1. Check that the retarded potentials of an oscillating dipole (Eqs.
11.12 and 11.17) satisfy the Lorenz gauge condition. Do not use
approximation 3. Reference equation 11.12 ... Reference equation 11.17
... Get solution
2. Equation 11.14 can be expressed in “coordinate-free” form by writing ... Do so, and likewise for Eqs. 11.17, 11.18. 11.19, and 11.21. Reference equation 11.14 ... Reference equation 11.17 ... Reference equation 11.18 ... Reference equation 11.19 ... Reference equation 11.21 ... Get solution
3. Find the radiation resistance of the wire joining the two ends of the dipole. (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 R = 790 (d/λ)2 Ω, where λ is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm), should you worry about the radiative contribution to the total resistance? Get solution
4. A rotating electric dipole can be thought of as the superposition of two oscillating dipoles, one along the x axis and the other along the y axis (Fig. 11.7), with the latter out of phase by 90˚: ... Using the principle of superposition and Eqs. 11.18 and 11.19 (perhaps in the form suggested by Prob. 11.2), find the fields of a rotating dipole. Also find the Poynting vector and the intensity of the radiation. Sketch the intensity profile as a function of the polar angle θ, and calculate the total power radiated. Does the answer seem reasonable? (Note that power, being quadratic in the fields, does not satisfy the superposition principle. In this instance, however, it seems to. How do you account for this?) Reference equation 11.18 ... Reference equation 11.19 ... Reference prob 11.2 Equation 11.14 can be expressed in “coordinate-free” form by writing ... Do so, and likewise for Eqs. 11.17, 11.18. 11.19, and 11.21. Get solution
5. Calculate the electric and magnetic fields of an oscillating magnetic dipole without using approximation 3. [Do they look familiar? Compare Prob. 9.35.] Find the Poynting vector, and show that the intensity of the radiation is exactly the same as we got using approximation 3. Reference prob.9.35 Suppose ... (This is, incidentally, the simplest possible spherical wave. For notational convenience, let (kr − ωt) ≡ u in your calculations.) (a) Show that E obeys all four of Maxwell’s equations, in vacuum, and find the associated magnetic field. (b) Calculate the Poynting vector. Average S over a full cycle to get the intensity vector I. (Does it point in the expected direction? Does it fall off like r−2, as it should?) (c) Integrate I · da over a spherical surface to determine the total power radiated. Get solution
6. Find the radiation resistance (Prob. 11.3) for the oscillating magnetic dipole in Fig. 11.8. Express your answer in terms of λ and b, and compare the radiation resistance of the electric dipole. figure 11.8 ... Reference Prob. 11.3 Find the radiation resistance of the wire joining the two ends of the dipole. (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 R = 790 (d/λ)2 Ω, where λ is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm), should you worry about the radiative contribution to the total resistance? Get solution
7. Use the “duality” transformation of Prob. 7.64, together with the fields of an oscillating electric dipole (Eqs. 11.18 and 11.19), to determine the fields that would be produced by an oscillating “Gilbert” magnetic dipole (composed of equal and opposite magnetic charges, instead of an electric current loop). Compare Eqs. 11.36 and 11.37, and comment on the result. Reference equation 11.36 and 11.37 ... Reference prob 7.64. (a) Show that Maxwell’s equations with magnetic charge (Eq. 7.44) are invariant under the duality transformation ... where ... is an arbitrary rotation angle in “E/B-space.” Charge and current densities transform in the same way as qe and qm. particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using α = 90?) write down the fields produced by the corresponding arrangement of magnetic charge.] (b) Show that the force law (Prob. 7.38) ... is also invariant under the duality transformation. Get solution
8. A parallel-plate capacitor C, with plate separation d, is given an initial charge (±)Q0. It is then connected to a resistor R, and discharges, Q(t) = ... (a) What fraction of its initial energy (Q20 /2C) does it radiate away? (b) If C = 1 pF, R = 1000 Ω, and d = 0.1 mm, what is the actual number? In electronics we don’t ordinarily worry about radiative losses; does that seem reasonable, in this case? Get solution
9. Apply Eqs. 11.59 and 11.60 to the rotating dipole of Prob. 11.4. Explain any apparent discrepancies with your previous answer. Reference equation 11.59 ... Reference equation 11.60 ... Reference Prob. 11.4. A rotating electric dipole can be thought of as the superposition of two oscillating dipoles, one along the x axis and the other along the y axis (Fig. 11.7), with the latter out of phase by 90˚: ... Using the principle of superposition and Eqs. 11.18 and 11.19 (perhaps in the form suggested by Prob. 11.2), find the fields of a rotating dipole. Also find the Poynting vector and the intensity of the radiation. Sketch the intensity profile as a function of the polar angle θ, and calculate the total power radiated. Does the answer seem reasonable? (Note that power, being quadratic in the fields, does not satisfy the superposition principle. In this instance, however, it seems to. How do you account for this?) Get solution
10. An insulating circular ring (radius b) lies in the xy plane, centered at the origin. It carries a linear charge density λ = λ0 sin φ, where λ0 is constant and φ is the usual azimuthal angle. The ring is now set spinning at a constant angular velocity ω about the z axis. Calculate the power radiated. Get solution
