6.md

title: Problem Set 6

(8.1)

{:.question} Find the electric field for an infinitesimal dipole radiator.

This problem is most naturally expressed in spherical coordinates. We know that the vector potential is

\begin{align*} A_r &= \mu_0 \frac{I_0 d e^{-i k r}}{4 \pi r} \cos \theta \\ A_\theta &= -\mu_0 \frac{I_0 d e^{-i k r}}{4 \pi r} \sin \theta \\ A_\phi &= 0 \end{align*}

So the electric field is "simply"

E = \frac{1}{i \omega \mu_0 \epsilon_0} \nabla ( \nabla \cdot A) - i \omega A

First let's find \nabla \cdot A.

$$ \begin{align*} \nabla \cdot A &= \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 A_r)

• \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} ( A_\theta \sin \theta )
• \frac{1}{r \sin \theta} \frac{\partial A_\phi}{\partial \phi} \ &= \frac{\mu_0 I_0 d}{4 \pi} \left( \frac{\cos \theta}{r^2} \frac{\partial}{\partial r} \left( r e^{-ikr} \right)

• \frac{e^{-ikr}}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \sin^2 \theta \right) \ &= \frac{\mu_0 I_0 d}{4 \pi} \left( \frac{\cos \theta}{r^2} \left( e^{-ikr} - r i k e^{-ikr} \right)
• \frac{e^{-ikr}}{r^2 \sin \theta} \left( 2 \sin \theta \cos \theta \right) \right) \ &= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{r^2} - \frac{ik}{r} - \frac{2}{r^2} \right) \ &= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{r^2} + \frac{ik}{r} \right) \end{align*} $$

For any function f(r, \theta, \phi) in spherical coordinates,

\nabla f = \frac{\partial f}{\partial r} \hat{r} + \frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\theta} + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \hat{\phi}

So the three components of \nabla (\nabla \cdot A) are

$$ \begin{align*} [\nabla (\nabla \cdot A)]_r &= \frac{\partial (\nabla \cdot A)}{\partial r} \ &= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta \left( \left( \frac{-2}{r^3} - \frac{ik}{r^2} \right) e^{-ikr} + \left( \frac{1}{r^2} + \frac{ik}{r} \right) (-ik) e^{-ikr} \right) \ &= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{-2}{r^3} - \frac{ik}{r^2} - \frac{ik}{r^2} + \frac{k^2}{r} \right) \ &= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right) \

[\nabla (\nabla \cdot A)]_\theta &= \frac{1}{r} \frac{\partial (\nabla \cdot A)}{\partial \theta} \ &= - \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \left( \frac{1}{r^3} + \frac{ik}{r^2} \right) \frac{\partial}{\partial \theta} \cos \theta\ &= \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \left( \frac{1}{r^3} + \frac{ik}{r^2} \right) \sin \theta \

[\nabla (\nabla \cdot A)]_\phi &= 0 \end{align*} $$

Putting these together (and recalling that k = \omega \sqrt{\mu_0 \epsilon_0}) we find that

$$ \begin{align*} E_r &= \frac{1}{i \omega \mu_0 \epsilon_0} \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right)

• i \omega \frac{\mu_0 I_0 d}{4 \pi} \frac{\cos \theta}{r} e^{-ikr} \ &= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{i \omega \mu_0 \epsilon_0} \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right) - \frac{i \omega}{r} \right) \ &= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3}