CHNG X: T TRNG A.TM TT L THUYT 1. nh lut Bi-Xava-Laplatx:

[ ]3

0 ,.4 r

rlIdBdrrr

= (10-1)

trong Bdr

l vc t cm ng t do phn t dng in lIdr

gy ra ti im M x nh bi bn knh vc t rr , 0 =4.10-7 H/m, l hng s t, l t thm ca mi trng. 2. Nguyn l chng cht t trng:

(c dng in) = BdB

rr(10-2)

v (10-3) =

=n

iiBB

1

rr

3. Vc t cng t trng:

0BHr

r= (10-4)

4. Cm ng t gy bi mt on dng in thng:

RIB

4)cos(cos 210 = (10-5)

trong R l khong cch t im mun tnh cm ng t n dng in. Trng hp dng in thng di v hn: 1= 0, 2= :

RIB

20= (10-6)

5. Cm ng t gy bi dng in trn ti mt im trn trc:

2/3220

)(2 hRISB+

=

(10-7)

trong R l bn knh dng in, S l din tch ca dng in, h l khong cch t tm dng in trn ti im mun tnh cm ng t 6. nh l Oxtrgratxki-Gaox i vi t trng:

=)(

0S

SdBrr

(10-8)

7. nh l Ampe v lu s ca t trng:

=

=)( 1C

n

iiIldH

rr (10-9)

8. Cng t trng: Bn trong mt cun dy in hnh xuyn:

80

RnIH2

= (10-10)

n l s vng ca cun dy in hnh xuyn, R l bn knh ca ng trn tm l tm ca dng in hnh xuyn i qua im mun tnh t trng. Bn trong mt ng dy in thng di v hn: H=n0I (10-11) n0 l s vng dy trn n v di ca ng dy. 9. Lc tc dng ca t trng ln dng in:

[ ]BlIdFd rrr ,= (10-12) 10. Cng ca lc t:

A = I(2m - 1m) (10-13) trong 1m v 2m l t thng gi qua din tch lc u v lc sau ca mch in.

B. BI TP V HNG DN GII 10.1 Hai dy dn thng di song song xuyn qua v vung gc vi mt

phng hnh v. Khong cch gia hai dy l 32cm, khong cch t dng in I1 n im M l 8cm, khong cch t dng in I2 n im N l 8cm. Dng in I2 c chiu nh hnh v v c cng l 5A.

M NI1 I2

+

Hnh 10-1a. Hi dng in I1 phi c chiu v cng l bao nhiu cm

ng t ti N bng khng? b. Xc nh vc t cm ng t ti im M trong trng hp dng in

I1 va tm c trn. Gii

a. Gi vc t cm ng t do dng in I1 gy ra ti N l NB1

r , vc t cm

ng t do dng in I2 gy ra ti N l , cm ng t ti N bng khng

c ngha l : NB 2

r

NB2r

NB1r

M NI1 I2

+NBr

= + = 0 NB1r

NB 2r

NN BB 21rr

= th I1 phi ngc chiu so vi I2 v

B1N = B2N (1) Hnh 10-1a Ta c:

1

17

1

101 10.22 R

IRIB N

==

81

217

10.4010.2

=I (2)

v:

2

27

2

202 10.22 R

IRIB N

==

227

10.810.2

=I (3)

Thay (2) v (3) vo (1):

)(25408 1

12 AIII ==

Vy I1 ngc chiu so vi I2, c ln bng 25A.

b. Gi vc t cm ng t do dng in I1 gy ra ti M l , vc t cm ng t do dng in I

MB1r

2 gy ra ti M l

MBr

MB2r

MB1r

M NI1 I2

+

Hnh 10-1b

M2Br

c xc nh nh hnh v, v ln:

1

17

1

101 10.22 r

IrIB M

==

v:

2

27

2

202 10.22 r

IrIB M

==

Vc t cm ng t tng hp ti M l MBr

c xc nh:

MMM BBB 21rrr

+= v BB1M>B2M nn vc t cm ng t tng hp ti M l c xc nh nh hnh v, ln:

MBr

)(10.610.405

10.82510.210.2 522

7

2

2

1

17 TrI

rIBM

=

=

=

A

D

Br

C

10.2 Mt dy dn c gp li thnh hnh tam gic vung cn ADC c AD=AC=10cm (hnh v). Khung dy c t trong 1 t trng u cm ng t B=0,01T. Cho dng in I=10A chy trong khung theo chiu CADC. Xc nh lc t tc dng ln cc cnh ca khung dy.

