Vật Lý A1-CHUONG 10

Chng X. T trng ca dng in khng i CHNG X. T TRNG CA DNG IN KHNG I Chng ny nghin cu t trng do dng in khng i gy ra, tc dng gia cc dng in, v tc dng ca t trng ln dng in. Nh , chng ta ta s hiu c nguyn tc hot ng ca cc dng c v thit b in da trn tnh cht t ca dng in. 1. TNG TC T CA DNG IN NH LUT AMPRE 1. Th nghim v tng tc t a. Tng tc t gia cc nam chm Th nghim chng t hai thanh nam chm c th ht nhau nu hai cc khc tn t gn nhau, hoc y nhau nu cc cc ca chng cng tn. Cc thanh nam chm li c th ht c cc vn st. Cc tnh cht ca nam chm c gi l t tnh. Tng tc gia cc nam chm c gi l tng tc t. b. Tng tc gia dng in vi nam chm

I S N S N S

I N

(a)

(b) Hnh 10-1.Tc dng ca dng in ln kim nam chm

(c)

Th nghim chng t dng in cng c t tnh nh nam chm, ngha l dng in c th ht hoc y nam chm v ngc li nam chm cng c th ht hoc y dng in. Tht vy, ta t mt kim nam chm gn mt dy dn, song song vi dy dn cha c dng in (Hnh 10-1a). Khi cho dng in chy qua, kim nam chm quay lch i so vi phng ban u (Hnh10-1b). Nu i chiu dng in, kim nam chm cng lch nhng theo chiu ngc li (Hnh 10-1c). Sau , ta t mt nam chm gn mt dy dn. Khi cho dng in chy qua dn, nam chm s ht hoc y dy dn tu theo chiu ca dng in. Thay dy dn bng mt cun dy dn c dng in chy qua, ta cng thu c kt qu tng t (Hnh 10-2). Khi ch c cc dng in vi nhau, chng cng tng tc vi nhau. Tht vy, th nghim chng t: hai dy dn thng song song nhau, gn nhau, khi trong chng c dng in cng chiu chy qua th chng ht nhau, khi trong chng c dng in chy ngc chiu nhau th chng y nhau (Hnh 10-3). Hai ng dy in cng ht nhau hoc y nhau tu theo

192

Chng X. T trng ca dng in khng i dng in hai u ca chng cng chiu hay ngc chiu nhau. Mi cun dy nh vy tng ng vi mt nam chm: u cun dy no m khi nhn vo, ta thy c dng in chy ngc chiu quay ca kim ng h th l cc bc (N) ca nam chm, cn ngc li th l cc nam S (Hnh 10-4). V th ngi ta gi ng dy c dng in l nam chm in.

o

NN S

S

Hnh 10-2 Nam chm tc dng ln dng in

Hnh10-3 Tc dng gia hai dng in

Hnh 10-4 Cc bc (N), cc nam (S) ca nam chm in

Kt lun Qua cc th nghim trn, ngi ta kt lun: tng tc gia cc dng in cng l tng tc t. 2. nh lut Ampe (Ampre) thun li cho vic xc nh lc t, Ampre a ra khi nim phn t dng in, gi tt l phn t dng. Phn t dng in l mt on rt ngn ca dng in. V mt ton r hc, ngi ta biu din n bng mt vect I dl nm ngay trn phn t dy dn, c phng chiu l phng chiu ca dng in, v c ln Idl (hnh 10-5a). Ta gi s xt hai dng in hnh dng bt k, c cng ln lt l I v I0. Trn hai r r dng in , ta ly hai phn t dng bt k I. dl v I 0dl0 (hnh 10-5b) c v tr tng ng l r r r r O v M. t r = OM v gi l gc gia phn t I. dl v vect r . V mt phng P cha r r r I. dl v im M. V php tuyn n i vi mt phng P ti M ( n phi c chiu sao cho ba r r r vect I. dl , r v n theo th t hp thnh mt tam din thun). Gi 0 l gc gia phn t r r dng I 0dl0 v n . T khi nim phn t dng, v vi cch b tr nh trn, nh lut thc nghim ca Ampre pht biu nh sau: r r T lc do phn t dng in I. dl tc dng ln phn t dng I 0dl0 l mt r vect dF r r C phng vung gc vi mt phng cha phn t dng I 0dl0 v php tuyn n , r r r C chiu sao cho ba vect I 0dl0 , n , v d F theo th t hp thnh mt tam din thun, C ln bng dF =

0 I .dl . sin .I 0 .dl0 . sin 0 4 r2

(10-1)

193

Chng X. T trng ca dng in khng i Vi: 0 l mt hng s gi l hng s t, trong h n v SI n c gi tr bng: Henry 0 = 4.10-7 H/m ( met )

(10-2)

l mt s khng th nguyn, ph thuc vo tnh cht ca mi trng bao quanh cc phn t dng, c gi l t thm ca mi trng hay l t thm t i ca mi trng so vi chn khng; n gin, ta gi l t thm ca mi trng. Vi chn khng =1, vi khng kh = 1+0,03.10-6, vi nc: = 1- 0,72.10-6 V ca khng kh gn bng 1 nn trong nhng trng hp khng yu cu chnh xc cao, c th coi khng kh c = 1, tc l c th coi cc th nghim v tng tc t tin hnh trong khng kh nh l c thc hin trong chn khng. Pht biu nh lut Ampre trn y cng c th biu din bng biu thc sau:r

r Tng tc gia phn t dng I. dl v r phn t dng I 0 dl0v c ln:

Hnh 10-5

I. dl

r I dl ( Idl r ) dF = 0 . 0 0 (10-3) 4 r3 r Mt cch tng t, lc dF' do phn t r r dng I 0dl0 tc dng ln phn t dng I. dl :r r r r 0 Idl ( I 0 dl0 r ' ) dF' = . 4 r3

r

r

r

(10-4)

' 0 I 0 .dl0 . sin 0 .I .dl . sin ' 4 r2 Trong cng thc ny: r r r Vect r ' = r , cng ln nhng ngc chiu vi r , r = r. r r ' Gc 0 l gc gia I 0 dl0 vi r ' , r r r Cn gc ' l gc gia vect I. dl vi vect tch I 0 dl0 r ' .

dF' =

(10-5)

Ch : Trong nh lut Ampre, phn t dng ng vai tr tng t nh in tch im trong nh lut Coulomb. nh lut Ampre l nh lut c bn ca tng tc t, cng nh nh lut Coulomb l nh lut c bn ca tng tc tnh in. r r Ta thy hai lc dF v dF' khng tun theo nh lut Newton III. nh lut Ampre pht biu i vi phn t dng in. Trong thc t, ta ch c cc dng in hu hn tng tc vi nhau. xc nh lc tc dng ca mt dng in ln mt dng in khc, ta tng hp cc lc do tt c cc phn t ca dng in ny tc dng ln tt c cc phn t ca dng in r r kia, ta s c F' = - F , tc l i vi tng tc gia hai dng in hu hn, nh lut Newon III vn nghim ng. Tht vy, cc tnh ton da vo nh lut Ampre i vi tng

194

Chng X. T trng ca dng in khng itc gia cc dng in hu hn u cho kt qu ph hp vi thc nghim v tho mn nh lut Newton th III. 2. T TRNG V VECT CM NG T, VECT CNG T TRNG I. Khi nim t trng Ta bit rng, hai dng in cch nhau mt khong no trong chn khng vn ht nhau hoc y nhau vi mt t lc no . Vy c cn mt mi trng no ng vai tr truyn lc tng tc t dng in ny ln dng in kia hay khng? Ngi ta lp lun tng t nh vi in trng v tha nhn rng: dng in to ra trong khng gian bao quanh n mt dng vt cht c bit, gi l t trng. Chnh thng qua t tng m t lc c truyn t dng in ny ti dng in khc. Tnh cht c bn ca t trng l n tc dng ln bt k dng in no t trong n. Nh , ta c th gii thch c s tng tc gia cc dng in nh sau: Khi c mt dng in I1, n to ra xung quanh n mt t trng. Nu t mt dng in I2 khc vo t trng ca I1, t trng ca I1 s tc dng ln dng in I2 mt lc, ngc li I2 cng to ra xung quanh n mt t trng, t trng ny cng tc dng ln I1 mt t lc. Kt qu l hai dng in ny tng tc nhau thng qua t trng ca chng. II. Cc i lng c trng cho t trngr Gi s ta xt t trng do phn t dng I. dl gy ra ti mt im M cch n mt on r (hnh 10-6). T biu thc nh lut Ampre v tng tc gia hai phn t dng in, ta c nhn xt: vect r r r 0 Idl r dB = . (10-6) 4 r3 ch ph thuc vo: r - Phn t dng in I. dl , l phn t gy ra t trng, r - Bn knh vect r v , tc l vo v tr ca im M r trong t trng ca I. dl , ti ta t phn t dng Hnh 10-6 r Cm ng t gy bi phn t dng in I 0dl0 , m khng ph thuc vo phn t dng in r I 0dl0 . r Vy vect dB c xc nh theo (10-6), l vect c trng v mt tc dng lc cho r t trng ti im M gy bi phn t dng I. dl , v c gi l vect cm ng t do phn t r dng I. dl gy ra ti M. Biu thc (10-6) c gi l nh lut Biot-Xavart-Laplace, c th pht biu nh sau: r r Vect cm ng t dB do phn t dng I. dl gy ra ti im M, cch n mt khong r l mt vect c: Idl sin (10-7) - ln dB = 0 . 4 r2 r r ( l gc gia vect I. dl v vect r )

1. Vect cm ng t

195

Chng X. T trng ca dng in khng ir - phng vung gc vi mt phng cha phn t dng in I. dl v im M; r r r - chiu sao cho ba vect I. dl , r v dB theo th t hp thnh tam din thun ; - gc ti im M. r I dl Ngi ta cng c th xc nh chiu ca vect

r dB bng qui tc vn nt chai nh sau: t ci vn nt chai theo phng ca dng in, nu quay ci vn nt chai sao cho n tin theo chiu ca dng in th chiu quay ca n s ch chiu ca vect r cm ng t ti im (hnh 10-7). dB Trong h n v SI, cm ng t c tnh bng n v Tesla ( k hiu l T), s c nh ngha sau ny, t cng thc (10-26) mc 3.

