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Physics / Introduction to Electrodynamics 4 / Chapter 9 / Problem 35P

Introduction to Electrodynamics | 4th Edition | ISBN: 9780321856562 | Authors: David J. Griffiths

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Chapter 9, Problem 35P

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QUESTION:

Problem 35P

Suppose

(This is, incidentally, the simplest possible spherical wave. For notational convenience, let (kr − ωt) ≡ u in your calculations.)

(a) Show that E obeys all four of Maxwell’s equations, in vacuum, and find the associated magnetic field.

(b) Calculate the Poynting vector. Average S over a full cycle to get the intensity vector I. (Does it point in the expected direction? Does it fall off like r−2, as it should?)

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QUESTION:

Problem 35P

Suppose

(This is, incidentally, the simplest possible spherical wave. For notational convenience, let (kr − ωt) ≡ u in your calculations.)

(a) Show that E obeys all four of Maxwell’s equations, in vacuum, and find the associated magnetic field.

(b) Calculate the Poynting vector. Average S over a full cycle to get the intensity vector I. (Does it point in the expected direction? Does it fall off like r−2, as it should?)

ANSWER:

Step 1 of 3

Part a

We are required to show that the given $E$ obeys Maxwell’s equations.

Gauss’ law

$\nabla{}\cdot{}E=\nabla{}\cdot{}[\frac{Asin\ \theta{}}{r}\{cos(kr-\omega{}t)-(1/kr)sin(kr-\omega{}t)\}]\ \widehat{\varphi{}}$

$=0$

Faraday’s law

$\nabla{}\times{}E=\nabla{}\times{}[\frac{Asin\ \theta{}}{r}\{cos(kr-\omega{}t)-(1/kr)sin(kr-\omega{}t)\}]\ \widehat{\varphi{}}$

$=\frac{1}{rsin\theta{}}\frac{\partial{}}{\partial{}\theta{}}(sin\theta{}E{\ }_{\varphi{}})\ \widehat{r}-\frac{1}{r}\frac{\partial{}}{\partial{}r}(rE{\ }_{\varphi{}})\ \widehat{\theta{}}$

$=\frac{1}{rsin\theta{}}\frac{\partial{}}{\partial{}\theta{}}[E{\ }_{0}\frac{si{n}^{2}\theta{}}{r}(cos\ u-\frac{sin\ u}{kr})]\widehat{r}-\frac{1}{r}\frac{\partial{}}{\partial{}r}[E{\ }_{0}sin\theta{}(cosu-\frac{sin\ u}{kr})]\ \widehat{\theta{}}$

$=\frac{1}{rsin\ \theta{}}\frac{E{\ }_{0}}{r}2sin\theta{}cos\theta{}(cos\ u-\frac{sin\ u}{kr})\ \widehat{r}-\frac{1}{r}E{\ }_{0}sin\ \theta{}(-ksin\ u+\frac{sin\ u}{k{r}^{2}}-\frac{cos\ u}{r})\ \widehat{\theta{}}$

Now, $\nabla{}\times{}E=\ -\frac{\partial{}B}{\partial{}t}$

Therefore,

$B=\frac{2E{\ }_{0}cos\ \theta{}}{\omega{}{r}^{2}}\ (sin\ u+\frac{cos\ u}{kr})\ \widehat{r}+\frac{E{\ }_{0}\ sin\theta{}}{\omega{}r}(-kcos\ u+\frac{cos\ u}{k{r}^{2}}+\frac{sin\ u}{r})\ \widehat{\theta{}}$

This is the required equation of magnetic field.

Gauss’ law of magnetism

$\nabla{}\cdot{}B=\frac{1}{{r}^{2}}\frac{\partial{}}{\partial{}r}({r}^{2}B{\ }_{r})+\frac{1}{rsin\ \theta{}}\frac{\partial{}}{\partial{}\theta{}}(sin\ \theta{}\ B{\ }_{\theta{}})$

$\nabla{}\cdot{}B=\frac{1}{{r}^{2}}\frac{\partial{}}{\partial{}r}[\frac{2E{\ }_{0}cos\ \theta{}}{\omega{}}(sin\ u+\frac{cos\ u}{kr})]+\frac{1}{r\ sin\ \theta{}}\frac{\partial{}}{\partial{}\theta{}}[\frac{E{\ }_{0}si{n}^{2}\theta{}}{\omega{}r}(-k\ cos\ u+\frac{cos\ u}{k{r}^{2}}+\frac{sin\ u}{r})]$

$=\frac{1}{{r}^{2}}\frac{2E{\ }_{0}cos\ \theta{}}{\omega{}}(k\ cos\ u-\frac{cos\ u}{k{r}^{2}}-\frac{sin\ u}{r})+\frac{1}{rsin\ \theta{}}\frac{2E{\ }_{0}sin\ \theta{}\ cos\ \theta{}}{\omega{}r}(-kcos\ u+\frac{cos\ u}{k{r}^{2}}+\frac{sin\ u}{r})$

$=0$

So, $\nabla{}\cdot{}B=0$

Ampere’s law

$\nabla{}\times{}B=\frac{1}{r}[\frac{\partial{}}{\partial{}r}(rB{\ }_{\theta{}})-\frac{\partial{}B{\ }_{r}}{\partial{}\theta{}}]\ \widehat{\varphi{}}$

$=\frac{1}{r}\{\frac{\partial{}}{\partial{}r}[\frac{E{\ }_{0}sin\ \theta{}}{\omega{}}(-k\ cos\ u+\frac{cos\ u}{k{r}^{2}}+\frac{sin\ u}{r})]-\frac{\partial{}}{\partial{}\theta{}}[\frac{2E{\ }_{0}cos\ \theta{}}{\omega{}{r}^{2}}\ (sin\ u+\frac{cos\ u}{kr})]\}\ \widehat{\varphi{}}$

$=\frac{k}{\omega{}}\frac{E{\ }_{0}\ sin\ \theta{}}{r}(k\ sin\ u+\frac{cos\ u}{r})\ \widehat{\varphi{}}$

$=\frac{1}{c}\frac{E{\ }_{0}sin\ \theta{}}{r}(k\ sin\ u+\frac{cos\ u}{r})\ \widehat{\varphi{}}$

Now, $\frac{1}{{c}^{2}}\frac{\partial{}E}{\partial{}t}=\frac{1}{{c}^{2}}\frac{\partial{}}{\partial{}t}[\frac{Asin\ \theta{}}{r}\{cos(kr-\omega{}t)-(1/kr)sin(kr-\omega{}t)\}]\ \widehat{\varphi{}}$