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Homework Help: Maxwell's equations

  1. Apr 9, 2005 #1
    I want to make sure that I understand this good.
    Given E and B are possible in a region of free space (J=0) only if [tex] \triangledown \times E=0 [/tex] and [tex] \triangledown \cdot B = 0 [/tex]
     
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  3. Apr 9, 2005 #2

    dextercioby

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    That [itex] \vec{B} [/itex] needs to be stationary (time independent)...Else [itex] \vec{E} [/itex] would not be a purely potential-derived field.

    Daniel.
     
  4. Apr 9, 2005 #3
    I have this two equations:

    [tex]E=3\sin(3z-6t) \vec{k} [/tex]
    [tex]B=- \frac{1}{15} \sin(3z-6t) \vec{j}[/tex]

    So what should I do first?
     
  5. Apr 9, 2005 #4

    dextercioby

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    Verify whether such a field configuration satisfies the eqn-s

    [tex] \nabla\cdot\vec{B}=0 [/tex]

    [tex] \nabla\times\vec{E}=-\frac{\partial \vec{B}}{\partial t} [/tex]

    Isn't this what u're supposed to do?:confused:

    Daniel.
     
  6. Apr 9, 2005 #5

    jtbell

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    That depends on the question you're supposed to be answering. :confused:
     
  7. Apr 9, 2005 #6

    robphy

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    "Free space" means [tex]\rho={\color{red}0}[/tex] and [tex]\vec\jmath={\color{red}\vec 0}[/tex].
    So, in two of Maxwell's Equations, this means that
    [tex] \nabla \cdot \vec E=\rho ={\color{red}\ 0} [/tex] and
    [tex] \nabla \times \vec B=\vec \jmath + \frac{\partial \vec E}{\partial t}={\color{red}\ \vec 0} + \frac{\partial \vec E}{\partial t} [/tex].
    Of course, [from the other two equations] we must always have
    [tex] \nabla \cdot \vec B= 0 [/tex] and
    [tex] \nabla \times \vec E= -\frac{\partial \vec B}{\partial t} [/tex]

    [Insert your own conventional constants]
     
    Last edited: Apr 9, 2005
  8. Apr 9, 2005 #7
    What?! When we were taught Maxwell's equations in free space, we were told that: [tex]\nabla \cdot \vec E=\frac{\rho}{\epsilon_{0}}[/tex], as free space meant in air not in any sort of medium.
     
  9. Apr 9, 2005 #8

    robphy

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    So, maybe term I should have used is "source-free".
     
  10. Apr 9, 2005 #9
    [tex] \nabla\cdot\vec{B}=0 [/tex] I verified that

    [tex]\nabla\times\vec{E}=\left|\begin{array}{ccc}\vec{i }& \vec{j} &\vec{k}\\\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ 0 & 0 & 3\sin (3z-6t) \end{array} \right| [/tex]


    [tex] \triangledown \times E [/tex] gave me 0 and [tex] -\frac{\partial \vec{B}}{\partial t} =- \frac{2}{5} \cos(3z-6t) \vec{j}
    [/tex]

    I think this wrong?
     
    Last edited: Apr 9, 2005
  11. Apr 9, 2005 #10
    robert,

    You say:

    divB = 0, I agree

    curlE = 0, I agree

    -dB/dt = (-2/5)sin(3z - 6t)j, I think you've got a mistake here
     
  12. Apr 9, 2005 #11

    dextercioby

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    Should be a "cos" there.

    Daniel.
     
  13. Apr 9, 2005 #12
    [tex] \frac{\partial \vec{B}}{\partial t} =-\frac{2}{5} \cos(3z-6t) \vec{j}[/tex]
    So [tex] \nabla \times \vec E= -\frac{\partial \vec B}{\partial t} [/tex]
    are nor equal and they are not possible in region of space?
     
  14. Apr 9, 2005 #13

    dextercioby

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    That's that.It is not possible.They should identically solve every equation from Maxwell's system...

    Daniel.
     
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