Tài Liệu Điện Quang .-Hoàng Trọng

I HC QUC GIA H NI

TRNG I HC KHOA HC T NHIN

VT L I CNG 2

IN QUANG

H Ni 31/12/2013

I HC QUC GIA H NI

TRNG I HC KHOA HC T NHIN

VT L I CNG 2

IN QUANG

SINH VIN : HONG VN TRNG

LP : K54 a l

IN THOI : 0974 971 149

MAIL : [email protected]

H Ni 31/12/2013

Li chia s

in Quang l mn hc thuc "Khi kin thc c bn chung ca nhm ngnh" v cc ngnh o to ca trng HKHTN, HQGHN u hc mn hc ny.

Ni dung ca in Quang gm 2 phn chnh c lin quan cht ch vi nhau:

+ in hc (bao gm in v T)

+ Quang hc (tnh cht sng v tnh cht ht ca nh sng)

Trong phn in hc l nhng ni dung c bn v trng in v trng t. Trc ht, cn nm c cc khi nim, hiu v nh cc nh lut gii quyt cc bi tp lin quan y l iu kin cn. tip thu kin thc ca mn hc bt kh khn th cn phi xem li nhng kin thc lin quan ti mn Ton nh: o hm vi phn, tch phn hm mt bin (mn Gii tch 1 gio trnh Ton hc cao cp tp 2), tch phn hm nhiu bin (mn Gii tch 2 gio trnh Ton hc cao cp tp 3), ton t rot, ton t div,

Cc nh lut pht biu cho in tch im, ht c bn, yu t dng,.nhng bi ton li cn tm gi tr tng hp. V th, trong qu trnh vn dng l thuyt vo bi tp th thng gp kh khn do lin quan n php tnh tch phn m c th l i tm biu thc di du tch phn.

Bn cht ca php ly tch phn ch l php cng: cng v s cc s hng trong

mi s hng c gi tr v cng nh:

n

1i

in

b

a

)f(xn

ablimdxf(x)

Ta s dng tch phn khi c mt hoc nhiu yu t bin i, v d in trng do cc in tch v tr khc nhau l khc nhau, cc in tch phn b lin tc. Mun tm biu thc di du tch phn th phi xc nh gi tr (v d in trng) do mt yu t vi phn v cng nh gy nn sau ly tch phn cng cc gi tr li vi nhau. Ngoi ra, biu thc di du tch phn c th xut hin vct tc l chiu ca cc vct thay i theo tng yu t vi phn. Nu c 1 yu t bin i th ta c tch phn ca hm 1 bin, nu c 2 yu t bin i th ta c tch phn mt, nu c 2 yu t bin i m bin s ny l hm ca bin s kia th c th a v tch phn ng,

Ngoi ra, mt s nh l v nh lut cn c th hin di dng vi phn v lin quan ti ton t rot, ton t div: nh l O G trong in trng, t trng; nh lut Faraday v hin tng cm ng in t, nh lut Ampere v dng ton phn, cc phng trnh ca Maxwell. V vy cn phi hiu c rot v div:

S

dlE

limErot L0S

(trong S l din tch gii hn bi ng cong kn L)

V

dSE

limEdiv S0V

(trong V l th tch gii hn bi mt kn S)

Mt vn kh khn na l khi xc nh chiu theo quy tc bn tay phi th thng b gng tay. Cc bn c th thay bng quy tc vn inh c cho d tng tng, vi quy c: vn inh c xoay theo chiu kim ng h th inh c s chuyn ng tnh tin v pha trc v ngc li.

Phn Quang hc cng cha nhiu ni dung tng ng vi lch s pht trin ca n. Trong khun kh ca mn hc v cng phc v cho thi kt thc mn, chng ta nn tp trung vo bn cht sng ca nh sng (giao thoa, nhiu x, phn cc) v bn cht ht ca nh sng (hiu ng quang in, hiu ng Compton). Nhn chung, bi tp trong phn ny d hn phn in nhng l thuyt cn phi nh th kh nhiu.

File ny c cu trc nh sau: i vi mi phn th u tin l tm tt l thuyt vi cc cng thc hay s dng c bao quanh bi vin mu , sau l p dng l thuyt vo gii mt s bi ton lin quan. Cui file l mt s thi v kim tra. Cn hiu v ghi nh nhng ni dung l thuyt c bn, cch xy dng cng thc,v rt c th chng s xut hin trong bi thi kt thc mn hc.

Trn y l cht kin thc t i m mnh mun chia s cng cc bn. Do hn ch nhn thc v mn hc nn c th cn ni dung no vit cha ng hoc cha y , rt mong cc bn thng cm v gp mnh chnh sa thm.

Cc bn c iu g thc mc xin gi v a ch: [email protected]

Hoc ng kin ln page: THI HUS KHTN H NI ca trang web facebook.com cng trao i v tho lun.

Hong Vn Trng

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Hong Vn Trng 0974.971.149 1

MC LC

PHN I: IN HC .................................................................................................................. 7 A. L THUYT ......................................................................................................................... 7

1. in trng ........................................................................................................................ 7 a. in tch ......................................................................................................................... 7 b. nh lut Coulomb ......................................................................................................... 7 c. in trng ..................................................................................................................... 8 d. ng sc in trng ................................................................................................... 8 e. Nguyn l chng cht in trng .................................................................................. 8

2. nh l Ostrogradski Gauss (O G) .............................................................................. 9 a. Thng lng in trng ................................................................................................ 9 b. nh l O G ................................................................................................................. 9

3. in th .............................................................................................................................. 9 a. Cng ca lc tnh in.................................................................................................... 9 b. Tnh cht th ca trng tnh in ............................................................................... 10 c. Th nng ca mt in tch trong in trng ............................................................. 10 d. in th - Hiu in th ................................................................................................ 10 e. Mt ng th ................................................................................................................. 11 g. Mi lin h gia in th v cng in trng ..................................................... 11

4. Nng lng in trng ................................................................................................... 12 a. Nng lng tnh in ca vt dn ................................................................................ 12 b. Nng lng ca t in ................................................................................................ 12 c. Nng lng v mt nng lng in trng ........................................................... 13

5. Dng in ......................................................................................................................... 13 a. Mt dng in .......................................................................................................... 13 b. Phng trnh lin tc .................................................................................................... 14 c. Lc l ............................................................................................................................ 14

6. T trng ......................................................................................................................... 14 a. nh lut Ampere v tng tc t gia hai yu t dng c bn .................................. 14 b. T trng ...................................................................................................................... 15 c. nh lut Biot Savart Laplace ................................................................................ 15 d. Lc tc dng ca t trng ln dng in ................................................................... 15

7. nh lut Ampere v dng ton phn v ng dng ......................................................... 16 a. nh lut Ampere v dng ton phn ........................................................................... 16 b. ng dng ca nh lut Ampere v dng ton phn .................................................... 16 c. nh l Ostrogradski Gauss trong t trng ............................................................ 17

8. Lc Lorentz Hiu ng Hall ........................................................................................... 17 a. Lc Lorentz ................................................................................................................... 17 b. Hiu ng Hall ............................................................................................................... 17

9. Cc nh lut v cm ng in t .................................................................................... 18 a. nh lut Faraday ........................................................................................................ 18 b. nh lut Lenz .............................................................................................................. 18

10. Hin tng h cm v hin tng t cm ...................................................................... 18 a. Hin tng h cm ....................................................................................................... 18 b. Hin tng t cm ........................................................................................................ 19

11. Nng lng t trng .................................................................................................... 19 a. Nng lng t trng ................................................................................................... 19 b. Mt nng lng t trng ....................................................................................... 19

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12. H phng trnh Maxwell v h qu ............................................................................. 19 B. BI TP .............................................................................................................................. 21

Bi 1: in trng ca in tch im ................................................................................. 21 Bi 2: in th ca in tch im ....................................................................................... 22 Bi 3: in trng ca lng cc in ................................................................................ 22 Bi 4: in th ca lng cc in ...................................................................................... 24 Bi 5: in trng v in th ca dy dn thng di tch in .......................................... 24

Trng hp 1: M nm trn ng trung trc ca si dy v cch trung im ca si dy

mt khong l z. Xt trng hp c bit khi a ....................................................... 24 Trng hp 2: M nm trn ng thng AB v cch trung im O mt khong l z...... 27 Trng hp 3: M nm trn ng thng vung goc vi mt u ca si dy v cch u si dy mt khong l z .................................................................................................... 29 Trng hp 4: M nm trn ng thng vung goc vi si dy ti im cch u si dy mt khong l h, M cch si dy mt khong l z. y l trng hp tng qut cho 3 trng hp trn. ............................................................................................................ 31

Bi 6: in trng v in th ca vng dy trn tch in ................................................ 33

Trng hp 1: Vng dy trn tm O bn knh R, tch in q, mt in di l . im M nm trn trc ca vng dy v cch O mt khong l z. Xt trng hp c bit khi z bng 0. .............................................................................................................................. 33 Trng hp 2: Mt ng hnh tr rt mng co chiu cao h, tch in dng vi mt

in mt l . im M nm trn trc ca hnh tr v cch tm ca mt y mt khong l z .................................................................................................................................... 35

Trng hp 3: Bn cu rng co bn knh R, tch in dng vi mt in mt . im M nm ti tm ca bn cu..................................................................................... 36

Bi 7: in trng v in th ca a trn tch in .......................................................... 39

Trng hp 1: a trn tm O bn knh R, tch in dng vi mt in mt . im M nm trn trc ca a v cch O mt khong l z. Xt trng hp c bit khi z >> R

v khi R . ................................................................................................................... 39 Trng hp 2: Hnh tr c bn knh R, tch in dng vi mt in khi l . im M nm trn trc ca khi tr v cch tm ca mt y mt khong l z. ....................... 41

Bi 8: Xc nh vc t cm ng t B do mt dy dn thng di c dng I chy qua ti im M cch dy mt khong l z. ................................................................................. 44

Bi 9: Xc nh vct cm ng t B do mt dng in trn tm O bn knh R, cng I gy ra ti im M nm trn trc ca vng dy trn v cch O mt khong l z. Tm cm ng t ti tm O ca vng dy trn. ................................................................................ 46

Bi 10: Na vng dy dn in bn knh R = 0,49m v khi lng m = 250g, c dng in

I = 25A chy qua (hnh v). Hi cn mt t trng B c hng v ln nh th no na vng dy trn l lng trong khng gian. ............................................................. 47

