Question 9.35: Suppose E(r, θ, φ, t) = A sin θ/ r [cos (kr − ωt) − (1/kr) s...

Suppose E (r, \theta, \phi, t)=A \frac{\sin \theta}{r}[\cos (k r-\omega t)-(1 / k r) \sin (k r-\omega t)] \hat{ \phi }, \quad \text { with } \frac{\omega}{k}=c .

(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?)

(c) Integrate I · da over a spherical surface to determine the total power radiated .

\left[\text { Answer: } 4 \pi A^{2} / 3 \mu_{0} c\right]
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(a) (i) Gauss’s law: \nabla \cdot E =\frac{1}{r \sin \theta} \frac{\partial E_{\phi}}{\partial \phi}=0 .

(ii) Faraday’s law:

-\frac{\partial B }{\partial t}= \nabla \times E =\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta E_{\phi}\right) \hat{ r }-\frac{1}{r} \frac{\partial}{\partial r}\left(r E_{\phi}\right) \hat{ \theta }

 

=\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left[E_{0} \frac{\sin ^{2} \theta}{r}\left(\cos u-\frac{1}{k r} \sin u\right)\right] \hat{ r }-\frac{1}{r} \frac{\partial}{\partial r}\left[E_{0} \sin \theta\left(\cos u-\frac{1}{k r} \sin u\right)\right] \hat{ \theta }.

\text { But } \frac{\partial}{\partial r} \cos u=-k \sin u ; \frac{\partial}{\partial r} \sin u=k \cos u .

=\frac{1}{r \sin \theta} \frac{E_{0}}{r} 2 \sin \theta \cos \theta\left(\cos u-\frac{1}{k r} \sin u\right) \hat{ r }-\frac{1}{r} E_{0} \sin \theta\left(-k \sin u+\frac{1}{k r^{2}} \sin u-\frac{1}{r} \cos u\right) \hat{ \theta }.

Integrating with respect to t, and noting that \int \cos u d t=-\frac{1}{\omega} \sin u \text { and } \int \sin u d t=\frac{1}{\omega} \cos u . , we obtain

B =\frac{2 E_{0} \cos \theta}{\omega r^{2}}\left(\sin u+\frac{1}{k r} \cos u\right) \hat{ r }+\frac{E_{0} \sin \theta}{\omega r}\left(-k \cos u+\frac{1}{k r^{2}} \cos u+\frac{1}{r} \sin u\right) \hat{ \theta } .

(iii) Divergence of B:

\nabla \cdot B =\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} B_{r}\right)+\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta B_{\theta}\right)

 

=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left[\frac{2 E_{0} \cos \theta}{\omega}\left(\sin u+\frac{1}{k r} \cos u\right)\right]+\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left[\frac{E_{0} \sin ^{2} \theta}{\omega r}\left(-k \cos u+\frac{1}{k r^{2}} \cos u+\frac{1}{r} \sin u\right)\right]

 

=\frac{1}{r^{2}} \frac{2 E_{0} \cos \theta}{\omega}\left(k \cos u-\frac{1}{k r^{2}} \cos u-\frac{1}{r} \sin u\right) +\frac{1}{r \sin \theta} \frac{2 E_{0} \sin \theta \cos \theta}{\omega r}\left(-k \cos u+\frac{1}{k r^{2}} \cos u+\frac{1}{r} \sin u\right)

 

=\frac{2 E_{0} \cos \theta}{\omega r^{2}}\left(k \cos u-\frac{1}{k r^{2}} \cos u-\frac{1}{r} \sin u-k \cos u+\frac{1}{k r^{2}} \cos u+\frac{1}{r} \sin u\right)=0.

(iv) \text { Ampére }/Maxwell:

\nabla \times B =\frac{1}{r}\left[\frac{\partial}{\partial r}\left(r B_{\theta}\right)-\frac{\partial B_{r}}{\partial \theta}\right] \hat{ \phi }

 

=\frac{1}{r}\left\{\frac{\partial}{\partial r}\left[\frac{E_{0} \sin \theta}{\omega}\left(-k \cos u+\frac{1}{k r^{2}} \cos u+\frac{1}{r} \sin u\right)\right]-\frac{\partial}{\partial \theta}\left[\frac{2 E_{0} \cos \theta}{\omega r^{2}}\left(\sin u+\frac{1}{k r} \cos u\right)\right]\right\} \hat{\phi}

 

=\frac{E_{0} \sin \theta}{\omega r}\left(k^{2} \sin u-\frac{2}{k r^{3}} \cos u-\frac{1}{r^{2}} \sin u-\frac{1}{r^{2}} \sin u+\frac{k}{r} \cos u+\frac{2}{r^{2}} \sin u+\frac{2}{k r^{3}} \cos u\right) \hat{\phi}

 

=\frac{k}{\omega} \frac{E_{0} \sin \theta}{r}\left(k \sin u+\frac{1}{r} \cos u\right) \hat{\phi}=\frac{1}{c} \frac{E_{0} \sin \theta}{r}\left(k \sin u+\frac{1}{r} \cos u\right) \hat{\phi}.

