In spherical coordinates, let $\hat a_\theta$ and $\hat a_\phi$ denote unit vectors along the $\theta$ and $\phi$ directions. $$ E=\frac{100}{r}\sin\theta\cos(\omega t-\beta r)\,\hat a_\theta\ \text{V/m} $$ and \[ H=\frac{0.265}{r}\sin\theta\cos(\omega t-\beta r)\,\hat a_\phi\ \text{A/m} \

Question

In spherical coordinates, let $\hat a_\theta$ and $\hat a_\phi$ denote unit vectors along the $\theta$ and $\phi$ directions. $$ E=\frac{100}{r}\sin\theta\cos(\omega t-\beta r)\,\hat a_\theta\ \text{V/m} $$ and $$ H=\frac{0.265}{r}\sin\theta\cos(\omega t-\beta r)\,\hat a_\phi\ \text{A/m} $$ represent the electric and magnetic field components of the EM wave at large distances $r$ from a dipole antenna in free space. The average power crossing the hemispherical shell located at $r=1\text{ km}$, $0\le \theta\le \pi/2$ is

Accepted Answer

$1.06\pi$. The average power density is given by the time-average Poynting vector, $\langle Sangle=	frac12 E_0H_0$. Substituting the given field amplitudes at $r=1$ km and integrating over the hemisphere yields the total radiated power. The result is $1.06\pi$ W.

Solution

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