Hin tng sng in t hoc dng in cao tn khi truyn trong vt dnin tt ch tp trung mt lp mng b mt gi l hiu ng b mt hay

hiu ng skin

i lng c trng cho hiu ng b mt l thm su ca trng hay

dy lp skin , l khong cch sng in ti t b mt vo su

Br

Br

cBr

cBr

Thp

Cu


72/108

71

bn trong vt dn m ti bin ca Er

v Hr

gim i e = 2,718... ln

so vi gi tr ti b mt.

Theo (3.25) v (3.26) ta c

z0mm

z0mm

eHH

eEE

=

= (3.27)

Trong :

Em0 v Hm0 l bin ca Er

v Hr

ti b mt vt dn (z = 0). Theo nh

ngha thm su ca trng ta c

ee

E

E

m

0m == (3.28)

Suy ra

=

=

=

00

2

2

11

(3.29)

Nhn xt:

Trong cng thc (3.29), v l cc tham sin ca vt dn in.

thm su ca trng t l nghch vi cn bc hai ca tn s v in

dn sut ca vt dn. Chng hn Ag, Cu, Al ... c thm su ca

trng rt b c = 0,5 m di sng v tuyn f = 106 Hz. Do cc

kim loi ny dng lm mn chn sng in t rt tt.

Do c h/ bm nn dng in cao tn c cng phn b khng u

trong cng mt tit din ngang ca dy dn, do trkhng cng khng

u nhau tng ng. tin tnh ton ngi ta a ra khi nim tr

khng mt ring ca vt dn

Trkhng mt ring ca vt dn, k hiu ZS, l t sin p ca trng ri

trn mt n v chiu di theo chiu dng in v gi tr dng in chy

qua mt n v chiu rng t vung gc vi n


73/108

72

Xt vt dn phng, rng v hn v b dy ln. Chn h to Decac c

trc z trng vi phng truyn sng, mt phng vt dn trng vi mt phng

xOy.

Gi sEr

Ox. Theo nh lut Ohm ta c:

( )

+

==== +

iE

dzeEdzJSdJI 0mzi

00m

0x

S

rr

(3.30)

Lu : Tch phn (3.30) c ly t 0 , mt d b dy vt dn l huhn nhng dng in cao tn ch chy trn lp b mt rt mng nn b dy vt

dn c th xem l v hn.

Cng in trng Er

ti b mt vt dn bng in p ri trn mt n

v chiu di dc theo chiu dng in nn ta c

( ) ( ) SS0

0m

0mS iRi12

i1

i1

EE

IU

Z +=+

=+

=

+

==

do =

(3.31)

Trong :

=

2R 0S l in trngmt ring ca vt dn. (3.32)

x

y

Er

Jr

O

r


74/108

73

RS chnh l nguyn nhn lm tn hao sng in t trong vt dn. Nng

lng sng in t bin thnh nhit nng t nng vt dn.

S l phn khng ca trkhng mt ring ca vt dn ZS.

Nhn xt: Biu thc (3.32) cho thy rng mun gim tn hao nng lng

sng in t truyn dc vt dn cn phi s dng cc kim loi dn in tt nh

Au, Ag, Cu ...

3.4. Sphn cc ca sng phng

Sng in t c cc vector Er

v Hr

dao ng theo phng xc nh gi l

sng phn cc. Ngc li nu cc vector Er

v Hr

dao ng theo mi phng

ngu nhin gi l sng khng phn cc.

Sng in t phng c nhiu dng phn cc nh: phn cc elip, phn cc

trn v phn cc thng.

3.4.1. Phn cc elip

Trong qu trnh truyn sng nu ngn ca vector Er

vch mt hnh elip

trong khng gian gi l sng phn cc elip. Sng phn cc elip chnh l tng

hp ca 2 sng thnh phn cng tn s, cng phng truyn, nhng phng ca

Er

vung gc nhau.

