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
Plugging in the average power radiated by the electric dipole,
, with p0 = q0 d, we get
, we can substitute
. The coefficient
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.
For the magnetic dipole,
. Again setting this equal to
, we get
for , we get
. The coefficient
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.
The LARMOR FORMULA says
This is expended for a time t = v0/a, giving a total
The initial kinetic energy
Thus the fraction lost to EM radiation is
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!).
For a half-wave antenna
Charge conservation requires
For a 1-dimensional wire with a current I(z,t)
flowing in the direction, the same logic demands
ANSWER: The length is simple: where m. Thus . The current we can get from the power, using = 104 W. Thus or or .