Ngan Hang Cau Hoi Tdt0141040

8/8/2019 Ngan Hang Cau Hoi Tdt0141040

1/17

TRNG I HC S PHM K THUT TP.HCM

KHOA: IN T

B MN: C C K THUT IN T

Tn hc phn: TRNG IN T M hc phn:0141040

S VHT:3

Trnh o to:i hc

A - NGN HNG CU HI KIM TRA NH GI KIU T LUN.

Chng 1: M U

Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng 1

Cc i lng vect c trng cho trng in t :

E: vect cng in trng

H: vect cng t trng

D:vect in cm

B: vect t cm

J: vect mt dng in dn

H phng trnh Maxwell :

RotH = J +t

D

rotE = -t

B

divB = 0

divD =

D=E; B=H; J=E

Bi ton 1:cho in trng E ,tm t trng H = ?

Bi ton 2: cho t trng H,tm in trng E = ?

Bi ton 3: cho in trng E ,tm ?=

Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 1

Mc tiu kim tra nh gi Ni dung

Mc Nh cc kin thc cn nh :

RotH = J +t

D

rotE = -t

B

divB = 0

divD =

1

Biu mu 3a


2/17

D=E; B=H; J=E

Mc Hiu c cc kinthc hc

Hiu cc ngha ca h phng trnh Maxwell:

a) 2 phng trnh (1) v (2) nu ln mi quan h khng kht giatrng in bin thin v trng t bin thin

b) 2 phng trnh (3) v (4) nu ln dng hnh hc ca trng in

v trng tc) C 4 phng trnh nu ln mi quan h khng kht gia trngin t v mi trng cht

Kh nng vn dng cc kinthc hc

cc kin thc m sinh vin phi bit vn dng :

sinh vin phi bit cch tnh cc ton t vect nh grad,div,rot,divgrad trong cc h trc ta khc nhau bng cch s dngbng cc ton t vect c cho trc.

Kh nng tng hp: Bi ton 1:cho in trng E ,tm t trng H = ?

Bi ton 2: cho t trng H,tm in trng E = ?

Bi ton 3: cho in trng E ,tm ?=

Ngn hng cu hi v p n chi tit chng 1

tt Loi Ni dung im1 Cu hi

Cho trng in ( ) sin.cos.1

2ee

rE

r += .Hy tnh =?1

p n Theo phng trinh Maxwell,ta c

= divD =div(

E)=

divETrong h trc ta tr ta c:

divE =

= 0coscos

33=+

rr

vy 0=

1

2 Cu hi Trong mi trng =const, =const, =0, c trng inzetbyaxE .cos.sin.sin =

1.Tm H =?2.CMR : ..222 =+ba

2,5

p n Ta c :B

rotEt

=

m

1

2

Z

E

r

E

rr

rE zr

+

+

)(


3/17

cos ( cos sin cos sin )yz x y x yEE

rotE e e t b by axe a ax byey x

= =

sin( cos sin . sin cos . )y x

tB rotEdt a ax by e b ax by e

= =

0

sin( cos sin . sin cos . )y x

B tH a ax by e b ax by e

= =

Ta c :

( 0)D D

rotH J J E t t

= + = = =

2 2

0

( ) sin sin sin .y x z zH H a b

rotH e ax by t ex y

+= =

M :

0

0

sin sin cos .

sin sin sin .

z

z

D E ax by t eD

ax by t et

= =

=

2 2 2

0 0

DrotH a b

t

= + =

1.5

3 Cu hi Trong mi trng =const, =const, =0, c trng tzetbyaxH .cos.sin.sin =

1.Tm E =?2.CMR : ..222 =+ba

2,5

p n ( 0)D DrotH J J E t t

= + = = =

m sin cos cos . cos sin cos .x yrotH b ax by t e a ax by t e =

suy ra:

0 0

1 sin( cos sin . sin cos . )y x

D tE rotHdt a ax by e b ax by e

= = =

1

Ta c :

0

0

sin sin cos .

sin sin sin .