Gi s c 2 sng phng nh sau:

( )

( )+=

=

ztcosEjE

ztcosEiE

my2

mx1rr

rr

(3.33)

Sng tng hp c dng

=

+

2

mymx

21

2

my

2

2

mx

1 sinEE

EEcos2

E

E

E

E (3.34)

y l phng trnh m tng elip trong mt phng to (E1, E2). Trc

ln ca elip hp vi trc Ox mt gc c tnh theo:

= cosEE

EE22tg

2my

2mx

mymx (3.35)


75/108

74

Trong : Emx > Emy

Trong qu trnh truyn sng theo trc z, ngn ca vector Er

tng hp vch

nn mt ng elip xon trong khng gian

3.4.2. Phn cc trn

Nu 2 sng thnh phn c bin bng nhau: Emx = Emy = Em v lch pha

nhau mt gc2

= . Suy ra 1sin 2 = , 0cos = v phng trnh (3.34) tr

thnh2m

22

21 EEE =+ (3.36)

y l phng trnh m tng trn trong mt phng to (E1, E2).Trong qu trnh truyn sng theo trc z, ngn ca vector E

r

tng hp vch nn

mt ng trn xon trong khng gian, gi l sng phn cc trn.

Nu nhn theo chiu truyn sng vector Er

tng hp quay thun chiu kim

ng h, ta c sng phn cc trn quay phi. Nu nhn theo chiu truyn sng

vector Er

tng hp quay ngc chiu kim ng h, ta c sng phn cc trn

quay tri. Chiu quay ca vector Er

tng hp ph thuc vo du ca gc lch

pha2

3.4.3. Phn cc thng (tuyn tnh)

Trong qu trnh truyn sng theo trc z, vector Er

lun hng song song

theo mt ng thng gi l sng phn cc thng hay sng phn cc tuyn tnh.

trng hp ny gc lch pha ca 2 sng thnh phn c gi tr = 0, , 2, ...Suy ra sin = 0, cos = 1 v phng trnh (3.34) trthnh

0E

E

E

E2

my

2

mx

1 =

+ (3.37)

Hay


76/108

75

1

mx

my

2 EE

EE = (3.38)

y l phng trnh m tng thng i qua gc to hp vi trc Ox

mt gc c tnh theo

mx

my

E

Etg = (3.39)

Nhn xt: Tu thuc vo hng ca vector Er

ngi ta cn phn thnh 2

trng hp phn cc ngang v phn cc ng.

3.5. Sphn x v khc x ca sng phng

Mc tiu phn ny nghin cu qui lut ca sng phn x v khc x ti mt

phng phn cch rng v hn gia 2 mi trng c tham sin khc nhau.

n gin ta ch xt i vi sng phng ti phn cc thng ngang v ng.

3.5.1. Sng ti phn cc ngang

Nu vector Er

ca sng ti vung gc vi mt phng ti, gi l sng phn

cc ngang. Trong trng hp ny vector Er

ca sng ti s song song vi mt

phng phn cch 2 mi trng. Tm qui lut ca sng phn x v khc x ?

Chn h to Decac c mt xOy mt phng phn cch 2 mi trng,

trc z trng vi php tuyn ca mt phng phn cch 2 mi trng. Hai mi

trng l in mi c cc tham sin 1, 1, 2, 2 tng ng.

V sng ti l sng phng truyn theo phng zt, lp vi php tuyn z mt

gc t nn c th quay trc to quanh trc z cho trc x ca n ch phng

ca vector Er

ca sng ti. Ti mt phng phn cch s c sng phn x li mi

x

y

Er

Emx

EmyO


77/108

76

trng 1 vi gc phn xphn x truyn theo hng zpx, cn sng khc x ti mt

phng phn cch vi gc khc xi vo mi trng 2 theo phng zkx. Theo

h.v nhn thy rng Er

ca sng ti, sng phn x v sng khc x ch c 1

thnh phn theo trc x, cn Hr

ca cc sng trn c 2 thnh phn theo trc y v

z. p dng cc biu thc (3.4) v (3.5) ta c:

