Công thức và Lý thuyết Vật lý 11 NC

8/3/2019 Cng thc v L thuyt Vt l 11 NC

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TRNG NH HUNG T: 0908346838 TRANG 1

21F

21F

12F

q1.q2 >0

r21

F

12F

r

q1.q2 < 0

TM TT CNG THC V L THUYT HON CHNH

CHNG I. IN TCH IN TRNGI. Cach nhiem ien.C 3 cch nhim in mt vt: C xt, tip xc ,hng ngII.nh lut Cu lng:

Lc tng tc gia 2 in tch im q1; q2t cch nhau mt khong r trong mi trng c hng

s in mi l 12 21;F F c:- im t: trn 2 in tch.- Phng: ng ni 2 in tch.- Chiu: + Hng ra xa nhau nu q1.q2 > 0 (q1; q2 cng du)

+ Hng vo nhaunu q1.q2 < 0 (q1; q2 tri du)

- ln: 221

.r

qqkF

; k = 9.1092

2

.N m

C

(ghi ch: F l lc tnh in)

- Biu din:

3.Vt dn in, in mi:+ Vt (cht) c nhiu in tch t do dn in

+ Vt (cht) c cha t in tch t do cch in. (in mi)

4.nh lut bo ton in tch:Trong 1 h c lp v in (h khng trao i in tch vi cc h

khc) th tng i s cc in tch trong h l 1 hng s

III. in trng+ Khi nim:L mi trngtn ti xung quanh in tch v tc dng lc ln in tch khc

t trong n.+ Cng in trng: L i lng c trng cho in trng v kh nng tc dng lc.

EqFq

FE

. n v: E(V/m)

q > 0 : F

cng phng, cng chiu vi E

.

q < 0 : F

cng phng, ngc chiu viE

.

+ng sc in trng: L ng c v trong in trng sao cho hng ca tip tynti bt k im no trn ng cng trng vi hng ca vc t CT ti im .

Tnh cht ca ng sc:- Qua mi im trong .trng ta ch c th v

c 1 v ch 1 ng sc in trng.- Cc ng sc in l cc ng cong khngkn,n xut pht t cc in tch dng,tn cng ccin tch m.

- Cc ng sc in khng bao gi ct nhau.-Ni no c CT ln hn th cc ng sc v

mau v ngc li+in trng u:

- C vc t CT ti mi im u bng nhau.- Cc ng sc ca in trng u l cc ng thng song song cch u nhau

+ Vct cng in trng Edo 1 in tch im Q gy ra ti mt im Mcch Q mt

on r c: - im t: Ti M.


2/21


r r

- Phng: ng ni M v Q- Chiu: Hng ra xa Q nu Q > 0

Hng vo Q nu Q 0 q < 0 ME


3/21


* Ghi ch: cng thc chung cho 3 phn 6, 7, 8:

.

MN MN M N MN

AU V V E d

q

V. Vt dn trong in trng- Khi vt dn t trong in trng m khng c dng in chy trong vt th ta gi l vt dn

cn bng in (vdcb)+ Bn trong vdcb cng in trng bng khng. + Mt ngoi vdcb: cng in trng c phng vung gc vi mt ngoi+ in th ti mi im trn vdcb bng nhau+ in tch ch phn b mt ngoi ca vt,s phn b l khng u (tp trung ch li nhn)

VI. in mi trong in trng- Khi t mt khi in mi trong in trng th nguyn t ca cht in mi c ko dn

ra mt cht v chia lm 2 u mang in tch tri du (in mi b phn cc).Kt qu l trong khiin mi hnh thnh nn mt in trng ph ngc chiu vi in trng ngoi

VII. T in- nh ngha: H 2 vt dn t gn nhau, mi vt l 1 bn t. Khong khng gian gia 2 bn

l chn khng hay in miT in phngc 2 bn t l 2 tm kim loi phng c kch thc ln ,t i din nhau, song

song vi nhau- in dung ca t :L i lng c trng cho kh nng tch in ca t

QC

U (n v l F.)

Cng thc tnh in dung ca t in phng:

d

SC

.4.10.9

.9

. Vi S l phn din tch i din gia 2 bn.

Ghi ch :Vi mi mt t in c 1 hiu in th gii hn nht nh, nu khi s dng m tvo 2 bn t ht ln hn ht gii hn th in mi gia 2 bn b nh thng.

- Ghp t in song song, ni tip

GHP NI TIP GHP SONG SONGCch mc : Bn th hai ca t 1 ni vi bn th nht

ca t 2, c th tip tcBn th nht ca t 1 ni vi bn thnht ca t 2, 3, 4

in tch QB = Q1 = Q2= = Qn QB = Q1 + Q2+ + QnHiu in th UB = U1 + U2+ + Un UB = U1 = U2= = Unin dung

n21B C

1...

C

1

C

1

C

1

CB = C1 + C2+ + Cn

Ghi ch CB < C1, C2 Cn CB > C1, C2, C3

- Nng lng ca t in:2 2. .

2 2 2

QU C U QW

C

- Nng lng in trng: Nng lng ca t in chnh l nng lng ca in trngtrong t in.

T in phng

2

9

. .

9.10 .8.

E VW

vi V=S.d l th tch khong khng gian gia 2 bn t in phng

Mt nng lng in trng:2

8

W Ew

V k


4/21

TRNG NH HUNG T: 0908346838 TRANG 4CHNG II. DNG IN KHNG I

I. DNG IN Dng in l dng cc in tch (cc ht ti in) dichuyn c hngChiu quy c ca dng in l chiu dch chuyn c hng ca cc in tch dng. Dng in c:

* tc dng t (c trng) (Chiu quy c I)* tc dng nhit, tc dng ho hc tu theo mi trng.

