PHYS 110B - HW #7 Fall 2005, Solutions by David Pace Equations referenced as Eq. # are from Griffiths Problem statements are paraphrased

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1 PHYS B - HW #7 Fall 5, Solutions by David Pae Equations efeened as Eq. # ae fom Giffiths Poblem statements ae paaphased [.] Poblem.4 fom Giffiths Show that Eq..4, V, t an be witten as Eq..44, V, t q t v + v t q R v sin θ whee R vt is the veto fom the pesent loation of the patile to the obsevation point do not onfuse this symbol with R. Sine this elates to the situation of example.3 page 433 in Giffiths, the veloity is a onstant. The angle θ is that between the vetos R and v. Refeene figue.9 fo this geomety. It is also woth noting that at non-elativisti veloities whee v, the sala potential in simplifies to the well-known eletostati value. Solution Notiing how simila and ae, the best method fo showing thei equivalene seems to be ewiting the squae oot tem in. This will equie solving fo the dependene in tems of the pesent position of the patile sine does not appea in the final fom. The veloity is onstant in this poblem, so the etaded veloity witten in the oiginal solution is equivalent to the veloity at pesent time. Begin by solving fo v using the fat that R + vt as detemined fom the poblem statement. The magnitude of is, v R + vt v 3 R v + v t 4 R + vt R + vt 5 R + t R v + v t 6 R + t R v + v t 7 whee the value of is witten beause that is what appeas in the elevant equation. Now the squae oot tem in an be simplified. I ll just take the inside of the squae oot hee fo laity. t v + v t t v + v R + t R v + v t t 8 4 t t v + v + v R + t R v + v t t 9 4 t t R v + v t + R v + v t + R + t R v + v t t v R + t R v + v t t 4 t tr v v t + R v + v tr v + v 4 t + R + tr v + v t 4 t v R tv R v v 4 t + t v R v + R v R Piking up fom, R v + R v R R v os θ + R v R 3 v R os θ + v 4

2 v R sin θ + v R v v sin θ + v 5 6 [.] Poblem.7 fom Giffiths Show that Eq..7 is oet, R v sin θ Plugging this bak into gives, 7 R R u Eq..7 V, t q R v sin θ 8 q R v sin θ 9 and then use this esult to deive Eq..63, A q R R v 3 [ R R v v + R a + R ] v + R a v Solution Equating 6 and 7, Fom Giffiths, wt t t Eq..33 R wt Eq R R R R t t R 8 9 Using the above as the stating point, wite R, R R R t t 4 Take the non-etaded time deivative of R, R R R R + R R 5 R R 6 The time deivative of the ight hand side of 4 is, t t t t 7 whee the last step makes use of R t t, whih follows fom and 3. Witing out the left side of 9 gives, R R R wt 3 R wt 3 R wt 3 R v 33

3 3 Putting 33 bak into 9 podues, R v R v R R v R R R R 36 R R R v 37 and it emains to show that the denominato above is equal to R u whee u ˆR v Eq..64. R u R ˆR v 38 R R v 39 It will be easie to ay out the algeba in this fom and then eplae the R u tem in the final expession. A q [ v q [ a R u + v ] R u 47 R u + v R u ] R u 48 [ q ] R a R u R u v R u R u 49 q R u [ R a v ] R u 5 R R u 4 At this point it is wothwhile to wite out the time deivative in 5. Now fo the deivation of equation.63. Begin with the equation fo the veto potential as witten in Eq..4 the Liénad-Wiehet veto potential, A, t v V, t 4 Taking the time deivative of this gives, A [ v [ v ] V V + v V + v V ] 4 43 The geneal fom of the sala potential is given by Eq..39, V, t q R R v 44 The fist tem is, R R u R R v 5 R R v R v t t 5 53 The seond tem is solved in 33, and the thid tem is solved in equations 4 though 48. The esult is, R u v v R a 54 + v R a 55 q R u 45 whee the last step utilizes 39. Put this sala potential into ou pesent veto potential expession, A [ ] v q R u + v q R u 46 + v R a + v R R a R u Plug the esult of 57 bak into