11. A current I (t) flows around the circular ring in Fig. 11.8. Derive the general formula for the power radiated (analogous to Eq. 11.60), expressing your answer in terms of the magnetic dipole moment, m(t), of the loop. Figure 11.8 ... Equation 11.60 ... Get solution
12. An electron is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of the potential energy lost is radiated away? Get solution
13. A positive charge q is fired head-on at a distant positive charge Q (which is held stationary), with an initial velocity v0. It comes in, decelerates to v = 0, and returns out to infinity. What fraction of its initial energy ...is radiated away? Assume v0 ≪ c, and that you can safely ignore the effect of radiative losses on the motion of the particle. Get solution
14. In Bohr’s theory of hydrogen, the electron in its ground state was supposed to travel in a circle of radius 5 × 10−11m, held in orbit by the Coulomb attraction of the proton. According to classical electrodynamics, this electron should radiate, and hence spiral in to the nucleus. Show that v ≪ c for most of the trip (so you can use the Larmor formula), and calculate the lifespan of Bohr’s atom. (Assume each revolution is essentially circular.) Get solution
15. Find the angle θmax at which the maximum radiation is emitted, in Ex. 11.3 (Fig. 11.13). Show that for ultrarelativistic speeds (v close to c), ... ... What is the intensity of the radiation in this maximal direction (in the ultrarelativistic case), in proportion to the same quantity for a particle instantaneously at rest? Give your answer in terms of γ . Figure 11.13 ... Get solution
16. In Ex. 11.3 we assumed the velocity and acceleration were (instantaneously, at least) collinear. Carry out the same analysis for the case where they are perpendicular. Choose your axes so that v lies along the z axis and a along the x axis (Fig. 11.14), so that ... Check that P is consistent with the Liénard formula ... Figure 11.14 and 11.5 ... For relativistic velocities (β ≈ 1) the radiation is again sharply peaked in the forward direction (Fig. 11.15). The most important application of these formulas is to circular motion—in this case the radiation is called synchrotron radiation. For a relativistic electron, the radiation sweeps around like a locomotive’s headlight as the particle moves.] Get solution
17. (a) A particle of charge q moves in a circle of radius R at a constant speed v. To sustain the motion, you must, of course, provide a centripetal force mv2/R; what additional force (Fe) must you exert, in order to counteract the radiation reaction? [It’s easiest to express the answer in terms of the instantaneous velocity v.] What power (Pe) does this extra force deliver? Compare Pe with the power radiated (use the Larmor formula). (b) Repeat part (a) for a particle in simple harmonic motion with amplitude A and angular frequency ... Explain the discrepancy. (c) Consider the case of a particle in free fall (constant acceleration g). What is the radiation reaction force? What is the power radiated? Comment on these results. Get solution
18. A point charge q, of mass m, is attached to a spring of constant k. At time t = 0 it is given a kick, so its initial energy is...Now it oscillates, gradually radiating away this energy. (a) Confirm that the total energy radiated is equal to U0. Assume the radiation damping is small, so you can write the equation of motion as ... and the solution as ... with...and γ ≪ω0 (drop γ2 in comparison to ω2 0, and when you average over a complete cycle, ignore the change in e−γ t ). (b) Suppose now we have two such oscillators, and we start them off with identical kicks. Regardless of their relative positions and orientations, the total energy radiated must be 2U0. But what if they are right on top of each other, so it’s equivalent to a single oscillator with twice the charge; the Larmor formula says that the power radiated is four times as great, suggesting that the total will be 4U0. Find the error in this reasoning, and show that the total is actually 2U0, as it should be.13 Get solution
19. With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Reference equation 11.80 ... Get solution
20. Deduce Eq. 11.100 from Eq. 11.99. Here are three methods: (a) Use the Abraham-Lorentz formula to determine the radiation reaction on each end of the dumbbell; add this to the interaction term (Eq. 11.99). (b) Method (a) has the defect that it uses the Abraham-Lorentz formula—the very thing that we were trying to derive. To avoid this, let F(q) be the total d-independent part of the self-force on a charge q. Then ... where Fint is the interaction part (Eq. 11.99), and F(q/2) is the self-force on each end. Now, F(q) must be proportional to q2, since the field is proportional to q and the force is qE. So F(q/2) = (1/4)F(q). Take it from there. (c) Smear out the charge along a strip of length L oriented perpendicular to the motion (the charge density, then, is λ = q/L); find the cumulative interaction force for all pairs of segments, using Eq. 11.99 (with the correspondence q/2 → λ dy1, at one end and q/2 → λ dy2 at the other). Make sure you don’t count the same pair twice. Equation 11.100 ... Equation 11.99 ... Get solution
21. An electric dipole rotates at constant angular velocity ω in the x y plane. (The charges, ±q, are at ... the magnitude of the dipole moment is p = 2qR.) (a) Find the interaction term in the self-torque (analogous to Eq. 11.99). Assume the motion is nonrelativistic (ωR ≪ c). (b) Use the method of Prob. 11.20(a) to obtain the total radiation reaction torque on this system. (c) Check that this result is consistent with the power radiated (Eq. 11.60). Reference equation 11.60 ... Reference equation 11.99 ... Reference prob 11.20 Deduce Eq. 11.100 from Eq. 11.99. Here are three methods: (a) Use the Abraham-Lorentz formula to determine the radiation reaction on each end of the dumbbell; add this to the interaction term (Eq. 11.99). Get solution
