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Expert-verified Found in: Page 432 ### Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562 # Suppose ${\mathbit{E}}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{,}}{\mathbit{\theta }}{\mathbf{,}}{\mathbit{\varphi }}{\mathbf{,}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{A}}{\mathbf{}}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{r}}\left[cos\left(kr-\omega t\right)-\frac{1}{kr}sin\left(kr-\omega t\right)\right]{\mathbf{}}\stackrel{\mathbf{⏜}}{\mathbf{\varphi }\mathbf{}\mathbf{}}$ (This is, incidentally, the simplest possible spherical wave. For notational convenience, let$\left(kr-\omega t\right){\mathbf{\equiv }}{\mathbit{u}}$ in your calculations.) (a) Show that ${\mathbit{E}}$ 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 . (Does it point in the expected direction? Does it fall off like ${{\mathbit{r}}}^{\mathbf{-}\mathbf{2}}$, as it should?) (c) Integrate over a spherical surface to determine the total power radiated. [Answer:${\mathbf{4}}{\mathbit{\pi }}{{\mathbit{A}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{3}}{{\mathbit{\mu }}}_{{\mathbf{0}}}{\mathbit{c}}$]

(a)

The value of divergence of electric field of Maxwell’s equation is $\nabla .E=0$ .

The value of curl of electric field is $\nabla ×E=\frac{1}{r\mathrm{sin}\theta }\frac{\partial }{\partial \theta }\left(\mathrm{sin}\theta {E}_{\varphi }\right)\stackrel{⏜}{r}-\frac{1}{r}\frac{\partial }{\partial \theta }\left(r{E}_{\varphi }\right)\stackrel{⏜}{\theta }$

The value of magnetic field is$B=\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(\mathrm{sin}u+\frac{1}{kr}\mathrm{cos}u\right)\stackrel{⏜}{r}+\frac{A\mathrm{sin}\theta }{\omega r}\left(-k\mathrm{cos}u\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)\stackrel{⏜}{\theta }$.

The value of Gauss law of magnetism is $\nabla .B=0$.

The value of Ampere’s law.is $\frac{1}{{c}^{2}}\frac{\partial E}{\partial t}=\nabla ×B$.

(b)

The value of Intensity vector is $I=\frac{{A}^{2}{\mathrm{sin}}^{2}\theta }{2{\mu }_{0}c{r}^{2}}\stackrel{⏜}{r}$and the pointing vector S over the full

$cycleisS=\frac{{A}^{2}\mathrm{sin}\theta }{{\mu }_{0}\omega {r}^{2}}\left\{\begin{array}{c}\frac{2\mathrm{cos}\theta }{r}\left[\mathrm{sin}u\mathrm{cos}u+\frac{1}{kr}\left({\mathrm{cos}}^{2}u-{\mathrm{sin}}^{2}u\right)-\frac{1}{{k}^{2}{r}^{2}}\mathrm{sin}u\mathrm{cos}u\right]\theta \\ -\mathrm{sin}\theta \left(\begin{array}{c}-k{\mathrm{cos}}^{2}u+\frac{1}{k{r}^{2}}{\mathrm{cos}}^{2}u+\frac{1}{r}\mathrm{sin}u\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\mathrm{cos}u\\ -\frac{1}{{k}^{2}{r}^{3}}\mathrm{sin}u\mathrm{cos}u-\frac{1}{k{r}^{2}}{\mathrm{sin}}^{2}u\end{array}\right)\stackrel{⏜}{r}\end{array}\right\}$

(c) The value of total power radiated is$\frac{4\mathrm{\pi }}{3}\frac{{A}^{2}}{{\mu }_{0}c}$.

See the step by step solution

## Step 1: Write the given data from the question.

Consider this, incidentally, the simplest possible spherical wave. For notational convenience, let $\left(kr-\omega t\right)\equiv u$ in your calculations.

## Determine the formula of divergence of electric field of Maxwell’s equation, curl of electric field, magnetic field, Gauss law of magnetism, Ampere’s law, Intensity vector and total power radiated.