Cng dng in l i lng cho bit mnh ca dng in c tnh bi:q: in lng di chuyn qua cc tit din thng ca vt dnt: thi gian di chuyn(t0: I l cng tc thi)

Dng in c chiu v cng khng thay i theo thi gian c gi l dng in khng i (cng gi l dng ip mt chiu).

Cng ca dng in ny c th tnh bi:

qI =

t

trong q l in lng dch chuyn qua tit din thng ca vt dn trong thi gian t.Ghi ch:

a) Cng dng in khng i c o bng ampe k (hay miliampe k, . . . ) mc xen vomch in (mc ni tip).

b) Vi bn cht dng in v nh ngha ca cng dng in nh trn ta suy ra: * cng dng in c gi tr nh nhau ti mi im trn mch khng phn nhnh. * cng mch chnh bng tng cng cc mch r.

II. NH LUT M I VI ON MCH CH C IN TR1) nh lut:

Cng dng in chy qua on mch c c in tr R:- t l thun vi hiu in th hai u on mch.- t l nghch vi in tr.

R

UI (A)

Nu c R v I, c th tnh hiu in th nh sau :UAB = VA - VB = I.R ; I.R: gi l gim th ( st th hay st p) trn in tr.

Cng thc ca nh lut m cng cho php tnh in tr:I

UR ()

2) c tuyn V - A (vn - ampe) l th biu din I theo U cn gi l ng c trng vn - ampe.

i vi vt dn kim loi (hay hp kim) nhit nht nhc tuyn V A l on

ng thng qua gc cc trc: R c gi tr khng ph thuc U.(vt dn tuntheo nh lut m).Ghi ch :Nhc li kt qu tm hiu lp 9.

a) in tr mc ni tip:in tr tng ng c tnh bi:

Rm = Rl + R2+ R3+ + RnIm = Il = I2 = I3 = = InUm = Ul + U2+ U3+ + Un

b) in tr mc song song:

qI =

t

AI

RI

UA B

I

O U

R1 R2 R3 Rn

mm

m

UI =

R


5/21

TRNG NH HUNG T: 0908346838 TRANG 5in tr tng ng c anh bi:

1 2 3

1

m nR

1 1 1 1=

R R R R

Im = Il + I2 + + InUm = Ul = U2 = U3= =Un

c) in tr ca dy ng cht tit din u:: in tr sut (m)

SlR l: chiu di dy dn (m)

S: tit din dy dn (m2)

III. NGUN IN: Ngun in l thit b to ra v duy tr hiu in th duy tr dng in. Mi ngun in u c

hai cc, cc dng (+) v cc m (-). n gin ho ta coi bn trong ngun in c lc l lm di chuyn cc ht ti in (lectron; Ion)

gi cho:* mt cc lun tha lectron (cc m).* mt cc lun thiu lectron hoc tha t lectron hn bn kia (cc dng).

Khi ni hai cc ca ngun in bng vt dn kim loi th cclectron t cc (-) di chuyn qua vt dn v cc (+).Bn trong ngun, cc lectron do tc dng ca lc l di chuyn t cc(+) sang cc (-). Lc l thc hin cng (chng li cng cn ca trngtnh in). Cng ny c gi l cng ca ngun in. i lng c trng cho kh nng thc hin cng ca ngun in

gi l sut in ng E c tnh bi:q

A (n v

ca El V)trong : A l cng ca lc l lm di chuyn in tch t cc ny sang cc kia. ca ngun in.

|q| l ln ca in tch di chuyn.Ngoi ra, cc vt dn cu to thnh ngun in cng c in tr gi l in tr trong r ca ngunin.IV. PIN V ACQUY1. Pin in ho:

Khi nhng mt thanh kim loi vo mt cht in phn th giakim loi v cht in phn hnh thnh mt hiu in th in ho.Khi hai kim loi nhng vo cht in phn th cc hiu in th

in ho ca chng khc nhau nn gia chng tn ti mt hiu in th xc nh. l c s chto pn in ho. Pin in ho c ch to u tin l pin Vn-ta (Volta) gm mt thanh Zn v mt thanh Cu

nhng vo dung dch H2SO4 long.Chnh lch gia cc hiu in th in ho l sut in ng ca pin: E = 1,2V. 2. Acquy

Acquy n gin v cng c ch to u tin l acquy ch (cngi l acquy axit phn bit vi acquy kim ch to ra v sau)gm:

* cc (+) bng PbO2* cc (-) bng Pb

nhng vo dung dch H2SO4 long.Do tc dng ca axit, hai cc ca acquy tch in tri du v hot

ng nh pin in ho c sut in ng khong 2V.

RnR3R2R1mm

m

UI =

R


6/21

TRNG NH HUNG T: 0908346838 TRANG 6 Khi hot ng cc bn cc ca acquy b bin i v tr thnh ging nhau (c lp PbSO4 Ph bn

ngoi). Acquy khng cn pht in c. Lc phi mc acquy vo mt ngun in phc hi ccbn cc ban u (np in).

Do acquy c th s dng nhiu ln. Mi acquy c th cung cp mt in lng ln nht gi l dung lng v thng tnh bng n v

ampe-gi (Ah).1Ah = 3600C

IN NNG V CNG SUT IN - NH LUT JUN LENXI. CNG V CNG SUT CA DNG IN CHY QUA MT ON MCH1. Cng:Cng ca dng in l cng ca lc in thc hin khi lm di chuyn cc in tch t do trongonmch.Cng ny chnh l in nng m on mch tiu th v c tnh bi:

A = U.q = U.I.t (J)U : hiu in th (V)I : cng dng in (A); q : in lng (C); t : thi gian (s)

2 .Cng sutCng sut ca dng in c trng cho tc thc hin cng ca n. y cng chnh l cng sut intiu th bi on mch.