Sng ti

t1

t1

zikmz1my11

zikmx11

eHkHjH

eEiE

+=

=

rrr

rr

(3.40)

Sng phn x

px1

px1

zikmz1my11

zikmx11

eHkHjH

eEiE

+=

=

rrr

rr

(3.41)

Sng khc x

kx2

kx2

zikmz2my22

zikmx22

eHkHjH

eEiE

+=

=

rrr

rr

(3.42)

Trong :

01011k = v 02022k = l s sng ca mi trng 1 v 2 tng

ng. Cc phng truyn sng zt, zpx v zkx biu din qua x, y, z nh sau:

+==

+=

coszsinyzcoszsinyz

coszsinyz

kx

pxpxpx

ttt

(3.43)


78/108

77

V cc mi trng u l in mi nn p dng iu kin bin cho E

r

v H

r

ti mt phng phn cch xOy (z = 0) ta c:

my22my1my11

mx22mx1mx11

HHHHH

EEEEE

==+=

==+= (3.44)

Thay cc biu thc (3.40) - (3.43) vo (3.44) v cho z = 0 ta c:

=

=+

sinyikmy2

sinyikmy1

sinyikmy1

sinyikmx2

sinyikmx1

sinyikmx1

2px1t1

2px1t1

eHeHeH

eEeEeE

(3.45)

(3.45) lun tho mn y ta li c:

==

=

=+

sinyiksinyiksinyik

my2my1my1

mx2mx1mx1

2px1t1 eee

HHH

EEE

(3.46)

T biu thc cui ca (3.46) suy ra:

pxt = (3.47)

= sinksink 2t1 (3.48)

Nhn xt:

(3.47) m tnh lut phn x sng in t ti mt phng phn cch.

(3.48) m tnh lut khc x sng in t.

t

px

t

1Er

1Er

1Hr

1Hr

zpx

zt

zk

y

z

2Er

2Hr

O


79/108

78

011n = v 022n = (3.49)

ln lt l chit sut ca mi trng 1 v 2. Gi s1 = 2 = th nh lut khc

x ca sng in t phng c dng ging nh trong quang hc

= sinnsinn 2t1 (3.50)

m t gia cc bin phc ca sng ti, sng phn x v sng khc x

ngi ta a ra khi nim h s phn x v h s khc x.

H s phn x (reflective modulus) l t s gia bin phc ca sng

phn x v sng ti tnh cho Er

, k hiu R. H s khc x (refractive modulus)

l t s gia bin phc ca sng khc x v sng ti tnh cho Er

, k hiu T.

i vi sng phn cc ngang ta c:

m1

m1

ng

E

ER

= v

m1

m2

ng

E

ET

= (3.51)

Theo hvi vi sng phn cc ngang ta c:

==

==

==

cosHH,cosHH

cosHH,EE

EE,EE

m2my2tm1my1

tm1my1mx2m2

mx1m1mx1m1

(3.52)

v

2

m2m2

1

m1m1

1

m1m1

ZE

H

Z

EH

ZE

H

=

=

=

(3.53)


80/108

79

Trong :01

011Z

= v

02

022Z

= l tr sng ca mi trng 1 v 2

tng ng. Thay cc biu thc (3.52) v (3.53) vo (3.46) ri chia c 2 v ca

chng cho m1E

ta c

( )2

ng

1

tng

ngng

Zcos

TZ

cosR1

TR1

=

=+

(3.54)

Suy ra:

+

=

+

=

cosZcosZ

cosZ2T

cosZcosZ

cosZcosZR

1t2

t2ng

1t2

1t2ng

(3.55)

(3.55) gi l cng thc Fresnel

Gc khc xc th tnh c qua gc ti t theo nh lut khc x (3.48)

nh sau:

t2

2

1

2

t

2

1 sin1sink

k1cos

= (3.56)