M

r Xc nh vect dB theo qui tc vn nt chai T nh lut Ampre (10-3) v nh lut Biot-Savart-Laplace (10-6) ta suy ra lc do r r phn t dng I. dl tc dng ln phn t dng I 0 dl0 c xc nh bng cng thc: r r r dF = I 0 dl0 dB (10-8)

Hnh 10-7

2. Nguyn l chng cht t trng Ging nh in trng, t trng cng tun theo nguyn l chng cht: Vect cm r ng t B do mt dng in chy trong mt dy dn di hu hn gy ra ti mt im M bng r tng hp cc vect cm ng t dB do tt c cc phn t dng ca dng in gy ra ti im c xt. Tc l:

r B =

( Ca dong )

dB

r

(Tch phn ly theo c dng in)

(10-9)

Nu t trng do nhiu dng in gy ra th theo nguyn l chng cht t trng: Vect cm ng t ti mt im M trong t trng do nhiu dng in gy ra bng tng hp cc vct cm ng t do tt c cc dng in gy ra ti im . n r r r r r B = B1 + B2 +......+ Bn = Bi (10-10)i =1

3. Vect cng t trng r r Ngoi vect cm ng t B ngi ta cn a ra vect cng t trng H , c nh ngha bi biu thc sau: r r B H = (10-11)

0

nh ngha ny ch ng i vi mi trng ng nht v ng hng. Theo (10-6), r r vect B ph thuc bc nht vo do theo (10-11), H khng ph thuc vo . iu r c ngha l vect H c trng cho t trng do ring dng in gy ra v khng ph thuc vo tnh cht ca mi trng cha dng in. Do cng t trng khng bin i t ngt khi chuyn t mi trng ny sang mi trng khc (c khc nhau). V l , cc

196

Chng X. T trng ca dng in khng i r ng sc ca vect H i lin tc t mi trng ny sang mi trng khc c t thm khc nhau. Trong h n v SI, n v cng t trng lAmpe (k hiu l A/m). n v met

ny s c nh ngha mc di y (10-15). r r III. Xc nh vect cm ng t B v cng t trng H r r Sau y ta s xt mt vi v d tnh cm ng t B v vct H . 1. T trng ca dng in thng Xt mt on dy dn thng AB, c dng in I chy qua (Hnh 10-8). Hy xc nh r r vct cm ng t B v vct H do dng in gy ra ti mt im M nm cch dng in mt khong a. Ta tng tng chia AB thnh nhng phn t nh, c chiu di dl. r r Theo nh lut Biot-Savart-Laplace, vect cm ng t dB do phn t dng I. dl gy r ra ti im M c phng vung gc vi mt phng cha M v I. dl ( mt phng hnh v) v c ln: Idl sin dB = 0 . 4 r2

B I

2

a dl1A

r B

M

r d

r Theo nguyn l chng cht t trng, vect B do dng in trong on mch AB gy ra ti M bng tng r hp cc vect dB do tt c cc phn t dng ca on AB gy ra: r r B = dB (10-12) AB r V trong trng hp ny, tt c cc vect dB c cng phng chiu (vung gc vi mt phng hnh v v r r hng vo), nn B cng c phng chiu nh dB v c ln:B=

Hnh 10.8 xc nh vect cm ng t ca dng in thng Thay sin .dl rd v r = c:a sin 2

O Idl . sin r2 AB 4

vo biu thc di du tch phn trn y, ta

I B= 0 4a

1

sin d

Sau khi thc hin tch phn, ta c: B = 0 I(cos 1 -cos 2 ) (10-13) 4a Nu dng in thng di v hn, ta c: 1 = 0, 2 = , v t (10-13) ta tnh c: B= v suy ra: 197

0 I 2a

(10-14)

Chng X. T trng ca dng in khng iI (10-15) 2 a Trong h n v SI, ngi ta da vo cng thc (10-15) nh ngha n v ca cng t trng l A/m. Trong cng thc (10-15), nu cho I = 1A, chu vi ng trn bn knh a bng 2 a =1mt th: 1 Ampe A H= =1 1met m Vy ta c nh ngha nh sau: Ampe trn mt l cng t trng sinh ra trong chn khng bi mt dng in c cng 1 ampe, chy qua mt dy dn thng di v hn, tit din trn, ti cc im ca mt ng trn ng trc vi dy v c chu vi bng 1 mt.

H=

r r Ta hy xc nh vct B v H do dng in cng I chy trong dy dn hnh trn bn knh R gy ra ti im M nm trn trc ca dng in, cch tm O ca dng in mt khong h (Hnh 10-9). Ta c nhn xt, do tnh i xng ca dng in trn, bao g cng c th chn c nhng cp phn t dl1 v dl2 c chiu di bng nhau v nm i xng vi nhau qua tm O r r r r ca vng trn. Do cc vect cm ng t dB1 v dB2 do hai phn t dng I . dl1 v I . dl2

2. Dng in trn

gy ra ti M s i xng vi nhau qua trc ca dng in. Do tng hp hai vect ny ta r c 1 vect dB12 : r r r dB12 = dB1 + dB2 r nm trn trc ca dng in, do vect B do c dng in gy ra ti M cng nm trn trc y. Ta suy ra: cm ng t tng hp do c dng in trn gy ra ti M:r dB12

B

=

r dB1

r dB2

ca dong dien

dB

n

,

trong

dBn =

M r

r pm r S

r I . dl1

R I

r I . dl2

dBn l hnh chiu ca dB ln trc ca dng in do mt phn t dng I.dl gy ra ti M, l gc r gia dB vi trc ca dng in. Trong R = , sin =1, cos= . Do : r 2 2 . 0 I . dl RI .R 3 = o 3 B= dl r 4r ca dong dien ca dong dien 4 . 0 . R.I . 0 I 2 R= ( R 2 ) 3 4 r 2 r 3 . 0 I . S B= 2 ( R 2 + h 2 ) 3/ 2

0 Idl sin . cos 4 r2 r

Hnh 10-9 xc nh cm ng t gy bi dng in trn

=

Trong S = .R2 l din tch bao bi dng in trn; r = ( R2 + h2 )1/2. r Gi vect S l vect nm trn trc ca dng in, c cng bng S, c chiu l chiu tin ca ci vn nt chai khi ta quay cn ca n theo chiu ca dng in.

