THE UNIVERSITY OF BRITISH COLUMBIA

Physics 401 Assignment # 12:

SOLUTIONS:

Wed. 29 Mar. 2006 - finish by Wed. 5 Apr.
1. (p. 450, Problem 11.3) - Radiation Resistance of a Cell Phone:   Find the radiation resistance of the wire joining the two ends of the oscillating electric dipole described in Section 11.1.2. (This is the resistance that would give the same average power loss - to heat - as the oscillating dipole in fact puts out in the form of radiation.) Show that , where is the wavelength of the radiation. For the wires in an ordinary cell phone (say, d=5 cm), should you worry about the radiation contribution to the total resistance? Does it matter whether your cell phone uses the 900 MHz band or the 1.9 GHz band?1

ANSWER For a simple resistor R driven by a power supply that moves charge back and forth in an oscillation , we have a current and a power which averages to . Thus we can associate . Plugging in the average power radiated by the electric dipole, , with p0 = q0 d, we get . Since , we can substitute to get . The coefficient  N-A-2m-s-1, whose units are equivalent to W/A-2 or , leaving
.
For 900 or 1900 MHz, we have  cm or 15.8 cm, respectively,2 giving radiation resistances of or , respectively. Neither is a huge resistance, but both are certainly larger than that of the wires in your cell phone. The power is thus used quite efficiently. (Very little goes into useless heat; almost all is transmitted!) This is even more true of the higher frequency band: whatever transmission intensity is required, it can be realized with a smaller I.

2. (p. 454, Problem 11.6) - Radiation Resistance of a Magnetic Dipole Antenna:   Find the radiation resistance for the oscillating magnetic dipole shown in Fig. 11.8. Express your answer in terms of and b, and compare the radiation resistance of the electric dipole.3

ANSWER For the magnetic dipole, where . Again setting this equal to , we get . Again substituting for , we get . The coefficient , so
.
In this case, for a given frequency, the radiation resistance increases as the square of the area of the loop. For  cm or 15.8 cm, a 2.5 cm radius loop would have or , respectively. Thus the magnetic dipole antenna is similar to the electric dipole antenna at this size and frequency, but is much more strongly size- and frequency-dependent.

3. (p. 464, Problem 11.13) - Nonrelativistic Bremsstrahlung Radiation:
1. Suppose an electron decelerates at a constant rate a from some initial velocity v0 down to zero. What fraction of its initial kinetic energy is lost to EM radiation? (The rest is absorbed by whatever mechanism keeps the acceleration constant.) Assume (nonrelativistic case) so that the Larmor formula can be used.4

ANSWER The LARMOR FORMULA says . This is expended for a time t = v0/a, giving a total radiated energy . The initial kinetic energy . Thus the fraction lost to EM radiation is or
.

2. To get a sense of the numbers involved, suppose the initial velocity is thermal5 (around 105 m/s) and the distance over which the electron decelerates to rest is 30 Å. What can you conclude about radiation losses for electrons in an ordinary conductor?

ANSWER Using v02 = 2 a d with v0 = 105 m/s and d = 3 x 10-9 m, we have a = v02/2d = 1010/6 x 10-9 = 1.67 x 1018 m/s2. With q = -1.602 x 10-19 C and m = 0.911 x 10-30 kg, we get . (Not much!) The true picture is much stranger, of course; electrons are not localized point charges following classical trajectories, they are described by extended wavefunctions and do not radiate at all in things like atoms (luckily!).

4. Half-Wave Antenna:   Consider a half-wave linear antenna of length , with current , where .
1. Show that the linear charge density is , (i.e. the charge density is maximum at the times when the current is zero.)

ANSWER For a half-wave antenna and . Charge conservation requires . For a 1-dimensional wire with a current I(z,t) flowing in the direction, the same logic demands where . Thus . Integrating,

where and , giving .
Note that is also maximum at the places where the current is zero.

2. If an FM radio station broadcasts at a frequency of 10 MHz with a power of 10 kW from a half-wave antenna, how long must the antenna be? What is the current?

ANSWER The length is simple: where  m. Thus . The current we can get from the power, using = 104 W. Thus or or .

Jess H. Brewer
2006-04-02