Bi 11: Mt dy cp ng trc c ng knh trong d1 = 2mm v ngoi bc ch ng knh

d2 = 8 cm, gia li v v bc l cht in mi c hng s in mi = 3. Trong li v v bc tch in tri du nhau vi mt in di = 3,14.10-4 C/m. Hy xc nh cng in trng ti cc im cch trc mt khong: ............................................. 48

(a) r1 = 3 cm. ..................................................................................................................... 48 (b) r2 = 10 cm. ................................................................................................................... 48

Bi 12: Cho qu cu khng dn in tm O, bn knh R = 15 cm c tch in u vi

mt in tch khi = 1,699.10-7 C/m3, c t trong chn khng. ........................ 50 (1) Xc nh cng in trng ti im M cch tm O mt on: (a) r1 = 10 cm; (b) r2 = 30 cm. ........................................................................................................................ 50 (2) Ly in th ti v cng bng 0. Xc nh in th ti P cch tm 20 cm. ................ 50

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Bi 13: Mt qu cu kim loi tm O, bn knh R = 15 cm. Ly in th ti v cng bng 0, tch in cho qu cu n hiu in th 1500V. Hy xc nh: ...................................... 53

(a) in tch v mt in tch trn mt qu cu. .......................................................... 53 (b) Cng in trng, hiu in th ti cc im M v N ln lt cch tm O mt khong l 5 cm v 45 cm. ................................................................................................. 53 (c) Mt in trng ti cc im M, N. ....................................................................... 53

Bi 14: Mt dng in thng di v hn c dng in khng i 1A chy qua. Mt khung dy hnh ch nht ABCD t trong mt phng i qua dng in. Cho cnh AB = 30cm, BC = 20cm. on AB song song vi dng in, cch dng in 10cm. Hy xc nh t thng i qua cun dy. Cho hng s t thm ca mi trng bng 1. ............................ 54

Bi 15: Mt dy tch in lin tc nm dc theo trc Ox t im x = x0 n +. Mt in tch di trn dy l 0. Tnh cng in trng v in th ti gc ta O. ... 56

Bi 16: Mt thanh dn hnh tr, khi lng 0.72 kg, bn knh tit din 6 cm, c dng in I = 48A chy qua theo chiu mi tn, nm trn hai thanh ray c di L = 45 cm t song song v cch nhau mt khong d = 12 cm. Ton b h c t trong mt t trng u c ln 0.24 T, hng vung gc vi mt phng cha thanh dn v thanh ray. Thanh dn ng yn mt u ca ray v bt u ln khng trt theo ray. Tnh tc ca thanh dn ti thi im ri khi u kia ca ray. .......................................... 57

Bi 17: Mt dy dn c un nh hnh v, c dng I = 5A chy qua. Bn knh cung trn l R = 3 cm. Xc nh ln v hng ca cm ng t ti tm ca cung trn. ............. 58

Bi 18: Mt solenoid vi n = 400 vng/m c dng in bin thin I = 30(1 e-1.6t) A chy qua. Mt cun dy c tng cng N = 250 vng, bn knh 6cm c t ng trc vo trong lng ca solenoid. Tm sut in ng cm ng xut hin trong cun dy. ......... 59

Bi 19: Mt thanh di 14 cm c tch in u, c din tch tng cng l 22 C. Xc nh cng in trng v in th ti im nm trn trc thanh, cch trung im ca thanh mt khong l 36 cm....................................................................................... 60

Bi 20: Mt thanh dn in c mt khi lng l 0.04 kg/m, c treo bng hai si dy dn mm cho dng in I chy qua, t trong t trng Bin = 3,6 T, hng vung gc vo trong mt phng. Dng in I phi c hng v ln nh th no khng c sc cng trn cc dy treo. ..................................................................................................... 62

Bi 21: Mt dy dn gm vng dy trn c bn knh R v hai on dy thng, di, nm trong cng mt mt phng. Dy dn c dng in I = 7A chy qua theo chiu mi tn (hnh v). Tm biu thc ca vct cm ng t ti tm ca vng dy. ........................... 63

Bi 22: Mt cun c 15 vng dy, bn knh R = 10 cm, c cun quanh mt solenoid c bn knh 2 cm v n = 1000 vng/m. Dng in chy trong solenoid theo chiu mi tn (hnh v) v bin thin theo quy lut I = 5sin(120t) A. Tm biu thc ca sut in ng cm ng trong cun c 15 vng dy. .............................................................................. 64

Bi 23: Mt qu cu c, bn knh 40 cm, tch in u trong ton b th tch vi in tch

tng cng l +26C. Tm ln v hng ca cng in trng ti nhng v tr cch tm qu cu mt khong: ........................................................................................ 65

(a) 0 cm. ............................................................................................................................ 65 (b) 10 cm. .......................................................................................................................... 65

(c) 40 cm. .......................................................................................................................... 65 (d) 60 cm. .......................................................................................................................... 65

Bi 24: Mt thanh c di L nm dc theo trc x (hnh v). u bn tri ca thanh c

t ti gc ta . Thanh c tch in khng u vi mt in tch di l = .x ( l mt hng s dng). ............................................................................................... 67

(a) n v ca l g........................................................................................................ 67 (b) Tm in th ti im M cch gc ta mt khong d............................................. 67

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Bi 25: Bn dy dn thng song song di v hn c cng dng in I = 5A (hnh v). Cc dng in A v B hng vung gc vo trong mt phng hnh v. Cc dng C v D hng vung gc ra bn ngoi mt phng hnh v. Tm ln v hng ca cm ng t

B ti im M nm tm hnh vung c cnh 0,2m. ..................................................... 68 Bi 26: Mt qu cu khng dn in ng knh 8 cm, tch in u trong ton b th tch

vi in tch tng cng l +5,7C. Tnh in tch cha trong cc mt cu ng tm vi qu cu c bn knh. ........................................................................................................ 69

(a) r1 = 2 cm. ..................................................................................................................... 69

(b) r2 = 6 cm. ..................................................................................................................... 69 Tm ln v hng ca cng in trng ti cc mt cu ng tm . ............... 69

Bi 27: Tnh cng in trng v in th ti im P nm trn trc ca bn vnh khn

tch in u vi mt in tch mt (hnh v). ........................................................ 71 Bi 28: Mt vt dn hnh tr di v hn, bn knh R, c dng in I chy qua (hnh v) vi

mt dng J khng u trn tit din vt dn, J = br (vi b l hng s v r l khong

cch tnh t trc ca hnh tr). Tm ln cm ng t B ti nhng im nm cch trc hnh tr mt khong: ....................................................................................................... 73

(a) r1 < R. ...................................................................................................................... 73

(b) r2 > R. ..................................................................................................................... 73 Bi 29: Thanh dn c th trt khng ma st trn hai thanh ray song song, t cch nhau

mt khong l. Ton b h c t trong t trng u B hng vung gc vo trong

mt phng hnh v. Mt lc khng i c ln 1NFapp lm thanh dn trt u

sang phi vi tc 2m/s. B qua lc ma st. ............................................................... 74

(a) Tnh cng dng in chy trong in tr R = 8. ............................................... 74 (b) Tnh cng sut ta ra trn in tr R. ......................................................................... 74

Bi 30: Mt electrn chuyn ng trn qu o trn (hnh v) c ng nng E = 22,5 eV (1eV = 1,6.10-19 J), cm ng t B = 4,55.10-4 T. ............................................................ 75

(a) Tnh bn knh qu o in t, bit khi lng electrn m = 9,1.10-31 kg v in tch q = 1,6.10-19 C. .................................................................................................................. 75 (b) Chu k chuyn ng ca electrn. .............................................................................. 75

Bi 31: Mt si dy thng t nm ngang c dng I = 28A. Hi chiu v ln ca t trng bng bao nhiu n gy ra mt lc cn bng vi trng lng ca si dy. Cho bit khi lng trn mt n v chiu di ca si dy l: m/L = 46,6 g/m. .................... 77

Bi 32: Mt dy dn thng c tch thnh hai na vng trn nh nhau, c dng I chy qua. Xc nh cng t trng ti tm O ca vng trn. .......................................... 77

Bi 33: Tnh vct cm ng t ti tm C ca hnh c dng di y (hnh v) khi c dng I chy qua. ......................................................................................................................... 78

Bi 34: Hai vng dy dn mt ln mt nh t song song vi nhau (hnh v). Trong vng ln c dng I ang tng. Hi: .......................................................................................... 79

(a) Chiu ca dng in cm ng trong cun nh. ........................................................... 79 (b) Chiu ca lc tc dng ln cun nh. ......................................................................... 79

PHN II: QUANG HC .......................................................................................................... 81 A. L THUYT ....................................................................................................................... 81

1. C s quang hnh hc ...................................................................................................... 81 2. Giao thoa nh sng ........................................................................................................... 82

a. nh sng l mt song ................................................................................................... 82 b. Cng sng .............................................................................................................. 82 c. Giao thoa vn khng nh x ........................................................................................ 83 d. Giao thoa vn nh x .................................................................................................. 85 e. Cc giao thoa k ........................................................................................................... 88

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3. Nhiu x nh sng ............................................................................................................ 90 a. Nguyn l Hugen Fresnel .......................................................................................... 90 b. Nhiu x Fresnel ........................................................................................................... 90 c. Nhiu x Fraunhofer ..................................................................................................... 92

4. Phn cc nh sng ............................................................................................................ 95 5. Lng t nh sng ........................................................................................................... 97

a. Bc x nhit .................................................................................................................. 97 b. Cc nh lut bc x ca vt en tuyt i .................................................................. 97 c. Thuyt lng t nh sng ca Einstein ........................................................................ 98 d. Hin tng quang in ................................................................................................. 99 e. Hiu ng Compton ........................................................................................................ 99

B. BI TP ............................................................................................................................ 100 Bi 1: Chiu chm nh sng trng xung bn mng c chit sut n = 1,33 trong khng kh

vi gc ti 60o, nh sng c bc sng 550nm phn x cho cng cc i vi bc giao thoa bng 2. Hy xc nh b dy ca bn mng. Ngoi nh sng trn cn nh sng n sc no khi phn x cng cho cng cc i. ................................................... 100

Bi 2: Trn b mt ca mt quang c lm bng thy tinh c chit sut n = 1,7 ngi ta ph

mt lp trong sut c chit sut n1 = 1,7 . Hy xc nh b dy ti thiu ca lp trong

sut nh sng c bc sng 550nm khng b phn x. Coi nh sng chiu vung gc.101 Bi 3: Mt cu ca thu knh phng li tip xc vi bn thy tinh. Bn knh cong ca thu

knh l R = 100cm. Chiu chm sng n sc c bc sng = 0,5m ti vung gc vi bn thy tnh sao cho vn Newton xut hin mt trn mt cong ca thu knh. Cho bit chit sut ca vt liu lm thu knh l n1 = 1,5 v chit sut ca thy tinh l n3 = 1,7. ................................................................................................................................. 102

a) Hy xc nh bn knh ca vn ti th 5. ................................................................... 102 b) Khng gian gia thu knh v bn cha y sulphua cacbon c chit sut n2 = 1,63. Hy xc nh bn knh ca vn ti th 5. .............................................................. 102