198

Chng X. T trng ca dng in khng i

r r r Nh vy vect B v S cng chiu nhau. Khi c th biu din vect B nh sau: r r . 0 I . S B = . 2( R 2 + h 2 ) 3/ 2Ti tm ca dng in, h = 0 do : r r . 0 I . S B = . 2 R 3 c tng cho tnh cht t ca dng in trn, ngi ta a ra vect mmen t ca r dng in trn pm , c xc nh bi biu thc: r r (10- 16) pm = I. S , r r trong S l vec t din tch ca dng in c xc nh nh trn. Theo , pm cng r r hng vi B . Khi , vect B c xc nh bi: r . 0 r B = (10-17) . pm 2 2 3/ 2 2 (R + h ) r V th vect pm c trng cho tnh cht t ca dng in trn.3. T trng gy bi ht in tch chuyn ng r Trong mc 2.2 ta bit phn t dng I dl gy ra t trng c vect cm ng t r r r r 0 Idl r dB c xc nh bi nh lut Biot-Savart-Laplace: dB = . . 4 r3 Gi dS l din tch y ca phn t dng di dl, th tch ca phn t dng l dV=dS.dl, no l mt ht in trong phn t dng, s ht trong c phn t dng l n=nodV. Ch cc mi lin h bit: I=jdSn, v j= noqv, ta c th vit: r r r Idl =jdSn. dl =noqvdSn. dl r r V vn tc chuyn ng c hng ca ht in dng v cng chiu vi dl nn ta c r r th hon v v vi dl v c th vit: r r r r Idl = no.qdSn.dl v = no.q.dV. v = n.q v , trong dV=dS.dl l th tch ca phn t dng, n=nodV l s ht in trong th tch dV ca phn t dng. r r Tm li, ta thu c: Idl = n.q v . (10-18) T , ta tm c vect cm ng t gy bi mt ht in chuyn ng vi vn tc: r r r r r r dB 1 0 Idl r 1 0 n.q.v r . Bq = = . = n n 4 n 4 r3 r3 r r r q .v r Bq = 0 . (10-19) 4 r3 r r r r Nu q> 0, vect Bq c chiu sao cho 3 vect v ,r , Bq theo th t hp thnh tam r r din thun (Hnh 10-10a). Nu q0, b) q 0 nu < 900, d m < 0 nu >900, d m =0 nu =900. r r Mt khc, Bn= B.cos l hnh chiu ca vect B ln phng ca php tuyn n , do cng c th vit li (10-23) nh sau: r r (10-24) d m = B.dS cos = Bn.dS = B. dSB

2. T thng

tnh t thng qua din tch S hu hn, ta chia din tch thnh nhng phn t v r cng nh dS sao cho c th coi mi phn t l phng v trn , vect B khng i, khi r r t thng qua dS l d m = B. dS , v t thng gi qua ton b din tch S s c tnh bng tng ca cc t thng gi qua tt c cc phn t din tch c chia t din tch S y: r r (10-25) m = B.dS(S)

Nu S l mt mt phng vung gc vi cc ng cm ng t ( = 0) v t trng l r u ( B = const ) th ta c:

m =

( s)

B.dS

=B

(S)

dS = B.S

(10-26)

201


Trong h n v SI, n v ca t thng l Vbe, k hiu l Wb. n v Vbe s c nh ngha chng cm ng in t (chng11). T n v Vbe, ngi ta nh ngha n v cm ng t Tesla nh sau. Trong cng thc (10-26), nu m =1Wb, S = 1m2, = 0 th:

m 1Wb Wb = = 1 2 = 1Tesla (T) (10-26b) 2 S m 1m Vy: Tesla (T) l cm ng t ca mt t trng u gi qua mi mt vung din tch phng vung gc vi cc ng sc ca n mt t thng u 1Wb.Bn =3. Tinh cht xoy ca t trng Nghin cu t ph ca t trng cc dng in, ta thy cc ng cm ng t l cc ng cong kn. Theo nh ngha tng qut, mt trng c cc ng sc khp kn c gi l mt trng xoy. Vy t trng l mt trng xoy, hay nh ngi ta thng ni, t trng c tnh cht xoy. 4. nh l Oxtrogratxki - Gauss i vi t trng Pht biu nh l : T thng ton phn gi qua mt kn bt k lun lun bng khng. Cng thc biu din nh l O-G nh sau: r r B. dS =0 (10-27) (S)