Nu 2 mi trng l in mi c 1 = 2 = th (3.55) c vit li

t2

2

12t1

t1ng

t2

2

12t1

t2

2

12t1

ng

sin1cos

cos2T

sin1cos

sin1cos

R

+

=

+

=

(3.57)

3.5.2. Sng ti phn cc ng

Nu vector Er

ca sng ti nm trong mt phng ti, gi l sng phn cc

ng. Trong trng hp ny vector Hr

ca sng ti s song song vi mt phng

phn cch 2 mi trng. Tm qui lut ca sng phn x v khc x ?


81/108

80

Chn h to Decac c mt xOy mt phng phn cch 2 mi trng,

trc z trng vi php tuyn ca mt phng phn cch 2 mi trng v trc x ch

phng ca vector Hr

ca sng ti.

Theo h.v nhn thy rng Hr

ca sng ti, sng phn x v sng khc x

ch c 1 thnh phn theo trc x, cn Er

ca cc sng trn c 2 thnh phn theo

trc y v z. Tin hnh tng t nhi vi sng phn cc ngang ta c:

+

=

+

=

cosZcosZcosZ2

T

cosZcosZ

cosZcosZR

2t1

t2

2t1

2t1

(3.58)

T v R lin h vi nhau theo cng thc:

2

1 Z

ZTR1 =+ (3.59)

Nu 2 mi trng l in mi c 1 = 2 = th (3.58) c vit li

t2

2

11t2

t1

t2

2

11t2

t2

2

11t2

sin1cos

cos2T

sin1cos

sin1cos

R

+

=

+

=

(3.60)

px

t

1Er

1Er

1Hr

1Hr

zpx

zt

zk

y

z

2E

r

2Hr

O


82/108

81

3.5.3. Sng ti vung gc vi mt phng phn cch

Khi sng ti vung gc vi mt phng phn cch 2 mi trng, tc l t =

0, theo nh lut khc x ta c cos= 1 v do gc khc x= 0. H s khc

x v h s phn x trong cc biu thc ca (3.55) v (3.58) c dng n gin

nh sau:

21

2

21

21

12

2ng

12

12ng

ZZZ2

T,ZZZZ

R

ZZZ2

T,ZZZZ

R

+=

+

=

+=

+

=

(3.61)

3.5.4. Sphn x ton phn

Nu mi trng 1 c chit sut ln hn mi trng 2 n1 > n2, theo (3.50) ta

c:

t

2

1 sinnn

sin = (3.62)

c ngha l > t. Khi ta s c gc ti gii hn 0 < 0 0 th sng khc x khng i vo mi trng 2 m quay

trli mi trng 1 (ng vi > 2

), gi l hin tng phn x ton phn. Gc

0 gi l gc gii hn c xc nh theo cng thc:

1

20 n

narcsin= (3.64)

Hin tng phn x ton phn c ng dng truyn nh sng trong si

quang.


83/108

82

3.5.5. Skhc x ton phn

Nu sng ti truyn n mt phng phn cch vo mi trng 2 m khng

phn x trli mi trng 1 gi l s khc x ton phn. Trong trng hp ny

h s phn x bng 0. Gc ti ng vi hin tng khc x ton phn gi l gc

Brewster, k hiu l b. T (3.55) v (3.58) ta c gc Brewster i vi 2 trng

hp phn cc ngang v ng ca sng ti nh sau:

0sin1ZcosZ0R

0sin1ZcosZ0R

b2

2

12b1

b2

2

11b2ng

=

=

=

=

(3.65)

Nhn xt:

- 2 phng trnh trong (3.65) khng th c nghim ng thi, tc l ch c

1 trong 2 trng hp xy ra hin tng khc x ton phn. LT v TN ch ra

rng ch c sng phn cc ng mi c hin tng khc x ton phn v gc

Brewster bc xc nh nh sau:

2

1btg = (3.66)

- Cc kt qu nhn c i vi sng phn x v khc x ti mt phng

phn cch 2 mi trng l in mi cng ng i vi cc mi trng bt k c

in dn sut 0. Khi cc cng thc Fresnel trong (3.55) v (3.58) ch cn

thay = P v Z = ZP.