Ta c : .A

P U It

(W)

3. nh lut Jun - Len-x:Nu on mch ch c in tr thun R, cng ca lc in ch lm tng ni nng ca vt dn. Kt qu lvt dn nng ln v to nhit.Kt hp vi nh lut m ta c:

22. .

U A Q R I t t

R (J)

4. o cng sut in v in nng tiu th bi mt on mch

Ta dng mt ampe - k o cng dng in v mt vn - k o hiu in th. Cng sut tiu thc tnh hi:

P = U.I (W)-Ngi ta ch to ra ot-k cho bit P nh lch ca kim ch th.- Trong thc t ta c cng t in (my m in nng) cho bit cng dng in tc in nng tiu thtnh ra kwh. (1kwh = 3,6.10

6J)

II. CNG V CNG SUT CA NGUN IN1. CngCng ca ngun in l cng ca lc l khi lm di chuyn cc in tch gia hai cc duy tr hiu inth ngun. y cng l in nng sn ra trong ton mch.

Ta c : ItqA (J) : sut in ng (V)I: cng dng in (A)q : in tch (C)

2. Cng sut

Ta c : It

AP . (W)

III. CNG V CNG SUT CA CC DNG C TIU TH IN

Hai loi dng c tiu th in:

I

UA B

* dng c to nhit* my thu in


7/21

TRNG NH HUNG T: 0908346838 TRANG 71. Cng v cng sut ca dng c to nhit:

- Cng (in nng tiu th):2

2. .U

A R I t t R

(nh lut Jun - Len-x)

- Cng sut :2

2.U

P R IR

2. Cng v cng sut ca my thu ina) Sut phn in

- My thu in c cng dng chuyn ho in nng thnh cc dng nng lng khc khng phi l ninng (c nng; ho nng ; . . ).Lng in nng ny (A) t l vi in lng truyn qua my thu in.

tIqA pp ...

p : c trng cho kh nng bin i in nng thnh c nng, ho nng, .. . ca my thu in v gi l

sut phn in.-Ngoi ra cng c mt phn in nng m my thu in nhn t dng in c chuyn thnh nhit vmy c in tr trong rp.

tIrQ p ..2

- Vy cng m dng in thc hin cho my thu in tc l in nng tiu th bi my thu in l:tIrtIQAA pp ....

2 - Suy ra cng sut ca my thu in:

2.. IrIt

AP pp p .I: cng sut c ch; pr .I

2: cng sut hao ph (to nhit)

b) Hiu sut ca my thu in

Tng qut : H(%) = =

Vi my thu in ta c:

U

Ir

UtIU

tIH

ppp .1

..

..

Ghi ch : Trn cc dng c tiu th in c ghi hai chi s: (V d: 100W-220V)* P: cng sut nh mc.* U: hiu in th nh mc.

NH LUT M TON MCH, CC LOI ON MCHI. NH LUT M TON MCH1. Cng dng in trong mch kn:

- t l thun vi sut in ng ca ngun in- t l nghch vi in tr ton phn ca mch.

RrI

Ghi ch:

* C th vit : IrUIrR AB ).(

Nu I = 0 (mch h) hoc r


8/21

TRNG NH HUNG T: 0908346838 TRANG 8* Nu mch ngoi c my thu in ( p ;rP) th nh lut m tr thnh:

p

p

rrRI

* Hiu sut ca ngun in:

rR

RIrU

P

P

A

AH

tp

ich

tp

ich

1

II. NH LUT M I VI CC LOI MCH IN1. nh lut Ohm cha ngun (my pht):

Rr

UI AB

i vi ngun in :dng in i vo cc m v i ra t cc dng.UAB: tnh theo chiu dng in i t A n B qua mch (UAB = - UBA).

2. nh lut Ohm cho on mch cha my thu in:

Rr

UI

p

pAB

i vi my thu p : dng in i vo cc dngv i ra t cc m.

UAB: tnh theo chiu dng in i t A n B qua mch.3. Cng thc tng qut ca nh lut Ohm cho on mch gm my pht v thu ghp ni tip:

p

pAB

rrR

UI

Ch :

UAB: Dng in i t A n B (Nu dng in i ngc li l: -UAB) : ngun in (my pht) ; p : my thu. I > 0: Chiu dng in cng chiu chn.

I < 0: Chiu dng in ngc chiu chn. R: Tng in tr cc mch ngoi.

r: Tng in tr trong ca cc b ngun my pht.rp: Tng in tr trong ca cc b ngun my thu.

4. Mc ngun in thnh b:a. Mc ni tip:

nb

n

rrr

...

...

21

21

ch :Nu c n ngun ging nhau.

nrr

n

b

b

b. Mc xung i:

21

21

rrrb

b

c. Mc song song( cc ngun ging nhau).

1 ,r 2,r2 3 ,r3 n ,rn

b ,rb

1,r1 2,r2

1,r1 2,r2 ,r

,r

A B ,r R

I

A B ,r RI

A B ,r R

Iprp

AB

,r

R

I p,rp


9/21


nrrb

b

/

d. Mc hn hp i xng(cc ngun ging nhau).m:l s ngun trong mt dy (hng ngang).n:l s dy (hng dc).

Tng s ngun trong b ngun:N = n.m

Ch ng III. DNG IN TRONG CC MI TRNGI. H thng kin thc trong ch ng

1. Dng in trong kim loi- Cc tnh cht in ca kim loi c th gii thch c da trn s c mt ca cc electron t do trongkim loi. Dng in trong kim loi l dng dch chuyn c hng ca cc lectron t do.- Trong chuyn ng, cc lectron t do lun lun va chm vi cc ion dao ng quanh v tr cn bng cc nt mng v truyn mt phn ng nng cho chng. S va chm ny l nguyn nhn gy ra intr ca dy dnx kim loi v tc dng nhit. in tr sut ca kim loi tng theo nhit .- Hin t ng khi nhit h xung di nhit Tc no , in tr ca kim loi (hay hp kim) gimt ngt n gi tr bng khng, l hin t ng siu dn.