Write the formula of electric field is,

${\mathbf{\nabla }}{\mathbf{.}}{\mathbit{E}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left({r}^{2}{E}_{r}\right){\mathbf{+}}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\theta }}\left(sin\theta {E}_{\theta }\right){\mathbf{+}}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\frac{\mathbf{\partial }\mathbf{E}}{\mathbf{\partial }\mathbf{\varphi }}$ …… (1)

Here,role="math" localid="1658485458510" ${\mathbf{E}}$ is the electric field component of a spherical wave, role="math" localid="1658485452985" ${{\mathbit{E}}}_{{\mathbf{r}}}$is electric field component of a spherical wave, ${\mathbit{r}}$ is radius, ${{\mathbf{E}}}_{{\mathbf{\theta }}}$electric field component of a spherical wave and${{\mathbf{E}}}_{{\mathbf{\varphi }}}$is electric field component of a spherical wave.

Write the formula of curl of electric field.

${\mathbf{\nabla }}{\mathbf{×}}{\mathbf{E}}{\mathbf{=}}\frac{\mathbf{-}\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{r}}$ …… (2)

Here, ${\mathbf{B}}$is the magnetic field strength and${\mathbf{r}}$is radius.

Write the formula of magnetic field.

${\mathbit{B}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\theta }}\left[\frac{Asi{n}^{2}\theta }{r}\int cosu-\frac{1}{kr}\int sinu\right]\stackrel{\mathbf{⏜}}{\mathbf{r}}\phantom{\rule{0ex}{0ex}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left[Asin\theta \left(\int cosu-\frac{1}{kr}\int sinu\right)\right]{\mathbf{}}\stackrel{\mathbf{⏜}}{\mathbf{\theta }}{\mathbf{}}{\mathbf{}}$ …… (3)

Here, ${\mathbf{r}}$ is radius,${\mathbf{k}}$represent the wave number and ${\mathbf{A}}$ is constant.

Write the formula of Gauss law of magnetism.

${\mathbf{\nabla }}{\mathbf{.}}{\mathbf{B}}{\mathbf{=}}{\mathbf{0}}$ …… (4)

Here, is the magnetic field strength.

Write the formula of Ampere’s law.

${\mathbf{\nabla }}{\mathbf{×}}{\mathbf{B}}{\mathbf{=}}{\mathbf{\mu \sigma E}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\frac{\mathbf{\partial }\mathbf{E}}{\mathbf{\partial }\mathbf{t}}$ …… (5)

Here, ${\mathbf{E}}$ is the electric field component of a spherical wave, ${\mathbit{\mu }}$ is permeability, ${\mathbit{c}}$denotes the speed of light.

Write the formula of intensity vector.

${\mathbf{I}}{\mathbf{=}}\left(S\right)$ …… (6)

Here, S is Poynting vector.

Write the formula of total power radiated.

${\mathbf{P}}{\mathbf{=}}{\mathbf{\int }}{\mathbf{I}}{\mathbf{.}}{\mathbf{da}}$ …… (7)

Here,I is intensity vector.

## Step 3: (a) Determine the electric field of Maxwell’s equation.

From Gauss’s law,

$\nabla .E=\frac{{p}_{f}}{{\epsilon }_{0}}$

Here, ${\rho }_{f}$ is the free charge density.

Determine the divergence of electric field is,

Substitute .

As there is no free charge density here, therefore $\nabla .E=0$.

Hence, Gauss’s law is obeyed.

Determine the curl of electric field is,

role="math" localid="1658487525637" $\nabla ×E=\frac{1}{r\mathrm{sin}\theta }\left[\frac{\partial }{\partial \theta }\left(\mathrm{sin}\theta {E}_{\varphi }\right)-\frac{\partial {E}_{\theta }}{\partial \varphi }\right]\stackrel{⏜}{r}+\frac{1}{r}\left[\frac{\partial }{\partial r}\right]\left(r{E}_{\varphi }\right)\stackrel{⏜}{\varphi }+\frac{1}{r}\left[\frac{\partial }{\partial r}\left(r{E}_{\theta }\right)-\frac{\partial {E}_{r}}{\partial \theta }\right]\stackrel{⏜}{\varphi }$

Therefore, the value of curl of electric field is

$\nabla ×E=\frac{1}{r\mathrm{sin}\theta }\left[\frac{\partial }{\partial \theta }\left(\mathrm{sin}\theta {E}_{\varphi }\right)-\frac{\partial {E}_{\theta }}{\partial \varphi }\right]\stackrel{⏜}{r}+\frac{1}{r}\frac{\partial }{\partial r}\left(r{E}_{\varphi }\right)\stackrel{⏜}{\varphi }$

Substitute $\frac{-\partial B}{\partial t}$for $\nabla ×E$.