Bi 4: Hai tm thy tinh di 120 mm c mt u chm nhau cn u kia cch nhau 48 m

to thnh nm khng kh. Chiu chm nh sng n sc c bc sng = 0,48 m xung vung gc vi mt di ca nm. Hy xc nh: .............................................. 104

a) Khong vn. ................................................................................................................ 104 b) S vn giao thoa quan st c. ................................................................................. 104

Bi 6: Mt ngun sng im pht nh sng n sc bc sng = 0,5m c t trn trc vung gc i qua tm ca l trn truyn sng, bn knh r = 1 mm, cch l trn mt khong a = 1m. Hy xc nh khong cch b t mn n im quan st i vi im l trn cha ng 3 i Fresnel. .............................................................................. 105

Bi 7: Chiu chm nh sng n sc bc sng = 0,44 m ti vung gc vi khe hp b rng a. Trn mn quan st t cch khe hp 1m ngi ta o c khong cch t cc tiu nhiu x th 2 n cc i chnh gia l 50cm. Hy xc nh: ............................. 106

a) Gc nhiu x ng vi cc tiu th 2. ......................................................................... 106 b) B rng a ca khe hp. ............................................................................................... 106

Bi 8: Trong mt th nghim nhiu x ca sng phng qua mt khe hp di v hn, b rng khe l a = 1200nm, khong cch t mn n khe hp l 1m, nh sng c bc sng 600nm. Ly chnh gia mn lm gc, hy xc nh v tr gc v v tr trn mn ca cc i ph v cc tiu th nht (v pha gc m) trong cc trng hp khi chm sng ti:107

a) Vung gc vi khe hp. .............................................................................................. 108 b) To vi php tuyn ca khe hp mt gc 30o. ........................................................... 108

Bi 9: Chiu chm sng n sc song song bc sng = 0,55 m ti hai khe hp ging nhau c b rng a = 0,25mm; khong cch hai khe l d = 1,55mm. Mn quan st cch mt phng cha hai khe on D = 1m........................................................................... 109

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a) Xc nh khong cch gia cc cc i giao thoa. ..................................................... 109 b) C bao nhiu vn sng quan st c trong cc i trung tm ca bao hnh nhiu x. ........................................................................................................................................ 109

Bi 10: Chiu mt chm sng n sc bc sng 600nm ti vung gc vi mt cch t c hng s (chu k) l d = 1900nm v s khe l N = 104. Sau cch t t mt thu knh hi t, mn quan st t mt phng tiu din ca thu knh. Hy xc nh: .................... 111

a) V tr v b rng gc ca vch quang ph bc 2. ....................................................... 111 b) Trn mn quan st c bao nhiu vch quang ph. ................................................. 111

Bi 11: Chiu chm nh sng pht ra t ngun Natri ti vung gc vi cch t c cc thng s nh sau: hng s d = 1900 nm v s khe N = 104. Natri c hai nh sng n sc bc sng 589 nm v 589,59 nm. Hy xc nh: .................................................... 113

a) Khong cch gc gia hai vch quang ph bc 2 ca hai nh sng trn. ................... 113 b) Cch t c phn bit c hai vch quang ph bc 1 ca hai nh sng trn khng. Ti sao? ................................................................................................................................. 113

Bi 12: Chiu chm nh sng t nhin c cng Io ti h gm knh phn tch A v knh phn cc P. Hy xc nh gc gia hai quang trc ca hai knh P v A nh sng i qua h I = Io/8. B qua hin tng hp th nh sng khi qua hai knh. ........................ 114

Bi 13: Mt ngi thng thng nhy cm nht i vi nh sng c bc sng = 550 nm. Hy xc nh nhit ca mt hc en tuyt i mt ngi nhn r nht nh sng do n pht ra. ........................................................................................................ 114

Bi 14: Ph bc x ca mt tri cc i bc sng max = 480 nm. Coi b mt ca mt tri nh vt en tuyt i. Hy xc nh nhit b mt v nng sut bc x ton phn

ca mt tri. Cho h s Stefan Boltzmann = 5,67.10-8 W/m2K4. H s dch chuyn Wien b = 2898 m.K. ................................................................................................... 115

Bi 15: Mt ngun sng im cng sut 3W pht nh sng n sc bc sng 589 nm. Hy xc nh s photon i qua tit din 1cm2 theo phng vung gc vi phng truyn, cch ngun 1,75 m. ....................................................................................................... 115

Bi 16: Mt photon nng lng 150 keV tn x n hi trn electron t do ng yn di

gc tn x 90o. Cho bc sng Compton = 2,42.10-12 m, h = 6,625.10-34 J.s, c = 3.108 m/s. Hy xc nh:......................................................................................................... 116

a) Nng lng photon tn x. ......................................................................................... 116 b) ng nng v vn tc ca electron Compton (sau tn x). ........................................ 116

Bi 17: Mt photon nng lng 58 keV tn x n hi trn electron t do ng yn, sau tn x bc sng photon tng ln 25%. Hy xc nh: ....................................................... 117

a) Gc tn x. .................................................................................................................. 117 b) Bc sng v nng lng photon tn x. ................................................................... 117 Cho h = 6,625.10-34 J.s, c = 3.108 m/s, k = 2,43.10-12 m v ly 1eV = 1,6.10-19J. .......... 117

MT S THI V KIM TRA .................................................................................. 119 1. thi cui k h nm 2013 ........................................................................................... 119 2. thi cui k II nm hc 2012 2013 ......................................................................... 120 3. thi cui k h nm 2012 ........................................................................................... 122

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PHN I: IN HC

A. L THUYT

1. in trng

a. in tch

- L mt trong nhng thuc tnh vt l c bn ca vt cht.

- nh lut bo ton in tch: Trong mt h c lp, in tch c bo ton.

- in tch nguyn t: L in tch nh nht c bit trong t nhin. ln:

e = 1,6. 10-19 C (in tch ca electrn)

- Phn loi vt cht theo tnh cht dn in:

+ Vt dn in: l vt cht m cc in tch c th chuyn ng t do trong ton b th tch ca vt.

+ Vt cch in: l vt cht trong cc in tch khng th chuyn ng t do.

+ Vt bn dn: l vt cht ch dn in c mt iu kin xc nh.

b. nh lut Coulomb

- Th hin mi quan h tng tc gia hai in tch im ng yn.

- Ni dung:

Lc tng tc gia hai in tch im c phng nm trn ng thng ni hai in tch, c chiu l lc ht nu hai in tch tri du v l lc y nu hai in tch cng du, c ln t l thun vi tch ln ca hai in tch v t l nghch vi bnh phng khong cch gia chng.

- Biu thc:

Nr

r.

r

qqkF

2

21 (I.1)

Trong :

F l lc tng tc gia hai in tch im, c n v N.

q1, q2 c n v C.

r l khong cch gia hai in tch im, c n v m. Vct r c hng t in tch gy nn tc dng n in tch b tc dng.

k l hng s tnh in,0 l hng s in:

2

29

0

2

212

0C

Nm8,99.10

4

1k;

Nm

C8,85.10

- Lc Coulomb tun theo nguyn l chng cht:

n

1i

iFF

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c. in trng

- Khi nim: in trng l mi trng vt cht tn ti xung quanh in tch v tc dng lc in ln in tch khc t trong n.

- Vct cng in trng: Cng in trng ti mt im l mt i lng vct, c o bng lc tc dng ca in trng ln mt n v in tch dng ti im .

C

N

r

r.

r4

q

q

FE

2

00 (I.2)

+ Hng ca E trng vi hng ca lc tc dng F ln q0

E hng ra ngoi in tch im nu q > 0.

E hng vo trong in tch im nu q < 0.

+ ln:

200 r4

q

q

FE

Vc t r hng t in tch q n im cn kho st in trng.

d. ng sc in trng

- ng sc in trng l ng m tip tuyn ti mi im ca n trng vi phng ca vct cng in trng ti im . Chiu ca ng sc l chiu ca

vct cng in trng E .

- Quy c: S ng sc i qua mt n v din tch t vung gc vi ng

sc th bng ln ca vct cng in trng E .

- Tp hp cc ng sc in trng gi l in ph.

Tnh cht:

+ S ng sc i qua mt n v din tch vung gc vi chng th t l vi ln ca vect cng in trng ti .

+ Hai ng sc bt k khng th ct nhau.

+ ng sc b gin on ti ni c in tch. Cc ng sc i ra t in tch dng v i vo in tch m.

+ Cc ng sc n vung gc vi b mt ca vt dn trong iu kin tnh in.

e. Nguyn l chng cht in trng

- Cng in trng do h cc in tch im gy ra ti mt im bng tng vct cng in trng do tng in tch im gy ra ti im .

C

NEE

n

1i

i (I.3)

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2. nh l Ostrogradski Gauss (O G)

a. Thng lng in trng

- Thng lng in trng gi qua yu t din tch dS l tch v hng gia E

v dS.

E.dS.cosdS.Ed

Trong o:

+ l gc gia E v dS.

+ dSl vct php tuyn ca yu t din tch dS.

+ ln ca vct dSbng ln din tch ca yu t dS.

- Thng lng in trng gi qua mt kn S l:

S

E dS.E (I.4)

Trong trng hp ny, dSl hng ca php tuyn ngoi ca mt S ti yu t dS.

b. nh l O G

Thit lp mi quan h gia in tch v in trng.

- Ni dung: Thng lng in trng gi qua mt kn S bng tng i s cc in

tch cha trong mt kn chia cho hng s in 0.

- Biu thc:

n

1i

i

0S

E q1

dS.E

(I.5)

- Dng vi phn ca nh l (nu in tch phn b lin tc trong th tch ca vt):

0

),,(),,(Ediv

zyxzyx

( l mt in tch khi ti im c ta (x, y, z))

3. in th

a. Cng ca lc tnh in

- Cng cn thit a in tch th q0 t M n N l:

NM0

0

N

M

0MNr

1

r

1

4

qqdlEqA

(I.6)

rM v rN ln lt l khong cch t in tch q ti M v N.

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- Cng ny khng ph thuc vo dng ng cong dch chuyn m ch ph thuc vo v tr im u v im cui.

