9-Magnetostatics Part 1

Embed Size (px)

Citation preview

  • 8/7/2019 9-Magnetostatics Part 1

    1/14

    28 February 2011 1

    ElectromagneticsElectromagnetics TheoryTheory

    Magnetostatics: Part 1

    Outline:

    1. Biot -Savarts Law

    2. Amperes Circuit Law

    3. Magnetic Flux Density Maxwells Equations

    4. Maxwells Equation for Static Fields

    5. Magnetic Scalar and Vector Potentials

    6. Electric Boundary Conditions

    7. Capacitance

  • 8/7/2019 9-Magnetostatics Part 1

    2/14

    28 February 2011 2

    OverviewOverview

    Terms Electric Magnetic

    Field Intensity E H

    Flux Density D B

    RelationD = IE

    I = permittivity

    B=QH

    Q = permeability

    Maxwells Equation

    Basic Laws

    Flux

    !v

    !

    E

    Dv

    V

    JH

    B

    !v

    ! 0

    ! SD d] ! SB d]

    !

    !

    encl

    r

    Qd

    r

    QQ

    SD

    aF2

    21

    4TI

    !

    v!

    encl

    R

    I

    R

    I

    l

    al2

    0

    4T

    Q

    Electrostatic static

    electric fields

    ag etostatic staticag etic fields

  • 8/7/2019 9-Magnetostatics Part 1

    3/14

    28 February 2011 3

    BiotBiot--SavartsSavarts LawLaw

    Biot-Savarts law states that the differential ma netic field intensity dH produced at apoint P, as shown in Fi ure P, by the differential current element I dl is proportional to

    the product I dl and the sine of the an le E between the element and the line joinin P

    to the element and is inversely proportional to the square of the distance R between P

    and the element.

    sin

    R

    ddH

    E

    w

    2

    sin

    R

    dlkIdH

    E

    !T4

    1!k

    24

    sin

    R

    lI

    TE!

    3244 R

    dI

    R

    dId

    R

    TT

    RlalH

    v!

    v!

  • 8/7/2019 9-Magnetostatics Part 1

    4/14

    28 February 2011 4

    BiotBiot--SavartsSavarts LawLaw

    244 R

    d

    R

    dd

    R

    TTRlal

    H v!v!

    R= distance vector between dland point P

    dl = alon the direction of the currentI

    aR= points from the current elements to point P.

    Determinin the direction ofdH

    Conventional representation ofH (orI)

    (a) out of the pa e and

    (b) into the pa e.

    aR

  • 8/7/2019 9-Magnetostatics Part 1

    5/14

    28 February 2011 5

    BiotBiot--SavartsSavarts LawLaw

    !v

    R

    R

    v24T

    aJH

    v!

    s

    S

    24T

    aKH

    Volume currentLine current

    v

    !

    L

    R

    R

    dI

    24T

    al

    Surface current

  • 8/7/2019 9-Magnetostatics Part 1

    6/14

    28 February 2011 6

    BiotBiot--SavartsSavarts LawLaw

    To determine the field due to a straightline current carryin filamentaryconductor of finite len th AB as in the fi ure shown.

    JEETVaH )cos(cos

    412 !

    I

    Tri o Identities

    UU

    UU

    UUU

    UU

    sin

    1csc

    csccot1

    csccot

    tan

    1

    cot

    22

    2

    !

    !

    !

    !

    d

    d

    ? A J

    J

    V

    VT

    V

    V

    V

    T

    aH

    aRl

    aaRal

    RlH

    2/322

    3

    4

    ,

    ,

    4

    z

    dzI

    dzd

    zanddzdBut

    Idd

    zz

    J

    E

    E

    J

    E

    E

    EETV

    EVEEV

    T

    EEVEV

    aH

    aH

    !

    !

    !!

    2

    1

    2

    1

    sin4

    cos

    cos

    4

    ,cos,cot

    33

    22

    2

    dI

    ec

    decI

    decdzzLetting

  • 8/7/2019 9-Magnetostatics Part 1

    7/14

    28 February 2011 7

    BiotBiot--SavartsSavarts LawLaw

    Special Case 1: For a semi-infinite conductor (with respect to P), A(0, 0, 0) and B(0, 0, g)and E1 = 90r while E2 = 0r

    JTV

    aH )90cos0(cos4

    rrI

    JTV

    a4

    I!

    Special Case 2: For an infinite conductor (with respect to P), A(0, 0, -g) and B(0, 0, g)

    and E1 = 90r while E2 = 0r

    ? A JV

    arr 0cos0cos4

    I

    JTV

    a2

    I!