22. A particle of mass m and charge q is attached to a spring with force constant k, hanging from the ceiling (Fig. 11.18). Its equilibrium position is a distance h above the floor. It is pulled down a distance d below equilibrium and released, at time t = 0. (a) Under the usual assumptions (d ≪ λ ≪ h), calculate the intensity of the radiation hitting the floor, as a function of the distance R from the point directly below q. [Note: The intensity here is the average power per unit area of floor.] ... At what R is the radiation most intense? Neglect the radiative damping of the oscillator. (b) As a check on your formula, assume the floor is of infinite extent, and calculate the average energy per unit time striking the entire floor. Is it what you’d expect? (c) Because it is losing energy in the form of radiation, the amplitude of the oscillation will gradually decrease. After what time τ has the amplitude been reduced to d/e? (Assume the fraction of the total energy lost in one cycle is very small.) Get solution
23. A radio tower rises to 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 ω, with a total radiated power P (that’s 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 level 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. (a) In terms of the variables given (not all of which may be relevant), find the formula for the intensity of the radiation at ground level, a distance R from the base of the tower. You may assume that b ≪ c/ ω ≪ 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.] (b) 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? (c) 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 radioemission limit is 200 microwatts/cm2. Is KRUD in compliance? Get solution
24. As a model for electric quadrupole radiation, consider two oppositely oriented oscillating electric dipoles, separated by a distance d, as shown in Fig. 11.19. Use the results of Sect. 11.1.2 for the potentials of each dipole, but note that they are not located at the origin. Keeping only the terms of first order in d: ... (a) Find the scalar and vector potentials. (b) Find the electric and magnetic fields. (c) Find the Poynting vector and the power radiated. Sketch the intensity profile as a function of θ. Get solution
25. As you know, the magnetic north pole of the earth does not coincide with the geographic north pole—in fact, it’s off by about 11? Relative to the fixed axis of rotation, therefore, the magnetic dipole moment of the earth is changing with time, and the earth must be giving off magnetic dipole radiation. (a) Find the formula for the total power radiated, in terms of the following parameters: Ψ (the angle between the geographic and magnetic north poles), M (the magnitude of the earth’s magnetic dipole moment), and ω (the angular velocity of rotation of the earth). [Hint: refer to Prob. 11.4 or Prob. 11.11.] (b) 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. (c) Find the power radiated. (d) Pulsars are thought to be rotating neutron stars, with a typical radius of 10 km, a rotational period of 10−3 s, and a surface magnetic field of 108 T. What sort of radiated power would you expect from such a star?20 Reference Prob. 11.4 A rotating electric dipole can be thought of as the superposition of two oscillating dipoles, one along the x axis and the other along the y axis (Fig. 11.7), with the latter out of phase by 90˚: ... Using the principle of superposition and Eqs. 11.18 and 11.19 (perhaps in the form suggested by Prob. 11.2), find the fields of a rotating dipole. Also find the Poynting vector and the intensity of the radiation. Sketch the intensity profile as a function of the polar angle θ, and calculate the total power radiated. Does the answer seem reasonable? (Note that power, being quadratic in the fields, does not satisfy the superposition principle. In this instance, however, it seems to. How do you account for this?) Reference Prob. 11.11 A current I (t) flows around the circular ring in Fig. 11.8. Derive the general formula for the power radiated (analogous to Eq. 11.60), expressing your answer in terms of the magnetic dipole moment, m(t), of the loop. Figure 11.8 ... Equation 11.60 ... Get solution
26. An ideal electric dipole is situated at the origin; its dipole moment points in the ... direction, and is quadratic in time: ... Where ... is a constant. (a) Use the method of Section 11.1.2 to determine the (exact) electric and magnetic fields, for all r > 0 (there’s also a delta- unction term at the origin, but we’re not concerned with that). (b) Calculate the power, P(r, t), passing through a sphere of radius r . (c) Find the total power radiated (Eq. 11.2), and check that your answer is consistent with Eq. 11.60.21 Reference equation 11.60 ... Reference equation 11.2 ... Get solution
27. In Section 11.2.1 we calculated the energy per unit time radiated by a (nonrelativistic) point charge—the Larmor formula. In the same spirit: (a) Calculate the momentum per unit time radiated. (b) Calculate the angular momentum per unit time radiated. Get solution
28. Suppose the (electrically neutral) yz plane carries a time-dependent but uniform surface current ... (a) Find the electric and magnetic fields at a height x above the plane if (i) a constant current is turned on at t = 0: ... (ii) a linearly increasing current is turned on at t = 0: ... (b) Show that the retarded vector potential can be written in the form ... and from this determine E and B. (c) Show that the total power radiated per unit area of surface is ... Explain what you mean by “radiation,” in this case, given that the source is not localized.22 Get solution