- Nu c n in tch th cng dch chuyn s l:

n

1i NiMi0

0iMN

r

1

r

1

4

qqA

(I.7)

b. Tnh cht th ca trng tnh in

- Nu mt mi trng c cng dch chuyn in tch trn mt ng cong khp kn bng 0 th mi trng c tnh cht th.

0dl.Eqdl.FA 0

0dl.E (I.8)

- Gi tr dl.E gi l lu s ca vct cng in trng. Lu s ca vct cng in trng dc theo ng cong kn th bng 0.

0Erot

c. Th nng ca mt in tch trong in trng

- Th nng ca mt in tch q0 ti mt im trong in trng l mt gi tr bng cng ca lc tnh in dch chuyn in tch t im ang xt ra xa v cng.

- Th nng ca mt in tch im q0 trong in trng do q gy ra:

r4

qqW

0

0

(I.9)

- Nu h c n in tch th:

n

1i i0

0in

1i

ir4

qqWW

(I.10)

d. in th - Hiu in th

- in th ti mt im c tnh bng cng ca lc tnh in dch chuyn mt n v in tch dng t im ra xa v cng.

- in th ti mt im cch in tch q mt khong r, in tch th q0 l:

r4

q

q

Wdl.EV

00M

(I.11)

- Nu c nhiu in tch qi th:

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Hong Vn Trng 0974.971.149 11

n

1i i0

in

1i

ir4

qVV

(I.12)

- Do cng dch chuyn in tch q0 t M n N:

)V(VqA NM0MN

- Hiu in th:

Hiu in th gia hai im M v N trong in trng l mt i lng v tr s bng cng ca lc tnh in dch chuyn mt n v in tch dng t M n N.

NM0

MNMN VV

q

AU (I.13)

0V

AV MM

e. Mt ng th

- Mt ng th l tp hp nhng im c cng mt in th.

- Mt ng th ca in trng do in tch im gy ra l cc mt cu c tm nm ti ni c in tch.

- Tnh cht:

+ Cc mt ng th khng ct nhau.

+ Cng ca lc tnh in dch chuyn mt in tch trn mt ng th bng 0.

+ Vct cng in trng ti mi im trn mt ng th th vung gc vi mt ng th ti im .

g. Mi lin h gia in th v cng in trng

- Tnh in th t cng in trng:

M

M dl.EV (I.14)

Trong dl l vi phn ca yu t chiu di theo ng dch chuyn.

- Tnh cng in trng t in th:

Vc t cng in trng ti mt im bt k trong in trng bng v ngc du vi gradient ca in th ti im .

VgradE (I.15)

Ex = Vx ; Ey = Vy ; Ez = Vz

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4. Nng lng in trng

a. Nng lng tnh in ca vt dn

- Tnh cht ca vt dn:

+ Vct cng in trng iE ti mi im bn trong vt dn bng 0.

+ Thnh phn tip tuyn tE ca vct cng in trng ti mi im trn b mt vt dn bng 0.

+ in trng E ch cn li thnh phn vung gc: nEE

+ Mt vt dn l mt mt ng th. Vt dn cn bng tnh in l mt khi ng th.

+ Mt vt dn khc nm trong mt vt dn rng s khng b nh hng ca in trng ngoi.

+ S phn b in tch trn b mt vt dn ch ph thuc vo hnh dng ca vt dn . in tch q ch phn b trn b mt ca vt dn.

- Nng lng ca vt dn tch in;

2

2

CV2

1

C

q

2

1qV

2

1W (I.16)

Trong C l in dung ca vt dn. in dung ca vt dn bng s in tch cn truyn cho vt dn in th ca n tng ln 1V.

+ Nu vt dn l mt cu bn knh R th: R4C 0

+ T in phng: d

SC 0 (S din tch mt mt t, d khong cch)

+ T in tr:

1

2

0

R

Rln

2C

l (l chiu di t tr; R1, R2 bn knh

trong v ngoi)

+ T in cu: d

SC 0 (S din tch mt mt cu, d khong cch gia

hai bn t)

b. Nng lng ca t in

- T in l mt h gm hai vt dn c t rt gn nhau. Cc vt dn to nn t c gi l cc bn t.

- in dung ca t:

U

qC (I.17)

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Trong , q l in tch trn mt bn t v U l hiu in th gia hai bn t.

- Ghp t:

+ Ghp ni tip:

n

1i iC

1

C

1

+ Ghp song song:

n

1i

iCC

- Nng lng ca t tch in:

2

2

CU2

1

C

q

2

1qU

2

1W (I.18)

c. Nng lng v mt nng lng in trng

- Nng lng in trng:

.VE2

1W 20 (I.19)

Trong , V l th tch phn khng gian c in trng.

- Mt nng lng in trng:

2

0E2

1

V

Ww (I.20)

5. Dng in

a. Mt dng in

- Dng in: l dng chuyn di c hng ca cc ht mang in tch.

- Cng dng in l lng in tch chuyn qua trong mt n v thi gian.

dt

dqI

- Chiu ca dng in c quy c l chiu chuyn ng ca cc ht mang in tch dng.

- Vct mt dng in:

Vct mt dng in ti mi im c ln bng cng dng dI chuyn qua yu t mt dSn t vung gc vi hng chuyn ng ca cc ht mang in ti chia cho dSn:

ndS

dIj (I.21)

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+ Hng ca vct j trng vi hng chuyn ng ca cc ht mang in tch

dng.

- Do , cng dng qua mt mt c tit din S l:

s

dS.jI (I.22)

b. Phng trnh lin tc

Mt kn S nm trong mi trng c dng in chy qua th:

dVt

dVjdiv

dt

dqdS.j

V VS

t

jdiv

(I.23)

- i vi dng dng th:

0jdiv

c. Lc l

Lc l c c trng bng cng m n thc hin c a mt n v in tch dng i c theo mch in.

Vq

A (I.24)

6. T trng

a. nh lut Ampere v tng tc t gia hai yu t dng c bn

- Yu t dng c bn: l mt on v cng ngn ca dy dn c dng chy qua

(k hiu l dlI ) trong dlc chiu trng vi chiu dng in.

- nh lut Ampere: lc t do yu t dng 11dlI tc dng ln yu t

dng 22 dlI cng t trong chn khng l mt i lng vct 12dF c:

+ Phng vung gc vi mt phng cha yu t 22 dlI v php tuyn n ( n c

chiu sao cho 1dl , r , n to thnh mt tam din thun).

+ C chiu sao cho 2dl , n , 12dF to thnh mt tam din thun.

+ ln:

2111222

12r

sindl.IsindlIkdF

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Trong :

n;dlI

r;dlI

222

111

; k l h s t l: 4

k 0

Tng qut: 311220

12r

]rdl[IdlI.

4

dF

(I.25)

b. T trng

- nh ngha: T trng l mi trng vt cht tn ti xung quanh in tch chuyn ng v tc dng lc t ln in tch khc chuyn ng trong n.

- Cm ng t do mt in tch q chuyn ng vi vn tc v l:

30

r

]rv[q.

4

B

(I.26)

Trong , r l khong cch t q n im cn xc nh cm ng t.

c. nh lut Biot Savart Laplace

Vct cm ng t dB do yu t dng dlI gy ra ti im P, cch yu t dng mt khong r l mt i lng vct c:

+ Gc ti P.

+ Phng vung gc vi mt phng cha phn t dng in dlI v P.

+ Chiu sao cho dl , r , dB to thnh mt tam din thun.

+ ln:

20

r

sinIdl.

4

dB vi r,dlI

Tng qut:

30

r

]rdl[I.

4

dB

(I.27)

d. Lc tc dng ca t trng ln dng in

- Lc Ampere: lc t tc dng ln dng in I bng lc tc dng ln cc ht e chuyn ng c hng trong dy dn.

- nh lut Ampere v lc tc dng ln yu t dng dlI :

]Bdl[IdF (I.28)

Vy lc tc dng ln c dng in l: dng

dFF

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7. nh lut Ampere v dng ton phn v ng dng

a. nh lut Ampere v dng ton phn

- Lu s ca vc t cm ng t theo ng cong kn L:

L

dlBL (I.29)

- Ni dung nh lut: lu s ca vct cm ng t theo ng cong kn bng tch

0 vi tng i s cc dng in xuyn qua mt S gii hn bi ng cong kn .

n

1k

k0

L

IdlB (I.30)

+ Nu trong din tch gii hn bi ng cong kn L c mt dng l j th:

S

dSjI

jBrot 0 (I.31)

b. ng dng ca nh lut Ampere v dng ton phn

- T trng do mt dy in thng co kch thc ngang ng k, bn knh tit din l R.

+ Ti im cch trc ng dy mt khong r < R:

B c phng l tip tuyn ca ng trn bn knh r.

B c chiu c xc nh theo quy tc vn inh c.

ln: 20

R2

I.rB (I.32)

+ Ti im cch trc ng dy mt khong r > R:

Chiu c xc nh theo quy tc vn inh c.

ln: R2

IB 0 (I.33)

- T trng ca mt ng dy in thng (Solenoid)

B c phng song song vi trc ca ng dy.


ln: n.IB 0 (I.34)

(vi n s vng dy trn mt n v di)

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- T trng trong cun dy hnh xuyn (Toroid)

B c phng tip tuyn vi trc ca li.


ln: n.IB 0 (I.35)

(vi n s vng dy trn mt n v di)

c. nh l Ostrogradski Gauss trong t trng

Thng lng t trng qua mt kn bt k th bng 0.

0dSBS

B (I.36)

0Bdiv (I.37)

8. Lc Lorentz Hiu ng Hall

a. Lc Lorentz

- Lc t tc dng ln in tch q chuyn ng vi vn tc v:

]Bv[qF (I.38)

ln:

sinqvBF

+ Lc Lorentz khng sinh cng do vF

+ Khng th thay i nng lng ca cc ht in tch bng cch tc dng ln n mt t trng c nh.

+ Nu c ng thi E v B tc ng th:

]Bv[qEqF (I.39)

b. Hiu ng Hall

- Mt bn kim loi c dng mt chiu chy qua c t trong t trng B vung gc vi mt bn ca bn. Quan st thy c hiu in th gia mt trn v di ca bn.

bjBRU HH (I.40)

Trong o:

+ RH l h s Hall: ne

1RH

+ j l ln mt dng in

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+ B l ln ca cm ng t.

+ b l rng ca bn.

9. Cc nh lut v cm ng in t

a. nh lut Faraday

Sut in ng cm ng trong mt vng dy dn bng v tri du vi tc bin thin theo thi gian ca t thng gi qua vng dy .