3.6. iu kin bin gn ng Leontovic

Xt sng phng khc x ti mt phng phn cch 2 mi trng tin mi

(mi trng 1) vo mi trng c in dn sut ln 2 (mi trng 2), ta c:

2E212P1 tghaykk


84/108

83

t

2E2

1 sintg

sin

(3.68)

Nh vy: vi mi gc ti t khi tho mn iu kin (3.67) th gc khc x

0, c ngha l sng khc x truyn vo mi trng c in dn sut ln theo

phng php tuyn vi mt phng phn cch 2 mi trng khng ph thuc vo

gc ti t.

Nu chn trc z trng vi phng php tuyn ca mt phng phn cch th

Er

v Hr

ca sng khc x trong mi trng 2 c dng:

( ) ( )

==

=

2022P02

202

EkHZkE

HHr

rr

rr

r

r

(3.69)

Trong :

- 0r

l vector n v tip tuyn vi mt phng phn cch 2 mi trng

- H2, E2 l cc thnh phn tip tuyn ca Hr

v Er

ca sng khc xst

mt phng phn cch

Theo iu kin bin tng qut ti mt phng phn cch ta c:

=

=

21

21

HH

EE (3.70)

Suy ra:

= 12P1 HZE (3.71)

(3.71) m t quan h gia cc thnh phn tip tuyn ca Hr

v Er

ca sng in

t phng truyn t mi trng in mi qua mi trng dn in c in dn

sut ln, gi l iu kin bin gn ng Leontovic. Trong thc tiu kin

bin gn ng Leontovic c ng dng tnh tn hao ca sng in t truyn

dc b mt cc kim loi dn in tt.

3.7. Sng phng trong mi trng khng ng hng

3.7.1. Mi trng khng ng hng


85/108

84

Mi trng ng hng c cc tham s in t , , l cc hng s;

Er

//Dr

; Br

//Hr

theo cc phng trnh vt cht:

ED 0rr

= , HB 0rr

= (3.72)

Trong tn ngoi cc mi trng ng hng cn c cc mi trng khng

ng hng, theo cc hng khc nhau cc tham sin t, c gi tr

khc nhau. , c biu din di dng tensor t thm t

v tensor in

thm t

nh sau:

=

=

zzzyzx

yzyyyx

xzxyxx

zzzyzx

yzyyyx

xzxyxx

,tt

(3.73)

Cc phng trnh vt cht trong mi trng khng ng hng s l:

EDr

t

r

= , HBr

t

r

= (3.74)

Hay:

zzzyzyxzxz

zyzyyyxyxy

zxzyxyxxxx

zzzyzyxzxz

zyzyyyxyxy

zxzyxyxxxx

HHHB

HHHB

HHHB

EEED

EEED

EEED

++=

++=

++=

++=

++=

++=

(3.75)

Nhn xt:

- (3.75) cho thy rng Er

# Dr

; Br

# Hr

- Trong thc t khng tn ti cc mi trng m c, u l tensor, chc cc mi trng khng ng hng nh sau:

Mi trng c , l hng s v t thm l tensor t

, gi l mi trng

khng ng hng t quay. Th d: ferrite b t ho bi t trng khng i l

mi trng t quay i vi sng in t, c ng dng trong k thut siu cao

tn lm cc tbiu khin s truyn sng.


86/108

85

Mi trng c , l hng s v in thm l tensor t

, gi l mi

trng khng ng hng in quay. Th d: cht kh b ion ho (plasma) di

tc dng ca t trng khng i l mi trng in quay i vi sng in t.