2. Dng in trong cht in phn

- Dng in trong cht in phn l dng chuyn dch c h ng ca cc ion d ng v catt v ion m

v ant. Cc ion trong cht in phn xut hin l do s phn li ca cc phn t cht tan trong mitr ng dung mi.Khi n cc in cc th cc ion s trao i lectron vi cc in cc ri c gii phng ra , hoc

tham gia cc phn ng ph. Mt trong cc phn ng ph l phn ng cc dng tan, phn ng ny xyra trong cc bnh in phn c ant l kim loi m mui cu n c mt trong dung dch in phn.- nh lut Fa-ra-y v in phn.

Khi l ng m ca cht c gii phng ra cc in cc t l vi ng lng gamn

Aca cht v

vi in l ng q i qua dung dch in phn. ( q=It )

Biu thc ca nh lut Fa-ra-y:It

n

A

Fm

1

vi F 96500 (C/mol)3. Dng in trong cht kh

- Dng in trong cht kh l dng chuyn dch c h ng ca cc ion d ng v catt, cc ion m vlectron v ant.

Khi c ng in tr ng trong cht kh cn yu, mun c cc ion v lectron dn in trong cht khcn phi c tc nhn ion ho (ngn la, tia la in....). Cn khi cng in tr ng trong cht kh mnh th c xy ra s ion ho do va chm lm cho s in tch t do (ion v lectron) trong cht khtng vt ln (s phng in t lc).

S ph thuc ca c ng dng in trong cht kh vo hiu in th gia ant v catt c dngphc tp, khng tun theo nh lut m (tr hiu in th rt thp).

- Tia la in v h quang in l hai dng phng in trong khng kh iu kin th ng.

nmrr

m

b

b

r, r,

r, r,


10/21

TRNG NH HUNG T: 0908346838 TRANG 10C ch ca tia la in l s ion ho do va chm khi c ng in tr ng trong khng kh ln hn

3.105 (V/m)- Khi p sut trong cht kh ch cn vo khong t 1 n 0,01mmHg, trong ng phng in c s phngin thnh min: ngay phn mt catt c min ti catt, phn cn li ca ng cho n ant l ctsng ant.Khi p sut trong ng gim d i 10-3mmHg th min ti catt s chim ton b ng, lc ta c tia

catt. Tia catt l dng lectron pht ra t catt bay trong chn khng t do.4. Dng in trong chn khng

- Dng in trong chn khng l dng chuyn dch c h ng ca cc lectron bt ra t catt b nungnng do tc dng ca in tr ng.c im ca dng in trong chn khng l n ch chy theo mt chiu nht nh t ant sang catt.

5. Dng in trong bn dn

- Dng in trong bn dn tinh khit l dng dch chuyn c h ng ca cc lectron t do v l trng.Tu theo loi tp cht pha vo bn dn tinh khit, m bn dn thuc mt trong hai loi l bn dn loi

n v bn dn loi p. Dng in trong bn dn loi n ch yu l dng lectron, cn trong bn dn loi pch yu l dng cc l trng.Lp tip xc gia hai loi bn dn p v n (lp tip xc p n) c tnh dn in ch yu theo mt

chiu nht nh t p sang n.

Ch ng IV. T TRNGI. T TRNG1. Tng tac tTng tac gia nam cham vi nam cham, gia dong ien vi nam cham va gia dong ien vi dongien eu goi la tng tac t. Lc tng tac trong cac trng hp o goi la lc t.2. T trng- Khai niem t trng: Xung quanh thanh nam cham hay xung quanh dong ien co t trng.

Tong quat: Xung quanh ien tch chuyen ong co t trng.- Tnh chat c ban cua t trng: Gay ra lc t tac dung len mot nam cham hay mot dong ien at

trong no.- Cam ng t: e ac trng cho t trng ve mat gay ra lc t, ngi ta a vao mot ai lng vect

goi la cam ng t va k hieu la B

.

Phng cua nam cham th nam can bang tai mot iem trong t trng la phng cua vect cam ngt B

cua t trng tai iem o. Ta quy c lay chieu t cc Nam sang cc Bac cua nam cham th la

chieu cua B

.3. ng sc tng sc t la ng c ve sao cho hng cua tiep tuyen tai bat k iem nao tren ng cung

trung vi hng cua vect cam ng t tai iem o.4. Cac tnh chat cua ng sc t:- Tai moi iem trong t trng, co the ve c mot ng sc t i qua va ch mot ma thoi.- Cac ng sc t la nhng ng cong kn. Trong trng hp nam cham, ngoai nam cham cacng sc t i ra t cc Bac, i vao cc Nam cua nam cham.


11/21

TRNG NH HUNG T: 0908346838 TRANG 11- Cac ng sc t khong cat nhau.- Ni nao cam ng t ln hn th cac ng sct o ve mau hn (day hn), ni nao cam ngt nho hn th cac ng sc t o ve tha hn.5. T trng eu

Mot t trng ma cam ng t tai moi iem eu bang nhau goi la t trng eu.II. PHNG, CHIEU VA O LN CUA LC T TAC DUNG LEN DAY DAN MANG DONG

IEN1. Phng : Lc t tac dung len oan dong ien co phng vuong goc vi mat phang cha oan dongien va cam ng tai iem khao sat .2. Chieu lc t : Quy tac ban tay traiQuy tac ban tay trai : at ban tay trai duoi thang e cac ng cam ng t xuyen vao long ban tay vachieu t co tay en ngon tay trung vi chieu dong ien. Khi o ngon tay cai choai ra 90 o se ch chieucua lc t tac dung len oan day dan.3. o ln (nh luat Am-pe). Lc t tac dung len oan dong ien cng o I, co chieu dai hp vit trng eu B mot goc sinBIF