4 4 A [ q R R a v + v ] R R a u R u 58 [ q R R R R u R u u a v u R + v R a ] 59 qr [ R u a R u v ] R v v + v + R a v 6 R u 3 qr R u 3 qr R u 3 R q R u 3 q R R v 3 [ R u a v ] R + v + R a v [ R u R a [ R u R a v + R ] v + R a v v + R ] v + R a v [ R R v v + R a + R v + R a v ] whee in the last line I have eplaed the R u tems. This final esult mathes as intended. [3.] Poblem.9 fom Giffiths The eleti field is, E, t v / λ dx v sin θ/ 3/ ˆR R 67 a Use Eq..68, E, t q v / ˆR 4πɛ o v sin θ/ 3/ R 65 whee the λ is also a onstant and may be taken outside the integal. It emains to set the limits of this integal and pefom the mathematial steps, all of whih have been employed peviously. to solve fo the eleti field a distane d away fom an infinite wie that is aying a unifom line hage λ whih moves down the wie with onstant speed v. b Use Eq..69, B v E 66 to solve fo the magneti field due to this wie. Solution Let the wie lie along the x axis. This poblem quikly edues to a fom simila to hapte in Giffiths. Notie that fo the linea hage density the q in the expession fo the eleti field may be taken in infinitesimal piees, dq λ dx. Fom the geomety of the poblem we know that the x omponent of the eleti field will be zeo. Choosing any loation along the wie, the symmety will ause the x omponents of the eleti field due to that pat of the wie to the left to anel those x omponents due to that pat on the ight. This geomety is simila to figue.9 in Giffiths, but with the distane fom the line to the obsevation point set at a length d. The y omponent will be all that emains of the eleti field and this is eoded though ˆR sin θ ŷ. Futhemoe, sin θ d/r so we an make the eplaement /R sin θ/d. The integal is now, E, t λ v / sin θ sin θ dx d v sin θ/ 3/ ŷ 68 It appeas as though onveting the integal in tems of the angle θ is the best method. This is one again aom-

5 5 plished fom geometial onens, tan θ d x 69 x d tan θ d ot θ 7 Making these substitutions leads to, E, t λ v / d dz v z ŷ 76 3/ dx d s θ dθ 7 The integand simplifies, d sin dθ 7 θ sin 3 θ dx d v sin θ/ ŷ sin θ dθ 3/ d v sin θ/ ŷ 3/ The integal beomes, E, t λ v / π sin θ dθ d v sin θ/ 3/ ŷ whee the limits follow fom the natue of the infinite wie. When the hage density is all the way to the left, the angle is zeo. As the hage density gets all the way to the ight this angle goes to π. This integal may be solved with the following substitution, Let z os θ dz sin θ dθ sin θ z 75 whee the limits have hanged aoding to the substitution. This poblem is being solved in Catesian oodinates though it lealy has ylindial symmety. We use this to ou advantage by laiming that the veto natue of ou solution is simply known fom geomety. I keep the ŷ dietion hee beause that will failitate the oss podut neessay fo finding the magneti field. Sine the linea hage density is unifom it is known that the eleti field lines will be dieted along the adial veto. E, t λ v / d Hee is an integal solution we an use, dz v + v z ŷ 3/ 77 dx x a + x 3/ a a + x 78 Using ou integal solution in the eleti field expession gives, E, t λ v / d v z v + v z / ŷ 79 λ z d v + v z / ŷ 8 λ d v / + v v / ŷ 8 + v λ d ŷ λ πɛ o d ŷ 8

