Chapter 11: Q23P (page 497)

A radio tower rises to height h above flat horizontal ground. At the top is a magnetic dipole antenna, of radius b, with its axis vertical. FM station KRUD broadcasts from this antenna at (angular) frequency $ω$ , with a total radiated power P (that’s averaged, of course, over a full cycle). Neighbors have complained about problems they attribute to excessive radiation from the tower—interference with their stereo systems, mechanical garage doors opening and closing mysteriously, and a variety of suspicious medical problems. But the city engineer who measured the radiation level at the base of the tower found it to be well below the accepted standard. You have been hired by the Neighborhood Association to assess the engineer’s report.
(a) In terms of the variables given (not all of which may be relevant), find the formula for the intensity of the radiation at ground level, a distance R from the base of the tower. You may assume that $b ≪ c / ω ≪ h$ . [Note: We are interested only in the magnitude of the radiation, not in its direction—when measurements are taken, the detector will be aimed directly at the antenna.]
(b) How far from the base of the tower should the engineer have made the measurement? What is the formula for the intensity at this location?
(c) KRUD’s actual power output is 35 kilowatts, its frequency is 90 MHz, the antenna’s radius is 6 cm, and the height of the tower is 200 m. The city’s radio-emission limit is 200 microwatts $/ {cm}^{2}$ . Is KRUD in compliance?

Short Answer

Expert verified

(a) The formula for the intensity of the radiation at ground level, a distance R from the base of the tower is $I = \frac{3 R^{2}}{8 π {(h^{2} + R^{2})}^{2}}$ .

(b) The observation made at the location is $h = R$ and it corresponds to $I = \frac{3 }{32 π R^{2}}$ .

Step by step solution

Expression for the flux intensity of the magnetic dipole:

Write the expression for the magnitude of the intensity of radiation.

$I = < S >$ …… (1)

Here, $<S>$ is the Poynting vector which is given as:

$<S> = (\frac{μ_{0} m_{0}^{2} ω^{4}}{32 π^{2} c^{3}}) \frac{\sin^{2} θ}{r^{2}} \hat{r}$

Here, $μ_{0}$ is the permeability of magnetic field in free space, $m_{0}$ is the maximum value of the magnetic dipole moment, $ω$ is the angular frequency, c is the speed of light, r is the shortest distance from the source towards the observer, and $θ$ is the angle made by the displacement vector r with the vertical.

Determine the formula for the intensity of the radiation at ground level:

(a)

Draw the given situation.

From the above figure, the data is observed as,

$\begin{array}{rcl} r^{2} & = & R^{2} + h^{2} \\ \sin^{2} θ & = & \frac{R^{2}}{r^{2}} \end{array}$

Write the expression for the total radiated power.

$ = \frac{μ_{0} m_{0}^{2} ω^{4}}{32 π c^{3}}$ …… (2)

Substitute $<S> = (\frac{μ_{0} m_{0}^{2} ω^{4}}{32 π^{2} c^{3}}) \frac{\sin^{2} θ}{r^{2}} \hat{r}$ in equation (1).

$I = (\frac{μ_{0} m_{0}^{2} ω^{4}}{32 π^{2} c^{3}}) \frac{\sin^{2} θ}{r^{2}} I = (\frac{μ_{0} m_{0}^{2} ω^{4}}{32 π^{2} c^{3}}) \frac{(\frac{R^{2}}{r^{2}})}{r^{2}} I = (\frac{μ_{0} m_{0}^{2} ω^{4}}{32 π^{2} c^{3}}) (\frac{R^{2}}{r^{2}} \times \frac{1}{r^{2}}) I = (\frac{μ_{0} m_{0}^{2} ω^{4}}{32 π^{2} c^{3}}) (\frac{R^{2}}{{(h^{2} + R^{2})}^{2}}) . . . . . . . . (3)$

From equations (2) and (3),

$I = \frac{1}{32} (12) (\frac{R^{2}}{{(h^{2} + R^{2})}^{2}}) I = \frac{3 R^{2}}{8 π {(h^{2} + R^{2})}^{2}} . . . . . . . . . (4)$

Therefore, the formula for the intensity of the radiation at ground level, a distance R from the base of the tower, is

I = \frac{3 <P> R^{2}}{8 π {(h^{2} + R^{2})}^{2}}

Determine the distance from the base of a tower and formula for the intensity at the location:

(b)

The measurement are taken by the engineer for the maximum intensity. The first derivative the magnitude of the intensity of flux will corresponds to zero. Hence, from equation (4),

$\frac{\partial I}{\partial R} = 0 \frac{3 }{8 π} \frac{\partial}{\partial R} \frac{R^{2}}{{(R^{2} + h^{2})}^{2}} = 0 \frac{2 R}{{(R^{2} + h^{2})}^{2}} - \frac{4 R^{3}}{{(R^{2} + h^{2})}^{3}} = 0$

On further solving, the above equation becomes,

$\frac{2 R^{2}}{R^{2} + h^{2}} = 1 2 R^{2} = R^{2} + h^{2} R^{2} = h^{2} h = R$

Substitute the value of R in equation (4).

$I = \frac{3 R^{2}}{8 π {(R^{2} + R^{2})}^{2}} I = \frac{3 R^{2}}{8 π {(2 R^{2})}^{2}} I = \frac{3 R^{2}}{8 π (4 R^{4})} I = \frac{3 }{32 π R^{2}} . . . . . . . . (5)$ …… (5)

Therefore, the observation made at the location is $h = R$ and it corresponds to $I = \frac{3 }{32 π R^{2}}$ .

Determine the compliance of KRUD:

(c)

Re-write the equation (5) in terms of h.

$I = \frac{3 }{32 π h^{2}}$

Substitute $P = 35 kW$ and $h = 200 m$ in the above expression.

$I = \frac{3 (35 kW \times \frac{10^{3} W}{1 kW})}{32 π {(200 m)}^{2}} I = 0.02611 W / m^{2} \times (\frac{10^{- 6} μW}{1 W}) (\frac{1 m^{2}}{10^{4} {cm}^{2}}) I = 2.611 μW / {cm}^{2}$

The value of city’s radio emission limit is $200 \frac{μW}{{cm}^{2}}$ ,the KRUD is in compliance with the city’s radio emission limits as the value is $2.611 \frac{μW}{{cm}^{2}}$ .

Therefore, Yes, the KRUD is in compliance.