B o ln cua cam ng t . Trong he SI, n v cua cam ng t la tesla, k hieu la T.III. NGUYEN LY CHONG CHAT T TRNGGia s ta co he n nam cham( hay dong ien ). Tai iem M, T trng ch cua nam cham th nhat la

1B , ch cua nam cham th hai la

2B , , ch cua nam cham th n la nB . Goi B la t trng cua he tai

M th: nBBBB ...21

T TRNG CUA DONG IEN CHAY TRONG DAY DAN CO HINH DANG ACBIET

1. T trng cua dong ien chay trong day dan thang dai

Vect cam ng t B

tai mot iem c xac nh:- iem at tai iem ang xet.- Phng tiep tuyen vi ng sc t tai iem ang xet- Chieu c xac nh theo quy tac nam tay phai

- o lnr

IB 710.2

2. T trng cua dong ien chay trong day dan uon thanh vong tronVect cam ng t tai tam vong day c xac nh:- Phng vuong goc vi mat phang vong day- Chieu la chieu cua ng sc t: Khum ban tay phai theo vong dy cua khung day sao cho chieu t

co tay en cac ngon tay trung vi chieu cua dong ien trong khung , ngon tay cai choay ra ch chieung sc t xuyen qua mat phang dong ien

- o lnR

NIB 7102

R: Ban knh cua khung day danI: Cng o dong ienN: So vong day

3. T trng cua dong ien chay trong ong day danT trng trong ong day la t trng eu. Vect cam ng tB

c xac nh

- Phng song song vi truc ong day

B


12/21

TRNG NH HUNG T: 0908346838 TRANG 12- Chieu la chieu cua ng sc t

- o ln nIB 710.4

Nn : So vong day tren 1m

N la so vong day, la chieu dai ong day

TNG TAC GIA HAI DONG IEN THANG SONG SONG. LC LORENX

1. Lc tng tac gia hai day dan song song mang dong ien co:- iem at tai trung iem cua oan day ang xet- Phng nam trong mat phang hnh ve va vuong goc vi day dan- Chieu hng vao nhau neu 2 dong ien cung chieu, hng ra xa nhau neu hai dong

ien ngc chieu.

- o ln : r

IIF 21710.2 l: Chieu dai oan day dan, r Khoang cach gia hai day dan

2. Lc Lorenx co:- iem at tai ien tch chuyen ong- Phng vuong goc vi mat phang cha vect van toc cua hat mang ien va vect cam

ng t tai iem ang xet- Chieu tuan theo quy tac ban tay trai: at ban tay trai duoi thang e cac ng cam ng t xuyen

vao long ban tay va chieu t co tay en ngon tay trung vi chieu dong ien. Khi o ngon tay cai choaira 90o se ch chieu cua lc Lo-ren-x neu hat mang ien dng va neu hat mang ien am th chieungc lai- o ln cua lc Lorenx vBSinqf : Goc tao bi B,v

KHUNG DAY MANG DONG IEN AT TRONG T TRNG EU1. Trng hp ng sc t nam trong mat phang khung dayXet mot khung day mang dong ien at trong t trng eu B

nam trong mat phang khung day.- Canh AB, DC song song vi ng sc t nen len lc t tac dung lenchung bang khong- Goi 1F

, 2F

la lc t tac dung len cac canh DA va BC.

Theo cong thc Ampe ta thay 1F

, 2F

co- iem at tai trung iem cua moi canh- phng vuong goc vi mat phang hnh ve- chieu nh hnh ve(Ngc chieu nhau)- o ln F1 = F2

Vay: Khung day chu tac dung cua mot ngau lc. Ngau lc nay lam chokhung day quay ve v tr can bang ben2. Trng hp ng sc t vuong goc vi mat phang khung dayXet mot khung day mang dong ien at trong t trng eu B

vuong goc vi

mat phang khung day.- Goi 1F

, 2F

, 3F

, 4F

la lc t tac dung len cac canh AB, BC, CD, DA

Theo cong thc Ampe ta thay 31 FF

, 42 FF

Vay: Khung day chu tac dung cua cac cap lc can bang. Cac lc nay khunglam quay khung.c. Momen ngau lc t tac dung len khung day mang dong ien.

N Q

PM

I1

I2

B

F

C

D

A

B

I

.

2F

1F

+B

1F

2F

3F

4F

A B

D C


13/21

TRNG NH HUNG T: 0908346838 TRANG 13Xet mot khung day mang dong ien at trong t trng eu B

nam trong mat phang khung day

Tong quatVi )n,B(

Ch ng V. CM NG IN T1. T thng qua din tch S:

= BS.cos ; Li (Wb)

Vi L la o t cam cua cuon day VnL 27104 (H)

Nn : so vong day tren mot n v chieu dai

2. Sut in ng cm ng trong mch in kn:

tc

(V)

- ln sut in ng cm ng trong mt on dy chuyn ng:

sinBlvc (V) ),( vB

- Sut in ng t cm:t

iLc

(V) (dau tr ac trng cho nh luat Lenx)

3. Nng l ng t tr ng trong ng dy: 2

2

1LiW (J)

4. Mt nng l ng t tr ng: 27108

1Bw

(J/m3)

Chng VI. KHC X NH SNGI. Hien tng khuc xa anh sangHien tng khuc xa anh sang la hien tng khi anh sang truyen qua mat phan cach gia hai moitrng trong suot, tia sang b be gay khuc (oi hng ot ngot) mat phan cach.

2. nh luat khuc xa anh sang+ Tia khuc xa nam trong mat phang ti va ben kia phap tuyen so vi tia ti. (Hnh 33)+ oi vi mot cap moi trng trong suot nhat nh th t so gia sin cua

goc ti (sini) vi sin cua goc khuc xa (sinr) luon luon la mot so khong oi.So khong oi nay phu thuoc vao ban chat cua hai moi trng va c goila chiet suat t oi cua moi trng cha tia khuc xa (moi trng 2) oi vimoi trng cha tia ti (moi trng 1); k hieu la n21.