6 6 This solution is time-independent. It also mathes the solution we would have found fo a unifomly haged wie in the eletostati ase hene the laim that this poblem follows those of hapte in the text. b The magneti field is found using this solution fo the eleti field. Sine the veloity is also a onstant we have, B vˆx λ πɛ o d ŷ λv πɛ o d ẑ 83 This an be witten in a moe familia fom noting /µ o ɛ o. B µ oλv πd ẑ 84 whih is again the ommon stati esult. The magneti field still foms loops aound the wie. The z dietion is not unique in this ase and as we otate aound the wie the dietion of the magneti field will otate aodingly. [4.] Poblem.6 fom Giffiths A patile of hage q ests at the oigin. A patile of hage q is moving along the z axis with a onstant veloity v. a Solve fo the foe on q due to q at time t. This is given by F t when the seond patile is at z vt. b Solve fo F t and omment on whethe Newton s thid law equal and opposite foes holds in this poblem. Solve fo the linea momentum pt in the EM fields. Giffiths Hint: Ignoe onstant tems beause they ae not elevant to the following pat of this poblem. You answe should be, pt µ oq q 4πt ẑ 85 d Show that the sum of the foes found in the pevious pats of this poblem is equal to the negative time deivative of the momentum we have also found. Povide a physial intepetation of this esult. In othe wods, show that, F + F d pt 86 dt and disuss the physial amifiations of this expession. Solution a In this pat, we want to detemine the foe on the moving patile due to the patile that is fixed at the oigin. The fixed patile establishes an eleti field though whih patile moves. The foe is given by, F t q E 87 whee E must be evaluated at the pesent position of patile. This is solved as, q F q z ẑ 88 q q v t ẑ 89 Thee is no teatment of etaded potentials o etaded times in this solution beause we assume that the eleti field of patile has aleady been established thoughout all of spae. b Now we onside the evese situation of pat a. We an follow the same poedue as in that pat. Detemine the eleti and magneti fields geneated by patile and then use F q E + v B whee v v. A lot of time is immediately saved by notiing that the expession fo the magneti field of patile has been given in Giffiths example.4 as Eq..69, B v E 9 whih gives the magneti field due to patile. Notie that the magneti field of a moving haged patile is eveywhee pependiula to its eleti field. At the oigin whee patile ests, the eleti field due to patile is dieted along the z-axis. The veloity of patile whih ould also be onsideed the negative veloity of patile in the efeene fame of patile is also dieted along the z-axis. The tem v E is zeo in this ase. The foe is alulated using the eleti field of patile as given by Eq..68 still fom example.4, E, t q v / ˆR 4πɛ o v sin θ 3/ R 9 whee θ is the angle between R the veto dieted fom the field soue s pesent loation to the obsevation point and v the veloity of patile. In this poblem θ 8 o beause v and R ae antipaallel. The eleti field of patile is, E q v / 4πɛ o v ẑ 3/ vt 9 q v / v t ẑ 93

7 7 The foe may now be alulated, F q q v / v t ẑ 94 and in this ase we see that Newton s thid law does not hold beause the foes found in pats a and b ae not equal and opposite. The linea momentum stoed in the EM fields is given by Eq. 8.9, p em µ o ɛ o S dτ 95 whee this is an integal ove all spae. In this poblem, the momentum is due to the total fields. Patile s eleti field adds to that of patile. We have, we an ewite the spheial angle as, sin θ a 99 p em µ o ɛ o µ o E + E B dτ 96 sin θ a R ɛ o E B + E B dτ 97 The E B is onstant in time. An explanation of this popety is as follows: Patile is not aeleating, so this tem of the Poynting veto epesents the enegy flux density that it aies along with it. Sine none of the enegy is lost though adiation, the amount of enegy flux tagging along with patile is onstant in time. This enegy flux density has some value that is non-zeo, but this value is a onstant and Giffiths has asked us to ignoe it in this alulation. The time dependent momentum is given by, p em ɛ o E B dτ 98 This expession may at fist lead us to believe that the est of the alulation will be simple. Thee is going to be a lot of algeba and integation, howeve, beause B must be detemined using the expession fo E as in 9. The integation is taken ove all spae, so we begin with 9. The angle θ in this expession is that between the v and R vetos. In ode to pefom the integation we need to ewite this angle in tems of the θ, the spheial oodinate angle with the pime added in ode to keep tak of whih angle we ae onsideing. Looking at the image below, sin θ R whee θ is one of the spheial oodinates we will integate ove. The eleti field of patile is, E q q q v v sin θ R v v sin θ R 3/ v v sin θ R 3/ 3/ ˆR R t v R R 3 t v R 3 4 whee this uses R t v and ˆR t v/r. The denominato an be simplified, v sin θ R The eleti field is, 3/ R 3 R v sin θ 3/ 5 E q v / 4πɛ o R v / sin θ t v 6 3/

