- #1

EmilyRuck

- 136

- 6

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