THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 401 Assignment # 11:
 
RADIATION 1
 
SOLUTIONS:
 
Wed. 22 Mar. 2006 - finish by Wed. 29 Mar.

1. (p. 449, Problem 11.2) - Electric Dipolar Radiation: Equation (11.14),
   
   $$V(r,\theta,t) = - {p_0 \omega \over 4\pi\epsz c} \left(\c . . . . . . ght) \sin\left[\omega\left(t - {r \over c}\right)\right] \, ,$$
   
   
   can be expressed in "coordinate-free" form by writing $p_0\cos\theta = \Vec{p}_0\cdot\Hat{r}$.  Do so . . .
   ANSWER:  . . . and similarly for Equation (11.17),
   
   $$\Vec{A}(r,\theta,t) = - {\muz p_0 \omega \over 4\pi r} \sin\left[\omega\left(t - {r \over c}\right)\right] \Hat{z} \, ,$$
   
   
   ANSWER:  . . . Equation (11.18),
   
   $$\Vec{E}(r,\theta,t) = - {\muz p_0 \omega^2 \over 4\pi} \l . . . . . . ft[\omega\left(t - {r \over c}\right)\right] \Hat{\theta} \, ,$$
   
   
   ANSWER: 
   . . . Equation (11.19),
   
   $$\Vec{B}(r,\theta,t) = - {\muz p_0 \omega^2 \over 4\pi c} . . . . . . left[\omega\left(t - {r \over c}\right)\right] \Hat{\phi} \, ,$$
   
   
   ANSWER: 
   . . . and Equation (11.21),
   
   $$\langle \Vec{S} \rangle = \left( \muz p_0^2 \omega^4 \over 32\pi^2 c \right) {\sin^2\theta\over r^2} \Hat{r} \, .$$
   
   
   ANSWER:  As always,
   
   $$\Vec{S} \equiv {1\over\muz}\Vec{E}\times\Vec{B} = {\omega^2 \over 4\pi} \cdot {\muz \omega^2 \over 4\pi c} \cdot$$
   
   
   
   
   $$\cdot {[(\Hat{r} \times \Vec{p}_0) \times \Hat{r}] \times . . . . . . ^2} \; \cos^2\left[\omega\left(t - {r \over c}\right)\right]$$
   
   
   
   
   $$\hbox{\rm or} \quad \langle \Vec{S} \rangle = {\muz \omeg . . . . . . at{r} \times \Vec{p}_0\vert^2 \over r^2\right) \, \Hat{r} \; .$$
   
   

2. Atomic Dipoles: Explain why you can safely assume ${\displaystyle {\Vec{m}_0 \over c} \ll \Vec{p}_0 }$ for an atom with magnetic dipole moment $\Vec{m}_0$ and electric dipole moment $\Vec{p}_0$, assuming typical values of relevant physical quantities.   ANSWER:  Using Bohr's model of the H atom, we have an electron ( $m_e \simeq 0.911\times10^{-30}$ kg, $e \simeq 1.6\times10^{-19}$ C) orbiting a heavy nucleus at radius $a_0 \simeq 0.53\times10^{-10}$ m, with angular momentum $L = m_e a_0^2 \omega = \hbar \simeq 1.05\times10^{-34}$ kgm²/s. Thus $\omega = \hbar/m_e a_0^2 \simeq 4.13\times10^{16}$ s^-1. The amplitude of the electric dipole moment (in the plane of the orbit) is thus $p_0 = e a_0 \simeq 0.85\times10^{-29}$ Cm. The magnetic dipole moment is $m_0 = \pi a_0^2 (e\omega/2\pi) \simeq 0.927\times10^{-23}$ m²C/s, and $m_0/c \simeq 3.09\times10^{-30}$ 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 $\omega$, 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 $b \ll c/\omega \ll h$. [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
      

      $$\langle \Vec{S} \rangle = \left(\muz m_0^2 \omega^4 \over 32\pi^2c^3\right) {\sin^2\theta \over r^2} \, \Hat{r} \, .$$
      
      
      The total power radiated, Eq. (11.40), is
      
      $$\langle P \rangle = \left(\muz m_0^2 \omega^4 \over 12\pi c^3\right) \, .$$
      
      
      
      
      $$\hbox{\rm Thus} \qquad \langle \Vec{S} \rangle = \left( . . . . . . gle P \rangle\sin^2\theta\over 8\pi r^2\right) \, \Hat{r} \, .$$
      
