1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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: Apr 16, 2013
  4. Apr 16, 2013 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Consider the Electric Field E(t,x,y,z) = Acos(ky-wt)k
Loading...