Hnh 10-2

Gii

82

Cc lc tc dng vo tng cnh ca khung dy c phng chiu nh hnh v. V ln: v cc cnh ca khung dy u vung gc vi vc t cm ng t B

r do ta p

dng cng thc:

CAFr

DCFr

ADFr

I

A

D

Br

C

F = IBl FAD = FCA = IBa

= 10.10-2.10-1 = 10-2(N) )(10.41,110.22 22 NIBaFDC

===

10.3 Mt dy dn c un thnh hnh ch nht c cc cnh a = 16cm, b = 30cm, c dng in cng I = 6A chy qua. Xc nh vc t cng t trng ti tm ca khung dy.

Gii Vc t 1H

r, , , 2Hr

3Hr

4Hr

ln lt do cc on dy dn mang dng in DA, AB, BC, CD gy ra ti tm O ca hnh ch nht ABCD c phng vung gc vi mt phng hnh v, chiu hng vo trong, c ln c xc nh theo cng thc:

)cos(cos4 21

=r

IH

Mt khc:

22131cos2.

24 ba

abI

bIHH

+===

2

1

r Hr

I

D

B

C

A

Hnh 10-3

)/(3)10.30()10.16(

10.1610.30.14,3

62222

2

2 mA+

=

22142'cos2.

24 ba

baI

aIHH

+===

)/(5,10)10.30()10.16(

10.3010.16.14,3

62222

2

2 mA+

=

Hr

cng phng cng chiu vi cc vc t thnh phn, ln ca ti tm O ca hnh ch nht ABCD l: H

r

H = H1 + H2 + H3 + H4 = 2(3 +10,5) = 27(A/m)

83

10.4 Mt dy dn c un thnh hnh tam gic u mi cnh a = 50cm. Trong dy dn c dng in cng I = 3,14A chy qua. Xc nh vc t cng t trng ti tm ca khung dy.

Gii Vc t , , ln lt do cc on dy dn mang dng in CA, AB, BC gy ra ti tm O ca hnh tam gic ABC c phng vung gc vi mt phng hnh v, chiu hng vo trong, c ln c xc nh theo cng thc:

1Hr

2Hr

3Hr

)cos(cos4 21

=r

IH

trong

63ar =

Mt khc:

cos2.4321 rIHHH ===

I BC

A

Hr

Hnh 10-4

)/(3

610.50.3.14,3.4

14,32

mA==

Gi l vc t cng t trng tng hp ti tm O ca tam gic, ta c:

0Hr

3210 HHHHrrrr

++= V 3 vc t , , cng phng cng chiu nn 1H

r2Hr

3Hr

0Hr

cng phng cng chiu vi cc vc t thnh phn, ln ca 0H

rti tm O ca tam

gic ABC l: H0 = 3H1 = 3.3 = 9(A/m)

10.5 Mt khung dy trn bn knh R = 5cm, Khung gm 12 vng dy,

trong mi vng dy c dng in cng I = 0,5A. Xc nh cm ng t ti tm ca khung dy.

Gii p dng cng thc:

84

RnI

RnIB 70 10.2

2==

)(10.54,705,0

5,0.1210.2 57 T ==

Vy cm ng t do khung dy mang dng in gy ra ti tm ca khung l: B = 7,54.10-5T 10.6 Mt dy dn di, an gia c un li thnh mt hnh vng trn

nh hnh v. Bn knh vng trn dy dn l R = 6cm. Trong dy dn c dng in cng I = 3,75A chy qua. Xc nh vc t cm ng t ti tm ca vng dy.