29. Use the duality transformation (Prob. 7.64) to construct the electric and magnetic fields of a magnetic monopole qm in arbitrary motion, and find the “Larmor formula” for the power radiated Reference prob 7.64 iv class="question"> (a) Show that Maxwell’s equations with magnetic charge (Eq. 7.44) are invariant under the duality transformation ... where ... is an arbitrary rotation angle in “E/B-space.” Charge and current densities transform in the same way as qe and qm. particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using α = 90˚) write down the fields produced by the corresponding arrangement of magnetic charge.] (b) Show that the force law (Prob. 7.38) ... is also invariant under the duality transformation. Get solution
30. Assuming you exclude the runaway solution in Prob. 11.19, calculate (a) the work done by the external force, (b) the final kinetic energy (assume the initial kinetic energy was zero), (c) the total energy radiated. Check that energy is conserved in this process. Reference prob 11.19 With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Get solution
31. (a) Repeat Prob. 11.19, but this time let the external force be a Dirac delta function: F(t) = kδ(t) (for some constant k).25 [Note that the acceleration is now discontinuous at t = 0 (though the velocity must still be continuous); use the method of Prob. 11.19 a) to show that ; Δa = −k/mτ . In this problem there are only two intervals to consider: (i) t 0, and (ii) t > 0.] (b) As in Prob. 11.30, check that energy is conserved in this process. Reference prob 11.19 With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Reference prob 11.30 Assuming you exclude the runaway solution in Prob. 11.19, calculate (a) the work done by the external force, (b) the final kinetic energy (assume the initial kinetic energy was zero), (c) the total energy radiated. Check that energy is conserved in this process. Get solution
32. A charged particle, traveling in from −∞along the x axis, encounters a rectangular potential energy barrier ... Show that, because of the radiation reaction, it is possible for the particle to tunnel through the barrier—that is, even if the incident kinetic energy is less than U0, the particle can pass through.26 [Hint: Your task is to solve the equation ... subject to the force ... Refer to Probs. 11.19 and 11.31, but notice that this time the force is a specified function of x, not t. There are three regions to consider: (i) x 0, (ii) 0 x L, (iii) x > L. Find the general solution for a(t), v(t), and x(t) in each region, exclude the runaway in region (iii), and impose the appropriate boundary conditions at x = 0 and x = L. Show that the final velocity (v f ) is related to the time T spent traversing the barrier by the equation ... and the initial velocity (at x = −∞) is ... To simplify these results (since all we’re looking for is a specific example), suppose the final kinetic energy is half the barrier height. Show that in this case ... In particular, if you choose ... the initial kinetic energy is (8/9)U0, and the particle makes it through, even though it didn’t have sufficient energy to get over the barrier!] Reference prob 11.19With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Reference prob 11.31 (a) Repeat Prob. 11.19, but this time let the external force be a Dirac delta function: F(t) = kδ(t) (for some constant k).25 [Note that the acceleration is now discontinuous at t = 0 (though the velocity must still be continuous); use the method of Prob. 11.19 a) to show that ; Δa = −k/mτ . In this problem there are only two intervals to consider: (i) t 0, and (ii) t > 0.] (b) As in Prob. 11.30, check that energy is conserved in this process. Get solution
33. (a) Find the radiation reaction force on a particle moving with arbitrary velocity in a straight line, by reconstructing the argument in Sect. 11.2.3 without assuming v(tr ) = 0. (b) Show that this result is consistent (in the sense of Eq. 11.78) with the power radiated by such a particle (Eq. 11.75). Equation 11.78 ... Equation 11.75 ... Get solution
34. a) Does a particle in hyperbolic motion (Eq. 10.52) radiate? (Use the exact formula (Eq. 11.75) to calculate the power radiated.) (b) Does a particle in hyperbolic motion experience a radiation reaction? (Use the exact formula (Prob. 11.33) to determine the reaction force.) [Comment: These famous questions carry important implications for the principle of equivalence.27] Reference equation 11.75 ... Reference equation 10.52 ... Reference prob 11.33 Get solution
35. Use the result of Prob. 10.34 to determine the power radiated by an ideal electric dipole, p(t), at the origin. Check that your answer is consistent with Eq. 11.22, in the case of sinusoidal time dependence, and with Prob. 11.26, in the case of quadratic time dependence. the result of Prob. 10.34 ... Reference equation 11.22 ... Reference problem 11.26 Get solution
2. Equation 11.14 can be expressed in “coordinate-free” form by writing ... Do so, and likewise for Eqs. 11.17, 11.18. 11.19, and 11.21. Reference equation 11.14 ... Reference equation 11.17 ... Reference equation 11.18 ... Reference equation 11.19 ... Reference equation 11.21 ... Get solution
3. Find the radiation resistance of the wire joining the two ends of the dipole. (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 R = 790 (d/λ)2 Ω, where λ is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm), should you worry about the radiative contribution to the total resistance? Get solution
