(b) (0, 3) L1 (−3, 0) L3 Figure P3.55: Problem 3.55. Solution: (a) ♥ Z C Bdℓ= Z...

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Transcript of (b) (0, 3) L1 (−3, 0) L3 Figure P3.55: Problem 3.55. Solution: (a) ♥ Z C Bdℓ= Z...

  • Problem 3.37 For each of the following scalar fields, obtain an analytical solution for ∇T and generate a corresponding arrow representation.

    (a) T = 10+ x, for −10≤ x ≤ 10 (b) T = x2, for −10≤ x ≤ 10 (c) T = 100+ xy, for −10≤ x ≤ 10 (d) T = x2y2, for −10≤ x,y ≤ 10 (e) T = 20+ x+ y, for −10≤ x,y ≤ 10 (f) T = 1+sin(πx/3), for −10≤ x ≤ 10 (g) T = 1+cos(πx/3), for −10≤ x ≤ 10

    (h) T = 15+ r cosφ , for {

    0≤ r ≤ 10 0≤ φ ≤ 2π.

    (i) T = 15+ r cos2 φ , for {

    0≤ r ≤ 10 0≤ φ ≤ 2π.

    Solution:

    (a)

    T = 10+ x

    ∇T = x̂ ∂T ∂x

    + ŷ ∂T ∂y

    + ẑ ∂T ∂ z

    = x̂.

    The direction of∇T displays the fact thatT increases linearly withx only.

  • (b)

    T = x2

    ∇T = x̂ ∂T ∂x

    + ŷ ∂T ∂y

    + ẑ ∂T ∂ z

    = x̂2x.

  • (c)

    T = 100+ xy

    ∇T = x̂ ∂T ∂x

    + ŷ ∂T ∂y

    + ẑ ∂T ∂ z

    = x̂y+ ŷx.

    The magnitude of the gradient increases monotonically withx.

  • (d)

    T = x2y2

    ∇T = x̂ ∂T ∂x

    + ŷ ∂T ∂y

    + ẑ ∂T ∂ z

    = x̂2xy2 + ŷ2x2y.

  • (e)

    T = 20+ x+ y

    ∇T = x̂ ∂T ∂x

    + ŷ ∂T ∂y

    + ẑ ∂T ∂ z

    = x̂+ ŷ.

  • (f)

    T = 1+sin(πx/3)

    ∇T = x̂ ∂T ∂x

    + ŷ ∂T ∂y

    + ẑ ∂T ∂ z

    = x̂ 2π 6

    cos(πx/3).

  • (g)

    T = 1+cos(πx/3)

    ∇T = x̂ ∂T ∂x

    + ŷ ∂T ∂y

    + ẑ ∂T ∂ z

    = −x̂ 2π 6

    sin(πx/3).

  • (h)

    T = 15+ r cosθ

    ∇T = r̂ ∂T ∂ r

    + φ̂φφ 1 r

    ∂T ∂φ

    = r̂ cosφ − φ̂φφsinφ = x̂.

  • (i)

    T = 15+ r cos2 θ

    ∇T = r̂ ∂T ∂ r

    + φ̂φφ 1 r

    ∂T ∂φ

    = r̂ cos2 φ − φ̂φφ2sinφ cosφ .

  • Problem 3.44 Each of the following vector fields is displayed in Fig. P3.44 in the form of a vector representation. Determine∇ ·A analytically and then compare the result with your expectations on the basis of the displayed pattern.

    (a) A = −x̂cosxsiny+ ŷsinxcosy, for −π ≤ x,y ≤ π

    Figure P3.44(a)

    Solution:

    A = −x̂cosxsiny+ ŷsinxcosy

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    = ∂ ∂x

    (−cosxsiny)+ ∂ ∂y

    (−sinxcosy)

    = sinxsiny−sinxsiny = 0

    Yes,A is divergenceless everywhere.

  • (b) A = −x̂sin2y+ ŷcos2x, for −π ≤ x,y ≤ π

    Figure P3.44(b)

    Solution:

    A = −x̂sin2y+ ŷcos2x

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    = ∂ ∂x

    (−sin2y)+ ∂ ∂y

    (cos2x) = 0

    Yes,A is divergenceless everywhere.

  • (c) A = −x̂xy+ ŷy2, for −10≤ x,y ≤ 10

    Figure P3.44(c)

    Solution:

    A = −x̂xy+ ŷy2

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    = ∂ ∂x

    (−xy)+ ∂ ∂y

    (y2) = −y+2y = y

    NO, A is not divergenceless everywhere. It is divergenceless only aty = 0.

  • (d) A = −x̂cosx+ ŷsiny, for −π ≤ x,y ≤ π

    Figure P3.44(d)

    Solution:

    A = −x̂cosx+ ŷsiny

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    = ∂ ∂x

    (−cosx)+ ∂ ∂y

    (siny) = sinx+cosy

    NO, A is not divergenceless everywhere.

  • (e) A = x̂x, for −10≤ x ≤ 10

    Figure P3.44(e)

    Solution:

    A = x̂x

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    + ∂Az ∂ z

    = 1

    This indicates that the divergence ofA is the same at all points in the defined space. In other words, every small volume is a source of flux (more flux leaving thevolume than entering it), and the net generated flux is the same at all locations.

