Question

In spherical coordinates, let ${\hat{a}}_0,{\hat{a}}_{\varphi }$ denote unit vectors along the $\theta ,\varphi$ directions.

$\overrightarrow{E}=\frac{100}{r}sin\theta cos(\omega t-\beta r){\hat{a}}_{\theta }V/m$ and

$\overrightarrow{H}=\frac{0.265}{r}sin\theta cos(\omega t-\beta r){\hat{a}}_{\varphi }A/m$

represent the electric and magnetic field components of the EM wave at large distance r from a dipole antenna, in free space. The average power (W) crossing the hemispherical shall located at r = 1 km, $0\le \theta \le \pi /2$ is

• A. 55.5

Answer 1

The correct option is A 55.5
$\overrightarrow{E}=Re\left[\frac{100}{r}sin\theta e^{-j\beta r}{e}^{j\omega r}\right]{\hat{a}}_{\theta }V/m$

$\overrightarrow{H}=Re\left[\frac{0.265}{r}sin\theta e^{-j\beta r}{e}^{j\omega t}\right]{\hat{a}}_{\varphi }A/m$

Poynting vetor,

$\overrightarrow{P}=\frac{1}{2}Re[\overrightarrow{E}\times \overrightarrow{H}]=\frac{26.5}{2r^2}si{n}^2\theta {\hat{a}}_{r}W/m^2$

The average power crossing the given hemispherical shell is,

power = ${\int }_{\theta =0}^{\pi /2}{\int }_{\varphi =0}^{2\pi }\overrightarrow{P}\cdot r^{2}sin\theta d\theta d\varphi {\hat{a}}_r$

$=\frac{26.5}{2}\times 2\pi {\int }_{\theta =0}^{\pi /2}si{n}^3\theta d\theta$

$=\frac{26.5}{2}\times 2\pi \frac{2}{3}=55.5W$