z

z

B H ax by t e

Bax by t e

t

= =

=

M :2 2

0

( ) sin sin sin .y x z zE E a b

rotE e ax by t ex y

+= =

2 2 2

0 0

BrotE a b

t

= + =

1,5

4 Cu hi Trong mi trng = , = , =0, c trng in 23


4/17

( ) xezttzE 4.010.6cos.100),( 7 = 1.Tm H =?2.Trng in trn c tnh cht th hay khng?

p n Ta c:

( )74 cos 6 10 0.4xy yE

rotE e e t zz

= =

m :( )

( )

7

7

0

2sin 6 10 0.4

3

2sin 6 10 0.4

3

y

y

BrotE B rotEdt t z e

t

B H t z e

= = =

= =

1,5

V 0rotE nn trng in cho khng c tnh cht th 0,55 Cu hi Cho trng in 2 3r zE e .5r e .r.cos e .r = + + . Trng in trn

c tnh cht th hay khng?

1

p n ( ) ( )

23 2cos 0

r z r z z r

z

rE rE e E E E e E rotE e

r z z r r r r e e

= + + =

= + vy trng in khng c tnh cht th

1

Chng 2: TRNG IN T TNH

Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng 2

Cc i lng vect c trng cho trng in tnh:

D: vect in cm,E: vect cng in trng

Cc i lng vect c trng cho trng t tnh:

B: vect t cm

H: vect cng t trng

H phng trnh Maxwell ca trng in t tnh

trng in tnh

rotE = 0; divD = ;

E: vect cng in trng

D:vect in cm; D=E

trng t tnh

RotH = 0; divB = 0;

H: vect cng t trng

B: vect t cm; B=H

4


5/17

Bi ton 1: Tm in trng E , D , = ? bng phng php gii phng trnh Laplace-Poisson

Bi ton 2: Tm in trng E , D , = ? bng phng php s dng nh lut Gauss .

Cc mc tiu kim tra nh gi v dng cu hi gi chng 2

Mc tiu kim tra nh gi Ni dungMc Nh cc kin thc cn nh :

a) phng trnh Laplace-Poisson: = -/

b) nh lut Gauss: .s

D ds q=

Mc Hiu -sinh vin cn phi hiu : trng n tnh v trng t tnh l haimt ca trng n t tnh, chng hon ton c lp vi nhau.

-sinh vin phi hiu cc tnh cht ca trng in tnh,khi nim vnng lng trng in , in dung:

-nng lng trng in :21 1 1. . .

2 2 2EV

W D EdV QU C U= = =

-in dung :Q

CU

=


sinh vin phi bit vn dng phng trnh Laplace Poisson vnh lut Gauss tm trng in tnh..

Kh nng tng hp: Bi ton 1: Tm in trng E , D , = ? bng phng php giiphng trnh Laplace-Poisson?

Bi ton 2: Tm in trng E , D , = ? bng phng php sdng nh lut Gauss ?

Ngn hng cu hi thi v p n chi tit chng 2

tt Loi Ni dung im1 Cu

hiCho qu cu bn knh a, mang in tch vi mt in tch khi =kR,tmi trng khng kh.Hy xc nh E , D , do qu cu ny gy ra bngphng php s dng nh lut Gauss? (bit rng th ti tm ca qu cu bng 0vmi trngtrongqu cu c =const).

2,5

pn

S dung h trc ta cu (HTTC), ta co: 0;0;0 ==

R0.5

Trng hp 1: R < aAp dung nh luat Gauss ta c:

4

1

2

1 ...4...4. RKDRqdsD

s

==

1

5


6/17

U

d

x0

2

12

1.

.RK

ERKD ==

The

.3

..

...

3

0

2

011

RKdR

RKdRE

RR

=== Trng hp 2: R > aAp dung nh luat Gauss ta c:

42

22 ...4...4. aKDRqdsD

s

==

02

4

22

4

2.

..

R

aKE

R

aKD ==

The

+=+== aR

aKaKdREdREdRE

R

a

aR 11.

.3

.)..(.