4. A rotating electric dipole can be thought of as the superposition of two oscillating dipoles, one along the x axis and the other along the y axis (Fig. 11.7), with the latter out of phase by 90˚: ... Using the principle of superposition and Eqs. 11.18 and 11.19 (perhaps in the form suggested by Prob. 11.2), find the fields of a rotating dipole. Also find the Poynting vector and the intensity of the radiation. Sketch the intensity profile as a function of the polar angle θ, and calculate the total power radiated. Does the answer seem reasonable? (Note that power, being quadratic in the fields, does not satisfy the superposition principle. In this instance, however, it seems to. How do you account for this?) Reference equation 11.18 ... Reference equation 11.19 ... Reference prob 11.2 Equation 11.14 can be expressed in “coordinate-free” form by writing ... Do so, and likewise for Eqs. 11.17, 11.18. 11.19, and 11.21. Get solution
5. Calculate the electric and magnetic fields of an oscillating magnetic dipole without using approximation 3. [Do they look familiar? Compare Prob. 9.35.] Find the Poynting vector, and show that the intensity of the radiation is exactly the same as we got using approximation 3. Reference prob.9.35 Suppose ... (This is, incidentally, the simplest possible spherical wave. For notational convenience, let (kr − ωt) ≡ u in your calculations.) (a) Show that E obeys all four of Maxwell’s equations, in vacuum, and find the associated magnetic field. (b) Calculate the Poynting vector. Average S over a full cycle to get the intensity vector I. (Does it point in the expected direction? Does it fall off like r−2, as it should?) (c) Integrate I · da over a spherical surface to determine the total power radiated. Get solution
6. Find the radiation resistance (Prob. 11.3) for the oscillating magnetic dipole in Fig. 11.8. Express your answer in terms of λ and b, and compare the radiation resistance of the electric dipole. figure 11.8 ... Reference Prob. 11.3 Find the radiation resistance of the wire joining the two ends of the dipole. (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 R = 790 (d/λ)2 Ω, where λ is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm), should you worry about the radiative contribution to the total resistance? Get solution
7. Use the “duality” transformation of Prob. 7.64, together with the fields of an oscillating electric dipole (Eqs. 11.18 and 11.19), to determine the fields that would be produced by an oscillating “Gilbert” magnetic dipole (composed of equal and opposite magnetic charges, instead of an electric current loop). Compare Eqs. 11.36 and 11.37, and comment on the result. Reference equation 11.36 and 11.37 ... Reference prob 7.64. (a) Show that Maxwell’s equations with magnetic charge (Eq. 7.44) are invariant under the duality transformation ... where ... is an arbitrary rotation angle in “E/B-space.” Charge and current densities transform in the same way as qe and qm. particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using α = 90?) write down the fields produced by the corresponding arrangement of magnetic charge.] (b) Show that the force law (Prob. 7.38) ... is also invariant under the duality transformation. Get solution
8. A parallel-plate capacitor C, with plate separation d, is given an initial charge (±)Q0. It is then connected to a resistor R, and discharges, Q(t) = ... (a) What fraction of its initial energy (Q20 /2C) does it radiate away? (b) If C = 1 pF, R = 1000 Ω, and d = 0.1 mm, what is the actual number? In electronics we don’t ordinarily worry about radiative losses; does that seem reasonable, in this case? Get solution
9. Apply Eqs. 11.59 and 11.60 to the rotating dipole of Prob. 11.4. Explain any apparent discrepancies with your previous answer. Reference equation 11.59 ... Reference equation 11.60 ... Reference Prob. 11.4. A rotating electric dipole can be thought of as the superposition of two oscillating dipoles, one along the x axis and the other along the y axis (Fig. 11.7), with the latter out of phase by 90˚: ... Using the principle of superposition and Eqs. 11.18 and 11.19 (perhaps in the form suggested by Prob. 11.2), find the fields of a rotating dipole. Also find the Poynting vector and the intensity of the radiation. Sketch the intensity profile as a function of the polar angle θ, and calculate the total power radiated. Does the answer seem reasonable? (Note that power, being quadratic in the fields, does not satisfy the superposition principle. In this instance, however, it seems to. How do you account for this?) Get solution
10. An insulating circular ring (radius b) lies in the xy plane, centered at the origin. It carries a linear charge density λ = λ0 sin φ, where λ0 is constant and φ is the usual azimuthal angle. The ring is now set spinning at a constant angular velocity ω about the z axis. Calculate the power radiated. Get solution
11. A current I (t) flows around the circular ring in Fig. 11.8. Derive the general formula for the power radiated (analogous to Eq. 11.60), expressing your answer in terms of the magnetic dipole moment, m(t), of the loop. Figure 11.8 ... Equation 11.60 ... Get solution
12. An electron is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of the potential energy lost is radiated away? Get solution
13. A positive charge q is fired head-on at a distant positive charge Q (which is held stationary), with an initial velocity v0. It comes in, decelerates to v = 0, and returns out to infinity. What fraction of its initial energy ...is radiated away? Assume v0 ≪ c, and that you can safely ignore the effect of radiative losses on the motion of the particle. Get solution