(V)dt

d B

(I.41)

Nu c N vng dy th:

(V)dt

dN B

(I.42)

b. nh lut Lenz

- Ni dung: Sut in ng cm ng lun to ra dng cm ng sao cho t trng m n sinh ra chng li s bin thin ca thng lng t trng sinh ra n.

- Cc cch co th to ra sut in ng cm ng nh sau, da vo (I.43):

+ Thay i cm ng t B

+ Thay i din tch A ca vng dy

+ Thay i gc gia vng dy v t trng.

(V)cosBdSdt

d

S

(I.43)

10. Hin tng h cm v hin tng t cm

a. Hin tng h cm

Khi t hai ng dy gn nhau th dng in trong cun th nht bin thin s to nn sut in ng cm ng cun th 2. Sut in ng ny t l vi tc bin thin ca t thng qua cun 2 do dng trong cun 1 gy ra.

21221 NM gi l h s h cm, c n v Henri

- Sut in ng cm ng trong cun 2 l:

dt

dN 2122

dt

dIM 1212 (I.44)

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b. Hin tng t cm

- L hin tng cm ng xy ra trong chnh cun dy c lp c dng xoay chiu.

- Dng in qua cun dy bin thin, t thng do dng sinh ra cng bin thin v sinh ra sut in ng cm ng c chiu chng li nguyn nhn sinh ra n.

L.IN

L: h s t cm c n v Henri, ph thuc vo thng s hnh hc ca si dy v li vt liu t trong lng cun dy.

dt

dIL (I.45)

11. Nng lng t trng

a. Nng lng t trng

2

B LI2

1W (I.46)

b. Mt nng lng t trng

0

2

BB

B

2

1

V

Ww (I.47)

Vi V l th tch trong lng cun cm.

12. H phng trnh Maxwell v h qu

a. S to thnh in trng do t trng bin thin. Phng trnh Maxwell Faraday

- Bt k mt t trng no bin i theo thi gian cng sinh ra mt in trng

xoy BE . Sut in ng bng lu s ca vct theo vng dy dn kn:

dlE B

Theo nh lut cm ng in t ca Faraday:

S

B dSBdt

ddlE

Lu s ca vct cng in trng xoy dc theo vng dy kn bt k bng v tri du vi tc bin thin theo thi gian ca t thng gi qua din tch gii hn bi vng dy kn o.

- Biu thc in trng trong trng hp tng qut:

t

BErot

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b. S to thnh t trng do in trng bin thin

- Dng in dch l dng in ng vi in trng bin i theo thi gian v phng din sinh ra t trng.

- nh l dng ton phn trong trng hp tng qut:

D0 IIdlB

Trong , I v ID ln lt l cng dng in dn, cng dng in dch.

Dng vi phn ca nh l:

t

EjBrot 00

Trong , j v t

Ej 0D

ln lt l mt dng in dn v mt dng in dch.

c. Cc phng trnh Maxwell

- in trng v t trng ng thi tn ti trong khng gian to thnh mt trng thng nht gi l in t trng. H qu chnh ca l thuyt Maxwell l kt lun v bn cht sng in t ca nh sng.

- Cc phng trnh Maxwell:

+ T trng bin thin theo thi gian sinh ra in trng xoy:

t

BErot

(I.48) (phng trnh Maxwell Faraday)

+ T thng gi qua mt kn bt k bng 0:

0Bdiv (I.49) (nh l O G trong t trng)

+ in trng xoy sinh ra dng in dch v t sinh ra t trng bin i:

t

EjBrot 00 (I.50) (phng trnh Maxwell Ampere)

+ in tch ngoi l ngun gc ca in trng:

0

Ediv (I.51) (nh l O G trong in trng)

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d. Sng in t

- T trng bin thin sinh ra in trng bin thin v in trng bin thin ny li sinh ra t trng bin thin. Kt qu l c mt h trng in t bin i lan truyn trong khng gian. l sng in t.

- Vn tc truyn song in t trong mi trng:

cv (I.52)

Trong : v ln lt l hng s in mi v hng s t thm ca mi trng,

(m/s)458792299

1c

00

l vn tc nh sng trong chn khng.

- Nng lng ca song in t. Vct Pointing (khng cn hc)

+ Mt nng lng ca sng in t:

0

22

0

0

22

0

BE

B

2

1E

2

1

+ Vct Pointing biu th nng lng sng in t truyn qua mt n v din tch trong mt n v thi gian:

BE

1S

0

(I.53)

ln trung bnh ca S :

0

00

0

2

02

00

BE

2

1

Bc

2

1Ec

2

1S (I.54)

Trong , E0 v B0 ln lt l gi tr cc i ca in trng v t trng.

B. BI TP

Bi 1: in trng cua in tich im

- Khi c 1 in tch im q: th cng in trng E ti im P cch in tch q mt khong l r c xc nh nh sau:

+ E c im t ti im P.

+ E c phng nm trn ng thng ni in tch q vi im P.

+ E c chiu t in tch q ti P nu q > 0 v c chiu ngc li nu q < 0.

+ ln ca E c xc nh theo cng thc (vi l hng s in mi ca mi trng):

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20r4

qE

C

N

Vy:

r

r

r4

qE

2

0 (I.50)

- Khi c n in tch im:

in trng tng cng v tr ang xt c xc nh theo nguyn l chng cht.

n21 E...EEE

Bi 2: in th cua in tich im

- Khi c 1 in tch im q: th in th V ti im P cch in tch q mt khong l r c xc nh nh sau:

- in th ti mt im c tnh bng cng ca lc tnh in dch chuyn mt n v in tch dng t im ra xa v cng. Th ti mt im l i lng v hng.

P

P dsEV vi ds l vi phn ca qung ng dch chuyn.

ds.s4

qV

r

2

0

P

vi s l khong cch t yu t ds ti in tch p.

1

r

1

4

q

s

1

4

qV

0r0

P

r.4

qV

0

P

Coulomb

JuneVol (I.51)

- Th ti im P quanh mt in tch dng th c gi tr dng, quanh mt in tch m th c gi tr m.

- Khi c n in tch im:

Ta c th tnh th tng cng mt im gy ra bi mt nhm in tch im nh nguyn l chng cht. Vi n in tch, th tng cng l:

n

1i i0

in

1i

ir4

qVV

Vol (I.52)

Bi 3: in trng cua lng cc in

- Lng cc in l mt h gm 2 in tch im c ln bng nhau v ngc nhau v du, t cch nhau mt khong l d.

In xong 31/12/2013

Hong Vn Trng 0974.971.149 23

- Tnh in trng ca lng cc in ti im M nm trn trc ca lng cc (ng thng ni hai in tch im) v cch trung im ca n mt khong l z:

in trng gy ra ti M do in tch dng v in tch m ca lng cc gy ra. Ap dng nguyn l chng cht in trng ta c:

)()( EEE

+ E c phng nm trn ng thng ni hai in tch.

+ E c chiu t in tch m n in tch dng.

+ V ln (vi trng hp trn hnh v):

22

2

0

2

0

2

0

)()(

2z

d1

2z

d1

z4

q

2

dz

q.

4

1

2

dz

q.

4

1

EEE

Khi im M cch rt xa trung im ca lng cc (z >> d) th 12z

d . Khai

trin biu thc trn theo cng thc Taylor Maclaurin ta c:

...z

d1

2z

d1

2

; ...z

d1

2z

d1

2

Do o:

z

2d.

z4

q

z

d1

z

d1

z4

qE

2

0

2

0

30 z

qd.

2

1E

C

N (I.53)

- Tch qd c gi l mmen ca lng cc in. Mmen ca lng cc in l mt i lng vect, c ln bng tch qd, c chiu hng dc theo trc ca lng cc t in tch m n in tch dng.

+ - q q

M d z

)(E )(E

In xong 31/12/2013

Hong Vn Trng 0974.971.149 24

- in trng gy ra bi lng cc in t l nghch vi lp phng khong cch t im cn xc nh n trung im ca lng cc.

Bi 4: in th cua lng cc in

Xt in th ti im M cch trung im ca lng cc mt on l z v cch in tch dng v m mt khong l z+ v z-

Theo nguyn l chng cht, in th ti M l tng in th do in tch dng v in tch m gy ra:

)()(0

)()(

2

1i

iz

q

z

q

4

1VVVV

)()(

)()(

0 .zz

zz

4

1V

Nu z >> d (d l khong cch gia hai in tch) th ta c mt cch gn ng:

2

)()(

)()(

z.zz

d.coszz

Do :

qdz4

cosV

2

0 Vol (I.54)

- Tch qd gi l m men ca lng cc in.

- in th M s khng thay i nu quay M quanh trc ca lng cc sao cho z

v khng thay i.

- in th bng 0 ti mi im nm trn mt phng vung gc vi trc ca lng cc ti trung im ca n.

Bi 5: in trng v in th cua dy dn thng di tich in

Si dy AB c di 2a, tch in dng vi mt din di .Tnh in trng ti im M.

Trng hp 1: M nm trn ng trung trc ca si dy v cch trung im ca

si dy mt khong l z. Xt trng hp c bit khi a

q q

M

d

z (+)

z()

+

z

In xong 31/12/2013

Hong Vn Trng 0974.971.149 25

in trng:

- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im c trong si dy.

- Chia si dy tch in thnh nhng yu t c di dx v cng nh (nh ti mc coi nh l mt in tch im) v cch trung im O ca si dy mt on l x. tnh in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx

gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.

Mt s bn thc mc: sao khng t vi phn l dx v khong cch t dx ti tm l d hoc t vi phn l dl v khong cch t dl ti tm l x. L do l trong khi nim ch c in trng do cc in tch im gy ra, v th ta phi chia si dy thnh v s cc in tch im v dx c chiu di rt nh, nh ti mc coi in tch trn n ging nh in tch im. on vi phn c chiu di rt nh th c th t l dx, dl, dk, nhng bin khi ly tch phn phi t tng ng vi n: t on vi phn l dx th khong cch t dx ti O l x, t on vi phn l dl th khong cch t n ti O l l.

Ch : nu y l mt hm ca x th dx v dy l hon ton khc nhau v: dy = ydx.

- Tnh in trng do yu t dx gy ra ti M:

+ C gc t ti M.

+ in tch ca yu t dx l: dq = .dx

+ in trng Ed c chiu nh hnh v, c phng hp vi ng trung trc ca

AB mt gc .