0

43

20 102

1

2 Cuhi

Cho mt t in phng nh hnh v, gia hai bn cc t l lp in mi c =3 0x. Hy xc nh cng in trng v th gia hai bn cc t(bit rng mi trngtrongt c =const).

2,5

pn

S dung h trc ta cc (HTTC), ta co:

0;0;0 =

=

zyx

0.5

Ap dung phng trnh Laplace-Poisson ta c:

0

02

2

2

..3

x

x=

=

10

20

.4

..3C

x

x+=

210

30 ..4

.CxC

x ++=

1

Ap dng iu kin b ,ta c:

=

=

=

=

UC

dUdC

d

U

2

0

201 .4

.

0)(

)0(

Uxd

Udx ++= .).4

.(

.4

.

0

20

0

30

cng in trng: ).4

.(

.4

..3

0

20

0

20

d

UdxgradE ==

1

6


7/17

b

ar

3 Cuhi

cho mt t in cu nh hnh v, mi trng gia hai bn cct c r = 5, ti bn knh R = a c mt phn b intch mt =const. Hy xc nh in dung v nng lngin trng ca t?

3

pn S dng HTTC, ta c : 0;0;0 == R 0.5

p dng nh lut Gauss ta c :22

1 ..4....4. aDRqdsDs

==

02

2

2

2

..5

..

R

aE

R

aD ==

=== ab

adR

R

adREU

a

b

a

b

11

.5

..

..5

..

0

2

02

2

1.5

in dung ca t :ab

UqC

11.20 0

==

nng lng in trng : )11(.5

...2

2

.

0

24

ab

aUqWE ==

1

4 Cuhi

Mt tr trn c bn knh a,mang in tch mt phn b u vi mt =const. Hy xc nh E , D , do mt tr ny gy ra(bit rng mitrngbn trongv bn ngoi c =const v (r0)=0).

3

pn S dng h trc ta tr ( HTTT),ta c :0;0;0 == zr 0.5

Trng hp 1: r > ap dng nh lut Gauss ta c:

aLDrLqdsD

s

...2.....2. 11 ==

.

..11 r

aE

r

aD ==

r

radr

r

adrE

r

r

r

r

011 ln

..

.

..

00

===

1.5

Trng hp 2 : r < a

p dng nh lut Gauss ta c : 0....20. 22 === DrLqdsDs

00 22 == ED

a

radr

r

adrE

a

r

a

r

012 ln

..

.

..

00

===

1

7


8/17

5 Cuhi

in tch phn b mt trn hai mt tr r=a va r=b >a c dng :

0

0

khi r a

akhi r b

b

= =

=hy xc nh E , D , trong cc min ?bit rng (a)=0

3

pn S dng HTTT,ta c : 0;0;0 == zr 0.5

Trng hp 1: r < a

p dng nh lut Gauss ta c: 0....20. 11 === DrLqdsDs

00 11 == ED

0.0 11

== r

drE

0.5

Trng hp 2: b > r > a

p dng nh lut Gauss ta c: aLDrLqdsDs

...2.....2. 022 ==

.

.. 02

02 r

aE

r

aD ==

r

aadrEdrE

r

a

rln

... 0202

===

1

Trng hp 3: r > b

p dng nh lut Gauss ta c: 0....20. 33 === DrLqdsDs

00 33 == ED

a

badrE

r

aln

.. 03

==

1

6 Cuhi Trong h trc ta tr tn ti hm th c dng : 0

1510.E .cos . r

r =

, trong E0l hng s. Hi hm th cho c tha mn phng trnh Laplacehay khng?

2,5

pn S dng

HTTT, ta c : 0;0;0 =

zr

hm th cho c tha mn phng trnh Laplace khi: 0=

0.5

m :

+

=

2

2

11

rrr

rr

=

rrE

rr

rr

115cos..10

130

=

rrE

r

115.cos..10

130

2

2

2

8


9/17

vy : 0= nn hm th cho tha mn phng trnh Laplace

7 Cuhi

Gia hai bn cc phng song song cch nhau khong cch x=d, c cng

in trng bin thin theo quy lut :2

x 0 2

xE e .E 1

d

=

ur uur.

a) Hy xc nh v hiu in th gia hai bn cc t (bit rng th ti dl th thp)

b) Nu t in trn c t tip vo hiu in th U1 th cng intrng thay i nh th no?