Tng ion ca kh quyn tri t cng l mi trng in quay i vi sng in

t, khi truyn sng v tuyn trong tng ion cn xt n tnh khng ng hng

ca n.

3.7.2. Tensor tthm v tensor in thm

Ferrite chnh l hp cht Fe3O4 v mt s oxide kim loi khc nh MnO,

MgO, NiO ... va c tnh cht in mi va c tnh cht st t, = 5 20, =

10-4 10-6 (m)-1. Khi khng c t trng khng i , 0Hr

= 0, ferrite biu hin

nh mt mi trng ng hng i vi s truyn sng in t. Khi c t

trng khng i, 0Hr

0, ferrite biu hin tnh cht ca mi trng khng ng

hng t quay i vi s truyn sng in t. Tensor t thm c dng nh

sau:

=

0

x

x

00

0ia

0iat

(3.76)

Trong :

Mme

Hme

a

ia

1

0

0

00

0

M

2M

2 00

yxxy

2M

20M

0yyxxx

=

=

=

==

===

(3.77)

Vi:

- e l in tch ca electron


87/108

86

- m0 l khi lng ca electron

- M l ln ca vector t ho ca ferrite

- l tn s ca sng in t

- M l tn s cng hng t quay

- 0 l hng s t

Kh b ion ho c mt s lng ln cc /tch t do gm electron v ion,

gi l mi trng plasma, c rt ln. Khi khng c t trng khng i , 0Hr

=

0, plasma biu hin nh mt mi trng ng hng i vi s truyn sng in

t. Khi c t trng khng i, 0Hr

0, plasma biu hin tnh cht ca mi

trng khng ng hng in quay i vi s truyn sng in t. Tensor

in thm c dng nh sau:

=

z

x

x

00

0ib

0ibt

(3.78)

Trong :

00

220

00

0

M

2

20

0zzz

2M

20M

0

yxxy

2M

2

20

0yyxxx

m

Ne

Hme

1

b

ib

1

=

=

==

=

==

===

(3.79)

Vi:

- M l tn s cng hng t quay

- e l in tch ca electron


88/108

87

- m0 l khi lng ca electron

- N l s electron trong 1 n v th tch

- 0 l hng sin

- 0 l hng s t

- l tn s ca sng in t

3.7.3. Sng phng trong ferrite b tho

Xt sng phng iu ho truyn dc theo phng ca vector t trng

khng i t ho vt liu ferrite rng v hn. Chn trc z trng vi phng

truyn sng v vector 0Hr

, s dng tensor t thm (3.76) v iu kin ngang

ca sng phng TEM (3.1) cho cc phng trnh Maxwell ta c:

0E

HiaHizE

HiaHizE

0H

Eiz

H

Eiz

H

z

xyx

x

yxx

y

z

yx

xy

=

+=

=

=

=

=

(3.80)

Nghim ca (3.80) c dng:

ikzmymx

ikz

mymx

eHjHiH

eEjEiE

+=

+=

rrr

rrr

(3.81)

Thay (3.81) vo (3.80) ta c:

ak 2x22 = (3.82)

Suy ra:


89/108

88

( )

( )ak

ak

x

x

=

+=

+

(3.83)

Khi vn tc pha v trsng c tnh theo cng thc:

( )

( )

=

+=

=

=

+=

=

+

+

+

aZ

aZ

a

1

kv

a

1

kv

xP

xP

x

ph

x

ph

(3.84)

Cc thnh phn ca Hr

v Er

ca sng phng trong ferrite b t ho:

++

+

++

+

++

=

=

=

xPy

yPx

xy

HZE

HZE

HiH

(3.85)

V

=

=

=

xPy

yPx

xy

HZE

HZE

HiH

(3.86)

Hay di dng vector:

( ) ( )

++

++

+

+

+

=

=

+=+

mxm

P

zktim

HH

kHZE

ejiiHHrrr

rrr

(3.87)

V


90/108

89

( ) ( )