14. In Bohr’s theory of hydrogen, the electron in its ground state was supposed to travel in a circle of radius 5 × 10−11m, held in orbit by the Coulomb attraction of the proton. According to classical electrodynamics, this electron should radiate, and hence spiral in to the nucleus. Show that v ≪ c for most of the trip (so you can use the Larmor formula), and calculate the lifespan of Bohr’s atom. (Assume each revolution is essentially circular.) Get solution
15. Find the angle θmax at which the maximum radiation is emitted, in Ex. 11.3 (Fig. 11.13). Show that for ultrarelativistic speeds (v close to c), ... ... What is the intensity of the radiation in this maximal direction (in the ultrarelativistic case), in proportion to the same quantity for a particle instantaneously at rest? Give your answer in terms of γ . Figure 11.13 ... Get solution
16. In Ex. 11.3 we assumed the velocity and acceleration were (instantaneously, at least) collinear. Carry out the same analysis for the case where they are perpendicular. Choose your axes so that v lies along the z axis and a along the x axis (Fig. 11.14), so that ... Check that P is consistent with the Liénard formula ... Figure 11.14 and 11.5 ... For relativistic velocities (β ≈ 1) the radiation is again sharply peaked in the forward direction (Fig. 11.15). The most important application of these formulas is to circular motion—in this case the radiation is called synchrotron radiation. For a relativistic electron, the radiation sweeps around like a locomotive’s headlight as the particle moves.] Get solution
17. (a) A particle of charge q moves in a circle of radius R at a constant speed v. To sustain the motion, you must, of course, provide a centripetal force mv2/R; what additional force (Fe) must you exert, in order to counteract the radiation reaction? [It’s easiest to express the answer in terms of the instantaneous velocity v.] What power (Pe) does this extra force deliver? Compare Pe with the power radiated (use the Larmor formula). (b) Repeat part (a) for a particle in simple harmonic motion with amplitude A and angular frequency ... Explain the discrepancy. (c) Consider the case of a particle in free fall (constant acceleration g). What is the radiation reaction force? What is the power radiated? Comment on these results. Get solution
18. A point charge q, of mass m, is attached to a spring of constant k. At time t = 0 it is given a kick, so its initial energy is...Now it oscillates, gradually radiating away this energy. (a) Confirm that the total energy radiated is equal to U0. Assume the radiation damping is small, so you can write the equation of motion as ... and the solution as ... with...and γ ≪ω0 (drop γ2 in comparison to ω2 0, and when you average over a complete cycle, ignore the change in e−γ t ). (b) Suppose now we have two such oscillators, and we start them off with identical kicks. Regardless of their relative positions and orientations, the total energy radiated must be 2U0. But what if they are right on top of each other, so it’s equivalent to a single oscillator with twice the charge; the Larmor formula says that the power radiated is four times as great, suggesting that the total will be 4U0. Find the error in this reasoning, and show that the total is actually 2U0, as it should be.13 Get solution
19. With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Reference equation 11.80 ... Get solution
20. Deduce Eq. 11.100 from Eq. 11.99. Here are three methods: (a) Use the Abraham-Lorentz formula to determine the radiation reaction on each end of the dumbbell; add this to the interaction term (Eq. 11.99). (b) Method (a) has the defect that it uses the Abraham-Lorentz formula—the very thing that we were trying to derive. To avoid this, let F(q) be the total d-independent part of the self-force on a charge q. Then ... where Fint is the interaction part (Eq. 11.99), and F(q/2) is the self-force on each end. Now, F(q) must be proportional to q2, since the field is proportional to q and the force is qE. So F(q/2) = (1/4)F(q). Take it from there. (c) Smear out the charge along a strip of length L oriented perpendicular to the motion (the charge density, then, is λ = q/L); find the cumulative interaction force for all pairs of segments, using Eq. 11.99 (with the correspondence q/2 → λ dy1, at one end and q/2 → λ dy2 at the other). Make sure you don’t count the same pair twice. Equation 11.100 ... Equation 11.99 ... Get solution
21. An electric dipole rotates at constant angular velocity ω in the x y plane. (The charges, ±q, are at ... the magnitude of the dipole moment is p = 2qR.) (a) Find the interaction term in the self-torque (analogous to Eq. 11.99). Assume the motion is nonrelativistic (ωR ≪ c). (b) Use the method of Prob. 11.20(a) to obtain the total radiation reaction torque on this system. (c) Check that this result is consistent with the power radiated (Eq. 11.60). Reference equation 11.60 ... Reference equation 11.99 ... Reference prob 11.20 Deduce Eq. 11.100 from Eq. 11.99. Here are three methods: (a) Use the Abraham-Lorentz formula to determine the radiation reaction on each end of the dumbbell; add this to the interaction term (Eq. 11.99). Get solution