+ ln ca Ed l:

220

2

0

zx

dx

4

1

r

dq

4

1dE

- in trng do c on dy gy ra ti M:

a

a

y

a

a

x

a

a

EdEdEdE

x

-a

a

M

dx

xEd

yEdEd

z O

r

A

B

In xong 31/12/2013

Hong Vn Trng 0974.971.149 26

Do tnh cht i xng ca si dy nn 0Eda

a

y

. Suy ra:

a

a

xEdE

+ E c phng v chiu trng vi xEd

+ V ln:

a

a

xdEE

3/2220

2222

022

x

xz

dx

4

z

xz

z

xz

dx

4

1

xz

zdEdEcosdE

a

a

3/2220

a

a

x

xz

dx

4

zdEE

Ta c: dcos

zdxz.tanx

2

Do o:

3/2

2

3

2

0

3/2232

0

3/22222

0

a

a

3/2220

cos

1z

d

cos

z

4

z

tan1z

d

cos

z

4

z

.tanzz

d

cos

z

4

z

xz

dx

4

zE

.sinz

2

1sin

z2

dcos

z4

2

dcosz4

z

.dcos

cos

z

4

z

00

000

-0

3

3

2

0

220 za

a.

z2

E

C

N (I.55)

- Khi a th a >> z, sin = 1 hoc

1

za

alim

22av in trng ti

M l:

z2

E

0

In xong 31/12/2013

Hong Vn Trng 0974.971.149 27

in th:

in th ti M do yu t vi phn dx gy ra c tnh bng cng m lc in trng ca dx dch chuyn mt n v in tch dng t M ra xa v cng.

in th ti M do dx gy ra l:

22

00 zx4

dx

r.4

dqdV

in th ti M do c on dy gy ra l: p dng nguyn l chng cht (i vi i lng v hng) ta c:

a

a22

0

n

1i

in zx4

dxdVVlimV

a

0

22

0022

0

zxxln2

zx

dx

4

2V

a

zlnzaaln2

V 22

0

z

zaaln.

2

V

22

0

Vol (I.56)

- Khi a th a >> z v

z

zaa 22

. Do , V

Trng hp 2: M nm trn ng thng AB v cch trung im O mt khong l z

in trng:

- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn si dy.

- Chia si dy tch in thnh nhng yu t c di dx v cng nh (nh ti mc coi nh l mt in tch im) v cch u A ca si dy mt on l x. tnh

z

A

B

-a

a

dx

x

M

Ed

In xong 31/12/2013

Hong Vn Trng 0974.971.149 28

in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx gy ra

ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.

- in trng do yu t dx gy ra ti M:

+ C gc t ti M.

+ in tch ca yu t dx l: dq = .dx

+ in trng gy ra ti M l Ed c phng trng vi ng thng AB v c chiu nh hnh v.

+ ln ca Ed l:

20

2

0 x

dx

4

x

dq

4

1dE


B

A

EdE

+ ln:

az

az0

az

az

2

0

B

Ax

1.

4

x

dx

4

dEE

az

1

az

1.

4

E

0

C

N (I.57)

in th:



x

dx

4

x.4

dx

x.4

dqdV

000


az

az00

xln4

x

dx

4

dVV

az

az

az

azln

4

V

0

Vol (I.58)

In xong 31/12/2013

Hong Vn Trng 0974.971.149 29

Trng hp 3: M nm trn ng thng vung gc vi mt u ca si dy v cch

u si dy mt khong l z

in trng:


- Chia si dy thnh nhng yu t vi phn dx v cng nh, cch A mt on l x. tnh in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx



+ C gc t ti M.

+ Ed c chiu nh hnh v, c phng hp vi ng thng AM mt gc .

+ ln ca Ed l:

22020 zxdx

4

1

r

dq

4

1dE

(v dq = .dx)


2a

0

y

2a

0

x

2a

0

EdEdEdE

+ V ln:

2a

0

2

0

22

0

2a

0

xx cos.xz

dx

4

1dE.cosdEE

a

M: dcos

zdxz.tanx

2

Do o:

A

dx

z

r

M

Ed

B

0

x

xEd

yEd

2a

In xong 31/12/2013

Hong Vn Trng 0974.971.149 30

220000

000

2222

0

x

z4az

2a

4

sin

z4

.sin

z4

dcosz4

tanzzcos

dz.cos

4

E

220000

000

2222

0

2a

0

22

0

2a

0

2a

0

yy

z4az

z

4

cos

z4

.cos

z4

dsinz4

tanzzcos

dz.sin

4

.sinzx

dx

4

1dE.sindEE

Du " " trong Ey th hin rng vct yE c chiu hng xung di so vi trc

tung Oy.

2

y

2

x EEE

)z(4az

z

)z(4az

4a.

4

E

222

2

222

2

0

z4

E

0

C

N (I.59)

+ Chiu ca E hp vi phng ngang mt gc sao cho: x

y

E

Etan

in th:



22

000 zx

dx

4

r.4

dx

r.4

dqdV


2a

0

22

0

2

022

0

zxxln4

zx

dx

4

dVV

a

In xong 31/12/2013

Hong Vn Trng 0974.971.149 31

z

z4a2aln

4

V

22

0

Vol (I.60)

Trng hp 4: M nm trn ng thng vung gc vi si dy ti im cch u si dy mt khong l h, M cch si dy mt khong l z. y l trng hp tng qut cho 3 trng hp trn.

in trng:


Chia si dy thnh nhng yu t vi phn dx v cng nh, cch H mt on l x. tnh in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx



+ C im t ti M.

+ Ed c chiu nh hnh v, c phng hp vi ng thng HM mt gc .

+ ln ca Ed l:

22020 zxdx

4

1

r

dq

4

1dE

(do dq = dx)


B

A

y

B

A

x

B

A

EdEdEdE

+ V ln:

a2h 2

22

0

B

A

xx cos.xz

dx

4

1dE.cosdEE

h

ah

h

z

dx

M

Ed

A

B

0

a

x

xEd

yEd

h

H

r

2

In xong 31/12/2013

Hong Vn Trng 0974.971.149 32

M:

dcos

zdxz.tanx

2

Do :

220000

000

2222

0

x

zh)(2a

h2a

z4

sin

z4

.sin

z4

dcosz4

cos

tanzzcos

zd

4

E

220000

000

2222

0

a2h

22

0

a2h

HA

yy

zh)(2a

z

z4

cos

z4

.cos

z4

dsinz4

sin

tanzzcos

zd

4

.sinzx

dx

4

1dE.sindEE

h

B

2

y

2

x EEE

cossinz4

E 22

0

z4

E

0

C

N (I.61)

+ Chiu ca E hp vi phng ngang mt gc sao cho: x

y

E

Etan

in th:



22

000 zx

dx

4

r.4

dx

r.4

dqdV


In xong 31/12/2013

Hong Vn Trng 0974.971.149 33

2ah

h

22

0

2ah

h22

0

zxxln4

zx

dx

4

dVV

22

22

0 zhh

zh)(2ah)(2aln

4

V

Vol (I.62)

Bi 6: in trng v in th cua vng dy trn tich in

Tnh in trng v in th ti im M.

Trng hp 1: Vng dy trn tm O bn knh R, tch in q, mt in di l . im M nm trn trc ca vng dy v cch O mt khong l z. Xt trng hp c

bit khi z bng 0.

in trng:

- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn vng dy trn.

Chia vng dy thnh nhng yu t vi phn dx v cng nh. tnh in trng ti M do vng dy gy ra, ta phi tnh in trng do tng yu t dx gy ra ti M l

cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.


+ C im t ti im M.

+ in trng gy ra ti M l Ed c phng v chiu nh hnh v.

+ ln ca Ed :

in tch ca yu t dx l: dq = .dx = .Rd

22020 zRRd

4

1

r

dq

4

1dE

- in trng do c vng dy gy ra ti M:

2

0

y

2

0

x

2

0

EdEdEdE

M

dx

z

R

xEd

EdyEd

O d

q

r

In xong 31/12/2013

Hong Vn Trng 0974.971.149 34

Do tnh cht i xng ca vng dy nn 0Ed2

0

x

. Suy ra: 2

0

yEdE

+ E c phng v chiu trng vi yEd

+ V ln:

2

0

2

0

22

0

2

0

y .coszR

Rd

4

1dE.cosdEE

+ Ta c:

22 zR

zcos

2

02222

0 zR

z

zR

Rd

4

1E

2/3220 zRRz

2

1E

C

N (I.63)

Khi z = 0 th ta c in trng ti tm ca vng dy trn tch in l: E = 0.

in th:



22

000 zR

d

4

R

r.4

Rd

r.4

dqdV


22

0

2

022

0 zR4

R.2

zR

d

4

RdVV

22

0 zR2

RV

Vol (I.64)

Khi z = 0 th in th ti tm ca vng dy trn tch in l: 02

V

In xong 31/12/2013

Hong Vn Trng 0974.971.149 35

Trng hp 2: Mt ng hnh tr rt mng c chiu cao h, tch in dng vi mt

in mt l . im M nm trn trc ca hnh tr v cch tm ca mt y mt khong l z

in trng:

- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn mt xung quanh ca hnh tr.

Chia ng tr thnh cc nhn trn c rng dx v cng nh. tnh in trng ti M do ng tr gy ra, ta phi tnh in trng do tng nhn trn c rng dx gy

ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.

- in trng do nhn trn co b rng dx gy ra ti M, tm vng trn cch M mt khong x l:

+ C im t ti im M.

+ Ed c phng v chiu nh hnh v.

+ ln ca Ed :

in tch ca nhn trn b rng dx l: dq = 2R.dx (R l bn knh ng tr)

Do o theo trng hp 1 ca bi ny ta c:

dx

xR

R.x

2

1dx

xR

2R.x

4

1dE

3/2220

3/2220

- in trng do c ng tr gy ra ti M:

z

h-z

EdEdE

O

y x

h

z

M

z

Ed

Hnh tr rng

In xong 31/12/2013

Hong Vn Trng 0974.971.149 36

+ E c phng v chiu trng vi Ed

+ V ln:

z

h-z

3/222

22

0

z

h-z

3/2220

z

h-z xR

)Rd(x

4

Rdx

xR

x

2

RdEE

z

hz

22

0 xR

1

2

RE

22220 zR

1

h)-z(R

1

2

RE

C

N (I.65)

in th:

in th ti M do v s cc nhn trn c b rng v cng nh dx gy ra.

in th ti M do mt nhn trn b rng dx gy ra l: theo trng hp 1 bi ny th:

dxxR2

Rdx

xR4

R2.dV

22

0

22

0

in th ti M do c ng tr gy ra l:

z

hz

22

0

z

h-z22

0

xRxln2

R

xR

dx

2

RdVV

22

22

0 h)-(zRh)(z

zRzln

2

RV

Vol (I.66)

Trng hp 3: Bn cu rng c bn knh R, tch in dng vi mt in mt . im M nm ti tm ca bn cu.

in trng:

- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn b mt ca bn cu.