Bieu thc: 21sin

sinn

r

i

+ Neu n21 > 1 th goc khuc xa nho hn goc ti. Ta noi moi trng (2)

chiet quang kem moi trng (1).

M : Momen ngau lc t (N.m)I: Cng o dong ien (A)B: T trng (T)S: Dien tch khung day(m2)

M = IBSsin

i

r

N

N

I

S

K

(1

(2


14/21

TRNG NH HUNG T: 0908346838 TRANG 14+ Neu n21 < 1 th goc khuc xa ln hn goc ti. Ta noi moi trng (2) chiet quang hn moi trng (1).+ Neu i = 0 th r = 0: tia sang chieu vuong goc vi mat phan cach se truyen thang.+ Neu chieu tia ti theo hng KI th tia khuc xa se i theo hng IS (theo nguyen l ve tnh thuan

nghch cua chieu truyen anh sang).

Do o, ta co12

21

1

nn .

3. Chiet suat tuyet oi Chiet suat tuyet oi cua mot moi trng la chiet suat cua no oi vi chan khong. V chiet suat cua khong kh xap x bang 1, nen khi khong can o chnh xac cao, ta co the coi chiet

suat cua mot chat oi vi khong kh bang chiet suat tuyet oi cua no. Gia chiet suat t oi n21 cua moi trng 2 oi vi moi trng 1 va cac chiet suat tuyet oi n2 va

n1 cua chung co he thc:1

221

n

nn

Ngoai ra, ngi ta a chng minh c rang:Chiet suat tuyet oi cua cac moi trng trong suot t le nghch vi van toc truyen anh sang trong

cac moi trng o:

2

1

1

2

v

v

n

n

Neu moi trng 1 la chan khong th ta co: n1 = 1 va v1 = c = 3.108 m/s

Ket qua la:2

n =2v

c hay v2 =2n

c .

V van toc truyen anh sang trong cac moi trng eu nho hn van toc truyen anh sang trong chankhong, nen chiet suat tuyet oi cua cac moi trng luon luon ln hn 1.

Y ngha cua chiet suat tuyet oi

Chiet suat tuyet oi cua moi trng trong suot cho biet van toc truyen anh sang trong moi trng onho hn van toc truyen anh sang trong chan khong bao nhieu lan.

HIEN TNG PHAN XA TOAN PHAN VA NHNG IEU KIEN E HIEN TNG XAYRA.

1. Hien tng phan xa toan phanHien tng phan xa toan phan la hien tng ma trong o ch ton tai tia phan xa ma khong co tia

khuc xa.

2. ieu kien e co hien tng phan xa toan phan Tia sang truyen theo chieu t moi trng co chiet suat ln sang moi

trng co chiet suat nho hn. (Hnh 34) Goc ti ln hn hoac bang goc gii han phan xa toan phan (i gh).

3. Phan biet phan xa toan phan va phan xa thong thngGiong nhau

Cung la hien tng phan xa, (tia sang b hat lai moi trng cu). Cung tuan theo nh luat phan xa anh sang .Khac nhau

Hien tng phan xa thong thng xay ra khi tia sang gap mot mat phan cach hai moi trng vakhong can them ieu kien g.

G

S

R

K

I

J

i i/

r

(Hnh 34)

H


15/21

TRNG NH HUNG T: 0908346838 TRANG 15Trong khi o, hien tng phan xa toan phan ch xay ra khi thoa man hai ieu kien tren. Trong phan xa toan phan, cng o chum tia phan xa bang cng o chum tia ti. Con trong phan

xa thong thng, cng o chum tia phan xa yeu hn chum tia ti.

4. Lang knh phan xa toan phanLang knh phan xa toan phan la mot khoi thuy tinh hnh lang tru co tiet dien thang la mot tam giac

vuong can

ng dungLang knh phan xa toan phan c dung thay gng phang trong mot so dung cu quang hoc (nhong nhom, knh tiem vong ).Co hai u iem la t le phan tram anh sang phan xa ln va khong can co lp ma nh gng phang.

Ch ng VII. MT V CC DNG C QUANG

Lng knh1. nh ngha

Lang knh la mot khoi chat trong suot hnh lang tru ng, co tietdien thang la mot hnh tam giac.

ng i cua tia sang n sac qua lang knh Ta ch khao sat ng i cua tia sang trong tiet dien thang ABC

cua lang knh.

Noi chung, cac tia sang khi qua lang knh b khuc xa va tia loluon b lech ve pha ay nhieu hn so vi tia ti.

Goc lech cua tia sang n sac khi i qua lang knhGoc lech D gia tia lo va tia ti la goc hp bi phng cua tia tiva tia lo, (xac nh theo goc nho gia hai ng thang).

2. Cc cng thc ca lng knh:

A'iiD

'rrA

'rsinn'isin

rsinnisin

iu kin c tia l

)sin(sin

2

0

0

Ani

ii

iAgh

Khi tia sng c gc lch cc tiu: r = r = A/2; i = i = (Dm + A)/2

Khi goc lech at cc tieu: Tia lo va tia ti oi xng nhau qua mat phangphan giac cua goc chiet quang A .

Khi goc lech at cc tieu Dmin :2

sin2

sin minA

nAD

THAU KNH MONG1. nh ngha

SR

I

Ji1 i2

r1 r2

A

B C

D


16/21


OF F/

(Hnh 36)

(a)

(b)

(c)

OF/ F

(Hnh 37)

(a)

(b)

(c)

Thau knh la mot khoi chattrong suot gii han bi haimat cong, thng la hai matcau. Mot trong hai mat co thela mat phang.

