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Vector potential components in small loop antenna

  1. Mar 17, 2014 #1
    Consider a small, thin loop in the [itex](x,y)[/itex] plane centered in the origin and with radius [itex]a[/itex]. We are interested in the vector potential [itex]\mathbf{A}[/itex] generated by the loop at a point [itex]P(r, \theta, \phi)[/itex], with [itex]2 \pi a \ll r[/itex], so at a great distance (moreover, [itex]a \ll \lambda[/itex]).

    We need two coordinates systems:

    - one for the source point, for which variables and unit vectors will be primed ([itex]r', \theta', \phi'[/itex]);
    - one for the observation point [itex]P(r,\theta,\phi)[/itex].

    Because of the use of the spherical coordinates, the unit vectors do not coincide in any case.

    A constant current [itex]I[/itex] flows through the loop and in any point we have only [itex]\mathbf{I} = I_{\phi} \mathbf{u}_{\phi'}[/itex], where [itex]\mathbf{u}_{\phi'}[/itex] is the unit vector in the [itex]\phi[/itex] direction.

    The vector potential [itex]\mathbf{A}[/itex] is computed as usual through the integral

    [itex]\mathbf{A} = \displaystyle \frac{\mu}{4 \pi} \int_{loop} \mathbf{I} \frac{e^{-j \beta R}}{R} dl'[/itex]

    but we have to express every [itex]\displaystyle \mathbf{I} \frac{e^{-j \beta R}}{R} dl'[/itex] in terms of the [itex](\mathbf{u}_r, \mathbf{u}_{\theta}, \mathbf{u}_{\phi})[/itex] unit vector, so the unprimed coordinate system. We will obtain a new vector with (in general) all the three components. [itex]R[/itex] is the distance between the position of the actual length element [itex]dl'[/itex] and [itex]P[/itex]: [itex]R = |\mathbf{r} - \mathbf{r'}|[/itex].

    But why the resulting [itex]\mathbf{A}[/itex] has only the [itex]\phi[/itex]-component?

    So, the question is: why a vector like [itex]\mathbf{I}[/itex], which has in its own primed coordinate system only a [itex]\phi'[/itex]-component, after the shown integration originates a vector potential [itex]\mathbf{A}[/itex] with only the same component, [itex]A_{\phi}[/itex]?

    This is the procedure followed in Balanis, Antenna Theory, Ch. 5.

    Thank you anyway!

    Emily
     
  2. jcsd
  3. Mar 18, 2014 #2

    jasonRF

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    The short answer is "symmetry", but that can often be said flippantly without really working it out. In all cases it should be clear that [itex]\mathbf{A}[/itex] can only have an x and a y component (the source azimuth unit vector is [itex]\mathbf{u}_\phi^\prime = \cos\phi^\prime \mathbf{u}_y - \sin \phi^\prime \mathbf{u}_x[/itex]; when it doubt use Cartesian vector components!) . However, it should be clear that the resulting vector potential must be invariant to rotations about the z axis. So if we find A in the x-z plane then we should be able to rotate the result to obtain A everywhere.

    Now, you should be able to convince yourself that [itex]\mathbf{A}(x,y=0,z)[/itex] should only have an x component: by either physical arguemtns or setting up the integral and looking at odd/even functions the x component of must vanish. By symmetry you should likewise have only an x component in the y-z plane, and for the arbitrary rotation about z, you should just have an azimuthal component of A.

    You can also just set up the full integral (use Cartesian vector components again; note it is fine to allow [itex]\mathbf{A} = \mathbf{u}_x A_x(\theta,\phi,r) + \mathbf{u}_y A_y(\theta,\phi,r) [/itex] ) and integrate to find that only an azimuth component survives.

    I hope that makes sense,

    jason
     
  4. Mar 18, 2014 #3

    jasonRF

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    I think if your current was not uniform (was a function of [itex]\phi[/itex]) then the symmetry argument would break down and you would have more than just teh azimuthal A.
     
  5. Mar 18, 2014 #4
    First of all thank you for your observations.
    Good idea switching to Cartesian coordinates: because of the position of the unit vectors [itex]\mathbf{u}_\phi^\prime[/itex], now we have evidence that [itex]\mathbf{A}[/itex] can only have (in Cartesian coordinates) components only parallel to the [itex](x,y)[/itex] plane. My problem now is that [itex]\mathbf{u}_\phi[/itex] (not primed) surely lays in this plane; but with a generic [itex]P(r, \theta, \phi)[/itex], [itex]\mathbf{u}_\theta[/itex] and [itex]\mathbf{u}_r[/itex] could have anyway some component in the same plane, so this still seems to prevent me saying that [itex]\mathbf{A}[/itex] has only a [itex]\phi[/itex] component.

    If it is possible, let's try by physical arguments.

    Ok, if in the [itex](x,z)[/itex] plane I have only an [itex]x[/itex] component, and [itex]\mathbf{A}[/itex] is invariant with respect to rotation about the [itex]z[/itex] axis, in the [itex](y,z)[/itex] plane again it should have only an [itex]x[/itex] component.
    But how could I determine that [itex]\mathbf{A}[/itex] in the [itex](x,z)[/itex] has only an [itex]x[/itex] component? Is it possible without integral, just by physical considerations?

    Probably yes!

    Sure, it begins to!

    Emily
     
  6. Mar 18, 2014 #5

    jasonRF

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    I'm sure you mean y component in the x-z plane. Anyway, consider a point at which to evaluate A, call it (x,z) of course. Not consider contributions to the x component of A (use the Cartesian vectors here!): in particular, what is the result when you add the contribution from the current elements located at [itex]\phi^\prime = \alpha[/itex] and [itex]\phi^\prime = -\alpha[/itex]? Draw the picture. I think you will see that the contributions from those two current elements are opposite in the x direction, so the sum is zero. Of course the contributions of those two current elements to the y component of A are equal and in the same direction.

    jason
     
  7. Mar 18, 2014 #6

    jasonRF

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    I didn't explain this well - sorry! Think of yourself sitting at the point at which you will evaluate A and looking toward the current loop. The current loop is symmetric, so if you cannot "see" the coordinate axes you cannot tell what direction you are looking. In all cases you see a small loop with the current nearest you flowing from left to right (current in the far edge of loop is right to left). If at your given location the vector potential is pointed toward your right, then if you move to another location and look toward the loop the vector potential will still be pointed toward your right. It couldn't be anything else, because the current loop you "see" is identical from your point of view. Thus, if you are at (x,y,z) = (10, 0, 10), and you only have a vector potential toward your right it is in the y direction. Now if you move to (x,y,z) = (0,10,10) the vector potential is still toward your right, which is in the negative x direction.

    jason
     
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