R

dx d

z M

Ed

In xong 31/12/2013

Hong Vn Trng 0974.971.149 37

Chia bn cu thnh cc vng nhn trn c rng dx v cng nh. tnh in trng ti M do bn cu rng gy ra, ta phi tnh in trng do tng vng nhn trn

c rng dx gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.

- in trng do vng nhn trn co b rng dx gy ra ti M, tm vng nhn trn cch M mt khong z v bn knh vng nhn l r:

+ C im t ti im M.


+ ln ca Ed :

in tch ca vng nhn trn b rng dx l: dq = 2r.dx

Do o theo trng hp 1 ca bi trc v in trng do vng dy trn tch in gy ra, ta c:

dx

zr

rz

2

1dx

zr

2r.z

4

1dE

3/2220

3/2220

- in trng do c bn cu gy ra ti M:

EdE

+ E c phng v chiu trng vi Ed

+ V ln:

dEE

Ta c:

Rcosz

Rsinr

Rddx

(vi d l gc tm m cung dx chn)

Do :

2

00

2

0

3/2222

3

0

2

0

3/222220

.d.cossin2

d

)cos(sinR

.cossinR

2

dcosRsinR

.R.RcosRsin

2

dEE

2

0

2

0

2

002

sin

2

)(sin.dsin

2

In xong 31/12/2013

Hong Vn Trng 0974.971.149 38

04

E

C

N (I.67)

Cch khc:

Ta c th khng cn chia thnh cc vng nhn trn c b rng dx. u tin tnh in trng do yu t vi phn din tch dS gy ra ti M, sau ly tch phn theo mt ca bn cu.

- in trng do yu t vi phn din tch dS gy ra c ln:

20 R

dq

4

1dE

M: .dx.rddq (vi l gc m dx chn tm ng trn bn knh r)

dd.sin.Rd.Rsin.Rddq 2

Do : dd..sin4

R

dq

4

1dE

0

2

0

Khi ly tch phn trn ton mt bn tr th s c c d v d cng bin thin. tnh tch phn 2 lp ny th phi chuyn v tch phn 1 lp bng cch p dng h qu ca nh l Fubini.

in trng do c bn cu gy ra ti M l:

0

2

00

2

0

2

00

S0S

4

.d.cossin4

2d.d.cossin

4

dd.cos.sin4

dE.cosE

in th:

in th ti M do v s cc vng nhn trn c b rng v cng nh dx gy ra.

in th ti M do mt vng nhn trn b rng dx gy ra l: theo trng hp 1 v in th ca vng dy trn tch in th:

dxxr2

rdx

xr4

r2.dV

22

0

22

0

M:

Rcosz

Rsinr

Rddx

nn ta c:

In xong 31/12/2013

Hong Vn Trng 0974.971.149 39

dsin

2

RRd

cosRsinR2

RsindV

02222

0

in th ti M do c bn tr rng gy ra l:

200

2

00

cos2

Rdsin

2

RdVV

02

RV Vol (I.68)

Bi 7: in trng v in th cua a trn tich in

Tnh in trng v in th ti im M.

Trng hp 1: a trn tm O bn knh R, tch in dng vi mt in mt . im M nm trn trc ca a v cch O mt khong l z. Xt trng hp c bit

khi z >> R v khi R .

in trng:

- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im nm trn din tch b mt ca a trn.

Chia a trn thnh cc yu t xuyn c rng dr v cng nh, bn knh r. tnh in trng ti M do a gy ra, ta phi tnh in trng do tng yu t xuyn c

b rng dr gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.

- in trng do yu t xuyn co b rng dr gy ra ti M, tm yu t xuyn cch M mt khong z l:

+ C im t ti im M.


+ ln ca Ed :

in tch ca yu t xuyn b rng dr l: dq = 2r.dr (r l bn knh yu t xuyn tng ng)

O d

M

z

xEd

EdyEd

R

q

dr r

In xong 31/12/2013

Hong Vn Trng 0974.971.149 40

Do o theo trng hp 1 ca bi trc v vng dy trn tch in ta c:

dr

zr

r.z

2

1dr

zr

2r.z

4

1dEdE

3/2220

3/2220

y

- in trng do c a trn gy ra ti M:

yx dEdEdEE

Do tnh cht i xng ca a trn m: 0dEx nn: ydEE

+ E c phng v chiu trng vi ydE

+ V ln:

R

0

3/222

22

00

3/2220

y

zr

)zd(r

4

zdr

zr

r

2

zdEE

R

R

0

220

R

0

220 zr

1

2

z

zr2

1

1

4

zE

220 zR

1

z

1

2

zE

220 zR

z1

2

E

C

N (I.69)

Khi z >> R th 1zR

z22

v in trng ti M l: 0E . Khi a

trn tch in coi nh mt in tch im gy ra mt in trng cch xa v cng.

Khi R th 0zR

z22

v in trng ti M l:

02

E

. Khi

a trn tch in coi nh mt mt phng rng v hn.

in th:

in th ti M do v s cc vng xuyn c b rng v cng nh dr gy ra.

in th ti M do mt vng xuyn b rng dr gy ra l: theo trng hp 1 ca bi trc v in th do vng dy trn tch in gy ra th:

drzr2

.rdr

zr4

r2.dV

22

0

22

0

In xong 31/12/2013

Hong Vn Trng 0974.971.149 41

in th ti M do c a trn gy ra l:

R

022

22

0

R

022

0 zr

)zd(r

4

zr

drr

2

dVV

R0

22

0

R

0

22

0

zr2

2

1

zr

4

zzR2

V 22

0

Vol (I.70)

Khi z >> R th zzR22 v in th ti M l: 0V . Khi a trn

tch in coi nh mt in tch im c th v cng bng 0.

Khi R th 22 zR v in th ti M l: V . Khi a trn

tch in coi nh mt mt phng rng v hn gy ra mt in th im rt gn b mt ca n.

Trng hp 2: Hnh tr c bn knh R, tch in dng vi mt in khi l . im M nm trn trc ca khi tr v cch tm ca mt y mt khong l z.

in trng:

- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im ca khi tr.

z

z

Hnh tr c

O

y x

h

M

Ed

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Hong Vn Trng 0974.971.149 42

Chia khi tr thnh cc a trn rt mng, c rng dx v cng nh, khong cch t M ti a trn l x. tnh in trng ti M do khi tr gy ra, ta phi tnh

in trng do tng a trn c b rng dx gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.

- in trng do a trn co b rng dx gy ra ti M:

+ C im t ti im M.


+ ln ca Ed :

in tch ca a trn b rng dx l: dq = .R2.dx

Do o theo trng hp 1 ca bi ny ta c:

220

y

xR

x1

2

.dxdEdE

- in trng do c khi tr gy ra ti M:

yx dEdEdEE

Do tnh cht i xng trc ca khi tr m: 0dEx nn: ydEE

+ E c phng v chiu trng vi dE

+ V ln:

z

hz22

22z

hz0

z

hz22

z

hz0

z

hz22

0

xR

)xd(R

2

1dx

2

dx

xR

xdx

2

dxxR

x1

2

dEE

22220

z

hz

22z

hz0

h)-(zRzRhzz2

xRx2

22220

h)-(zRzRh2

E

C

N (I.71)

in th:

in th ti M do v s cc a trn c b rng v cng nh dx gy ra.

in th ti M do mt a trn b rng dx gy ra l: theo trng hp 1 ca bi ny th ta c:

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Hong Vn Trng 0974.971.149 43

xxR2

.dxdV 22

0

in th ti M do c khi tr gy ra l:

dxxxR2

dVV

z

h-z

22

0

z

hz

z

h-z

22

0

z

h-z

z

h-z

22

0

x2

1dxxR

2

dxxdxxR

2

h2

1dxxR

2

z

h-z

22

0

Tnh: dxxRIz

h-z

22

t:

xvdxdv

xR

xduxRu

22

22

Suy ra:

zhz

222z

hz

22

z

h-z22

2z

hz

22

z

h-z22

2z

h-z

22z

hz

22

z

h-z22

222z

hz

22

z

h-z22

2z

hz

22

xRxlnRxRx2I

dxxR

RxRx2I

dxxR

RdxxRxRx

dxxR

RRxxRx

dxxR

xxRxI

22

2222222

h)-(zRh)(z

zRzlnRh)-(zRh)(zzRz

2

1I

h

h)-(zRh)(z

zRzlnRh)-(zRh)(zzRz

4

V

22

2222222

0

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Hong Vn Trng 0974.971.149 44

Vol (I.72)

(kt qu lng nhng qu, khng bit tnh ton c ng khng)

Bi 8: Xc nh vc t cm ng t B do mt dy dn thng di c dng I chy qua ti im M cch dy mt khong l z.

Cm ng t ti im M gy ra bi dng in thng AB c xc nh bng cm

ng t tng hp gy ra bi cc yu t dng in dlI trn dng in thng .

- Cm ng t ti M do yu t dng in dlI gy ra l mt i lng vct dB c xc nh theo nh lut Biot Savart Laplace:

+ C phng vung gc vi mt phng cha dlI v EM

+ C chiu sao cho dlI , EM v dB to thnh mt tam din thun.

Mun xc nh chiu ca dB ta s dng quy tc vn inh c (d s dng hn quy tc bn tay phi). t u nhn ca inh c ti im M sao cho inh c vung gc

vi mt phng cha dlI v EM (hy tng tng theo hnh hc khng gian). Xoay

inh c sao cho chiu quay ca inh c t vct dlI n vct EM th chiu chuyn

ng tnh tin ca inh c s l chiu ca vc t dB .

Khi ta xoay inh c m mun bit inh c s chuyn ng tnh tin theo hng no th da vo quy c: nu inh c quay cng chiu kim ng h (nhn t u to n u nhn ca inh) th inh c s chuyn ng tnh tin v pha trc v ngc li. Nh vy, trng hp trn ta xoay inh c cng chiu kim ng h nn inh c s chuyn ng tnh tin theo hng t ngoi xuyn vo trong trang giy.

d hnh dung chiu chuyn ng ca inh c cc bn lm th nghim sau: dng t vt xoy inh c vo mt cnh ca bng g. Khi ta xoay t vt theo chiu kim ng h th inh c s i su vo cnh ca. Mun ly inh c ra ta phi xoay t vt ngc chiu kim ng h.

2 B

I

z

dlI

1 A

E

M

dB

r

In xong 31/12/2013

Hong Vn Trng 0974.971.149 45

Lu : Trong thc t c mt vi inh c khi xoay ngc chiu kim ng h th n s chuyn ng tnh tin v pha trc v ngc li: van bnh gas, c gi cnh qut khi quay,. Nhng quy tc vn inh c m chng ta s dng ch tun theo quy c trn.

