- #1

CAF123

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## Homework Statement

A thin linear antenna of length ##d## is excited in such a way that it carries the sinusoidal current ##I(t) = I_0 \sin(2\pi z/d)e^{ i\omega t}##, where ##\omega = 2\pi c/d##.

(i) Beginning with the exact solution for the vector potential in Lorenz gauge, calculate the power radiated per unit solid angle (averaged over a periodic time of the oscillation) far from the antenna and plot the angular distribution as a function of ##\theta##.

[Hint: The power radiated into a solid angle ##d \Omega## in the direction ##\hat r## is ##dP = \langle \mathbf S \rangle \cdot \hat r dA## where ##dA = r^2d\Omega## and ##\langle \mathbf S \rangle## is the time-averaged Poynting vector. Use that ##\langle \mathbf S \rangle = \langle c \mathbf E \times \mathbf B \rangle = c \frac{1}{2} Re \mathbf E^* \times \mathbf B.] ##

(ii) Compare your result 2(i) with that obtained from the dipole approximation, and explain where and why this approximation fails in this case.

## Homework Equations

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Given in the question

## The Attempt at a Solution

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I believe the required exact expression of the vector potential is $$\mathbf A(\mathbf r, t) = \frac{1}{4\pi r c} \int d^3 r' \frac{J(\mathbf r', t - \frac{|\mathbf r - \mathbf r'|}{c})}{|\mathbf r - \mathbf r'|}$$ In the far field limit, ##r \gg d## and ##\lambda \ll r \Rightarrow kr \ll 1## I am really just looking for some hints to start, I could try expanding ##J## but not quite sure what this will look like. Thanks. I have other expressions for A but they involve dipole terms which is not what we have here.