8 8 We ae finally in a position to detemine the magneti field, B q v / R v / sin θ v t v 3/ 7 whee the angle is equivalent to the θ in the field expession. B µ oq v sin θ 4π v / R v / sin θ 3/ ˆφ 3 q v / R v / sin θ 3/ v sin θ ˆφ 8 whih uses v t v and inopoates the spheial oodinate vesion of the veloity ẑ os θ ˆ sin θ ˆθ, v v ẑ ˆ 9 vos θ ˆ sin θ ˆθ ˆ v sin θ ˆφ v sin θ ˆφ The eleti field due to the patile loated at the oigin is something we should all be able to immediately wite down fom memoy, E q ˆ 4 We may now etun to the alulation of the time dependent potion of the momentum. Beginning with the next few lines I will emove the pime notation fom the angula oodinates. The following integal is taken ove all spae in spheial oodinates. The diffeential volume element in spheial oodinates is dτ sin θ d dθ dφ. p em ɛ o E B dτ 5 ɛ o q ˆ µ oq v sin θ v / 4π R v / sin θ ˆφ dτ 6 3/ µ oq q v 6π µ oq q v 6π v v sin θ R v / sin θ ˆ ˆφ sin θ d dθ dφ 7 3/ sin θ R v / sin θ 3/ ˆθ d dθ dφ 8 Integations ove the unit vetos of uvilinea oodinates ae not simple if even possible, so we must onvet this ˆθ veto into Catesian oodinates. Apply the definition ˆθ os θ os φ ˆx + os θ sin φ ŷ sin θ ẑ. Doing so immediately demonstates that the x and y omponents of the momentum ae zeo. Sine the integation is taken ove all spae, this equies integation ove the ange φ π. Suh an integation taken ove a sine o osine funtion is zeo. The ˆθ dependene is emoved and leaves us at inluding a fato of π fom the φ integation, p em µ oq q v 6π π v π sin θ R v / sin θ sin θ ẑ d dθ 9 3/ µ oq q v 8π v π sin 3 θ R v / sin θ ẑ d dθ 3/

9 To ontinue onwad we need to deide whih integation to ay out next. Thee is no obvious eason to hoose one ove the othe at this juntion. The R tem has dependene within it, so we should expand this out befoe ontinuing. R R R t v t v vt os θ + v t whee this θ is that of the spheial oodinate system beause it involves. Now the momentum expession an be ewitten as, p em µ oq q v π v sin 3 θ 8π vt os θ + v t v / sin θ ẑ d dθ 3/ 9 µ oq q v 8π v π sin 3 θ { v sin θ vt os θ + v t ẑ d dθ 3 3/ and now the following esult fom an integal table enouages us to integate in the oodinate next, x dx bx + a a + bx + x 3/ 4a b a + bx + x 4 b + a 4a b a + b + + b + a 4a b 5 a + b + b 4a b + 4a 4a b a 6 b 4a b + 4 a 4a b 4a b b + 4 a 7 a b a b a + b a + b 8 This is applied to 3 with a v t, b vt os θ, and v / sin θ, esulting in, p em µ oq q v π v 8π sin 3 θ v / sin θ v t v / sin θ vt os θ ẑ dθ 9 µ oq q v 8π π v vt v / sin θ sin 3 θ dθ v / sin θ os θ The denominato hee should be simplified befoe we attempt the θ integation. ẑ 3 3 vt v / sin θ v / sin θ os θ vt v / sin θ v / sin θ os θ v / sin θ + os θ v / sin θ + os θ 3