  • (f) A = x̂xy2, for −10≤ x,y ≤ 10

    Figure P3.44(f)

    Solution:

    A = x̂xy2

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    + ∂Az ∂ z

    = y2

  • (g) A = x̂xy2 + ŷx2y, for −10≤ x,y ≤ 10

    Figure P3.44(g)

    Solution:

    A = x̂xy2 + ŷx2y

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    + ∂Az ∂ z

    = y2 + x2

  • (h) A = x̂sin (πx

    10

    )

    + ŷsin (πy

    10

    )

    , for −10≤ x,y ≤ 10

    Figure P3.44(h)

    Solution:

    A = x̂sin(πx/10)+ ŷsin(πy/10)

    ∇ ·A = ∂Ax ∂x

    + ∂Ay ∂y

    + ∂Az ∂ z

    = π 10

    [cos(πx/10)+cos(πy/10)]

  • (i) A = r̂ r + φ̂φφr cosφ , for {

    0≤ r ≤ 10 0≤ φ ≤ 2π.

    Figure P3.44(i)

    Solution:

    A = r̂ r + φ̂φφr cosφ

    ∇ ·A = 1 r

    ∂ ∂ r

    (rAr)+ 1 r

    ∂Aφ ∂φ

    + ∂Az ∂ z

    = 2−sinφ

  • (j) A = r̂ r2 + φ̂φφr2sinφ , for {

    0≤ r ≤ 10 0≤ φ ≤ 2π.

    Figure P3.44(j)

    Solution:

    A = r̂ r2 + φ̂φφr2sinφ

    ∇ ·A = 1 r

    ∂ ∂ r

    (rAr)+ 1 r

    ∂Aφ ∂φ

    + ∂Az ∂ z

    = 3r + r cosφ

  • Problem 3.46 For the vector fieldE = x̂xz − ŷyz2 − ẑxy, verify the divergence theorem by computing:

    (a) the total outward flux flowing through the surface of a cube centered at the origin and with sides equal to 2 units each and parallel to the Cartesian axes, and

    (b) the integral of∇ ·E over the cube’s volume. Solution:

    (a) For a cube, the closed surface integral has 6 sides:

    n

    E ·ds= Ftop+Fbottom+Fright +Fleft +Ffront +Fback,

    Ftop = ∫ 1

    x=−1

    ∫ 1

    y=−1

    (

    x̂xz− ŷyz2− ẑxy )∣

    z=1 · (ẑdy dx)

    = − ∫ 1

    x=−1

    ∫ 1

    y=−1 xy dy dx =

    (

    (

    x2y2

    4

    )∣

    1

    y=−1

    )∣

    1

    x=−1 = 0,

    Fbottom= ∫ 1

    x=−1

    ∫ 1

    y=−1

    (

    x̂xz− ŷyz2− ẑxy )∣

    z=−1 · (−ẑdy dx)

    = ∫ 1

    x=−1

    ∫ 1

    y=−1 xy dy dx =

    (

    (

    x2y2

    4

    )∣

    1

    y=−1

    )∣

    1

    x=−1 = 0,

    Fright = ∫ 1

    x=−1

    ∫ 1

    z=−1

    (

    x̂xz− ŷyz2− ẑxy )∣

    y=1 · (ŷ dz dx)

    = − ∫ 1

    x=−1

    ∫ 1

    z=−1 z2 dz dx = −

    (

    (

    xz3

    3

    )∣

    1

    z=−1

    )∣

    1

    x=−1 =

    −4 3

    ,

    Fleft = ∫ 1

    x=−1

    ∫ 1

    z=−1

    (

    x̂xz− ŷyz2− ẑxy )∣

    y=−1 · (−ŷ dz dx)

    = − ∫ 1

    x=−1

    ∫ 1

    z=−1 z2 dz dx = −

    (

    (

    xz3

    3

    )∣

    1

    z=−1

    )∣

    1

    x=−1 =

    −4 3

    ,

    Ffront = ∫ 1

    y=−1

    ∫ 1

    z=−1

    (

    x̂xz− ŷyz2− ẑxy )∣

    x=1 · (x̂ dz dy)

    = ∫ 1

    y=−1

    ∫ 1

    z=−1 z dz dy =

    (

    (

    yz2

    2

    )∣

    1

    z=−1

    )∣

    1

    y=−1 = 0,

  • Fback= ∫ 1

    y=−1

    ∫ 1

    z=−1

    (

    x̂xz− ŷyz2− ẑxy )∣

    x=−1 · (−x̂ dz dy)

    = ∫ 1

    y=−1

    ∫ 1

    z=−1 z dz dy =

    (

    (

    yz2

    2

    )∣

    1

    z=−1

    )∣

    1

    y=−1 = 0,

    n

    E ·ds= 0+0+ −4 3

    + −4 3

    +0+0 = −8 3

    .

    (b)

    ∫∫∫

    ∇·E dv = ∫ 1

    x=−1

    ∫ 1

    y=−1

    ∫ 1

    z=−1 ∇·(x̂xz− ŷyz2− ẑxy)dz dy dx

    = ∫ 1

    x=−1

    ∫ 1

    y=−1

    ∫ 1

    z=−1 (z− z2)dz dy dx

    =

    (

    (

    xy

    (

    z2

    2 − z

    3

    3

    ))∣

    1

    z=−1

    )∣

    1

    y=−1

    1

    x=−1

    = −8 3

    .

  • Problem 3.55 Verify Stokes’s theorem for the vector fieldB = (r̂ cosφ + φ̂φφsinφ) by evaluating:

    (a) n ∫

    C B · dℓℓℓ over the path comprising a quarter section of a circle, as shown in

    Fig. P3.55, and

    (b) ∫

    S (∇×××B) ·ds over the s