3

pn

S dung h trc ta cc ( HTTC), ta co:

0;0;0 =

=

zyx

Ta co : 20 2..)(

.d

xE

x

EdivEdivD

=

===

Hiu in th : )3

.(. 00

ddEdxEU

d

==

1

Khi at vao hieu ien the U1 th ta co:

20

2

2 ..2

d

xE

x=

=

12

2

0.

Cd

xE

x+=

212

30 ..3

.CxC

d

xE ++=

Ap dng iu kin b ,ta c:

=

=

==

12

1011 3

0)(

)0(

UCdUEC

d

U

110

2

3

0 ).3

(.3

.Ux

d

UE

d

xE++=

cng in trng )3

(. 102

20

d

UE

d

xEgradE ==

2

8 Cu

hi

Hai in cc phng cch nhau khong cch d=50mm c ni vi ngun c

hiu in th U=500V, gia hai in cc c in tch phn b di dng mt in tch khi : 0300. .x = . Hy tnh th , cng in trng E tiv tr x=25mm.

3

pn

S dung HTTC, ta co: 0;0;0 =

=

zyx

Ap dng phng trnh Laplace- Poisson : xx

.3002

2

==

1

9


10/17

12

150 Cxx

+=

213 .50 CxCx ++=

Ap dng iu kin b ,ta c :

=

=

=

=

UCd

UdC

d

U

2

21 50

0)(

)0(

Vy : Uxd

Udx ++= )50(50

23

V250500025.0)05.0

500)05.0(50()025.0.(50 23)025.0( ++=

cng in trng: )50(150 22d

UdxgradE ==

( ) )/(10000)05.0

500)05.0(50()025.0.(150

22025.0 mVE =

2

Chng 3: TRNG IN T DNG

Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong ch ng 3

Cc i lng vect c trng cho trng in dng trong vt dn :


,E: vect cng in trng

Cc i lng vect c trng cho trng t dng:

B: vect t cm

H: vect cng t trng

H phng trnh Maxwell ca trng in dng

rotE = 0

divJ = 0

J=E

H phng trnh Maxwell ca trng t dng

RotH = J

divB = 0

B=H

Bi ton 1:Tnh in tr cch in v dng in r ca t in, cp tr ng trc.Bi ton 2: Tnh t trng A,B,H trong v ngoi cp tr ng trc .




H phng trnh Maxwell ca trng in dng

10


11/17

rotE = 0

divJ = 0

J=E

H phng trnh Maxwell ca trng t dng

RotH = J

divB = 0

B=H

Phng trnh Laplace-Poisson dnh cho t th vect A :

A J= V

nh lut Amper-Maxwell :

L

Hdl I =

Mc Hiu sinh vin phi hiu cc tnh cht ca trng in dng:

-Tnh cht th:

rotE =0-Tnh tiu tn:

p=J.E ;

P = .V

pdV =U.I-Dng dn chy lin tc:

divJ =0sinh vin phi hiu cc tnh cht ca trng t dng,khi nim v

nng lng trng t, in cm

-nng lng trng t :21 1 1. . .

2 2 2MV

W B HdV I L I = = =

-in cm : LI

=


-sinh vin phi bit vn dng h phng trnh Maxwell tnh intr cch in v dng in r.

-sinh vin phi bit vn dng phng trnh Laplace-Poisson dnh cho

t th vect A v nh lut Amper-Maxwell tnh t trng A,B,Htrong v ngoi cp tr ng trc.