22. A particle of mass m and charge q is attached to a spring with force constant k, hanging from the ceiling (Fig. 11.18). Its equilibrium position is a distance h above the floor. It is pulled down a distance d below equilibrium and released, at time t = 0. (a) Under the usual assumptions (d ≪ λ ≪ h), calculate the intensity of the radiation hitting the floor, as a function of the distance R from the point directly below q. [Note: The intensity here is the average power per unit area of floor.] ... At what R is the radiation most intense? Neglect the radiative damping of the oscillator. (b) As a check on your formula, assume the floor is of infinite extent, and calculate the average energy per unit time striking the entire floor. Is it what you’d expect? (c) Because it is losing energy in the form of radiation, the amplitude of the oscillation will gradually decrease. After what time τ has the amplitude been reduced to d/e? (Assume the fraction of the total energy lost in one cycle is very small.) Get solution
23. A radio tower rises to 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 ω, with a total radiated power P (that’s 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 level 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. (a) In terms of the variables given (not all of which may be relevant), find the formula for the intensity of the radiation at ground level, a distance R from the base of the tower. You may assume that b ≪ c/ ω ≪ 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.] (b) 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? (c) 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 radioemission limit is 200 microwatts/cm2. Is KRUD in compliance? Get solution
24. As a model for electric quadrupole radiation, consider two oppositely oriented oscillating electric dipoles, separated by a distance d, as shown in Fig. 11.19. Use the results of Sect. 11.1.2 for the potentials of each dipole, but note that they are not located at the origin. Keeping only the terms of first order in d: ... (a) Find the scalar and vector potentials. (b) Find the electric and magnetic fields. (c) Find the Poynting vector and the power radiated. Sketch the intensity profile as a function of θ. Get solution
25. As you know, the magnetic north pole of the earth does not coincide with the geographic north pole—in fact, it’s off by about 11? Relative to the fixed axis of rotation, therefore, the magnetic dipole moment of the earth is changing with time, and the earth must be giving off magnetic dipole radiation. (a) Find the formula for the total power radiated, in terms of the following parameters: Ψ (the angle between the geographic and magnetic north poles), M (the magnitude of the earth’s magnetic dipole moment), and ω (the angular velocity of rotation of the earth). [Hint: refer to Prob. 11.4 or Prob. 11.11.] (b) 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. (c) Find the power radiated. (d) Pulsars are thought to be rotating neutron stars, with a typical radius of 10 km, a rotational period of 10−3 s, and a surface magnetic field of 108 T. What sort of radiated power would you expect from such a star?20 Reference Prob. 11.4 A rotating electric dipole can be thought of as the superposition of two oscillating dipoles, one along the x axis and the other along the y axis (Fig. 11.7), with the latter out of phase by 90˚: ... Using the principle of superposition and Eqs. 11.18 and 11.19 (perhaps in the form suggested by Prob. 11.2), find the fields of a rotating dipole. Also find the Poynting vector and the intensity of the radiation. Sketch the intensity profile as a function of the polar angle θ, and calculate the total power radiated. Does the answer seem reasonable? (Note that power, being quadratic in the fields, does not satisfy the superposition principle. In this instance, however, it seems to. How do you account for this?) Reference Prob. 11.11 A current I (t) flows around the circular ring in Fig. 11.8. Derive the general formula for the power radiated (analogous to Eq. 11.60), expressing your answer in terms of the magnetic dipole moment, m(t), of the loop. Figure 11.8 ... Equation 11.60 ... Get solution
26. An ideal electric dipole is situated at the origin; its dipole moment points in the ... direction, and is quadratic in time: ... Where ... is a constant. (a) Use the method of Section 11.1.2 to determine the (exact) electric and magnetic fields, for all r > 0 (there’s also a delta- unction term at the origin, but we’re not concerned with that). (b) Calculate the power, P(r, t), passing through a sphere of radius r . (c) Find the total power radiated (Eq. 11.2), and check that your answer is consistent with Eq. 11.60.21 Reference equation 11.60 ... Reference equation 11.2 ... Get solution
27. In Section 11.2.1 we calculated the energy per unit time radiated by a (nonrelativistic) point charge—the Larmor formula. In the same spirit: (a) Calculate the momentum per unit time radiated. (b) Calculate the angular momentum per unit time radiated. Get solution
28. Suppose the (electrically neutral) yz plane carries a time-dependent but uniform surface current ... (a) Find the electric and magnetic fields at a height x above the plane if (i) a constant current is turned on at t = 0: ... (ii) a linearly increasing current is turned on at t = 0: ... (b) Show that the retarded vector potential can be written in the form ... and from this determine E and B. (c) Show that the total power radiated per unit area of surface is ... Explain what you mean by “radiation,” in this case, given that the source is not localized.22 Get solution
29. Use the duality transformation (Prob. 7.64) to construct the electric and magnetic fields of a magnetic monopole qm in arbitrary motion, and find the “Larmor formula” for the power radiated Reference prob 7.64 iv class="question"> (a) Show that Maxwell’s equations with magnetic charge (Eq. 7.44) are invariant under the duality transformation ... where ... is an arbitrary rotation angle in “E/B-space.” Charge and current densities transform in the same way as qe and qm. particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using α = 90˚) write down the fields produced by the corresponding arrangement of magnetic charge.] (b) Show that the force law (Prob. 7.38) ... is also invariant under the duality transformation. Get solution
