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  • CHNG IX

    TRNG TNH T

  • I. Dng in: l dng chuyn di c hng ca cc

    ht mang in. Chiu ca dng in theo qui c

    l chiu chuyn ng ca cc ht mang in tch

    dng , hay ngc chiu vi chiu chuyn ng

    ca cc ht mang in tch m

    1. Cng dng in: CD qua din tch S l

    mt i lng c tr s bng in lng chuyn

    qua din tch trong mt n v thi gian .

    dq l in lng chuyn qua din tch S trong

    thi gian dt

    dt

    dqi

  • in lng chuyn qua din tch S trong khong

    thi gian t l:

    Nu phng chiu v cng ca dng in

    khng thay i theo thi gian th dng in c

    gi l dng in khng i v c k hiu l I.

    tt

    idtdqq00

  • 2. Vect mt dng in

    Vect mt dng in ti mt

    im M l vect c:

    * Gc ti M

    * C hng l hng chuyn ng ca ht in

    dng i qua im

    * C ln bng CD i qua mt n v din

    tch t vung gc vi hng

    ndS

    dIj

    +

    +

    + j

    ndS

  • Cng dng in qua din tch S

    dSn l hnh chiu ca dS ln mt

    phng thng gc vi

    Nu mi trng trong c dng in do cc ht

    mang in tch q vi mt n chuyn ng vi

    vn tc gy ra th:

    . . .cos .nS S S

    I dI j dS j dS j d S

    j

    nj

    dSn

    dS

    v

    j nqv

  • 3. Sut in ng ca ngun in

    S ca ngun in l mt i lng c gi tr

    bng cng ca lc in trng do ngun to ra

    lm dch chuyn in tch +1 mt vng quanh

    mch kn ca ngun :

    l in trng do ngun to ra (khng phi

    l trng tnh in)

    Nu in trng ny ch tn ti trn mt on s

    ca on mch th:

    dsE

    C

    .

    )(

    *

    *E

    dsE

    s

    .

    )(

    *

  • 4. nh lut Ohm dng vi phn

    Xt hai din tch nh dSn nm vung gc vi cc

    ng dng, v cch nhau mt khong nh dl.

    Gi V v V + dV l in th ti hai din tch y,

    dI l cng dng in chy qua chng. Theo

    L Ohm, ta c:

    M:

    :in dn xut ca mi trng

    RdVRdVVVdI //)(

    V V+dV

    dl

    dSn j

    dl

    dV

    dS

    dIj

    dSdl

    dVdI

    dS

    dlR

    n

    nn

    1

    .1

    .

    EEjEdl

    dV

    1

    1

  • V v lun cng phng chiu vi nhau nn:

    y l dng vi phn ca nh lut Ohm

    5. Phn t dng in:

    L mt on rt ngn ca dng in c biu

    din bng vect nm trn dy dn c

    phng chiu l phng chiu ca dng in v

    c ln bng Idl.

    j E

    j E

    dlI

  • III. T trng:

    1. Khi nim t trng: Vt l hin i cho rng

    bt k mt dng in no cng to ra khong

    khng gian xung quanh n mt dng vt cht

    gi l t trng. Biu hin v s tn ti ca TT

    l lc tc dng ln mt kim nam chm hay mt

    dng in khi ta t chng vo trong t trng.

    Tng tc gia nam chm vi nam chm, nam

    chm vi dng in, dng in vi dng in gi

    l tng tc t

  • 2. nh lut Ampere:

    Lc do phn t dng in tc dng ln

    phn t dng in t cch nhau mt on

    r l:

    0 = 4.10-7 l hng s t

    l t thm ca mi trng

    l vecto v t n

    1 1I dl

    1 2I dl

    2 2 1 1

    3

    0

    ( )

    4

    I dl I dl rd F k

    r

    k

    r1 1I dl 1 2I dl

  • 3.Vect cm ng t:

    Vect cm ng t do phn t dng in gy

    ra ti im M cch n mt khong r l :

    l vect v t PTD n im M

    dlI

    3.

    4 r

    rdlIBd o

    r

  • c :

    * Gc ti im M

    * Phng vung gc vi

    phn t dng in v im M

    * Chiu xc nh bng qui

    tc vn nt chai

    * ln :

    Biu thc trn c Biot Savart Laplace

    a ra t thc nghim, do cn gi l L Biot

    Savart Laplace

    Bd

    2

    sin

    4 r

    IdldB o

    O

    M P

    d B

    I dl

  • 3.Nguyn l chng cht t trng

    Vect cm ng t do mt dng in bt k gy ra ti im M:

    Vect cm ng t do nhiu dng in gy ra:

    Vect cm ng t l mt i lng vt l c trng cho t trng v mt tc dng lc.

    4.Vect CTT: Trong mi trng ng cht v ng hng vect CTT c nh ngha:

    BdB

    n

    i

    iBB

    1

    o

    BH

  • IV. ng dng

    1.Xc nh vect cm ng t do dng in thng c

    dng in khng i I chy qua.

