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Application of Maxwell-Ampere's law

  1. Jun 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Two equal and opposite charges (± q) start from positions ± y, respectively and move on the y axis with constant velocity u, so that they approach symmetrically the axis origin. Using generalized Ampere's law find the magnetic field in any given point at the xz plane.

    2. Relevant equations

    Maxwell-Ampere's law

    3. The attempt at a solution

    Lost!
     
  2. jcsd
  3. Jun 2, 2013 #2
    Sorry. You have to show some work. If you can show work, we're allowed to help, and I'd love to do some physics right now, because life is boring right now until 5:00. So, just show how you'd think it would work.
     
  4. Jun 2, 2013 #3

    rude man

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    A good place to start might be with: what's the generaslized Ampere's law? Note that current density = 0 everywhere on the xz plane.
     
  5. Jun 2, 2013 #4
    The displacement current will also be zero? (because from Gauss's law the surface integral of E.dS will be zero?). Therefore the magnetic field is zero?
     
  6. Jun 2, 2013 #5

    rude man

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    Well, D = εE so would you say E = 0 everywhere on the xz plane? And if not, is it time-changing?
     
  7. Jun 2, 2013 #6
    Well, yes the electric field is time changing, so I assume I can't use Gauss's law. How can I express the electric field as a function of time?
     
  8. Jun 2, 2013 #7

    rude man

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    If the charges were fixed at +/- y could you come up with the field at a point on the xz plane?
     
  9. Jun 2, 2013 #8
    If P(x,0,z) is a point on the xz plane, then the electric field on P due to the two charges, would be the vectorial sum of the electric fields due to each charge, i.e.

    [tex]\vec{E} = \frac{q \hat{r_1}}{4 \pi \epsilon_0 r_1^2}+\frac{q \hat{r_2}}{4 \pi \epsilon_0 r_2^2} =
    \frac{- q \hat{j}}{2 \pi \epsilon_0 r^2} [/tex]

    ?
     
  10. Jun 2, 2013 #9

    rude man

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    Your expression for E in terms of r1 and r2 is OK, but then the addition is not.
    As y → 0 the E field at (x,0,z) must go to zero, right?

    Express your r1 and r2 vectors in terms of x, y and z. You got the resultant unit vector right, it is j.
     
    Last edited: Jun 2, 2013
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