Kh nng tng hp: Bi ton 1:Tnh in tr cch in v dng in r ca t in, cptr ng trc

Bi ton 2: Tnh t trng A,B,H trong v ngoi cp tr ng trc



12/17

U

d

x0

tt Loi Ni dung im1 Cu hi Cho mt t in phng nh hnh v,gia hai bn cc t l lp in mi c

0

1

4x 8 =

+, r=3. Hy xc nh cng in trng ,in tr

cch in v mt in tch khi gia hai bn cc t?(bit rng dintch bn cc t l S ?

2,5

p nS dng HTTC, ta c: 0;0;0 =

=

zyx

0.5

Ta c : xtheoconstJdivJ ==0

( ) ( )ddJdxEUxJJEd

82.84 2

0 00

+==+==

dd

USSJI

dd

UJ ro

82

...

82

.2

02

0

+==

+=

1

( )

( )( )

( )ddxU

EDdd

xUE

82

84..3.

82

842

0

2 +

+==

+

+=

(

0

2

.

82

S

dd

I

UR

rocd

+==

dd

UdivD

82

.122

0

+==

1

2 Cu hi T in cu c bn knh trong a=1cm;bn knh ngoi b=5cm;gia 2 ct t

l lp in mi cC

R= , (C =10-4 s). Dong ien ro chay

qua lp ien moi co cng o I=0.2A, haytnh:hieu ien the gia hai cot tu, ien dan rocua tu.

2,5

p nS dng HTTC, ta c : 0;0;0 =

=

R

divJ=02

2( : )

K J K R J K K const J E

R CR = = = =

1

24 . 44

S

II JdS R J K K

= = = =

vy : ( ) ln 2564

b

a

I bU a EdR V

C a

= = = =

1

in dn r : G = I/U =7,82.10-4 S 0.5

12


13/17

b

a

2

1

I

3 Cu hi Mt dy dn c bn knh a,mang dng in vi mt 0jej z= , ttrong mi trng khng kh .Hy xc nh H , B , A do dy dn gy ra?Bit rng mi trng bn trong dy dn c r=3 v A(r=0)=0

2,5

p nS dng HTTT, ta co: 0;0;0 =

=

zr 0.5

r < ap dng nh lut Ampere ta c:

2

...2...2.

2

011

rjHrIdlH

C

==

00

10

1 .32

.

2

.

rjB

rjH ==

The4

...3

2

...32

00

0

00

011

rjdr

rjdrBA

rr ===

1

r > ap dng nh lut Ampere ta c :

2

...2...2.

2

022

ajHrIdlH

C

==

r

ajB

r

ajH

.2

..

.2

.2

00

2

2

0

2

==

ajajdrBdrBdrBA

r

a

ar

ln2

..

4

...3...

2

00

2

002

01

02

===

1

4 Cu hi Cho mot tru mang ien nh hnh ve. Tnh ien cam

tren mot n v dai cua day dan.

3

p nS dng HTTT, ta co: 0;0;0 =

=

zr 0.5

a < r < b

p dng nh lut Ampere ta c : 2 2. 2. . .C

H dl I r H I = =2

2 22 2

IIH B

r r

= =

2 22 2

2 2 2 2

1 .ln .2 .1 ln2 8 4M

V

I Ib bW B H dV a a

= = =

1

0 < r < a

p dng nh lut Ampere ta c:2

1 1 2. 2. . . .

C

rH dl I r H I

a= =

11 12 2

. ..

2 2

I rI rH B

a a

= =

1

13


14/17

2 24

1 11 1 1 2 4

1. .2 .1

2 8 4 16MV

I IaW B H dV

a

= = =

1 2 1 2

2

2( )ln

8 2M MW W bLI a

+= = +

0.5

Chng 4: TRNG IN T BIN THINCc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng 4

Cc i lng vect c trng cho trng in t bin thin:

E: vect cng in trng

H: vect cng t trng

D:vect in cm

B: vect t cm


H phng trnh Maxwell catrng in t bin thin:

RotH = J +t

D

rotE = -t

B

divB = 0

divD =

D=E; B=H; J=E

Bi ton 1:Tnh cc gi tr c trng ca sng nh : bc sng ,h s pha, tn s Bi ton 2:Xc nh sng in ,sng t . .




