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Homework Help: Consider the Electric Field E(t,x,y,z) = Acos(ky-wt)k

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the Electric Field [itex]E(t,x,y,z) = A\cos(ky-wt)\hat{k}[/itex]

    a) Find the magnetic field such that [itex]\partial_t B + \nabla \times E = 0 [/itex]
    b) Show that ##\nabla \cdot E = 0## and that ##\nabla \cdot B = 0##.

    3. The attempt at a solution

    So for part a, the curl of [itex]E = -\partial B / \partial t [/itex]

    Curl of E works out to be [itex]-Ak\sin(ky-wt)[/itex] I think but then I'm not sure how you get the magnetic field.

    For part b, div E = 0. I don't understand how to get this. I do the calculations and get [itex]-A*t*\sin(k*y - t*w) - A*y*\sin(k*y - t*w)[/itex]? Where am I going wrong? I don't know how to do the second part at all, div B = 0.

    Thanks in advanced
    Last edited by a moderator: Apr 16, 2013
  2. jcsd
  3. Apr 16, 2013 #2


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    The curl of a vector field is a vector field. What is the direction vector?

    Find B by solving the equation you are given:
    [tex]\nabla\times \vec{E}= -\frac{\partial\vec{B}}{\partial t}[/tex]

    div fi+ gj+ hk is f_x+ g_y+ h_z. n Here f and g are 0 while h is a function only of y and t, not z.

    Last edited by a moderator: Apr 16, 2013
  4. Apr 16, 2013 #3


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    Hi Smazmbazm! :smile:

    (i assume k is the unit vector in the z direction)
    in which direction? :wink:
    i'm not sure how you got that :confused:

    div is simply ∂/∂x of the x coordinate + … + …

    try again :smile:
  5. Apr 16, 2013 #4
    Right so for the first part [itex] \nabla \times E = [\partial / \partial y Acos(ky - wt) - 0]\hat{i} + [\partial / \partial x Acos(ky - wt) - 0]\hat{j} + [0 - 0]\hat{k} [/itex].

    [itex]\nabla \times E = -Aksin(ky - wt)\hat{i}[/itex], correct?

    For b, right. I think I confused myself. The definition of div is [itex]\partial f / \partial x + \partial g / \partial y + \partial h / \partial z [/itex]. Is that f, g and h referring to each direction? Like [itex]\hat{i}, \hat{j}, \hat{k}[/itex]? If so, that makes more sense

    Thanks guys
  6. Apr 16, 2013 #5


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    correct :smile:
    yup, that's div(f,g,h) :smile:
  7. Apr 16, 2013 #6
    Sorry the previous post should be [itex]\partial B / \partial t = -Aksin(ky - wt)\hat{i} [/itex] rather then [itex]\
    \nabla \times E = -Aksin(ky - wt)\hat{i} [/itex]

    So to get the electric field you integrate -Aksin(ky - wt) with respect to t? Not sure where to go from there..
  8. Apr 16, 2013 #7


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    Now I'm getting confused …

    isn't it the magnetic field that you want? :confused:

    Yes, integrate ∂B/∂t wrt t, and you get B. :smile:
  9. Apr 16, 2013 #8
    Yes, sorry, magnetic not electric. Ok, thanks tiny-tim
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