$\frac{-\partial B}{\partial t}=\frac{1}{r\mathrm{sin}\theta }\frac{\partial }{\partial \theta }\left[\frac{A{\mathrm{sin}}^{2}\theta }{r}\left(\mathrm{cos}u-\frac{1}{kr}\mathrm{sin}u\right)\right]\stackrel{⏜}{r}\phantom{\rule{0ex}{0ex}}-\frac{1}{r}\frac{\partial }{\partial t}\left[A\mathrm{sin}\theta \left(\mathrm{cos}u-\frac{1}{kr}-\frac{1}{kr}\mathrm{sin}u\right)\right]\stackrel{⏜}{\theta }$ …… (8)

Here,$u=\left(kr-\omega t\right)$.

Integrate equation (8),

$B=\frac{1}{r\mathrm{sin}\theta }\frac{\partial }{\partial \theta }\left[\frac{A{\mathrm{sin}}^{2}\theta }{r}\left(\int \mathrm{cos}u-\frac{1}{kr}\int \mathrm{sin}u\right)\right]\stackrel{⏜}{r}\phantom{\rule{0ex}{0ex}}-\frac{1}{r}\frac{\partial }{\partial \theta }\left[A\mathrm{sin}\theta \left(\int \mathrm{cos}u-\frac{1}{kr}\int \mathrm{sin}u\right)\right]\stackrel{⏜}{\theta }$ …… (9)

Substitute role="math" localid="1658488205000" $-\frac{1}{\omega }\mathrm{sin}u\int \mathrm{cos}udt$ for $\int \mathrm{cos}udt$ and $\frac{1}{\omega }\mathrm{cos}u$for $\int \mathrm{sin}udt$ into equation (9).

$B=\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(\mathrm{sin}u+\frac{1}{kr}\mathrm{cos}u\right)\stackrel{⏜}{r}+\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}-kcosu\frac{1}{k{r}^{2}}cosu+\frac{1}{r}sinu\right)\stackrel{⏜}{\theta }$

Therefore, the value of magnetic field is

$B=\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}sinu+\frac{1}{kr}cosu\right)\stackrel{⏜}{r}+\frac{A\mathrm{sin}\theta }{\omega r}-kcosu\frac{1}{k{r}^{2}}cosu+\frac{1}{r}sinu\right)\stackrel{⏜}{\theta }$

Determine the Gauss’s law of magnetism,

Substitute $\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(\mathrm{sin}u+\frac{1}{kr}\mathrm{cos}u\right)\stackrel{⏜}{r}+\frac{A\mathrm{sin}\theta }{\omega r}-k\mathrm{cos}u\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r\mathrm{sin}u}\right)\stackrel{⏜}{\theta }$ for B into equation (4).

$\nabla .B=\frac{1}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}B\right)+\frac{1}{r\mathrm{sin}\theta }\frac{\partial }{\partial \theta }\left(\mathrm{sin}\theta {B}_{\theta }\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{r}^{2}}\frac{\partial }{\partial r}\left[\frac{2A\mathrm{cos}\theta }{\omega r}\left(\mathrm{sin}u+\frac{1}{kr}\mathrm{cos}u\right)\right]\phantom{\rule{0ex}{0ex}}+\frac{1}{r\mathrm{sin}\theta }\frac{\partial }{\partial r}\left[\frac{A{\mathrm{sin}}^{2}\theta }{\omega r}\left(-k\mathrm{cos}u+\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{{r}^{2}}\frac{2A\mathrm{cos}\theta }{\omega }\left(k\mathrm{cos}u-\frac{1}{k{r}^{2}}\mathrm{cos}u-\frac{1}{r}\mathrm{sin}u\right)\phantom{\rule{0ex}{0ex}}+\frac{1}{r\mathrm{sin}\theta }\frac{2A\mathrm{sin}\theta \mathrm{cos}\theta }{\omega r}\left(-k\mathrm{cos}u+\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)$

Solve further as

$\nabla .B=\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(k\mathrm{cos}u-\frac{1}{k{r}^{2}}\mathrm{cos}u-\frac{1}{r}\mathrm{sin}u-k\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)\phantom{\rule{0ex}{0ex}}=0$

Hence, the Gauss law of magnetism is obeyed.

