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**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!!

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