3 Gabarito Prova 3 { Eletromagnetismo I { Noturno ?· 3a Gabarito Prova 3 { Eletromagnetismo I { Noturno…

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  • 3a Gabarito Prova 3 Eletromagnetismo I Noturno

    Q1 [4.0]

    (a) [0.5]

    ~B(t, r, ) ' 0k2c

    4

    eikr

    rr ~p = 0 k

    2 c

    4

    eikr

    rp0eitr

    (p0ze

    it)=0k

    2cp04

    ei(krt)

    r(z cos + sen ) z

    = 0k2cp0

    4

    ei(krt)

    rsen

    (b) [1.5]

    ~ ~B = 1r

    [1

    sen

    (B sen )

    r (rB)

    r

    ]= 0k

    2cp04

    1

    r

    [ei(krt)

    r2sen

    (sen2

    )

    r sen (ei(krt)

    )r

    ]

    i0k3cp0

    4

    ei(krt)

    rsen

    Entao

    ~E = c2

    (~ ~B)dt

    ~E = c20k

    3p04

    ei(krt)

    rsen

    ~E =0k

    3c3p04

    ei(krt)

    rsen

    (c) [1.0]

    ~Sp =1

    0~E ~B

    = 10

    0k3c3p0

    4

    ei(krt)

    rsen

    0k2cp0

    4

    ei(krt)

    rsen ( )

    =0k

    5c4p20162

    e2i(krt)

    r2sen2 r =

    00

    05c4p20

    162c5

    e2i(krt)

    r2sen2 r

    =4p20

    1620c3e2i(krt)

    r2sen2 r

    (d) [1.0]

    ~Spt =1

    T

    T0

    1

    0Re[ ~E]Re[ ~B]dt

    =4p20

    1620c3sen2

    r2r

    2

    2/0

    cos2(kr t)dt

    =4p20

    3220c3sen2

    r2r

    Q2 [3.0]

    1

  • a) [1,0] Condicoes de contorno

    ~Etang1 = ~Etang2 (E1p E1p) cos 1 = E2p cos 2

    ~Htang1 = ~Htang2 n11c

    (E1p + E1p) =

    n22c

    E2p

    Segue que

    E1p E1p =cos 2cos 1

    E2p; E1p + E1p =

    n21n12

    E2p

    entao

    E1p =

    n21n12

    cos 2cos 1n21n12

    + cos 2cos 1

    E1p; E2p =2

    cos 2cos 1

    + n21n12

    E1p

    Da

    rp =E1E1

    =

    n21n12

    cos 2cos 1n21n12

    + cos 2cos 1

    =

    n22

    cos 1 n11 cos 2n22

    cos 1 +n11

    cos 2

    e

    tp =E2E1

    =2

    cos 2cos 1

    + n21n12

    =2n11 cos 1

    n11

    cos 2 +n22

    cos 1

    b) [0,5]

    rp = 0n22

    cos 1 n11

    cos 2 = 0

    (n21n12

    )1 sin2 B

    1 n

    21

    n22sin2 B = 0(

    n21n12

    )2(n12n22

    )2sin2 B 1 +

    n21n22

    sin2 B = 0

    sin2 B =

    1(n21n12

    )2n21n22(n12n21

    )2c) [0,5]

    tp =2

    cos 2cos 1

    + n21n12

    =2

    1n22n1

    + n21n12=2n11n2

    Os coeficientes associados a conservacao de energia sao o R e T (e nao r e p).

    d) [0,5]

    T =~S2 kt~S1 kt

    =(2/v2)E

    2p2 cos 2

    (1/v1)E2p1 cos 1=2v11v2

    t2pcos 2cos 1

    =2v11v2

    (2n11n2

    )2 1n22n1

    =22v1n111v2n2

    n1n2

    =v2v1

    v1/c

    v2/c= 1

    2

  • e) [0,5]

    rs = 0n11

    cos 1 n22

    cos 2 = 0(n12n21

    )1 sin2 1

    1 n

    21

    n22sin2 1 = 0(

    n21n12

    )2(n12n22

    )2sin2 1 1 +

    n21n22

    sin2 1 = 0

    sin2 1 =

    1(n21n12

    )2n21n22(n21n12

    )2

    Q3 [3.0]

    a) [0,5]

    ~E = ~B

    t ey

    y[Eei(kyt)ez

    ]= ikEei(kyt)ex =

    ~B

    t

    ~B

    t= ikEei(kyt)ex ~B =

    k

    Eei(kyt)ex

    b) [0,5] Para que a onda possa se propagar, o vetor de onda precisa ser real; como

    k =

    2 2pc2

    entao e preciso que > p.

    c) [0,5]

    (k2) =

    (2 2pc2

    )

    2k

    k = 2

    c2

    k

    k= vgvf = c

    2

    d)

    ~S =1

    40

    [~E ~B + ~E ~B

    ]=

    1

    40

    [Eei(kyt)z k

    Eei(kyt)x+ Eei(kyt)z k

    Eei(kyt)x

    ]=

    1

    20

    k

    |E|2y

    e) [0,5] Se < p, entao,

    k = i

    2p 2

    c2=i

    ; c

    2p 2

    e~E = Eei(kyt)z = Eeiey/z

    3

  • f) [0,5]

    ~E = Eey/ei z ~B = i|k|Eey/zeitex

    ~S =1

    40

    [~E ~B + ~E ~B

    ]=

    1

    40

    [Eeitey/ez (i)

    |k|Eeitey/ex + E

    eitey/ez i|k|Eeitey/ex

    ]= ey

    1

    20

    |k||E|2e2y/[i+ i] = 0

    4

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