Chng minh : Ta hy tnh t thng qua mt mt kn S bt k t trong t trng (hnh 10-13). Theo qui c, i vi mt kn, ngi ta chn chiu dng ca php tuyn l chiu hng ra ngoi mt . V vy, t thng ng vi ng cm ng t i vo mt kn l m ( >900, do cos 0. Nu ht in mang in m (q< 0) th chiu ca FL ly ngcli vi trng hp q>0 (xem hnh 10-24). Theo (10-45) lc Lorentz vung gc vi vn tc chuyn ng ca ht nn cng thc hin bi lc ny lun bng khng.2. Chuyn ng ca ht in trong t trng u

r

209

Chng X. T trng ca dng in khng iTa xt chuyn ng ca ht vi vn tc v , c khi lng m, in tch q (q> 0), trong t trng u, khng i theo thi gian, c cm ng t B . V lc Lorentz lun vung gc vi r vect vn tc v v khng thc hin cng nn ng nng ca ht khng bin i, ln ca vn tc cng khng i, lc Lorentz ch lm cho phng ca vect vn tc thay i. Nh vy, lc Lorentz ng vai tr ca lc hng tm, ngha l:

r

r

mv2 FL =qvB.sin = RTa xt hai trng hp sau y :a. Vn tc v ca ht vung gc vi cm ng t B

(10-46)

r

r

V vn tc v ca ht vung gc vi cm ng t B nn lc Lorentz lm cho ht chuyn ng trong mt phng vung gc vi vect cm ng t B , c qu o trn (hnh 1025) bn knh R. T (10-46) ta suy ra:

r

r

r

FL = qvB= R=

mv2 , R(10-47)

mv qB

Chu k quay ca ht: 2R 2m T= = v qB V tn s quay: 2 qB = = T m

r v

r B

r v

r B

(10-48)

Hnh 10.25

(a)(10-49)

(b)

Chuyn ng ca ht in trong t trng, v B a) trng hp q> 0. b) trng hp q< 0

r

r

Cc biu thc (10-48), (10-49) chng t chu k v tn s quay (T, ) khng ph thuc vo bn knh R v vn tc v ca ht.b. Trng hp vect v hp vi vect B mt gc . Trong trng hp ny, c th phn r tch vect v thnh hai thnh phn:

r

r

thnh phn v vung gc vi B v thnh phn

r

r

r v // song song vi vect r r r v = v + v //

r B:

Thnh phn vung gc buc ht in chuyn ng theo qu o trn vi bn knh:

R=

mv qB

210

Chng X. T trng ca dng in khng iCn thnh phn song song v// c tc dng lm cho ht chuyn ng theo phng ca cm ng t B vi vn tc v//. Vy ht tham gia ng thi hai chuyn ng, kt qu l qu o ca ht l ng xon c, c bn knh nh (10-51), bc ca qu o xon c bng: (10-52) h = v//T Chuyn ng ca ht in trong t trng c nhiu ng dng: to ra vn tc rt ln ca ht in trong cc my gia tc ht (cyclotron) trong vic nghin cu ht nhn nguyn t v cc ht c bn v cc ng dng khc. My chn vn tc o t s e/m ca electron m Joseph Jonh Thomson to ra nm 1897 da trn s chuyn ng trong t trng ca cc ht in c vn tc khc nhau. Da trn hin tng chuyn ng ca ht in trong t trng, nm 1879, Edwin H.Hall ln u tin dng du ca hiu in th Hall xc nh du ca ht in chuyn ng to nn dng in v ng chng t rng cc ht in chuyn ng to nn dng in trong kim loi l cc ht mang in m. v.v...HNG DN HC CHNG X I. MC CH, YU CU

r

Sau nghi nghin cu chng ny, yu cu sinh vin: 1. Hiu c v nh cc nh lut: Ampre, Boit-Savart-Laplace, cc nh l: Oxtrogratxki-Gaux v t thng qua mt kn, nh l Ampre v dng in ton phn. 2.Vn dng c cc nh l v nh lut trn tnh c t trng gy bi: dng in thng, dng in trn, cun dy hnh xuyn, cun dy thng di, khung dy in kn... 3. Xc nh c t trng gy bi ht in chuyn ng v lc Lorentz tc dng ln ht in chuyn ng trong t trng.II. TM TT NI DUNG

1. Thc nghim xc nhn c lc tng tc gia cc dng in tng t nh tng tc gia cc nam chm. Lc ny c gi l lc t. Ampre da ra nh lut thc nghim: lc t dF do phn t dng I dl tc dng ln phn t dng I o dl o cch n mt khong r c xc nh bi tch vect kp (10-3):

r

r

r

r r r r o I 0 dl0 ( Idl r ) dF = 4 r3

(1)

trong , o l hng s t: o = 4.10-7H/m. 2. Dng in gy ra xung quanh n mt t trng, t trng truyn lc tng tc gia cc dng in, n tc dng ln bt k dng in no t trong n. i lng c trng cho t r r trng v mt tc dng lc l vect cm ng t B v cng t trng H . Phn t dng in Idl gy ra vect cm ng t dB ti im M cch n mt on r c xc nh bi nh lut Biot-Savart-Laplace (10-6): r r r 0 Idl r dB = (2) . 4 r3

r

r

211

Chng X. T trng ca dng in khng iNh vy, lc do phn t dng Idl tc dng ln phn t dng I o dl o biu din qua cm ng t l:

r

r

r r r dF = I o dl o dB

(3)