10 v / sin θ + os θ vt v / sin θ v / sin θ os θ v / sin θ + os θ vt v / sin θ v / sin θ sin θ 33 v / sin θ + os θ vt v / sin θ v / sin θ + sin θ 34 v / sin θ + os θ vt sin θ v / sin θ v / 35 vt sin θ v / vt sin + θ v / v / sin θ + os θ v / sin θ 36 os θ 37 v / sin θ We may finally etun to the deivation of the momentum, p em µ oq q v 8π µ oq q v 8π µ oq q 8πt π v sin 3 θ vt sin θ v / + os θ ẑ dθ 38 v / sin θ v π sin θ os θ sin vt v / θ + ẑ dθ 39 v / sin θ π sin θ + sin θ os θ ẑ dθ 4 v / sin θ We have two moe integals to solve. π sin θ dθ os θ π 4 The othe integal is moe involved, π 4 sin θ os θ v / sin θ 43 Begin with the substitution u os θ, leading to du sin θ dθ. It beomes neessay to apply sin θ os θ as well, π sin θ os θ v / sin θ θπ θ du dθ u dθ v os θ 44

11 u du 45 v + v u u du 46 v v + u Moving on the momentum tem, d dt pt d µ o q q ẑ 58 dt 4πt µo q q 4π t ẑ 59 v Applying the integal tables, leads to, v v v x dx a + x u du 47 v + u a + x u du + u v v + u v + v Inopoating these integal esults into 4, p em µ oq q + ẑ 5 8πt µ oq q 4πt ẑ 53 whee this esult is exatly what Giffiths told us to expet. d We ae asked to veify, Fo the foes, F + F d pt 54 dt F + F q q v t ẑ + q q v / v t ẑ 55 q q v t + v ẑ 56 q q t ẑ 57 Equating 57 and 6, µ oq q 4πt ẑ 6 q q t ẑ µ oq q 4πt ẑ 6 ɛ o µ o 6 ɛ o µ o 63 whee this expession is tue so we have veified the equality. In this ase we have the equivalent of an inelasti ollision. The appaent net foe of this system is ontained in the momentum hange of the EM fields. [5.] Poblem.3 fom Giffiths Calulate the adiation esistane of the wie onneting the hages that fom the dipole of setion.. in Giffiths. Radiation esistane is that value of esistane in the wie esulting in an amount of enegy dissipation as heat equal to the atual enegy output of the system due to adiation. Show, d R 79 Ω 64 λ whee λ is the adiation wavelength. Fo wies of length d 5 m e.g. in a adio, is it neessay to woy about the ontibution of adiation to thei esistane? Solution This poblem is solved by ompaing diffeent models fo enegy dissipation. The powe of a iuit is given by Eq. 7.7, P I R 65 whee I have hosen the fom that inludes a esistane fato beause that is what we seek in this poblem.

