ngan hang cau hoi tdt0141040
TRANSCRIPT
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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
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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
+
+
)(
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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
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( ) 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
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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
=
Kh nng vn dng cc kinthc hc
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
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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
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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
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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
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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
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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 :
J: vect mt dng in 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 .
Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 3
Mc tiu kim tra nh gi Ni dung
Mc Nh cc kin thc cn nh :
H phng trnh Maxwell ca trng in dng
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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
=
Kh nng vn dng cc kinthc hc
-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
Ngn hng cu hi v p n chi tit chng 311
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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
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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
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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
J: vect mt dng in dn
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 . .
Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 4
Mc tiu kim tra nh gi Ni dung
Mc Nh cc kin thc cn nh :
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
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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
= =
= =
Kh nng vn dng cc kinthc hc
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 . .
Ngn hng cu hi v p n chi tit chng 4
tt Loi Ni dung im
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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
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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