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Half wave antenna

1,444
2
Vector Potential

1. Homework Statement
Consider two half wave antennas each ahving current

[tex] I(z,t)=\hat{z} I_{0}\cos\omega t\sin k(\frac{d}{2}-|z|) [/tex]
where [itex] k=\omega/c[/itex]
Each antenna has length d and points in the z direction. Antenna 1 is at [itex] (\Delta/2,0,0)[/itex] and antenna two is at [itex](-\Delta/2,0,0)[/itex]

a) Find the vector potential A
b) Find the electric and magnetic field
c) Find [tex] dP/d\Omega[/tex]
d) Evalute [tex] dP/d\Omega [/tex] in the X Y plane when the antenna is seaparated by a distance lambda/2. Along what direction is the radiation preferentialy propagated?

2. Homework Equations
In CGS units so...
[tex] \vec{A}(\vec{r},t)=\frac{1}{c}\int \frac{\vec{J}(\vec{r},t_{r})}{|\vec{r}-\vec{r'}|} d\tau [/tex]


3. The Attempt at a Solution
So we need the current as a function of z' and the retarded time

[tex]I(z,t)=\hat{z} I_{0}\cos\omega t\sin k(\frac{d}{2}-|z|) [/tex]
[tex]I(z,t)=\hat{z} I_{0}\cos\omega (t-\frac{\mathcal{R}}{c})\sin k(\frac{d}{2}-|z|) [/tex]

where [tex]\mathcal{R}=\sqrt{z'^2+r^2-2z'r\cos\theta}[/tex]
since we want the fields far away (radiation zone), expand
[tex]\mathcal{R}\approx r\left(1-\frac{z'}{r}\cos\theta[/tex]
so then
[tex] \cos\omega (t-\frac{\mathcal{R}}{c})\approx\cos\omega\left(t-\frac{r}{c}\left(1-\frac{z'}{r}\cos\theta\right)\right)[/tex]
[tex] \cos\omega (t-\frac{\mathcal{R}}{c})\approx\cos\omega t \cos\frac{r}{c}\left(1-\frac{z'}{r}\cos\theta\right)+\sin\omega t\sin\frac{r}{c}\left(1-\frac{z'}{r}\cos\theta\right)[/tex]

So then A is calculated like this? make the approximation that [tex] \mathcal{R}\approx r [/tex]

[tex] \vec{A} = \hat{z}\frac{I_{0}}{rc}\int \sin k\left(\frac{d}{2}-|z|\right)\left(\cos\omega (t-\frac{\mathcal{R}}{c})\left(\cos\omega t \cos\frac{r}{c}\left(1-\frac{z'}{r}\cos\theta\right)+\sin\omega t\sin\frac{r}{c}\left(1-\frac{z'}{r}\cos\theta\right)\right) dz'[/tex]

Ok since there are two antennas how should the integration be performed...
should i do for each antenna separately? That is integrate one of them from [itex]\Delta[/itex] to [itex]\Delta+\frac{d}{2}[/itex] and one of them from [itex]\Delta[/itex] to [itex]\Delta-\frac{d}{2}[/itex] ?? And then add the two results?

Thanks for your help!!
 
Last edited:

Answers and Replies

clem
Science Advisor
1,308
15
Since it is a radiation problem, I think you are allowed to assume r>>r' and
kr>>1, which greatly simplifies the problem.
 

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