Ngi ta cn a ra vect cng t trng H c trng cho tc dng ca t r trng, trong trng hp mi trng ng nht v ng hng, lin h vi vect B theo biu r r thc: B =o H 3. T trng tun theo nguyn l chng cht: B = hay

r

r

(L)

r dB

r r B = Bii

T cng thc (2), ta tm c ln ca vect cm ng t B gy bi mt on dy dn in thng c dng in I ti im cch n mt on a bng:

r

B=Nu dng in thng di v hn th

o I (cos 1 - cos 2 ) 4a o I 2a

B=

suy ra

H=

I 2 a

Cng t (2) ta tnh c cm ng t do dng in trn cng I bn knh R gy ra ti im nm trn trc cch tm O mt khong h (11-17): r r . 0 . 0 I . S r B= = . pm 2 2 3/ 2 2 2 3/ 2 2 ( R + h ) 2 ( R + h ) r r trong p m = IS l mmen t ca dng in trn, c phng trng vi trc ng trn, c chiu trng vi chiu ca vect B . Nu cho h=0, ta tm c cm ng t B gy bi dng in trn ti tm O. 4. T (2), nu ch n mi lin h Idl = nq v , vi n l tng s ht in trong phn t dng Idl ta d dng tm c vect cm ng t do ht in q chuyn ng vi vn tc v gy ra ti im cch n mt on r (10-19): r r r 0 qv r Bq = 4 r3 5. biu din t trng mt cch trc quan, ngi ta a ra khi nim ng sc t trng (ng cm ng t). Khc vi ng sc ca trng tnh in, ng sc t l nhng ng cong kn. Do t thng qua mt kn S bng khng:

r

r

r

r

r

r r B.dS = 0 v suy ra

div B = 0.

r

l nh l O-G i vi t trng. nh l cho thy cc ng sc t l nhng ng cong kn. 6. Tnh cht xoy ca t trng cn c th hin nh l v dng in ton phn (nh l Ampre) (10-32): 212

Chng X. T trng ca dng in khng i r r H. dl =

( C)

Ik =1

n

k

trong ,

Ik =1

n

k

l tng i s cc dng in xuyn qua din tch gii hn bi ng

cong kn C. nh l Ampre gip tnh ton thun li cm ng t B v cng t trng H ti mt im bn trong ng dy in hnh xuyn:

B=o

nI 2 R

trong , n l tng s vng dy qun trn ng, R l bn knh ca vng trn tm O ca vng xuyn i qua im tnh B. T ta tnh c cm ng t gy bi ng dy thng di v hn c s vng dy trn mt n v di no:

B = onoI7. T biu thc (3) ta suy ra lc t dF tc dng ln phn t dng Idl t trong t trng c cm ng t B :

r

r

r

r r r dF = Idl BTa suy ra mt on dy dn di l c dng in I t trong t trng c cm ng t B r (trn l vect B = const) s chu tc dng mt lc t:

r r r F = I .l B l lc Ampre. T ta suy ra hai dng in I1, I2 song song nhau s ht nhau nu cng chiu, s y nhau nu ngc chiu. Lc do dng in ny tc dng ln mt on di l ca dng in kia l (10-39): F21 =

0 I1 I2.l = F12 2d

8. Mt khung dy in kn c dng in I t trong t trng B s s chu tc dng ca r mt mmen lc M (11-40): r r r M = Pm B r r trong , p m = IS l mmen t ca ca dng in I chy trong khung dy. Khung dy nh vy trong t trng B s c mt th nng: Wm= - Pm .B 9. Khi t thng qua mch in thay i, lc t thc hin mt cng:

r r

A= I(m2 - m1) =Im,trong , m l bin thin t thng gi qua din tch ca mch in c cng dng I khng i.

213


r B vi vn tc v s chu tc dng ca lc 10. Nu ht in q chuyn ng trong t trngLorentz:

r r r FL =q v B r r r Lc Lorentz FL vung gc vi v v B , nn cng ca lc ny bng khng, n ch lmi phng chuyn ng ca ht in, khng lm cho ng nng ca ht in thay i v r r ng vai tr ca lc hng tm. Nu t trng l u v vn tc v vung gc vi B th ht in s chuyn ng theo qu o trn trong mt phng vung gc vi B , cn nu v hp vi r r B mt gc th ht in s chuyn ng theo ng xon c c trc cng phng vi B , r r cng chiu vi B nu l gc nhn, ngc chiu vi B nu l gc t.III. CU HI N TP

r

r

1. Nu th nghim minh ho tng tc gia dng in v nam chm, gia dng in v dng in. 2. Pht biu nh lut Ampre, vit biu thc dB gy bi phn t dng Idl ti mt r im trong t trng ca n. Nu r phng chiu v ln ca dB . 3. Pht biu nguyn l chng cht t trng. p dng nguyn l ny nh th no tnh t trng gy bi cc dng in. 4. Tnh cm ng t B v cng t trng H gy bi dng in thng ni chung, dng in thng di v hn, bi dng in trn. 5. Xc nh cm ng t B gy bi in tch q chuyn ng vi vn tc v. 6. nh ngha ng sc t v t ph. Nu tnh cht ca ph ng sc t. V ph cc ng sc ca t trng gy bi mt vi dng in. 7. nh ngha t thng, rt ra nh l O-G i vi t trng. 8. Ti sao ni t trng c tnh cht xoy? Vit biu thc ton hc th hin tnh cht xoy ca t trng. r 9. nh ngha lu s ca ca vect cng t trng H . Thit lp nh l Ampre. Cho v d minh ho nh l ny. 10. ng dng nh l Ampre v dng in ton phn tnh cng t trng H (v tnh B) ti mt im bn trong cun dy hnh xuyn. T suy ra biu thc ca cng t trng H v cm ng t B gy bi ng dy in thng di v hn. 11. Vit biu thc lc Ampre ca t trng B tc dng ln phn t dng in Idl . Nu r phng chiu ln ca lc ny. 12. Tm lc tc dng gia hai dng in thng song song di v hn cng chiu v ngc chiu nhau. 13. Tnh cng ca t lc khi lm di chuyn mt mch in kn trong trong t trng. 14. Tm t lc tc dng ln ht in q chuyn ng trong t trng (lc Lorentz).

