THE UNIVERSITY OF BRITISH COLUMBIA

Physics 401 Assignment # 11:

SOLUTIONS:

Wed. 22 Mar. 2006 - finish by Wed. 29 Mar.
1. (p. 449, Problem 11.2) - Electric Dipolar Radiation: Equation (11.14),

can be expressed in "coordinate-free" form by writing .  Do so . . .
ANSWER . . . and similarly for Equation (11.17),

ANSWER . . . Equation (11.18),

. . . Equation (11.19),

. . . and Equation (11.21),

2. Atomic Dipoles: Explain why you can safely assume for an atom with magnetic dipole moment and electric dipole moment , assuming typical values of relevant physical quantities.   ANSWER Using Bohr's model of the H atom, we have an electron (  kg,  C) orbiting a heavy nucleus at radius  m, with angular momentum  kgm2/s. Thus  s-1. The amplitude of the electric dipole moment (in the plane of the orbit) is thus  Cm. The magnetic dipole moment is  m2C/s, and  Cm, which is a factor of smaller than p0.

3. (p. 473-474, Problem 11.22) - Broadcasting KRUD: A radio tower rises to a 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 , with a total radiated power P (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 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.
1. In terms of the variables given (not all of which may be relevant, of course) find the formula for the intensity of the radiation at ground level, a distance R away from the base of the tower. You may assume that . [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.]

ANSWER The average energy flux from a magnetic dipole antenna, Eq. (11.39), is

The total power radiated, Eq. (11.40), is

Here we have r2 = h2+R2 and , giving
.
The intensity at the base of the tower (R=0) is zero, Doh!

2. 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?   ANSWER At the point of highest intensity, of course. As for any extremum, this requires

or or .

3. 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?   ANSWER We don't need to know or b. All we need is  W and h=200 m. At the point of highest intensity (at ground level), R=h, we have

or
W/cm2, well within compliance.1

4. (p. 474, Problem 11.23) - Earth as a Pulsar: The magnetic north pole of the Earth does not coincide with the geographic North Pole - in fact, it's off by about 7 at present.2Relative to the fixed axis of rotation, therefore, the magnetic dipole moment vector of the Earth is changing with time, so the Earth must be giving off magnetic dipole radiation.
1. Find the formula for the total power radiated, in terms of the following parameters: (the angle between the geographic and magnetic north poles), M (the magnitude of the Earth's magnetic dipole moment), and (the angular velocity of rotation of the Earth). [Hint: refer to Prob. 11.4 or Prob. 11.12.]

ANSWER Problem 11.12 gives the total power radiated by a magnetic dipole generated by a time-varying current in a circular loop: . Problem 11.4 describes an electric dipole rotating about the axis as a superposition of two oscillating dipoles in the and directions, out of phase: . You are then invited to find the intensity as a function of the polar angle and calculate the total power radiated, explaining why the power seems to satisfy the superposition principle even though it is quadratic in the fields.
A more conventional way to represent precession of a dipole is to make the component real and the component imaginary: which amounts to the same thing as above. In the Earth's case it is only the transverse component that precesses; the axial component just adds a constant magnetic dipole field. Thus and we expect3
.

2. Using the fact that the Earth's magnetic field is about half a gauss at the Equator, estimate the magnetic dipole moment M of the Earth.   ANSWER From Eq. (5.87) on p. 246 we have the field of a static magnetic dipole:

which reduces to for (i.e. at the Equator). Thus or . (We neglect the tilt and the fact that the Earth's dipole is far from pointlike on the scale of RE.)