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An infinite charged line moving with velocity V and its energy current

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider an infinite charged line along the z-axis, with linear charge density .
    The charge moves uniformly with velocity v in the positive z direction.
    1. Give an expression for the electric and magnetic field.
    2. Give an expression for the energy
    flux density (or energy current).
    3. Why does the energy
    ow in the direction it does?
    4. Bonus question. Instead of a charged thin rod, one may imagine a
    neutral conductor, with positive ions, and negative electrons carrying the
    current. In this case there is no electric field, and thus not energy current.
    Explain the difference.


    2. Relevant equations
    If I am correct, the energy flux density or the energy current is the Poynting vector, which is 1/[itex]\mu0[/itex] E cross product B.

    For the E field and the B field, I somehow feel like it's just Gauss's law and Ampère's law. This is a rather important point though, as I'm not sure if I can just use them here. (In class we've already treated retarded potentials and such, but this question 'feels' like it is of less advanced material.)

    3. The attempt at a solution

    Alright, so I used Gauss's law, using a cylinder with radius r, to compute the E field. Similarly, I used a circle with radius r to compute the B field. E has a radial direction, and B is perpendicular to that, and I indicated that direction as the phi-hat direction, can I do that?
    Then, as E and B are mutually perpendicular, I could use the right hand rule to compute the direction of S. Actually doing the cross product might have gone wrong, I thought it was the length of E times the length of B times the sine of their angle, which is just E multiplied by B.
    I end up with
    2llhzrs.jpg

    Now, I have no idea how to explain the direction of the S vector. But most importantly, is what I have done up to this point correct? The bonus question I am not too worried about (as in, I don't really feel the urge to do it), as I don't know what to do there, but if it is not too hard, maybe someone could help me get started with that one too?

    Kind regards,
    Verdict
     
    Last edited: Mar 14, 2013
  2. jcsd
  3. Mar 14, 2013 #2

    mfb

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    Apart from the prefactor for S (where did the second 2 and the mu go?), I don't see an error so far. You should specify the velocity definition ("in positive z-direction"?).

    Concerning the bonus question, I don't see why someone would expect an energy flow there... the setup has a perfect symmetry.
     
  4. Mar 14, 2013 #3
    Ah oops. That went wrong in word, it should have said (2pi)^2. Should I specify the velocity definition in a separate line, or do you mean that it has to be reflected in the equation itself?

    And then, if it is indeed correct, explaining why the energy flow has the same direction as the flow of charge.. It just seems 'obvious' that this would be so, but that is hardly an explanation.
     
  5. Mar 14, 2013 #4

    mfb

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    I would do this, or call it vz.

    Energy flow: If we add some (hypothetical) start of the current flow, you have to accelerate particles against the electric field of the existing charges.
     
  6. Mar 14, 2013 #5
    Hm. Sure. And the direction of the acceleration is the direction of the energy flow?
     
  7. Mar 15, 2013 #6

    mfb

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    The direction of the particles, right.
    You need energy at the source, and send this along the line of moving charges.
     
  8. Mar 15, 2013 #7
    Hmm alright. Would you also agree with the explanation that in this case, the current flows in the Z direction, carrying positive charges and thus positive potential energy in this direction, and thus there is an energy flow in that direction?
     
  9. Mar 15, 2013 #8

    mfb

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    The charge type does not matter (S is proportional to lambda squared), the argument works for both charge types.
     
  10. Mar 15, 2013 #9
    Great, thanks a bunch
     
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