30. Assuming you exclude the runaway solution in Prob. 11.19, calculate (a) the work done by the external force, (b) the final kinetic energy (assume the initial kinetic energy was zero), (c) the total energy radiated. Check that energy is conserved in this process. Reference prob 11.19 With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Get solution
31. (a) Repeat Prob. 11.19, but this time let the external force be a Dirac delta function: F(t) = kδ(t) (for some constant k).25 [Note that the acceleration is now discontinuous at t = 0 (though the velocity must still be continuous); use the method of Prob. 11.19 a) to show that ; Δa = −k/mτ . In this problem there are only two intervals to consider: (i) t 0, and (ii) t > 0.] (b) As in Prob. 11.30, check that energy is conserved in this process. Reference prob 11.19 With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Reference prob 11.30 Assuming you exclude the runaway solution in Prob. 11.19, calculate (a) the work done by the external force, (b) the final kinetic energy (assume the initial kinetic energy was zero), (c) the total energy radiated. Check that energy is conserved in this process. Get solution
32. A charged particle, traveling in from −∞along the x axis, encounters a rectangular potential energy barrier ... Show that, because of the radiation reaction, it is possible for the particle to tunnel through the barrier—that is, even if the incident kinetic energy is less than U0, the particle can pass through.26 [Hint: Your task is to solve the equation ... subject to the force ... Refer to Probs. 11.19 and 11.31, but notice that this time the force is a specified function of x, not t. There are three regions to consider: (i) x 0, (ii) 0 x L, (iii) x > L. Find the general solution for a(t), v(t), and x(t) in each region, exclude the runaway in region (iii), and impose the appropriate boundary conditions at x = 0 and x = L. Show that the final velocity (v f ) is related to the time T spent traversing the barrier by the equation ... and the initial velocity (at x = −∞) is ... To simplify these results (since all we’re looking for is a specific example), suppose the final kinetic energy is half the barrier height. Show that in this case ... In particular, if you choose ... the initial kinetic energy is (8/9)U0, and the particle makes it through, even though it didn’t have sufficient energy to get over the barrier!] Reference prob 11.19With the inclusion of the radiation reaction force (Eq. 11.80), Newton’s second law for a charged particle becomes ... where F is the external force acting on the particle. (a) In contrast to the case of an uncharged particle (a = F/m), acceleration (like position and velocity) must now be a continuous function of time, even if the force changes abruptly. (Physically, the radiation reaction damps out any rapid change in a.) Prove that a is continuous at any time t, by integrating the equation of motion above from ... and taking the limit ... (b) A particle is subjected to a constant force F, beginning at time t = 0 and lasting until time T . Find the most general solution a(t) to the equation of motion in each of the three periods: (i) t 0; (ii) 0 t T ; (iii) t > T . (c) Impose the continuity condition (a) at t = 0 and t = T . Show that you can either eliminate the runaway in region (iii) or avoid preacceleration in region (i), but not both. (d) If you choose to eliminate the runaway, what is the acceleration as a function of time, in each interval? How about the velocity? (The latter must, of course, be continuous at t = 0 and t = T .) Assume the particle was originally at rest: v(−∞) = 0. (e) Plot a(t) and v(t), both for an uncharged particle and for a (nonrunaway) charged particle, subject to this force. Reference prob 11.31 (a) Repeat Prob. 11.19, but this time let the external force be a Dirac delta function: F(t) = kδ(t) (for some constant k).25 [Note that the acceleration is now discontinuous at t = 0 (though the velocity must still be continuous); use the method of Prob. 11.19 a) to show that ; Δa = −k/mτ . In this problem there are only two intervals to consider: (i) t 0, and (ii) t > 0.] (b) As in Prob. 11.30, check that energy is conserved in this process. Get solution
33. (a) Find the radiation reaction force on a particle moving with arbitrary velocity in a straight line, by reconstructing the argument in Sect. 11.2.3 without assuming v(tr ) = 0. (b) Show that this result is consistent (in the sense of Eq. 11.78) with the power radiated by such a particle (Eq. 11.75). Equation 11.78 ... Equation 11.75 ... Get solution
34. a) Does a particle in hyperbolic motion (Eq. 10.52) radiate? (Use the exact formula (Eq. 11.75) to calculate the power radiated.) (b) Does a particle in hyperbolic motion experience a radiation reaction? (Use the exact formula (Prob. 11.33) to determine the reaction force.) [Comment: These famous questions carry important implications for the principle of equivalence.27] Reference equation 11.75 ... Reference equation 10.52 ... Reference prob 11.33 Get solution
35. Use the result of Prob. 10.34 to determine the power radiated by an ideal electric dipole, p(t), at the origin. Check that your answer is consistent with Eq. 11.22, in the case of sinusoidal time dependence, and with Prob. 11.26, in the case of quadratic time dependence. the result of Prob. 10.34 ... Reference equation 11.22 ... Reference problem 11.26 Get solution