Determine the Ampere’s law,

As$\sigma =0$, therefore,

$\nabla ×B=\frac{1}{{c}^{2}}\frac{\partial E}{\partial t}\phantom{\rule{0ex}{0ex}}=\frac{1}{r}\left[\frac{\partial }{\partial r}\left(r{B}_{\theta }\right)-\frac{\partial B}{\partial \theta }\right]\varphi$

Substitute $\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(\mathrm{sin}u+\frac{1}{kr}\mathrm{cos}u\right)\stackrel{⏜}{r}+\frac{A\mathrm{sin}\theta }{\omega r}\left(-k\mathrm{cos}u\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)\stackrel{⏜}{B}$ for B .

$\nabla ×B=\frac{1}{r}\left\{\begin{array}{c}\frac{\partial }{\partial r}\left[\frac{A\mathrm{sin}\theta }{\omega }\left(-k\mathrm{cos}u+\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)\right]\\ -\frac{\partial }{\partial \theta }\left[\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(\mathrm{sin}u+\frac{1}{kr}\mathrm{cos}u\right)\right]\stackrel{⏜}{\varphi }\end{array}\right\}\phantom{\rule{0ex}{0ex}}=\frac{k}{\omega }\frac{A\mathrm{sin}\theta }{\omega }\left(k\mathrm{sin}u+\frac{1}{r}\mathrm{cos}u\right)\stackrel{⏜}{\varphi }\phantom{\rule{0ex}{0ex}}=\frac{A\mathrm{sin}\theta }{\omega }\left(k\mathrm{sin}u+\frac{1}{r}\mathrm{cos}u\right)\stackrel{⏜}{\varphi }$

Solve the term $\frac{1}{{c}^{2}}\frac{\partial E}{\partial t}$,

$\frac{1}{{c}^{2}}\frac{\partial E}{\partial t}=\frac{1}{{c}^{2}}\frac{A\mathrm{sin}\theta }{r}\left(\omega \mathrm{sin}u+\frac{\omega }{kr}\mathrm{cos}u\right)\stackrel{⏜}{\varphi }\phantom{\rule{0ex}{0ex}}=\frac{1}{{c}^{2}}\frac{\omega }{k}\frac{A\mathrm{sin}\theta }{r}\left(k\mathrm{sin}u+\frac{1}{r}\mathrm{cos}u\right)\stackrel{⏜}{\varphi }\phantom{\rule{0ex}{0ex}}=\frac{1}{c}\frac{A\mathrm{sin}\theta }{r}\left(k\mathrm{sin}u+\frac{1}{r}\mathrm{cos}u\right)\stackrel{⏜}{\varphi }\phantom{\rule{0ex}{0ex}}=\nabla ×B$

Hence, Ampere’s law is obeyed.

## Step 4: (b) Determine the Poynting vector and energy per unit time.

Determine the Poynting vector is given by the following equation.

$S=\frac{1}{{\mu }_{0}}\left(E×B\right)$ …… (10)

Substitute $\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(\mathrm{sin}u+\frac{1}{KR}\mathrm{cos}u\right)\stackrel{⏜}{r}+\frac{A\mathrm{sin}\theta }{\omega r}\left(-k\mathrm{cos}u\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)\stackrel{⏜}{\theta }$ for B and $\frac{A\mathrm{sin}\theta }{r}\left[\mathrm{cos}u-\frac{1}{kr}\mathrm{sin}u\right]\varphi$for E into above equation (10).