=

=

=

mxm

P

zktim

HH

kHZE

ejiiHH

rrr

rrr

(3.88)

Nhn xt:

- (3.85) v (3.87) m t sng phn cc trn quay phi

- (3.86) v (3.88) m t sng phn cc trn quay tri

Nh vy: khi sng phng truyn trong mi trng ferrite b t ho bi t

trng khng i, mi trng ny th hin cc tham sin t khc nhau i

vi sng phn cc trn quay phi v quay tri ng vi cc s sng k+ v k-; vn

tc pha vph+, vph

- v tr sng ZP+, ZP

- khc nhau. Do t thm ca mi

trng ferrite b t ho c gi tr khc nhau i vi sng phn cc trn quay

phi v quay tri nh sau:

a

a

x

x

=

+=

+

(3.89)

Nhn xt: khi sng phn cc thng truyn trong mi trng ferrite b t ho

dc theo t trng khng i 0Hr

hng theo trc z th vector Hr

ca sng in

t s quay i mt gc . Hin tng quay mt phng phn cc ca sng phn

cc thng truyn trong mi trng ferrite b t ho gi l h/ng Faraday. Gc

quay mt phng phn cc ca Hr

trong 1 n v chiu di trong ferrite gi l

hng s Faraday, k hiu l v c tnh theo cng thc:

( )aa22

kkxx +

=

=

+

(3.90)


91/108

90

Chng 4

NHIU X SNG IN T

4.1. Khi nim

Nu trong mi trng ng nht v ng hng c mt hay mt nhm vt

th m cc kch thc ca chng cbc sng ca sng in t th ti

c th xy ra hin tng sng phn x li mi trng, sng khc x truyn

vo cc vt th v si vng ca sng ti qua cc vt th lm cho cu

trc ca trng sng ti thay i. Hin tng trn gi l s nhiu x sng

in t ti cc v tr bt ng nht ca mi trng. Cc vt th ny gi lvt chng ngi, sng ti gi l sng scp, sng phn x gi l sng th

cp. Trng in t nhiu x ton phn l trng tng hp ca cc sng

scp, sng th cp v sng khc x

Mc tiu: xc nh trng th cp hoc trng ton phn ti mt im bt

k trong khng gian mi trng ng nht v ng hng ti thi im t

bt k khi bit cc tham sin v dng hnh hc ca vt chng ngi,v cu trc ca trng sng scp.

V vt chng ngi c dng hhc rt phc tp v nhng v tr khc nhau

so vi ngun scp, do bi ton nhiu x sng in t ch c th gii

gn ng. Trong thc t ngi ta thng dng cc i lng vt l nh tit

din phn x tng ng, tit din hp th ton phn ... c trng cho s

nhiu x sng in t. Vic gii chnh xc bi ton nhiu x sng in t ch c th thc hin i

vi vt chng ngi c dng hhc n gin nh htr trn nh di v hn,

hcu t rt xa ngun sng s cp, c ngha l cu trc ca ngun v

trng sng scp khng ph thuc vo vt chng ngi.

4.2. Nhiu x ca sng phng trn vt dn tr trn di v hn

4.2.1. Bi ton


92/108

91

- Gi s c mt vt dn in tt dng tr trn bn knh a di v hn t

trong kk v c sng phng iu ho truyn ti vung gc vi trc ca vt dn.

Xc nh trng th cp phn x t vt dn.

- Chn h to tr c trc z trng vi trc ca vt dn v sng phng iu

ho truyn dc theo trc Ox v vung gc vi trc ca vt dn. Khi s phn

cc ca sng ti c th xy ra 2 trng hp: tEr

// Oz v tEr

Oz. Nu sng ti l

sng phn cc thng bt k ca tEr

th n c xem nh l tng hp ca 2 trng

hp trn. Do vic gii bi ton nhiu x sng in

Documents

Ly Thuyet Truong Dien Tu - DHCN