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Question regarding current loops - electric dipole radiation

  1. Aug 3, 2015 #1

    fluidistic

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    1. The problem statement, all variables and given/known data; attempt and equations
    Many times I face problems with a wire loop with some current (which may or may not depend on time, which may or not move) "flowing" in it. And I'm asked to calculate the radiation due to it.
    So using the multipole expansion I know that the dominant terms in the radiation will be the electric dipole term, followed by the magnetic dipole term and then the electric quadrupole terms, assuming they are non-zero.
    So the first thing to do is to calculate the electric dipole term which is worth ##\vec p = \int \vec r \rho (\vec r ) dV##. And here is my question. Is this electric dipole term always worth 0, because there's no rho? Or must I imagine that there's a rho because there's a current?
    In an exam (where I had a circular current loop with ##I(t)=I_0 \cos (\omega t)##), I wrote down that rho=0 so that p(t)=0 and so that there's no radiation due to the electric dipole term. I then went on to calculate the magnetic dipole term (which was non zero) and calculated the radiation due to it. But I was awarded 0 credit whatsoever, because the professor did not buy my explanation for ##\vec p (t) =\vec 0## even though it is true that ##\vec p(t)=\vec 0##. He told me I could have used the continuity equation ##\nabla \cdot \vec J + \frac{\partial \rho}{\partial t}=0## to show that this moment vanishes (which really complicates things a lot).
    So I wonder whether I'm correct to assume that rho=0 or not. And if there's a quick way to show that ##\vec p (t)## vanishes for set up like these.



     
  2. jcsd
  3. Aug 3, 2015 #2

    mfb

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    You can have (net) charged objects moving around in the system, then you get a non-zero ρ. Otherwise this term will be zero. Your current loop has zero charge density everywhere if nothing else is specified.
     
  4. Aug 3, 2015 #3

    fluidistic

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    I am not so sure about this now. For instance consider the problem https://www.physicsforums.com/threads/given-a-current-calculate-the-charge-distribution.825923/, there's no mention of rho, only a mention of a current density J.
    And it turns out that rho is not worth 0, altough ##\frac{\partial \rho}{\partial t} = \text{constant}## so that ##\ddot {\vec p } (t)=0## so that there's no radiation due to the dipole term. But rho itself doesn't seem to be 0 even if there was no mention of rho.


    Edit: So my professor was maybe right not to buy my explanation for ##\vec p(t)=\vec 0##. Actually this dipole moment is likely not 0, but may depend linearly on time. But the second derivative of it with respect to time vanishes and the conclusion is the same, there's no radiation from that dipole term.
     
  5. Aug 3, 2015 #4

    mfb

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    Yes, because the problem statement gives a J with does not have zero divergence. If your problem is a current loop, this is not the case.
    The full problem statement would help if we are talking about a specific problem.
     
  6. Aug 3, 2015 #5

    fluidistic

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    Hmm but isn't that particular problem (the one I gave a link) a current loop problem? It's a closed loop with some current in it.
     
  7. Aug 3, 2015 #6

    mfb

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    This? No. The current loop would have the same current everywhere along a loop, this current density does not.
     
  8. Aug 3, 2015 #7

    fluidistic

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    Ok thanks a lot... Then I've no idea why my professor gave me no credit for a whole problem for the current loop ##I(t)=I_0 \cos ( \omega t)##, I stated that ##\vec p (t)=0## because rho =0. So that this dipole term did not contribute to the radiation. I had calculated the E and B radiation fields as well as the total power radiated due to a circular current loop with the current I just wrote. I did all good except for the justification that the dipole term does not radiate. And apparently rho=0 is a valid/solid justification... oh well, this hurts.
     
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