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Find the magnetic field in vacuum given an electric field

  1. Jun 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Given an electric field in vacuum:
    Screen Shot 2016-06-09 at 21.09.18.png
    Find the magnetic field under condition that |B| -> 0 as t -> infinity.
    2. Relevant equations
    I clearly have to use Maxwells equations to obtain the electric field here. Our equation for B has to satisfy them. More specifically Faraday's law and Ampere's law I think are most important here.
    3. The attempt at a solution
    I am going to use Faraday's law, take the - curl of the electric field and integrate it w.r.t to time. Curl of E will be E_0/(c*t^2) ( 1, 0 , 0).

    Integrate this with a minus sing w.r.t to time, and get E_0/(c*t)(1,0,0) + F(r), where F(r) is a function of r we need to find.

    Now, we also need B to satisfy Ampere's law and vanish at infinite time. But this field has zero curl, unless F(r) is non-zero, but how can F(r) be nonzero if B has to vanish at infinity? And even if it was non-zero, I have no idea how it could satisfy Ampere's law, as dE/dt if a function of t, but curl of B is not. And I can not find any other form of B except E_0/(c*t)(1,0,0) + F(r) as nothing else seems to satisfy Faraday's law. I am entirely confused.
     
    Last edited: Jun 9, 2016
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  3. Jun 11, 2016 #2

    mfb

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    E_0/(c*t)(1,0,0) goes to zero for infinite time, so F(r) can do that as well.
    Well, your B has to depend on time as well.
     
  4. Jun 11, 2016 #3

    TSny

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    But if F(r) is to go to zero for infinite time, then it appears that F(r) would depend on t. Then wouldn't Faraday's law no longer be satisfied for finite t?

    I can't find a solution to this problem.
     
  5. Jun 11, 2016 #4

    mfb

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    Wait, what exactly do you call F(r)? It is introduced outside an equation. Where is the corresponding Maxwell equation?
     
  6. Jun 12, 2016 #5
    F(r) is just an integration constant. I took an integral w.r.t to time, so I can add on any function of position to it, as long as b.c. are satisfied.
     
  7. Jun 12, 2016 #6

    mfb

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    So where do you solve the Maxwell equations now?
     
  8. Jun 14, 2016 #7
    What do you mean?
     
  9. Jun 14, 2016 #8

    mfb

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    You have an electric field, you want to find the magnetic field. You know that both together have to satisfy the Maxwell equations - that is the only connection between them (Faraday's law and Ampere's law are part of the Maxwell equations). You have to find a magnetic field such that electric plus magnetic field satisfy the equations. I don't understand how you could start at all without understanding that.
     
  10. Jun 15, 2016 #9
    I am trying to solve Maxwells equations. Maxwells equations are differential equations, so there is no one specific solution to them, fields are gauge invariant and what determines the solutions are boundary conditions. So I integrate the maxwells equations to find the solution as one solves a differential equation by integrating. F(r) is just an integration constant to satisfy the boundary conditions of the B field. Sorry if I want clear on that, but I hope now it will be more obvious what I am doing.
     
  11. Jun 15, 2016 #10

    mfb

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    Okay, where are those equations, before or after integration? That's what I am missing here.
     
  12. Jun 17, 2016 #11

    rude man

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    I used one of the Maxwell equations relating B to E. It's an ODE since the E field has a z component only. Blew up at t=0 though ...
    BTW is this in cgs or ? or are the units inconsistent in the given expression for E?
     
  13. Jun 18, 2016 #12

    nrqed

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    Note that the E field given also blows up for non zero y as t goes to zero. So there must be a condition on time that is implicitly assumed in the question (at least t>0)
     
  14. Jun 18, 2016 #13

    rude man

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    Agreed. Did you get an answer? Would like to compare, in private to avoid breaking pf's rules ...

    ... oh, and what about the dimensions discrepancy? Any comment?
     
  15. Jun 18, 2016 #14

    nrqed

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    Well, it would be much better if you posted here your steps, so that we (me or someone else) could check the steps.

    As for the units, there is something fishy from the very beginning as the E field given does not even have the units of an electric field. Is this a problem from a textbook or was it given by the prof?
     
  16. Jun 18, 2016 #15

    rude man

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    As they said in the old TV series The Prisoner: "That would be telling"! We are not supposed to provide OP's with turnkey answers.
    I agree, it's weird. But I've run into weird stuff with physicists using the cgs system (where there is no electric unit of any kind). Hard to believe my own physics intro course used cgs; thank goodness that seems to be waning or has waned.
     
  17. Jun 18, 2016 #16

    TSny

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    By choosing appropriate units for the constant ##E_0##, the overall units for the electric field will be correct for whatever system of units you have chosen to work in. So I don't see any problem with units.


    I think the OP's argument is correct for showing that there is no solution to this problem. To recap the OP's reasoning:

    (1) First note that ##\vec{\nabla} \times \vec{E} =\large \frac{E_0}{ct^2}## ##(1, 0, 0)##.

    (2) Integrating the vacuum Maxwell equation ##\vec{\nabla} \times \vec{E} = - \large \frac{\partial \vec{B}}{\partial t}## with respect to ##t## leads to ##\vec{B} = \large \frac{E_0}{ct}## ##(1, 0, 0) + \vec{f}(\vec{r})## where ##\vec{f}(\vec{r})## is an arbitrary vector function of position which is independent of time. ##\vec{f}(\vec{r})## is the "constant of integration" when integrating with respect to ##t##.

    (3) However, the boundary condition ##B \rightarrow 0## as ##t \rightarrow \infty## requires ##\vec{f}(\vec{r}) = \vec{0}## . So, ##\vec{B} = \large \frac{E_0}{ct}## ##(1, 0, 0)##.

    (4) It is impossible for ##\vec{B} = \large \frac{E_0}{ct}## ##(1, 0, 0)## to satisfy the vacuum Maxwell equation ##\vec{\nabla} \times \vec{B} = \large \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t}## since the left hand side is identically zero while the right hand side is nonzero.

    I think this argument is valid. Or am I overlooking something?
     
  18. Jun 19, 2016 #17

    rude man

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    Certainly misleading to call the constant E-something, don't you think? Let me restate E = k(0,0,y/t2). Then,
    in your point 4, why do you say the left-hand side is identically zero? ∇ x E = - dEz/dy = -k/t2 ≠ 0?

    Otherwise my derivation is same as yours I believe.
     
    Last edited: Jun 19, 2016
  19. Jun 19, 2016 #18

    TSny

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    Yes, a different symbol for the constant would be better.
    In point 4 we are looking at x B rather than x E. The result for x E was given in point 1 and used in point 2.
     
  20. Jun 19, 2016 #19

    rude man

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    Oops, I missed that. Right.
    I suppose the problem was conjured up from ∇ x E = - B/∂t alone, without regard to physical realizability. Thanks.
     
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