12 Solving fo the equivalent uent, I dq dt 66 d dt q o osωt Eq q o ω sinωt 68 This is an osillating paamete so it makes sense to solve fo the time-aveaged powe dissipation in the system. P I R 69 q oω sin ωtr 7 q oω R 7 whee the time aveaging does not affet the onstants and sin ωt /. The ω tem in 76 an now be eplaed, R µ od 6π µ oπ 3 4π λ 79 d 8 λ It emains to show that the pefato in 8 equals 79. µ o π 3 4π this is lose enough that we may laim 64 is tue. We might be tempted to hek whethe the units make sense. This an be poven quikly. In 8 the d/λ tem is unitless beause both vaiables ae in units of length and theefoe anel eah othe. Then, [Ω] [V ] [A] [J] [C][A] 83 The time-aveaged powe adiated fom the dipole is given. µ o π 3 [N] [m] [A] [s] [J] [C] [s] [s] 84 whee p o q o d. P µ op oω 4 π Eq.. 7 [J] [C] [s] [C] [J] [A][C] [Ω] 85 Equating 7 and 7, q oω R µ op oω 4 π R µ op oω 6πq o Radio waves may be teated with λ 3 m. Combining this with the given value of d.5 m gives, R µ oq od ω 6πq o µ od ω 6π Radiation is simply eletomagneti waves. We know that the solution needs to be in tems of the wavelength of these waves and only the angula fequeny tem in 76 an be ewitten in tems of wavelength. ω k k π λ ω k 4π 77 λ Ω 87 this is an insignifiant esistane and we ae justified in ignoing this ontibution when designing adios. [6.] Exat fields fo a pue dipole with time dependene. In the Loentz gauge i.e. using the etated potentials, deive the exat eleti and magneti fields assoiated with a pue point dipole at the oigin with dipole moment pt ptẑ. Using the fat that the uent density assoiated with time-hanging dipole moments is the polaization uent: J P /, whee P is the polaization dipole moment pe unit volume. Fo a point

13 3 dipole, the polaization is P ptδ 3. Using this you an show fo the speified point dipole that: A, t µ o 4π ṗt ẑ 88 Calulate B using B A in spheial oodinates. To get E, dont attempt to alulate the sala potential. Instead, use Faadays law to get E fom you expession fo B youll have to integate in time. In you esult, identify the adiation fa fields and show they yield the same answe as the deivation in letue, with pt p o osωt. Note that the only diffeene between you deivation hee and the one I did in letue is that you did not assume λ beause the objet in this poblem is a pue dipole, the fist two appoximations we made in letue ae exat hee these wee d and d λ, whee d is the size of the dipole. Solution The method fo solving this poblem is laid out fo us. Begin with the expession fo the etaded veto potential as given by Eq..9, A, t µ o J, t 4π R dτ 89 Rewite the uent density tem as the poblem statement suggests, J, t P 9 pt δ 3 ẑ 9 ṗt δ 3 ẑ 9 whee the dot ove the dipole moment veto indiates a time deivative this deivative taken with espet to the etaded potential, and the position veto is zeo beause the dipole is loated at the oigin. The veto R is dieted fom the soue to the obseve. The soue is at the oigin, so we have R this is the same as saying. Plae all this into the veto potential expession, A, t µ o 4π ṗt δ 3 dτ ẑ 93 µ o 4π ṗt ẑ 94 whee the / tem is emoved fom the integal eall that the integal is taken ove the pimed oodinates and in this ase R has no pimed oodinate dependene. We have detemined the veto potential and it mathes the given expession. This esult should be onveted into spheial oodinates using ẑ os θ ˆ sin θ ˆθ. A, t µ o 4π ṗt os θ ˆ sin θ ˆθ µ { o os θ ṗt ˆ sin θ ṗt ˆθ 4π Skipping to the non-zeo omponents of the ul, A µ { o 4π A θ A θ Continuing on, ˆφ 97 B A { µ o sin θ ṗt 4π θ os θ ṗt ˆφ 98 { µ o sin θ 4π ṗt ṗt sin θ ˆφ 99 The spatial deivative of the dipole moment tem is, ṗt ṗt ṗt pt t