\frac{1}{c^{2}} \frac{\partial E }{\partial t}=\frac{1}{c^{2}} \frac{E_{0} \sin \theta}{r}\left(\omega \sin u+\frac{\omega}{k r} \cos u\right) \hat{\phi}=\frac{1}{c^{2}} \frac{\omega}{k} \frac{E_{0} \sin \theta}{r}\left(k \sin u+\frac{1}{r} \cos u\right) \hat{\phi}

 

=\frac{1}{c} \frac{E_{0} \sin \theta}{r}\left(k \sin u+\frac{1}{r} \cos u\right) \hat{ \phi }= \nabla \times B .

(b) Poynting Vector:

S =\frac{1}{\mu_{0}}( E \times B )=\frac{E_{0} \sin \theta}{\mu_{0} r}\left(\cos u-\frac{1}{k r} \sin u\right)\left[\frac{2 E_{0} \cos \theta}{\omega r^{2}}\left(\sin u+\frac{1}{k r} \cos u\right)\right. \hat{\theta}

 

\left.+\frac{E_{0} \sin \theta}{\omega r}\left(-k \cos u+\frac{1}{k r^{2}} \cos u+\frac{1}{r} \sin u\right)(-\hat{ r })\right]

 

=\frac{E_{0}^{2} \sin \theta}{\mu_{0} \omega r^{2}}\left\{\frac{2 \cos \theta}{r}\left[\sin u \cos u+\frac{1}{k r}\left(\cos ^{2} u-\sin ^{2} u\right)-\frac{1}{k^{2} r^{2}} \sin u \cos u\right] \hat{ \theta }\right.

 

\left.-\sin \theta\left(-k \cos ^{2} u+\frac{1}{k r^{2}} \cos ^{2} u+\frac{1}{r} \sin u \cos u+\frac{1}{r} \sin u \cos u-\frac{1}{k^{2} r^{3}} \sin u \cos u-\frac{1}{k r^{2}} \sin ^{2} u\right) \hat{ r }\right\}

 

=  \begin{aligned}&\frac{E_{0}^{2} \sin \theta}{\mu_{0} \omega r^{2}}\left\{\frac{2 \cos \theta}{r}\left[\left(1-\frac{1}{k^{2} r^{2}}\right) \sin u \cos u+\frac{1}{k r}\left(\cos ^{2} u-\sin ^{2} u\right)\right] \hat{ \theta }\right. \\&\left.+\sin \theta\left[\left(-\frac{2}{r}+\frac{1}{k^{2} r^{3}}\right) \sin u \cos u+k \cos ^{2} u+\frac{1}{k r^{2}}\left(\sin ^{2} u-\cos ^{2} u\right)\right] \hat{ r }\right\}.\end{aligned}

Averaging over a full cycle, using \langle\sin u \cos u\rangle=0,\left\langle\sin ^{2} u\right\rangle=\left\langle\cos ^{2} u\right\rangle=\frac{1}{2} , we get the intensity: 

I =\langle S \rangle=\frac{E_{0}^{2} \sin \theta}{\mu_{0} \omega r^{2}}\left(\frac{k}{2} \sin \theta\right) \hat{ r }=\frac{E_{0}^{2} \sin ^{2} \theta}{2 \mu_{0} c r^{2}} \hat{ r } .

It points in the \hat{ r } direction, and falls off as 1 / r^{2}, as we would expect for a spherical wave.

\text { (c) } P=\int I \cdot d a =\frac{E_{0}^{2}}{2 \mu_{0} c} \int \frac{\sin ^{2} \theta}{r^{2}} r^{2} \sin \theta d \theta d \phi=\frac{E_{0}^{2}}{2 \mu_{0} c} 2 \pi \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4 \pi}{3} \frac{E_{0}^{2}}{\mu_{0} c} .

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