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 bc/ωh. [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/cm2. 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=3PR28πh2+R22.

(b) The observation made at the location ish=R and it corresponds to I=3P32πR2.

(c) Yes, the KRUD is in compliance with the value of city’s radio and the value is2.611μWcm2

Step by step solution

01

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=μ0m02ω432π2c3sin2θr2r^

Here,μ0 is the permeability of magnetic field in free space,m0 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, andθ is the angle made by the displacement vector r with the vertical.

02

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,

r2=R2+h2sin2θ=R2r2

Write the expression for the total radiated power.

P=μ0m02ω432πc3 …… (2)

Substitute S=μ0m02ω432π2c3sin2θr2r^in equation (1).

I=μ0m02ω432π2c3sin2θr2I=μ0m02ω432π2c3R2r2r2I=μ0m02ω432π2c3R2r2×1r2I=μ0m02ω432π2c3R2(h2+R2)2........(3)

From equations (2) and (3),

I=13212PR2h2+R22I=3PR28πh2+R22.........(4)

Therefore, the formula for the intensity of the radiation at ground level, a distance R from the base of the tower, is I=3PR28πh2+R22.
03

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),

IR=03P8πRR2R2+h22=02RR2+h22-4R3R2+h23=0

On further solving, the above equation becomes,

2R2R2+h2=12R2=R2+h2R2=h2h=R

Substitute the value of R in equation (4).

I=3PR28πR2+R22I=3PR28π2R22I=3PR28π4R4I=3P32πR2........(5) …… (5)

Therefore, the observation made at the location ish=R and it corresponds to I=3P32πR2.

04

Determine the compliance of KRUD:

(c)

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

I=3P32πh2

SubstituteP=35kW andh=200m in the above expression.

I=335kW×103W1kW32π200m2I=0.02611W/m2×10-6μW1W1m2104cm2I=2.611μW/cm2

The value of city’s radio emission limit is 200μWcm2,the KRUD is in compliance with the city’s radio emission limits as the value is 2.611μWcm2.

Therefore, Yes, the KRUD is in compliance.

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Most popular questions from this chapter

A charged particle, traveling in from along the x axis, encounters a rectangular potential energy barrier

U(x)={U0,if0<x<L,0,otherwise}

Show that, because of the radiation reaction, it is possible for the particle to tunnel through the barrier-that is, even if the incident kinetic energy is less thanU0, the particle can pass through 26.[Hint: Your task is to solve the equation

a=τa˙+Fm

Subject to the force

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Refer to Probs. 11.19 and 11.31, but notice that this time the force is a specified function ofx, nott. There are three regions to consider: (i)x<0, (ii) 0<x<L, (iii)x>L. Find the general solution fora(t), υ(t), andx(t)in each region, exclude the runaway in region (iii), and impose the appropriate boundary conditions at x=0andx=L. Show that the final velocity (υf)is related to the time spent traversing the barrier by the equation

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and the initial velocity (atx=) is

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To simplify these results (since all we're looking for is a specific example), suppose the final kinetic energy is half the barrier height. Show that in this case

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(a) Find the electric and magnetic fields at a height x above the plane if

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