+ ln:

20

EM

sinIdl.

4

dB vi EM,dlI

di ca yu t vi phn dl tnh theo gc l:

Do di dl v cng nh nn c th coi gc PKQ . Khi dl l cnh huyn

ca tam gic PKQ nn: sin

PQdl

Do gc d v cng nh nn c th coi PQ l cung trn ca ng trn tm M bn

knh r. Khi : PQ = r.d

Suy ra: sin

zd

sin

rddl

2 . M

sin

zEM

Vy: dsinz4

I

sin

zsin

sinIzd

EM

sinIdl

4

dB 0

2

22

2

0

- Cm ng t ti M do dng in thng gy ra l B :

dBB

+ B c im t ti M.

+ B c phng v chiu trng vi phng v chiu ca dB

+ ln:

2

1

2

1

0

0 cosz4

Idsin

z4

IdBB

210 coscosz4

IB (Tesla) (I.73)

K

dlI

d

M

Q

P

r

In xong 31/12/2013

Hong Vn Trng 0974.971.149 46

Khi dy dn di v hn th 1 00 v 2 1800. Khi , cm ng t do dy dn di v hn gy ra ti M cch dy mt khong z l:

z2

IB 0 (Tesla) (I.74)

Bi 9: Xc nh vct cm ng t B do mt dng in trn tm O bn knh R, cng I gy ra ti im M nm trn trc cua vng dy trn v cch O mt khong l z. Tm cm ng t ti tm O cua vng dy trn.

Cm ng t ti im M gy ra bi dng in trn c xc nh bng cm ng t

tng hp gy ra bi cc yu t dng in dlI trn dng in trn .

- Cm ng t ti M do yu t dng in dlI gy ra l mt i lng vct dB c xc nh theo nh lut Biot Savart Laplace:

+ C phng vung gc vi mt phng cha dlI v AM

+ C chiu sao cho dlI , AM v dB to thnh mt tam din thun.

+ ln:

20

AM

sinIdl.

4

dB vi 090AM,dlI

220

zR

Idl.

4

dB

- Cm ng t ti M do dng in trn gy ra l B :

yx dBdBdBB

Do tnh cht i xng ca dng in trn m 0dBx nn: ydBB

+ B c im t ti M.

+ B c phng vung gc vi vng in trn ti tm O.

dlI

R

O

z

M

dB

xdB

ydB

A

In xong 31/12/2013

Hong Vn Trng 0974.971.149 47

+ B c chiu trng vi chiu ca ydB t M ti O.

+ ln:

R2

0

R2

02

322

0

2222

0

R2

0

22

0y

dlzR4

RI

zR

R

zR

Idl

4

sinzR

Idl

4

dB.sindBB

R.2

zR4

RI

23

22

0

2

322

2

0

zR2

IRB

(Tesla) (I.75)

Khi z 0 th 323

22 RzR v cm ng t ti tm O ca dng in trn l:

2R

IB 0 (Tesla) (I.76)

Bi 10: Na vng dy dn in bn kinh R = 0,49m v khi lng m = 250g, c

dng in I = 25A chy qua (hnh v). Hi cn mt t trng B c hng v ln nh th no na vng dy trn l lng trong khng gian.

- Na vng dy l lng trong khng gian th n phi cn bng v lc, hay tng cc vct lc tc dng vo n bng 0.

Na vng dy c khi lng m nn n chu tc dng ca trng lc:

gmP (vi g l gia tc trng trng)

ln: P = mg = 10m (ly g 10 m/s2)

- Na vng dy l lng khi t trng tc dng lc t ln na vng dy v lc ny c ln bng trng lc P, c phng trng vi phng ca trng lc v c chiu

hng ln trn. Tc l F c phng v chiu nh hnh v:

I

P

F

dlI

I

O

M B

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Hong Vn Trng 0974.971.149 48

Gi s rng t trng ngoi l t trng u. Ti im M, cho dlI , B , F to

thnh mt tam din thun th B phi c phng vung gc vi mt phng giy v chiu i t ngoi xuyn vo trong t giy.

- Khi c t trng B th t trng s tc dng lc t ln na vng dy, lc t c ln l:

R

0

B.sindlIdFF vi 090B,dl

RIB.

- Vy khi na vng dy l lng th:

P-F

+ ln:

RIB.mg

PF

RI.

mgB (I.77)

(T)0,06549,0.14159,3.25

10.25,0B

Bi 11: Mt dy cp ng trc c ng kinh trong d1 = 2mm v ngoi bc ch

ng kinh d2 = 8 cm, gia li v v bc l cht in mi c hng s in mi

= 3. Trong li v v bc tich in tri du nhau vi mt in di = 3,14.10-4 C/m. Hy xc nh cng in trng ti cc im cch trc mt khong:

(a) r1 = 3 cm.

(b) r2 = 10 cm.

d2 d1 S1

S2

O

r2

r1 = 3

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Gi s dy cp ng trc rt di, li gia tch in dng v v bc tch in m.

Phn (a):

Xc nh cng in trng ti im M1 cch trc ca dy cp mt khong l r1 = 3cm.

Do c li gia v v bc u l cc vt dn nn cng in trng trong lng vt dn rng bng 0. V vy, im M1 nm gia hai vt dn nhng ch chu tc dng ca in trng do li gy ra.

Do tnh cht i xng trc ca dy cp nn cng in trng ti cc im nm trn mt tr c trc l trc ca dy cp th c ln bng nhau:

- Cng in trng 1E ti im M1:

+ C phng nm trn ng ni M1 vi tm O.

+ C chiu t O ti M1 (do li tch in dng).

+ ln:

Chn mt Gauss l mt cu S1, c trc l trc ca dy cp v cha im M1. p dng nh l O G trong in trng ta c: Thng lng in trng gi qua mt kn S1 bng tng i s cc in tch nm trong mt kn S1 chia cho hng s in mi.

Biu thc:

0

1

S

11

QdSE

1

(Q1 l tng i s in tch trong mt kn S1)

0

1

Squanh xung

1

QdSE

1

(v khng c in thng qua hai mt y)

0

1

Squanh xung

1

QdSE

1

(v E1 bng nhau ti mi im trn mt

xung quanh)

0

11

.hhr2E

(vi h l chiu di cp in)

01

1r2

E

(I.78)

7

12-

-4

1 10.28,68,85.103.3.2.3,14.0,0

3,14.10E

C

N

Phn (b):

Xc nh cng in trng ti im M2 cch trc ca dy cp mt khong l r1 = 10 cm.

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Hong Vn Trng 0974.971.149 50

Do c li gia v v bc u l cc vt dn nn cng in trng trong lng vt dn rng bng 0. V vy, im M2 nm ngoi v bc nn khng chu tc dng ca in trng do li gia gy ra. (ng sc b gin on ti ni c in tch)

Do tnh cht i xng trc ca dy cp nn cng in trng ti cc im nm trn cng mt tr c trc l trc ca dy cp th c ln bng nhau:

- Cng in trng 2E ti im M2:

+ C phng nm trn ng ni M2 vi tm O.

+ C chiu t M2 ti O (do v tch in m).

+ ln:

Chn mt Gauss l mt cu S2, c trc l trc ca dy cp v cha im M2. p dng nh l O G trong in trng ta c: Thng lng in trng gi qua mt kn S2 bng tng i s cc in tch nm trong mt kn S2 chia cho hng s in mi.

Biu thc:

0

2

S

22

QdSE

2

(Q2 l tng i s in tch trong mt

kn S2)

0

2

Squanh xung

2

QdSE

2

(v khng c in thng qua hai mt y)

0

2

Squanh xung

2

QdSE

2

(v E2 bng nhau ti mi im trn mt

xung quanh)

0

22

.hhr2E

(vi h l chiu di cp in)

02

2r2

E

(I.79)

7

12-

-4

2 10.88,18,85.103..2.3,14.0,1

3,14.10E

C

N

Bi 12: Cho qu cu khng dn in tm O, bn kinh R = 15 cm c tich in

u vi mt in tich khi = 1,699.10-7 C/m3, c t trong chn khng.

(1) Xc nh cng in trng ti im M cch tm O mt on: (a) r1 = 10 cm; (b) r2 = 30 cm.

(2) Ly in th ti v cng bng 0. Xc nh in th ti P cch tm 20 cm.

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Hong Vn Trng 0974.971.149 51

Phn (1a): Xc nh cng in trng 1E ti im M1 cch O mt khong 10 cm:

Do tnh cht i xng ca qu cu tch in nn cng in trng ti cc im nm trn cng mt mt cu tm O bn knh bt k th c ln bng nhau.

+ 1E c phng nm trn ng thng M1O.

+ 1E c chiu t O ti M1.

+ ln:

- Chn mt Gauss l mt cu S1 tm O bn knh r1. Ap dng nh l O G (i vi in trng) ta c: T thng gi qua mt cu kn S1 bng tng i s in tch nm

trong mt kn chia cho hng s in mi 0.

- Biu thc:

0

1

S

11

QdSE

1

(vi Q1 l tng in tch nm trong khi cu bn knh r1)

0

1

S

1

QdSE

1

0

1

S

1

QdSE

1

(do E1 bng nhau ti mi im trn mt cu S1)

0

3

12

11

r3

4.

r4E

0

11

3

r.E (I.80)

38,20103.8,85.10

0,1.3,14.1,699.10E

12-

-7

1

C

N

Phn (1b): Xc nh cng in trng 2E ti im M2 cch O mt khong 30 cm:

2E

1E

S2

S1

M2

O R r1

M1

r2

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Hong Vn Trng 0974.971.149 52

Do tnh cht i xng ca qu cu tch in nn cng in trng ti cc im nm trn cng mt mt cu tm O bn knh bt k th c ln bng nhau.

+ 2E c phng nm trn ng thng M2O.

+ 2E c chiu t O ti M2.

+ ln:

- Chn mt Gauss l mt cu S2 tm O bn knh r2. Ap dng nh l O G (i vi in trng) ta c: T thng gi qua mt cu kn S2 bng tng i s in tch nm

trong mt kn chia cho hng s in mi 0.

- Biu thc:

0

2

S

21

QdSE

2

(vi Q2 l tng in tch nm trong khi cu bn knh r2)

0

2

S

2

QdSE

2

0

2

S

2

QdSE

2

(do E2 bng nhau ti mi im trn mt cu S2)

0

3

2

22

R3

4.

r4E

220

3

2r3

R.E (I.81)

89,7

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