Gii

Ta coi nh ti tm O ca vng trn c t trng tng hp do dng in thng di v dng in trn gy ra. Gi vc t 1B

r, ln lt do cc dng in thng

v dng in trn gy ra ti tm O ca hnh trn, cc vc t ny c phng vung gc vi mt phng hnh v, hng ra ngoi, hng vo trong mt phng hnh v, c ln c xc nh:

2Br

1Br

2Br

II

Br

RI

RIB 701 10.22

==

v:

RI

RIB 702 10.22

==

V B2>B1 nn Br

cng phng cng chiu vi 2Br

, c ln:

Hnh 10-6

)1(10.2 712 ==

RIBBB

)(10.68,2)1(10.675,310.2 52

7 T ==

Vy vc t Br

c phng vung gc vi mt phng hnh v, chiu hng ra ngoi, c ln 2,68.10-5T.

85

10.7 Mt khung dy trn bn knh R = 10cm c dng in cng I = 1A chy qua. Xc nh vc t cm ng t ti: a. Mt im trn trc ca vng dy v cch tm O mt on h = 10cm. b. Tm O ca vng dy.

Gii

Khung dy in trn gy ra ti M v O cc vc t cm ng t ln lt l v MB

r0Br

c phng chiu c xc nh nh hnh v, ln:

0Br

M

MBr

IO

23

22

20

23

22

0

)(2)(2 hR

IR

hR

ISBM+

=+

=

[ ]

)(10.3,2)10()10(

)10.(1.10.2 6

23

2222

227

T

=+

=

RI

RIB 700 10.22

==

Hnh 10-7)(10.3,610

1.10.2 61

7

T

==

Vy: BM = 2,3.10-6 T B0 = 6,3.10-6 T

10.8 Mt khung dy dn hnh vung ABCD c cnh bng a. Trong khung

dy c dng in cng I chy qua. Khung dy c t trong mt t trng u c vc t cm ng t B

r song song vi cc cnh BC v AD

(hnh v). Tnh: Br

DA

CB

T T

a. M men ca lc t tc dng ln khung dy i vi trc T i qua tm hnh vung v song song vi cnh AB.

b. M men ca lc t tc dng ln khung dy i vi trc T bt k song song vi trc T.

Hnh 10-8

Hng dn

86

Lc t tc dng ln hai cnh AB v CD nh hnh v chng to thnh mt ngu lc, p dng cng thc tnh m men ca ngu lc:

CDFr

ABFr

Br

DA

CB

T T

MT = IBS = IBa2 V trc T song song vi trc T nn:

MT = MT = IBS = IBa2Vy m men ca lc t tc dng ln khung dy

i vi trcT v T bng nhau v bng IBa2. 10.9 Mt on dy dn thng di l = 10cm, c dng in cng I = 2A

chy qua. on dy dn chuyn ng vi vn tc v = 20cm/s trong mt t trng u c cm ng t B = 0,5T theo phng vung gc vi ng cm ng t. Dy dn chuyn ng theo chiu khin cho lc in t sinh cng cn. Tnh cng cn sau thi gian t = 10s.

Hng dn Khi on dy dn chuyn ng, lc t tc dng ln on dy dn c tnh theo cng thc:

F = IBl Theo bi cng ca lc F l cng cn, do : AC = -Fs = -IBlvt Thay cc gi tr vo ta c kt qu cn tm.

Kt qu: AC = -0,2J

r

DC

BA

10.10 Mt khung dy hnh vung ABCD chiu di mi cnh a =20 cm, c t gn dng in thng di v hn cng I=30A (hnh v) khung dy ABCD v dng in thng cng nm trong mt mt phng, cnh AD cch dng in thng mt on r = 1cm. Tnh t thng gi qua khung dy.

I

Hnh 10-10Hng dn Chia khung dy ABCD thnh nhng di hp c b rng dx cch dng in mt khong x. T trng do dng in gy ra ti dx l:

xI

xIB 701 10.22

==

x dx

r

DC

BA I

T thng gi qua din tch dS = adx l: dm = BdS = Bldx

+

=ar

rm x

dxIa710.2

87

r

arIa += ln10.2 7 Thay cc gi tr vo ta c:

rarIam

+= ln10.2 7

Kt qu: m = 13,2(Wb)

88

CHNG X: T TRNG A.TM TT L THUYT B. BI TP V HNG DN GII Gii Gii

Gii Gii Gii Gii Gii Hng dn