Thau knh mong la thauknh co khoang cach O1O2

cua hai chom cau rat nho so vi ban knh R1 va R2 cua cac mat cau.2. Phan loaiCo hai loai: Thau knh ra mong goi la thau knh hoi tu.

Thau knh ra day goi la thau knh phan k.ng thang noi tam hai chom cau goi la truc chnh cua thau knh.Coi O1 O2 O goi la quang tam cua thau knh.

3. Tieu iem chnh Vi thau knh hoi tu: Chum tia lo hoi tu tai iem F/ tren truc chnh. F/goi la tieu iem chnh cua thauknh hoi tu.

Vi thau knh phan k: Chum tia lo khong hoi tu thc s ma co ng keo dai cua chung cat nhau taiiem F/ tren truc chnh. F/goi la tieu iem chnh cua thau knh phan k .

Moi thau knh mong co hai tieu iem chnh nam oi xng nhau qua quang tam. Mot tieu iem goila tieu iem vat (F), tieu iem con lai goi la tieu iem anh (F/).

4. Tieu cKhoang cach f t quang tam en cac tieu iem chnh goi la tieu c cua thau knh: f = OF = OF/ .

5. Truc phu, cac tieu iem phu va tieu dien Moi ng thang i qua quang tam O nhng khong trung vi truc chnh eu goi la truc phu. Giao iem cua mot truc phu vi tieu dien goi la tieu iem phu ng vi truc phu o. Co vo so cac tieu iem phu, chung eu nam tren mot mat phang vuong goc vi truc chnh, tai tieu

iem chnh. Mat phang o goi la tieu dien cua thau knh. Moi thau knh co hai tieu dien nam hai benquang tam.

6. ng i cua cac tia sang qua thau knh hoi tuCac tia sang khi qua thau knh hoi tu se b khuc xa va lo ra khoi thau knh. Co 3 tia sang thng gap

(Hnh 36): Tia ti (a) song song vi truc chnh, cho tia lo i qua tieu iem

anh. Tia ti (b) i qua tieu iem vat, cho tia lo song song vi truc chnh.

Tia ti (c) i qua quang tam cho tia lo truyen thang.7. ng i cua cac tia sang qua thau knh phan kCac tia sang khi qua thau knh phan k se b khuc xa va lo ra khoi thau knh. Co 3 tia sang thng

gap (Hnh 37): Tia ti (a) song song vi truc chnh, cho tia lo co ng keo dai i qua tieu

iem anh. Tia ti (b) hng ti tieu iem vat, cho tia lo song song vi truc chnh. Tia ti (c) i qua quang tam cho tia lo truyen thang.

8. Qua trnh tao anh qua thau knh hoi tu


17/21

TRNG NH HUNG T: 0908346838 TRANG 17Vat that hoac ao thng cho anh that, ch co trng hp vat that nam trong khoang t O en F mi

cho anh ao.

9. Qua trnh tao anh qua thau knh phan kVat that hoac ao thng cho anh ao, ch co trng hp vat ao nam trong khoang t O en F mi cho

anh that.

10. Cong thc thau knh/

111

ddf

Cong thc nay dung c ca cho thau knh hoi tu va thau knh phan k.

11. o phong ai cua anh

o phong ai cua anh la t so chieu cao cua anh va chieu cao cua vat:d

d

AB

BAk

''

* k > 0 : Anh cung chieu vi vat.* k < 0 : Anh ngc chieu vi vat.Gia tr tuyet oi cua k cho biet o ln t oi cua anh so vi vat. Cong thc tnh o tu cua thau knh theo ban knh cong cua cac mat va chiet suat cua thau knh:

21

11)1(

1

RRn

fD .

Trong o, n la chiet suat t oi cua chat lam thau knh oi vi moi trng at thau knh. R1 va R2 laban knh hai mat cua thau knh vi qui c: Mat lom: R > 0 ; Mat loi: R < 0 ; Mat phang: R =

MT_CC TT CA MT

a/. nh nghave phng dien quang hnh hoc, mat giong nh mot may anh,cho mot anh that nho hn vat tren vong mac.b/. cau tao

thuy tinh the: Bo phan chnh: la mot thau knh hoi tu co tieu c f thay oi c vong mac: man anh, sat day mat ni tap trung cac te bao nhay sang dau cac day than kinhth giac. Tren vong mac co ien vang V rat nhay sang.

ac iem: d = OV = khong oi: e nhn vat cac khoang cach khac nhau (d thay oi) => fthay oi (mat phai ieu tiet )

d/. S ieu tiet cua mat iem cc vien Cv- iem cc can Cc S ieu tietS thay oi o cong cua thuy tinh the (va do o thay oi o tu hay tieu c cua no) e lam cho anhcua cac vat can quan sat hien len tren vong mac goi la s ieu tiet iem cc vien Cv


18/21

TRNG NH HUNG T: 0908346838 TRANG 18iem xa nhat tren truc chnh cua mat ma at vat tai o mat co the thay ro c ma khong canieu tiet ( f = fmax)

iem cc can Cciem gan nhat tren truc chnh cua mat ma at vat tai o mat co the thay ro c khi a ieu tiet toia ( f = fmin)Khoang cach t iem cc can Cc en cc vien Cv : Goi gii han thay ro cua mat- Mat thng : fmax = OV, OCc = = 25 cm; OCv =

e/. Goc trong vat va nang suat phan ly cua mat

Goc trong vat : tgAB

= goc trong vat ; AB: kch thc vat ; = AO = khoang cach t vat ti quang tam O cua mat .- Nang suat phan ly cua matLa goc trong vat nho nhat min gia hai iem A va B ma mat con co the phan biet c hai iem o.

min

1

1' 3500 rad- s lu anh tren vong macla thi gian 0,1s e vong mac hoi phuc lai sau khi tat anh sang kch thch.3. Cac tat cua mat Cach saa. Can thla mat khi khong ieu tiet co tieu iem nam trc vong mac .

