THE UNIVERSITY OF BRITISH COLUMBIA

*Physics 401 *
Assignment #
**11: **

** RADIATION ***1*

*SOLUTIONS:*

Wed. 22 Mar. 2006 - finish by Wed. 29 Mar.

Wed. 22 Mar. 2006 - finish by Wed. 29 Mar.

- (p. 449, Problem
**11.2**) -**Electric Dipolar Radiation:**Equation (11.14),

can be expressed in "coordinate-free" form by writing . Do so . . .**ANSWER**:

**ANSWER**:

**ANSWER**:

**ANSWER**:

**ANSWER**:*As always,*

**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 kgm*^{2}/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 m^{2}C/s, and Cm, which is a factor of smaller than*p*_{0}.- (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.- 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**r*^{2}=*h*^{2}+*R*^{2}and , giving

.

The intensity at the*base*of the tower (*R*=0) is zero,*Doh!* - 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 .* - 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?**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/cm^{2}, well within compliance.^{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
- (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.^{2}Relative 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.- 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 expect^{3}

. - 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*R*_{E}.) - Find the power radiated.
[Your answer should be several times 10
^{-5}W.]**ANSWER**:*The Earth's s*^{-1}. Plugging this, Am^{2}, and into

*gives .*

- Pulsars are thought to be rotating neutron stars,
with a typical radius of about km,
a typical surface magnetic field of T
and a variety of rotational periods
*T*; let's use s. What sort of radiated power would you expect from such a star? [See J.P. Ostriker and J.E. Gunn,*Astrophys. J.***157**, 1395 (1969).*Answer:*2 x 10^{36}W.]**ANSWER**:*Again we use to get = 10*^{27}Am^{2}. This is only some 8000 times bigger than the Earth's magnetic moment, but the frequency s^{-1}is a lot bigger, and is . . . well . . .**huge**. Thus (assuming the star's magnetic moment is perpendicular to its axis of rotation, which gives the biggest result) we get ! This is about a factor of two larger than the value predicted by Griffiths. No doubt this is because of the probability distribution of angles between the star's magnetic moment and its axis of rotation. If we assume all values of between 0 and are equally likely, then we should multiply our result by , giving . But this seems a little silly in two respects: first, we are just making an estimate for a "typical" neutron star. A 20% change of would have the same effect. Second, it seems improbable that the formation of neutron stars from supernovae of spinning suns would indiscriminately orient the star's magnetic moment relative to its axis or rotation; naively one might expect to be more likely, which would bias our estimate toward much smaller values of . A more realistic estimate would require a deeper knowledge of astrophysics than I (for one) possess.^{4}

- Find the formula for the total power radiated,
in terms of the following parameters:
(the angle between the geographic and magnetic north poles),

2006-03-30