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!

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]
     
  2. jcsd
  3. Apr 9, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar

    Staff: Mentor

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

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    "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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

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

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

Have something to add?



Similar Discussions: Maxwell's equations
  1. Maxwell's equations (Replies: 11)

  2. Maxwell's Equations (Replies: 2)

  3. Maxwell's Equation (Replies: 9)

  4. Maxwell's equations (Replies: 6)

  5. Maxwell's equations (Replies: 3)

Loading...