14 4 whee one again note that R. The double dot indiates a seond time deivative with espet to the etaded time. Continuing on with the alulation fo the magneti field. and 9 mathes Eq..9 the fa field expession so we an be onfident that we ae oet with ou magneti field solution. B { µ o sin θ 4π ṗt ṗt sin θ ˆφ 3 µ { o 4π sin θ pt sin θ ṗt ˆφ 4 The poblem statement suggests using Faaday s Law to solve fo the eleti field. This is an eo and we must atually use Ampee s Law to obtain the eleti field. Reall that Ampee s Law is, This is the exat expession fo the magneti field. We ould begin ou solution fo the eleti field, but it is wise to veify the answe we have so fa. Anothe pat of the poblem is to identify the adiation fields and show that they ae the same as those deived in lass. Let s do this fo the magneti field expession in 4 now, so that we an be sue it is oet befoe using it to find the eleti field. B µ o J + µo ɛ o E The adiation fields i.e. the fa fields ae given by the egion satisfying λ. As the name fa field implies, this is physially the egion of lage that is fa away fom the soue. All of the tems in the magneti field expession vay as / n whee n,. As we go to vey lage values of, both of the tems get vey small. The / tem deeases muh faste than the othe, howeve, so to fist ode only the / tem mattes in the adiation field. Keeping this tem and inseting the given pt p o osωt noting that this dipole moment must be evaluated at the etaded time, B µ o 4π { sin θ pt ˆφ 5 µ { o sin θ 4π p o osωt ˆφ 6 µ { o sin θ po ω osωt ˆφ 7 4π µ op o ω 4π µ op o ω 4π sin θ sin θ osωt ˆφ 8 os ω t ˆφ 9 The uent density tem is given in 9 and is zeo fo all points exept the oigin. Set this tem equal to zeo, but be awae that ou solution is diffeent at the oigin this is aeptable beause we ae only inteested in the adiation fields and the pefet dipole is unphysial so we ae not onened with the fields at the oigin. Ou method fom this point is to take the ul of the magneti field and then integate that expession in tems of the etaded time. We integate with espet to the etaded time beause that is how the exat answe is detemined. Any magneti field ating as a soue fo a time dependent eleti field is subjet to the etaded time teatment beause of the sepaation distane between this soue and eleti field being alulated. Keeping only the non-zeo tems of the ul of the magneti field gives,

15 5 B { sin θ θ sin θb φ ˆ + { B φ ˆθ µ { o sin θ 4π sin θ θ sin θ pt + sin θ ṗt µ { o sin θ pt + sin θ 4π sin θ θ ṗt ˆ ˆ + µ { o pt + ṗt 4π sin θ sin θ os θ ˆ sin θ µ { o pt os θ 4π + ṗt 3 µ { o os θ pt 4π + ṗt ˆ... sin θ pt sin θ pt + sin θ ṗt sin θ pt + sin θ pt + sin θ ṗt ṗt ˆθ + sin θ ṗt pt... ṗt pt sin θ ˆ + + sin θ + pt ˆθ ˆθ ˆθ ˆθ whee the spatial deivatives of the dipole moment tems have been solved as, and in simila fashion, pt pt... pt t... pt ṗt ṗt ṗt + ṗt + ṗt pt ṗt Inset the ul of the magneti field in 6 into Ampee s Law ealling that the uent density tem is zeo eveywhee exept at the oigin and we ae ignoing the oigin in this alulation,

16 6 µ o ṗt δ 3 E ẑ + µ o ɛ o µ { o os θ pt 4π + ṗt E { os θ pt + ṗt... ṗt pt sin θ ˆ + + sin θ + pt... ṗt pt sin θ ˆ + + sin θ + pt ˆθ ˆθ 3 4 E { os θ pt + ṗt... ṗt pt sin θ ˆ + + sin θ + pt ˆθ dt 5 { ṗt os θ + pt pt sin θ ˆ + + sin θ pt + ṗt ˆθ 6 Notie that the integation is taken with espet to the etaded time one again, beause we ae looking fo the exat solution. This integation amounts to emoving one of the time deivatives fom eah of the dipole tems beause these ae the only t dependenies in the expession. whee 3 mathes Eq..8, the fa field appoximation we alulated in letue. We see that 6, the exat solution fo the eleti due to this osillating dipole, is fa moe ompliated that the appoximate solutions deived in letue and Giffiths. Fom this expession, now only onside the one tem that has / dependene to solve fo the fa field appoximation, E pt sin θ sin θ p o osωt ˆθ 7 ˆθ 8 p oω sin θ os ω t ˆθ 9 µ op o ω sin θ 4π os ω t ˆθ 3