a)Cc i lng vect c trng cho trng in t bin thin:

b)S lan truyn ca sng in t phng trong mi trng in mi

l tngc)S lan truyn ca sng in t phng trong mi trng vt dn ltng

Mc Hiu H phng trnh Maxwell catrng in t bin thin:

RotH = J +t

D

14


15/17

rotE = -t

B

divB = 0

divD =

D=E; B=H; J=E

S lan truyn ca sng in t phng trong mi trng inmi l tng:

H s tt: 0 =

H s pha : .c

= =

2.

cc T

f

= = = ; v

= ;

2.f

=

= :tng tr sng

1sH e X E

=

uur urr; trong se l vect n v ch hng truyn

sng

= EXH :vect mt dng cng sut (vect Poyting )

S lan truyn ca sng in t phng trong mi trng vt dnl tng

H s tt , xuyn su :

2

1 2

= =

= =


sinh vin phi bit vn dng cc cng thc tnh khi sng in tphng trong mi trng in mi l tng xc nh sng in,sng t v tnh cc gi tr c trng ca sng nh : bc sng ,hs pha, tn s

Kh nng tng hp: Bi ton 1:Tnh cc gi tr c trng ca sng nh : bc sng ,hs pha, tn s

Bi ton 2:Xc nh sng in ,sng t . .


tt Loi Ni dung im

15


16/17

1 Cu hi Trong mi trng chn khng c mt sng in t phng lan truyntheo phng z ,vi cng t trng c dng:

( ) yeztCosztH .10.2.10.4),( 73 = . Hy xc nh : h spha ,bc sng ,cng in trng ?

2

p n Ta c : srad /10..2 7=

Tn s: Hzf 710.2

==

H s pha : mradc

/21,0. 00 ===

1

Bc sng : mf

cTc 30. ===

Tng tr sng: ==

120

0

0

Cng in trng:

( )7

1,51. 2 .10 0, 21 xz E H X e Cos t z e = =

1

2 Cu hi Cng in trng ca song in t phng lan truyn trong mitrng in mi khng tiu tn vi 0 = c dng :

( ) xeztCosztE ..4,010.6.10),( 7 = . Hy xc nh :tn s,bc sng ,vn tc truyn ,h s in mi tng i v cng ttrng ?

2,5

p nTa c : srad /10..6 7= Hzf 710

.2==

H s pha : mrad /.4,0 =

Bc sng: m5.2 ==

0.5

Vn tc truyn: smv /10.5,1 8==

4..

..2

00

2

00 ===

rr

1

Tng tr sng: ===

60

4

120

.0

0

r

Cng t trng:

( )71

0, 053. 6 .10 0, 4. . yS H e X E Cos t z e

= = uur uur ur r

1

3 Cu hi Song ien t phang lan truyen trong khong khtheo phng Z vi he so pha 30rad/m, bien o

cng o t trng la9

1(A/m) va theo hng

y. Hay tnh , f, ),( tzH , ),( tzE ?

2

16


17/17

p n 2. 2.( )

30 15m

= = =

883.10 15.10 ( )

/15

cf Hz

= =

1

( )

8 9

9

30( / ) . 30.3.10 9.10 ( / )1 1( / ) cos 9.10 30

9 9m yrad m c rad s

H A m H t z e

= = = == = +

1

B - HNG DN S DNG NGN HNG CU HI

- Thi im p dng:

- Phm vi cc trnh v loi hnh o to c th p dng: trnh i hc chnh qui

- Cch thc t hp cc cu hi thnh phn thnh cc thi.

- Cc hng dn cn thit khcNgn hng cu hi thi ny c thng qua b mn v nhm cn b ging dy hc phn.

Tp.HCM, ngy 9 thng 5 nm 2007

Ngi bin son(K v ghi r h tn, hc hm, hc v)

ThS Nguyn Ngc Hng

T trng b mn: KS Vi nh Phng

Cn b ging dy 1: ThS Nguyn Ngc Hng

Cn b ging dy 2 ThS Trng Vn Hin

17

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