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Homework Help: Vector potential due to oscillating dipole

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate the magnetic vector potential A at a point p located at a distance r from the axis of an oscillating dipole of length s.

    It is assumed that [tex]r\gg s[/tex] and that the current is the same throughout s.

    2. Relevant equations

    [tex]r=\sqrt{(x^2+(z-z')^2)},[/tex] where x,z is the horizontal and vertical coordinates of p, respectively, and z' is the vertical coordinate of the source point. The axis of the dipole lies on the z axis, and so x'=0. The problem is confined to the xz plane only.

    [tex]A=c\int^{s/2}_{-s/2}\frac{\exp(ikr)}{r}dz' \hat{z},[/tex]

    where c is a constant and [tex]k[/tex] is the wave number. The exponential comes from the fact that the current is a function of the retarded time, [tex][t]=t-r/c[/tex].

    3. The attempt at a solution

    I really don't know how to calculate this integral. Without the exponential I would've been fine, but now... lol wut? Are there perhaps some approximations, expansions, or variable changes that I could do? Any tips?

    If it is of any help, the answer is apparently the same answer as in the case of a current localized at the center of the dipole:

    [tex]A=d*\frac{\exp(i\omega[t])}{r}s \hat{z}[/tex]

    (d is a constant.)
     
    Last edited: Jan 12, 2010
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  3. Jan 13, 2010 #2

    gabbagabbahey

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    The problem statement you provided defines [itex]r[/itex] as the distance from the center of the dipole ([itex]z'=0[/itex] I presume) to the point [itex]p[/itex]...you'll want to use a different letter, like say, (capital) [itex]R[/itex] to represent the separation of the source point and field point:

    [tex]r=\sqrt{x^2+z^2}[/tex]

    [tex]R=\sqrt{(x^2+(z-z')^2)}[/tex]

    It seems like you are using [itex]c[/itex] to represent both a constant with units of Tesla-meters (in SI) and the speed of light....that's pretty confusing notation to me.

    Also, the current in an oscillating dipole varies with time, so why is there no [itex]t[/itex] in this expression?

    In SI units, the expression for the vector potential is something like

    [tex]\textbf{A}(\textbf{r},t)=\frac{\mu_0}{4\pi}\int \frac{I(t-\frac{R}{c})}{R}dz'\mathbf{\hat{z}}[/tex]
     
  4. Jan 13, 2010 #3
    Wow, that's a lot of annoying mistakes I did.

    Yes, my bad.

    Oh boy. Yes, that is definitely confusing. The factor c in front of the integral is not supposed to be the speed of light.

    I shouldn't have called c a constant, since there's supposed to be a time-dependent exponential included.

    Yes. In my case it's (yes, I even forgot the minus sign in the exponential)

    [tex]\frac{\mu_0I_0\exp(i\omega t)}{4\pi}\int_{-s/2}^{s/2}\frac{\exp(-ikR)}{R}dz'\hat{z},[/tex]

    where R is as you said, and [tex]I_0[/tex] is the amplitude of the current, which is independent of z'.

    I still don't know how to calculate the integral, though.
     
  5. Jan 13, 2010 #4

    gabbagabbahey

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    You'll probably want to use [itex]\omega/c[/itex] instead of [itex]k[/itex] here; as you'll want to use the approximation that [itex]s\ll \frac{\omega}{c}[/itex] at some point (basically that the dipole is very small compared to the wavelength of the radiation it produces)

    You also are given that [itex]s\ll r[/itex], and since [itex]|z'|\leq s[/itex], you know [itex]z'\ll r[/itex]....in order for this knowledge to be useful, you'll want to use the law of cosines to express [itex]R[/itex] in terms of [itex]z'[/itex], [itex]r[/itex] and the angle between the source and field point vectors (As measured from the center of the dipole). You can then Taylor expand both [itex]1/R[/itex] and [itex]\text{exp}(-i\omega R/c)[/itex] for small [itex]z'/r[/itex].
     
  6. Jan 14, 2010 #5
    Thank you very much!

    Doing as you say I get

    [tex]R=\sqrt{z'^2+r^2-2z'r\cos(\theta)}=r\sqrt{(z'/r)^2+1-2z'/r\cos(\theta)}.[/tex]

    Expanding [tex]1/R[/tex] and [tex]e^{-ikR}[/tex] around [tex]z'/r=0[/tex] yields [tex]1/r[/tex] and [tex]e^{-ikr}[/tex], respectively (the first order terms of both expansions contain factors [tex]z'/r^2[/tex] and therefore become insignificant).

    The integral then simply becomes

    [tex]\frac{\mu_0I_0e^{i\omega(t-kr)}}{4\pi r}\int dz' \hat{z},[/tex]

    since [tex]r[/tex] is independent of [tex]z'[/tex]. This gives me the correct answer.
     
    Last edited: Jan 14, 2010
  7. Jan 14, 2010 #6

    gabbagabbahey

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    Careful, you can definitely say terms with [itex]\left(\frac{z'}{r}\right)^2[/tex] are insignificant (the square of a very small number is an even smaller number), but that doesn't necessarily mean that [itex]\frac{z'}{r^2}[/itex] is negligible.....As an example, if [itex]s=10^{-6}[/itex] and [itex]r=10^{-4}[/itex], then [itex]|z'|\leq s\ll r[/itex] as demanded by the problem statement, but [itex]\frac{z'}{r^2}\leq\frac{s}{r^2}=100[/itex] won't be negligible.

    You will need to keep the first order terms in your expansions, and use your second approximation ([itex]s\ll \frac{\omega}{c}[/tex] )to later rid yourself of them.
     
  8. Jan 14, 2010 #7
    I'm not exactly having my finest moment as an aspiring physicist here. I blame it on simply rushing through the calculations since I'd really like to be doing something else for a little while. :cry:

    You're right. I didn't get any [itex]\frac{z'}{r^2}[/itex] terms, though; the expansions I did were incorrect. I did have to use the approximation [itex]s\ll \frac{\omega}{c}[/tex] in one of the expansions, though.

    Thanks again!
     
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