fmax < OC; OCc< ; OCv < => Dcan > Dthng- Sa tat : nhn xa c nh mat thng : phai eo mot thau knh phan ky sao cho anh vat

qua knh hien len iem cc vien cua mat.BAAB

knh

d )( VOCd

V

VOCddf

D11111

l = OO= khang cach t knh n mat, neu eo sat mat l =0 th fk = -OVb. Vien thLa mat khi khong ie tiet co tieu iem nam sau vong mac .fmax >OV; OCc > ; OCv : ao sau mat . => Dvien < DthngSa tat : 2 cach :+ eo mot thau knh hoi tu e nhn xa vo cc nh mat thng ma khong can ieu tiet(kho thc hien).+ eo mot thau knh hoi tu e nhn gan nh mat thng cch mt 25cm . (ay la cach thng dung )

BAABknh

25,0d )( COCd

C

COCddf

D11111

KNH LPa/. nh nhga:La mot dung cu quang hoc bo tr cho mat trong viec quang sat cac vat nho. No co tac dung lam tanggoc trong anh bang cach tao ra mot anh ao, ln hn vat va nam trong gii han nhn thay ro cua mat.b/. cau tao

Gom mot thau knh hoi tu co tieu c ngan(c vai cm)


19/21

TRNG NH HUNG T: 0908346838 TRANG 19c/. cach ngam chng

AB 1 1 2 2knhOk matO A B A B

d1 d1 d2 d2d1 < OF ; d1 nam trong gii han nhn ro cua mat: d1 + d1 = OKO ; d2 = OV

'

1 1

1 1 1

K f d d

Ngam chng cc canieu chnh e anh A1B1 la anh ao hiem ti CC : d1 = - (OCC - l)(l la khoang cach gia v tr at knh va mat)

BAAB knh

d )( COCd

C

COCdddf

D11111

Ngam chng CVieu chnh e anh A1B1 la anh ao hiem ti CV : d1

= - (OCV - l)BAAB knh

d )( VOCd

V

VOCdddf

D11111

d/. o boi giac cua knh lup* nh ngha:

o boi giac G cua mot dung cu quang hoc bo tr cho mat la t so gia goc trong anh cua mot vatqua dung cu quang hoc o vi goc trong trc tiep

0 cua vat o khi at vat tai iem cc can cua mat.

00 tan

tan

G (v goc va

0 rat nho)

Vi: 0AB

tg

* bi gic ca knh lp:

Gi l l khong cch t mt n knh v d l khongcch t nh AB n knh (d < 0), ta c :

A'B' A'B'tg

OA d '

suy ra:0

tg A'B' G .

tg AB d'

Hay:

G = k.d' +

(1)

k l phng i ca nh.

-Khi ngm chng cc cn: th d ' do :

d

dkG CC


20/21

TRNG NH HUNG T: 0908346838 TRANG 20-Khi ngm chng cc vin: th VOCd do :

V

VOC

d

dG

-Khi ngm chng v cc: nh AB v cc, khi AB ti CC nn:

AB ABtg

OF f

Suy ra:

G

f

Gc gi tr t 2,5 n 25.

khi ngam chng vo cc+ Mat khong phai ieu tiet

+ o boi giac cua knh lup khong phu thuoc vao v tr at mat.

Gia tr cua G

c ghi tren vanh knh: X2,5 ; X5.

Lu : - Vil l khong cch t mt ti knh lp th khi: 0 l < f GC > GVl = f GC = GVl > f GC < GV

- Trn vnh knh thng ghi gi tr25

( )G

f cm

V d: Ghi X10 th25

10 2, 5( )

G f cm f cm

KNH HIN VIa) nh ngha:Knh hin vi l mt dng c quang hc b tr cho mt lm tng gc trng

nh ca nhng vt nh, vi bi gic ln ln rt nhiu so vi bi gicca knh lp.

b) Cu to: C hai b phn chnh:

- Vt knh O1l mt thu knh hi t c tiu c rt ngn (vi mm), dng to ra mt nh tht rt ln ca vt cn quan st.

- Th knh O2cng l mt thu knh hi t c tiu c ngn (vi cm), dngnh mt knh lp quan st nh tht ni trn.

Hai knh c trc chnh trng nhau v khong cch gia chng khng i. B phn t sng dng chiu sng vt cn quan st.

d) bi gic ca knh khi ngm chng v cc:

- Ta c: 1 1 1 1

2 2 2

A B A Btg

O F f v tg =

AB

Do : 1 10 2

A Btg G x

tg AB f

(1)

Hay 1 2G k G

bi gic Gca knh hin vi trong trng hp ngm chng v cc bng tch ca phng i k1

ca nh A1B1qua vt knh vi bi gic G2ca th knh.


21/21


Hay1 2

.G

f .f

Vi: = /1 2F F gi l di quang hcca knh hin vi.

Ngi ta thng ly = 25cm

KNH THIN VNa) nh ngha:

Knh thin vn l dng c quang hc b tr cho mt lm tng gc trng nh ca nhng vt rt xa(cc thin th).

b) Cu to: C hai b phn chnh:- Vt knh O1: l mt thu knh hi t c tiu c di (vi m)

- Th knh O2: l mt thu knh hi t c tiu c ngn (vi cm)

Hai knh c lp cng trc, khong cch gia chng c th thay i c.

c) bi gic ca knh khi ngm chng v cc:- Trong cch ngm chng v cc,ngi quan st

iu chnh nh A1B2 v cc. Lc

1 1

2

A Btgf

v 1 101

A Btgf

Do , bi gic ca knh thin vn khi ngm chng v cc l :

1

0 2

ftgG

tg f

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Công thức và Lý thuyết Vật lý 11 NC