    To find unit vectoraJ is not always easy. A simple approach it to obtain it from:

    VJ aaa v! N where al= unit vector alon the line of currentaV = unit vector perpendicular to the line

    of current

  • 8/7/2019 9-Magnetostatics Part 1

    8/14

    28 February 2011 8

    ExampleExample

    The conductin trian ular loop in the fi ure shown carries a current of 10 A. FindH at

    (0, 0, 5).

  • 8/7/2019 9-Magnetostatics Part 1

    9/14

    28 February 2011 9

    Amperes Circuit LawAmperes Circuit Law

    Amperes circuit law states that the line inte ral ofH around a closedpath is the sameas the net current Ienc enclosed by the path.

    ! encldlHAmperes law is similar to Gausss law. Gausss law is applied when char e

    distribution is symmetrical while Amperes law is applied when current distributionis symmetrical.

    By applyin Stokess theorem to the left-hand side of the equation above:

    v!!SL

    enclddI SHlH

    But,

    S

    enclI SJ Therefore, JH !v

    Ma ells r

    ati

    Fr m here, it sh l be bserve thatvH =J { 0, that is a ma netostatic field is notconservative.

  • 8/7/2019 9-Magnetostatics Part 1

    10/14

    28 February 2011 10

    Magnetic Flux DensityMagnetic Flux Density Maxwells EquationMaxwells Equation

    The ma netic flux density B is similar to the electric flux density D. As D =I0E , thema netic flux density is related to the ma netic field intensity H accordin to:

    HQ! where Q0 = 4Tx 10-7 H/m

    Q0 is known as the permeability offree space.

    The ma netic flux throu h a surface S is iven by

    !S

    dSB] where the ma netic flux ] is in webers (Wb) andthe ma netic flux density is in Webers per square

    meter (WB/m2) or teslas (T).

  • 8/7/2019 9-Magnetostatics Part 1

    11/14

    28 February 2011 11

    Magnetic Flux DensityMagnetic Flux Density Maxwells EquationMaxwells Equation

    In an electrostatic field, the flux passin throu h a closed surface is the same as the char e

    enclosed; ] = Qencl = . Unlike electric flux lines, ma netic flux lines always close

    upon themselves as in (b). This is because it is notpossible to have isolatedmagnetic poles

    (or magnetic charges).An isolated ma netic char e does not exist.

    SD d

    By cuttin a ma net bar into

    smaller and smaller pieces, we

    would still et a bar with both

    north and south poles to ether.

  • 8/7/2019 9-Magnetostatics Part 1

    12/14

    28 February 2011 12

    Magnetic Flux DensityMagnetic Flux Density Maxwells EquationMaxwells Equation

    Thus, the total flux throu h a closed surface in a ma netic field can be written as:

    This equation is referred to as the law ofconservation ofthe magnetic flux orGausss

    law for magnetostatic fields.

    By applyin the diver ence theorem,

    ! 0SB d

    ! vS dvd BB

    0! BMa ells t

    ati

  • 8/7/2019 9-Magnetostatics Part 1

    13/14

    28 February 2011 13

    Maxwells Equation for Static FieldsMaxwells Equation for Static Fields

    Differential Form Integral Form Remarks

    Gausss law

    Non-existence of ma netic

    monopole

    Conservative nature ofelectrostatic field

    Amperes law

    vV! D

    0! B

    0!v E

    JH !v

    !S v

    vdvd VSD

    !S d 0SB

    !L S dd SJlH !S d 0lE

    We can summarize all of the four Maxwells equation as shown in the table below. Thechoice between differential or inte ral forms depend on the problem which we want to

    solve. A static field can only be electric or ma netic if it satisfies the correspondin

    Maxwells equations.

  • 8/7/2019 9-Magnetostatics Part 1

    14/14

    28 February 2011 14

    Magnetic Vector PotentialsMagnetic Vector Potentials

    Electric Magnetic

    Relations E = V(V/m) B = vA(Wb/m2)

    Point char e

    Line char e/current

    Surface

    char e/current

    Volume

    char e/current

    -

    Ma netic flux

    throu h a surface

    !

    !

    !

    v

    L

    R

    dv

    R

    d

    R

    Id

    T

    Q

    T

    Q

    T

    Q

    4

    4

    4

    JA

    KA

    lA

    ddd!

    d

    dd!

    d

    dd!

    d!

    v

    v

    S

    S

    L

    L

    vdV

    SdV

    ldV

    QV

    rr

    r

    r

    rr

    r

    r

    rr

    r

    r

    rr

    r

    V

    TI

    V

    TI

    V

    TI

    TI

    0

    0

    0

    0

    4

    1)(

    4

    1)(

    4

    1)(

    4)(

    ! L dlA]