      
      Here we have r² = h²+R² and $\sin^2\theta=R^2/r^2$, 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
      
      $$\DbyD{\vert\langle \Vec{S} \rangle\vert}{R} = 0 = {3\lang . . . . . . le \over 8\pi} \DbyD{}{R}\left(R^2 \over (h^2+R^2)^2\right)$$
      
      
      
      
      $$\hbox{\rm or} \qquad {2R \over (h^2+R^2)^2} -{4R^3 \over (h^2+R^2)^3} = 0$$
      
      
      or ${\displaystyle {2R^2 \over h^2+R^2} = 1}$ 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 $\omega$ or b. All we need is $\langle P \rangle = 3.5\times10^4$ W and h=200 m. At the point of highest intensity (at ground level), R=h, we have
      
      $$\vert\langle \Vec{S} \rangle\vert = {3\langle P \rangle h . . . . . . i (h^2+h^2)^2} = {3\times3.5\times10^4 \over 32\pi (200)^2}$$
      
      
      or  
      $= 2.611\;\mu$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$^\circ$ 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: $\Psi$ (the angle between the geographic and magnetic north poles), M (the magnitude of the Earth's magnetic dipole moment), and $\omega$ (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: $P_{\rm loop} = \muz \ddot{m}^2/6\pi c^3$. Problem 11.4 describes an electric dipole rotating about the $\Hat{z}$ axis as a superposition of two oscillating dipoles in the $\Hat{x}$ and $\Hat{y}$ directions, $\pi/2$ out of phase: $\Vec{p}(t) = p_0[\cos(\omega t)\Hat{x} + \sin(\omega t)\Hat{y}]$. You are then invited to find the intensity as a function of the polar angle $\theta$ 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 $\Hat{x}$ component real and the $\Hat{y}$ component imaginary: $\tilde{m}(t) = m_0 e^{i \omega t}$ which amounts to the same thing as above. In the Earth's case it is only the transverse component $M_\perp = M\sin\Psi \, e^{i\omega t}$ that precesses; the axial component $M_\parallel = M\cos\Psi$ just adds a constant magnetic dipole field. Thus $\ddot{m} = -\omega^2 M\sin\Psi \, e^{i\omega t}$ 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:
      
      $$\Vec{B}_{\rm dip} = {\muz\over4\pi} \left[3(\Vec{m}\cdot\Hat{r})\Hat{r} - \Vec{m} \over r^3\right]$$
      
      
      which reduces to $\Vec{B}_{\rm dip} = -\muz\Vec{m}/4\pi r^3$ for $\Hat{r} \perp \Vec{m}$ (i.e. at the Equator). Thus $M \approx 4\pi\times0.5\times10^{-4}\times(6.4\times10^6)^3/\muz$ or . (We neglect the $7^\circ$ tilt and the fact that the Earth's dipole is far from pointlike on the scale of R_E.)

   3. Find the power radiated. [Your answer should be several times 10^-5 W.]   ANSWER:  The Earth's $\omega = 2\pi/(24\times60\times60) \simeq 0.727\times10^{-4}$ s^-1. Plugging this, $M \approx 1.3\times10^{23}$ Am², and $\sin\Psi\approx\Psi = 0.122$ into
      
      $$P = {\muz \, \omega^4 \, M^2 \, \sin^2\Psi \over 6\pi c^3}$$
      
      
      gives .
       

   4. Pulsars are thought to be rotating neutron stars, with a typical radius of about $R\sim10$ km, a typical surface magnetic field of $B(R)\sim10^8$ T and a variety of rotational periods T; let's use $T\sim10^{-3}$ 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 10³⁶ W.]

      ANSWER:  Again we use $B_{\rm dip}^{\scriptscriptstyle\rm Equator} = \muz\, m /4\pi r^3$ to get $m \approx 4\pi\times10^{8}\times(10^4)^3/\muz$ = 10²⁷ Am². This is only some 8000 times bigger than the Earth's magnetic moment, but the frequency $\omega \approx 2\pi/10^{-3} = 6300$ s^-1 is a lot bigger, and $\omega^4$ 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 $\Psi$ between the star's magnetic moment and its axis of rotation. If we assume all values of $\Psi$ between 0 and $\pi$ are equally likely, then we should multiply our result by $\langle \sin^2\Psi \rangle = 1/2$, 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 $\omega$ 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 $\Psi \sim 0$ to be more likely, which would bias our estimate toward much smaller values of $\langle P_{\rm star} \rangle$. A more realistic estimate would require a deeper knowledge of astrophysics than I (for one) possess.⁴