r

r

r

214

Chng X. T trng ca dng in khng i15. Ht in q chuyn ng vi vn tc v c qu o nh th no trong t trng B = r r r r const? Xt trng hp v B , v trng hp v hp vi B mt gc .IV. BI TP 1. Mt dng in cng I = 6A chy trong mt dy dn in un thnh hnh vung

r

ABCD c cnh a = 10cm . Xc nh vect cm ng t B v cng t trng H ti tm O ca mch in . Chiu dng in ngc chiu kim ng h.p s:

B1=

o. . 4 .OM

(sin + sin )

;

Trong : OM = a/2

10 7.6 sin + sin = 1,69.10 5 T B1= 2 4 4 5.10 Vy v B = 4B1 = 6,67.10-5 T 6,67.10 5 = = 53,50 A / m. H= o 10 7.4

2. Mt dy dn ng knh d = 1mm qun thnh mt ng dy thng sao cho vect cm r ng t B trong ng c gi tr bng 3.10 -2T .Cng dng in chy trong ng dy bng 6A. Cun dy c my lp, bit rng cc vng dy qun st nhau. p s: p dng cng thc: B= o n0

Trong no l s vng qun trn mt n v di (tc l s vng qun trn mt di ca ng dy bng 1 m). T cng thc trn, ta rt ra: no =

B

o .

=

3.10 2 = 4000 vng / m 4 .10 7.6

Nu ng knh d ca si dy l 10-3 m th mi lp trn 1m s c:1 1 = 3 = 10 3 vng d 10

Vy s lp phi qun l:

4000 = 4 lp 1000

3. Mt dy dn c un thnh mt hnh tam gic u, mi cnh l a = 50cm. Dng r in chy trong dy dn c cng I=3,14 A. Tnh cng ca vect cm ng t B v

cng t trng H ti tm ca tam gic .p s:

B = 1,13.10 -5 T ; H = 9 A/m.

4. Mt dng in cng I chy trong mt dy dn un thnh hnh ch nht c cnh r l a v b. Xc nh cc vect B v H ti tm 0 ca hnh ch nht . Cho bit I=12A, a=16cm, b = 30cm . Chiu dng in ngc chiu kim ng h. p s:

B=

2 o 2 (b + a 2 ) = 68.10 -6 T 2 ab

215

Chng X. T trng ca dng in khng ir Chiu ca B v H vung gc vi mt hnh v v hng ra pha ngoi.

5. Cho hai dng in thng di v hn song song vi nhau t cch nhau 5cm, cng r ca hai dng in bng nhau v bng I=10A. Xc nh vect cm ng t B gy bi cc dng din ti mt im A nm gia hai dng in trong cc trng hp:

a) b)

Cc dng in chy cng chiu. Cc dng in chy ngc chiu nhau. a) B=0; b) B=1,6.10-4 T.

p s:

6. Mt ng dy in thng c qun bng mt si dy dn ng knh d=1mm, dng in chy trong dy dn l 4A. S lp qun trn ng dy l 3 lp. Tnh s vng dy qun trn r mt n v di ca ng. Tnh cng ca vect cm ng t B v cng t trng H bn trong ng. p s: n = 3000 vng/m;

B= 150,8.10-4 T;

H=12000A/m

7. Tm cng t trng ti mt im cch mt dy dn thng di v hn 2cm c dng in cng I=5A. p s:

H=

2a

=

5 = 39,8 A / m 2.3,14.2.10 2

8. Tm cng t trng ti tm mt dng in trn bn knh 1cm c dng in cng bng 1A.1 p s: H = = = 50 / m . 2 R 2.10 2

I1

M2

M1

A (h9) I1 A I2 B (h11)

B

M3

9. Hnh v (h 9) biu din tit din ca hai dy dn in thng di v hn c mang dng in I1, I2. Khong cch gia hai dy dn bng 10cm, I1=20A, I2=30A. Tm cng t trng gy bi cc dng I1 v I2 ti cc im M1, M2, M3. Cho bit AM1=2cm, AM2=4cm, BM3 =3cm. p s:

I3 C

H1=120 A/m;

H2=159A/m;

H3=135 A/m.

10. Gii bi tp trn, vi iu kin cc dng in I1 v I2 chy cng chiu. p s: H1=199A/m; H2=0A/m; H3=183 A/m. 11. Hnh v (h11) biu din tit din ca ba dng in thng di v hn.