$S=\frac{1}{{\mu }_{0}}\left[\frac{A\mathrm{sin}\theta }{r}\left[\mathrm{cos}u-\frac{1}{kr}\mathrm{sin}u\right]\stackrel{⏜}{\varphi }\right]x\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}\frac{2A\mathrm{cos}\theta }{\omega {r}^{2}}\left(\mathrm{sin}u+\frac{1}{kr}\mathrm{cos}u\right)\stackrel{⏜}{r}\\ +\frac{A\mathrm{sin}\theta }{\omega r}\left(-k\mathrm{cos}u+\frac{1}{k{r}^{2}}\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\right)\stackrel{⏜}{\theta }\end{array}\right]\phantom{\rule{0ex}{0ex}}=\frac{{A}^{2}\mathrm{sin}\theta }{{\mu }_{0}\omega {r}^{2}}\left\{\begin{array}{c}\frac{2\mathrm{cos}\theta }{r}\left[\mathrm{sin}u\mathrm{cos}u+\frac{1}{r}\left({\mathrm{cos}}^{2}u-{\mathrm{sin}}^{2}u\right)-\frac{1}{{k}^{2}{r}^{2}}\mathrm{sin}u\mathrm{cos}u\right]\stackrel{⏜}{\theta }\\ -\mathrm{sin}\left(\begin{array}{c}-k{\mathrm{cos}}^{2}u+\frac{1}{r}\mathrm{sin}u\mathrm{cos}u+\frac{1}{r}\mathrm{sin}u\mathrm{cos}u\\ -\frac{1}{{k}^{2}{r}^{2}}\mathrm{sin}u\mathrm{cos}u-\frac{1}{k{r}^{2}}{\mathrm{sin}}^{2}u\end{array}\right)\stackrel{⏜}{r}\end{array}\right\}\phantom{\rule{0ex}{0ex}}$

Average over a full cycle is,

$\left(\mathrm{sin}u\mathrm{cos}u\right)=0\phantom{\rule{0ex}{0ex}}\left({\mathrm{sin}}^{2}u\right)=\left({\mathrm{cos}}^{2}u\right)=\frac{1}{2}$

Determine the Intensity vector.

Substitute $\frac{{A}^{2}\mathrm{sin}}{{\mu }_{0}\omega {r}^{2}}\left(\frac{k}{2}\mathrm{sin}\theta \right)\stackrel{⏜}{r}$ for $\left(S\right)$into equation (6).

$I=\frac{{A}^{2}\mathrm{sin}}{{\mu }_{0}\omega {r}^{2}}\left(\frac{k}{2}\mathrm{sin}\theta \right)\stackrel{⏜}{r}\phantom{\rule{0ex}{0ex}}=\frac{{A}^{2}{\mathrm{sin}}^{2}\theta }{2{\mu }_{0}c{r}^{2}}\stackrel{⏜}{r}$

The intensity fluctuates as $\frac{1}{{r}^{2}}$ and faces in the direction of $\stackrel{⏜}{a}$. A spherical wave is predicted to behave in this way.

Therefore, the intensity vector is$\frac{{A}^{2}{\mathrm{sin}}^{2}\theta }{2{\mu }_{0}c{r}^{2}}\stackrel{⏜}{r}$.

## Step 5: (c) Determine the total power radiated.

Determine the total power radiated is,

Substitute $\frac{{A}^{2}{\mathrm{sin}}^{2}\theta }{2{\mu }_{0}c{r}^{2}}\stackrel{⏜}{r}$ for I.

role="math" localid="1658492229808" $P=\frac{{A}^{2}}{2{\mu }_{0}c}\int \frac{{\mathrm{sin}}^{2}\theta }{{r}^{2}}{r}^{2}\mathrm{sin}\theta d\theta d\varphi \phantom{\rule{0ex}{0ex}}=\frac{{A}^{2}}{2{\mu }_{0}c}2\mathrm{\pi }{\int }_{0}^{\mathrm{\pi }}{\mathrm{sin}}^{2}\mathrm{\theta d\theta }\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi }}{3}\frac{{\mathrm{A}}^{2}}{{\mathrm{\mu }}_{0}\mathrm{c}}$

Therefore, the value of total power radiated is $\frac{4\mathrm{\pi }}{3}\frac{{\mathrm{A}}^{2}}{{\mathrm{\mu }}_{0}\mathrm{c}}$ . ### Want to see more solutions like these? 