stunner5000pt
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Vector Potential
Consider two half wave antennas each ahving current
I(z,t)=\hat{z} I_{0}\cos\omega t\sin k(\frac{d}{2}-|z|)
where k=\omega/c
Each antenna has length d and points in the z direction. Antenna 1 is at (\Delta/2,0,0) and antenna two is at (-\Delta/2,0,0)
a) Find the vector potential A
b) Find the electric and magnetic field
c) Find dP/d\Omega
d) Evalute dP/d\Omega in the X Y plane when the antenna is seaparated by a distance lambda/2. Along what direction is the radiation preferentialy propagated?
In CGS units so...
\vec{A}(\vec{r},t)=\frac{1}{c}\int \frac{\vec{J}(\vec{r},t_{r})}{|\vec{r}-\vec{r'}|} d\tau
So we need the current as a function of z' and the retarded time
I(z,t)=\hat{z} I_{0}\cos\omega t\sin k(\frac{d}{2}-|z|)
I(z,t)=\hat{z} I_{0}\cos\omega (t-\frac{\mathcal{R}}{c})\sin k(\frac{d}{2}-|z|)
where \mathcal{R}=\sqrt{z'^2+r^2-2z'r\cos\theta}
since we want the fields far away (radiation zone), expand
\mathcal{R}\approx r\left(1-\frac{z'}{r}\cos\theta
so then
\cos\omega (t-\frac{\mathcal{R}}{c})\approx\cos\omega\left(t-\frac{r}{c}\left(1-\frac{z'}{r}\cos\theta\right)\right)
\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)
So then A is calculated like this? make the approximation that \mathcal{R}\approx r
\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'
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 \Delta to \Delta+\frac{d}{2} and one of them from \Delta to \Delta-\frac{d}{2} ?? And then add the two results?
Thanks for your help!
Homework Statement
Consider two half wave antennas each ahving current
I(z,t)=\hat{z} I_{0}\cos\omega t\sin k(\frac{d}{2}-|z|)
where k=\omega/c
Each antenna has length d and points in the z direction. Antenna 1 is at (\Delta/2,0,0) and antenna two is at (-\Delta/2,0,0)
a) Find the vector potential A
b) Find the electric and magnetic field
c) Find dP/d\Omega
d) Evalute dP/d\Omega in the X Y plane when the antenna is seaparated by a distance lambda/2. Along what direction is the radiation preferentialy propagated?
Homework Equations
In CGS units so...
\vec{A}(\vec{r},t)=\frac{1}{c}\int \frac{\vec{J}(\vec{r},t_{r})}{|\vec{r}-\vec{r'}|} d\tau
The Attempt at a Solution
So we need the current as a function of z' and the retarded time
I(z,t)=\hat{z} I_{0}\cos\omega t\sin k(\frac{d}{2}-|z|)
I(z,t)=\hat{z} I_{0}\cos\omega (t-\frac{\mathcal{R}}{c})\sin k(\frac{d}{2}-|z|)
where \mathcal{R}=\sqrt{z'^2+r^2-2z'r\cos\theta}
since we want the fields far away (radiation zone), expand
\mathcal{R}\approx r\left(1-\frac{z'}{r}\cos\theta
so then
\cos\omega (t-\frac{\mathcal{R}}{c})\approx\cos\omega\left(t-\frac{r}{c}\left(1-\frac{z'}{r}\cos\theta\right)\right)
\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)
So then A is calculated like this? make the approximation that \mathcal{R}\approx r
\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'
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 \Delta to \Delta+\frac{d}{2} and one of them from \Delta to \Delta-\frac{d}{2} ?? And then add the two results?
Thanks for your help!
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