Cho bit: AB = BC = 5cm, I1 = I2 = I v I3=2I. Tm mt im trn AC ti cng t trng gy bi ba dng in bng khng.p s: R rng l trn an BC, t trng tng hp gy bi ba dng in khng th r r r bng khng v ti c ba t trng H 1 , H 2 , H 3 u cng phng chiu. im M cn tm

ch c th nm trong an AB. t AM=x.Ta vit c: H1- H2 +H3 = 0;

I I 2I + =0 2x 2 (5 - x) 2 (10 - x)

216

Chng X. T trng ca dng in khng iPhp tnh cho ta: x=50 = 3,3cm 15

12. Cng bi ton trn, nu c ba dng in I1, I2, I3 u cng chiu. p s: Trong trng hp ny, im N cn tm khng th nm ngoi an AC v khi r r r H 1 + H 2 + H 3 lun lun khc khng. im N cn tm ch c th nm trn ng thng AC

trong cc khang AB hoc BC. t AN=x, ta vit c: r r r H 1 + H 2 + H 3 = 0, H1 = H 2 + H 3

I I 2 = + 2x 2 (5 - x) 2 (10 - x)Ta thu c mt phng trnh bc hai cho x, v c nghim bng: x1 =1,8cm ; x2 = 6,96cm.13. Hai dng in thng di v hn song song t cch nhau 5cm. Dng din chy trong cc dy cng chiu v c cng cng I1 = I2 =10A .Tm vect cng tH1 Hk H2 K

trng H gy bi hai dng in ti im K cch u mi dng 3cm (Hnh bt13).p s:2 H2 = H 12 + H 2 +2H1H2cos

a1M

a2N

(1) (2) (3)

Trong :

H1 = H2 =I/2 a

Hnh bt13

2 d2 = a12 + a 2 - 2a1a2cos =2a2-2a2 cos

Rt cos t (3) v H1, H2 t (2) v thay vo (1), ta c: H=I 2a 2

A

I

B

4a 2 d 2 = 58,68 A/mI D C

14. Cho hai dng in di v hn nm trong cng mt mt phng v vung gc vi nhau. Cng hai dng in u bng

5A v c chiu nh hnh bt14. Tm cng t trng H gy bi hai dng in ti cc im cch u hai dng 10cm .p s: HB=H1+H2= 2B

Hnh bt14

I

2a

=

2.5 =15,92A/m 2.3,14.10 1

L C B

T trng ti D c phng vung gc vi mt phng hnh v v c chiu hng vo pha trong hnh v, c ln bng: HD =15,92A/m, HC = HA =015. C mch in nh hnh v (Hnh bt15), dng in chy trong mch bng I =10A. Xc nh cm ng t B ti im O. Cho bit bn knh R ca cung trn bng R= 10cm v gc = 60 0 . 3 1 o I 6 p s: B= 4 12 R = 6,9.10 T = 6,9 T

2O

1 R

Hnh bt15

217

Chng X. T trng ca dng in khng i16. Ngi ta ni hai im A v B ca mt vng dy dn hnh trn vi hai cc ca mt ngun in. Phng ca cc dy ni i qua tm ca vng dy. B qua nh hng ca cc on dy ni. Xc nh cng t trng ti tm ca vng dy (Hnh bt16). p s:

H0=0.

17. Hai vng dy dn trn c tm trng nhau v c t sao cho trc ca chng vung gc vi nhau, bn knh mi vng dy bng R=2cm. Dng in chy qua chng c cng I1 = I2 =5A . Tm cng t trng ti tm ca cc vng dy .2 p s: H= H 12 + H 2 = 176 A/m.

A O BHnh bt16

I

18. Hai vng dy ging nhau bn knh r =10cm c t song song, trc trng nhau v mt phng ca chng cch nhau mt on a=20cm (hnh 18bt). Tnh cm ng t ti tm mi vng dy v ti im gia ca on thng ni tm ca chng trong hai trng hp:

+ _

a) Cc dng in chy trn cc vng dy bng nhau v cng chiu. b) Cc dng in chy trn cc vng dy bng nhau v ngc chiu.p s: a) Trng hp cc dng in cng chiu: Ti mt im bt k trn trc vng dy, ta c:01 B2 01 M a B2 M B1 02 b) B1 02 a)

B=

o

2 R2 + h2

(

R2

)

3/ 2

+

[R

R22

+ (a h )

2

]

3/ 2

T suy ra ti O1, h=0 ; ti O2, h=a.

Bo1 = Bo 2 =

o 1R22

R2 2 2 R R + a2

[

]

3/ 2

5 = 2,1.10 T

Ti M, h=a/2 ta c:

BM =

o2

.

(R

+a

2 3/ 2

)

= 1,35.10 5 T

c:

b) Trng hp cc dng in ngc chiu: Ti mt im bt k trn vng dy, ta R2 R2 B= o 2 R 2 + h 2 3 / 2 R 2 + (a h )2 3 / 2

(

)

[

]

T suy ra: Ti O1, h = 0,

Bo1 =

o 1

2 R R2 + a2

[

R2

]

= 1,7.10 5 T 3/ 2

r B01 hng cng chiu vi

r r r B1 . Ti 02,h = a, BO 2 hng cng chiu vi B2 .218

Chng X. T trng ca dng in khng iTi M, h = a/2, BM = 0.19. Xc nh cng in trng ti cc im nm bn trong v bn ngoi mt dy dn hnh tr c di v hn c dng in cng I chy qua. Cho bit bn knh tit din thng ca hnh tr l R. p s: H= .r. 2 2R

(H t l bc nht vi r), Vi 0

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