    Gii

    c phng vung gc vi mt phng

    hnh v v c ln

    X

    a

    x

    dx

    d B

    0 0

    2 2

    sin cos

    4 4

    Idx IdxdB

    r r

    B d B B dB

    M

    H

  • Ta c:

    Trng hp hai u dy di ra v cng

    2

    1

    2

    0 02 1

    ;cos

    cos

    cos (sin sin )4 4

    ar

    dx atg dx a

    I IB d

    a a

    01 2

    2 2

    IB

    a

  • Trng hp: H nm ngoi dng in thng

    Trng hp M nm trn ng ko di ca dng in th: B = 0

    2

    1

    0 02 1cos (sin sin )

    4 4

    I IB d

    a a

    H M

    M

  • 2. Mt on dy thng c dng in cng I,

    c un thnh mt cung trn AB, tm O bn

    knh R, gc AB = o. Xc nh vecto cm ng t

    ti O.

    Gii

    c phng vung gc vi

    mt phng hnh v c

    chiu hng vo v c

    ln

    0 02 2

    sin2

    4 4

    IdlIdl

    dBR R

    d B

    O

  • 0 0 0

    2 2

    0 0

    4 4

    4

    B d B

    I I RB dB dl

    R R

    IB

    R

  • 3. Xc nh vect cm ng t do dng in cng

    I chy trong dy dn un thnh vng trn bn

    knh R gy ra ti im M nm trn trc ca dng

    in v cch tm O ca n mt on h.

    Gii

    c phng chiu nh hnh v v c

    ln

    R

    M

    O

    r

    d B

    dl

    0

    24

    n t

    IdldB

    r

    B d B d B d B

    d B

  • nm trn trc, thng gc vi trc

    Do dng in trn i xng nn

    Nn

    Chiu ca l chiu tin ca nt chai khi xoay

    n theo chiu ca dng in

    ndB tdB

    0td B

    0 0 0

    32 32 2 2

    2

    0 0

    3 32 2 2 22 2

    cos

    24 4

    4

    2 2

    n nB d B B dB dB

    Idl IR IRRdl R

    r r rR h

    I R IS

    R h R h

    B

  • c trng cho tnh cht t ca dng in trn

    ngi ta a ra i lng momen t c nh

    ngha:

    l mt vecto nm trn trc ca dng in trn,

    c chiu l chiu tin ca nt chai khi xoay n

    theo chiu dng in v c ln bng din tch

    S ca dng in. Khi :

    mp IS

    S

    0 0

    3 32 2 2 22 22 2

    mI S pB

    R h R h

  • V d: Cho hai dng in thng di v hn cng

    dng in I nh hnh v. Xc nh B ti O

    B4 = 0, B6 = 0 v O nm trn ng

    ko di ca on (1) v (4) nn:

    I

    I

    R (6)

    (5)

    (4) (3)

    (2)

    (1)

    O

    1 2 3 4 5 6B B B B B B B

    1 2 3 5B B B B B

  • 00 01

    00

    2

    00 03

    00

    5

    (sin 0 sin 90 )4 4

    2

    4 8

    (sin 0 sin 90 )4 4

    2

    4 8

    IB

    R R

    II

    BR R

    IB

    R R

    II

    BR R

  • thng gc vi mt phng hnh v,

    hng ra, thng gc vi mt phng hnh v,

    hng vo nn

    1 2 3, ,B B B5B

    1 2 3 5

    0

    2

    B B B B B

    I

    R

  • Tnh B ti O

    O

    I

    I1 I2

    R

    (1)

    (2) (4)

    (3)

    A

    B

  • Ta c:

    thng gc vi mt phng hnh v v

    hng ra, thng gc vi mt phng hnh v v

    hng vo nn:

    1 2 3 4B B B B B

    1 2 3, ,B B B

    4B

    1 2 3 4

    01

    0

    3 1

    0 1 0 22 4

    (sin sin )4 cos 2

    (1 sin )4 cos

    2 (2 2 );

    4 4

    B B B B B

    IB

    R

    I

    R

    B B

    I IB B

    R R

  • Ta c

    r1 v r2 l in tr ca cc cung A2B v A4B , v

    in tr t l vi chiu di nn:

    1 1 2 2ABU r I r I

    1

    2

    2 1 2 4

    1 3

    2

    (2 2 )

    (2 2 ) 2

    r R

    r R

    I I R B B

    B B B

  • V. nh l Gauss ( i vi TT)

    1. ng sc t trng: l ng cong vch ra

    trong t trng sao cho tip tuyn ti mi im

    ca n trng vi phng ca vect cm ng t

    ti im y, chiu ca ng sc l chiu ca

    vect cm ng t.

    - ng sc TT l cc ng cong kn nn t

    trng l trng xoy.

    - Ngi ta qui c v s ng sc qua mt

    n v din tch nm vung gc vi phng ca

    TT t l vi ln ca vect cm ng t ti ni

    t din tch .

  • 2. T thng: T thng gi qua din

    tch dS l i lng:

    l vect c cng phng chiu

    vi php tuyn ca din tch ang xt v c

    ln bng chnh din tch dS

    T thng qua din tch dS v tr tuyt i t l vi

    s ng sc qua din tch dS.

    . . .cos

    n n

    d B d S B dS

    B dS BdS

    Sd

    dSn dS

    M

    Bn

  • T thng gi qua din tch S nm trong t

    trng bt k:

    Ta chia din tch S ra thnh cc phn t din tch

    dS VCB sao cho coi nh khng thay i trn

    din tch dS . T thng gi qua din tch S l:

    B

    )(S

    SdBd

  • 3.nh l Gauss:

    Theo quy c, i vi mt kn, ngi ta quy c

    chn chiu dng ca php tuyn hng ra pha

    ngoi mt . V vy, t thng ng vi ng sc

    i vo mt kn l m ( v > 90o), t thng ng

    vi