PHYS 110B - HW #7 Spring 2004, Solutions by David Pace Any referenced equations are from Griffiths Problem statements are paraphrased

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1 PHYS 0B - HW #7 Sping 2004, Solutions by David Pace Any efeenced euations ae fom Giffiths Poblem statements ae paaphased. Poblem 0.3 fom Giffiths A point chage,, moves in a loop of adius a. At time t 0 this chage is at the point a, 0) on the x axis; it moves with constant angula acceleation ω. Detemine the Liénad- Wiechet potentials fo points on the z axis. Solution The geneal foms of the Liénad-Wiechet potentials fo a chage ae, V, t) A, t) c Rc R v) µ o c v 4πRc R v) E ) v V, t) E ) c2 whee my R is euivalent to the lowe case scipt of Giffiths because making lowe case scipt s appeas to be beyond my ability. The paticle is confined to move in a cicle. Such a tajectoy may be witten as, wt) acosωt) ˆx + sinωt) ŷ 3) using the initial condition given in the poblem. The location of the paticle at the etaded time is found by eplacing t with t above. The vecto fom the souce i.e. the paticle at its etaded position) to the obsevation point is, R wt ) 4) z ẑ acosωt ) ˆx + sinωt ) ŷ 5) The magnitude of this vecto, which is needed to solve fo the potential in ), is, R R R R 6) z ẑ acosωt ) ˆx + sinωt ) ŷ) z ẑ acosωt ) ˆx + sinωt ) ŷ) z 2 + a 2 cos 2 ωt ) + sin 2 ωt )) 7) z 2 + a 2 8)

2 The expessions fo the potentials ae witten in tems of the velocity at the etaded time. Since wt ) epesents the position of the paticle at the etaded time the velocity is, vt ) wt ) acosωt ) ˆx + sinωt ) ŷ 9) a sinωt )ω) ˆx + cosωt )ω) ŷ 0) aω sinωt ) ˆx + cosωt ) ŷ ) Now compute the uantity R v, R v z ẑ acosωt ) ˆx + sinωt ) ŷ) aω sinωt ) ˆx + cosωt ) ŷ) 2) a 2 ω cosωt ) sinωt ) a 2 ω cosωt ) sinωt ) 3) 0 4) The scala potential can now be solved fo using ), V, t) The vecto potential follows by using 2), c c z 2 + a 2 ) 0) 5) z2 + a 2 6) A, t) aω sinωt ) ˆx + cosωt ) ŷ V, t) 7) aω sinωt ) ˆx + cosωt ) ŷ 8) z2 + a 2 aω sinωt ) ˆx + cosωt ) ŷ 4 πɛ o z2 + a 2 9) In this poblem the paticle tajectoy is geatly simplified by being constained to the cicula loop. This makes value of R constant and theefoe the etaded time is easily found. t t R c t z2 + a 2 c 20) The final fom of the vecto potential is, aω sin ω t )) z 2 +a 2 ˆx + cos ω t )) z 2 +a 2 ŷ c c A, t) 2) 4 πɛ o z2 + a 2 2

3 2. Poblem 0.4 fom Giffiths Show that, V, t) c c2 t v) 2 + v 2 ) 2 t 2 ) E ) can be witten as, V, t) R v2 sin 2 θ E ) whee R vt is the vecto fom the pesent location of the paticle to the obsevation point do not confuse this symbol with R). Since this elates to the situation of example 0.3 page 433 in Giffiths), the velocity is a constant. The angle θ is that between the vectos R and v. Refeence figue 0.9 fo this geomety. It is also woth noting that at nonelativistic velocities whee v 2, the scala potential in 22) simplifies to the well-known electostatic value. Solution Noticing how simila 22) and 23) ae, the best method fo showing thei euivalence seems to be ewiting the suae oot tem in 22). This will euie solving fo the dependence in tems of the pesent position of the paticle since does not appea in the final fom. The velocity is constant in this poblem, so the etaded velocity witten in the oiginal solution is euivalent to the velocity at pesent time. Begin by solving fo v using the fact that R + vt as detemined fom the poblem statement. v R + vt) v 24) R v + v 2 t 25) The magnitude of is, R + vt) R + vt) 26) R 2 + 2t R v + v 2 t 2 27) 2 R 2 + 2t R v + v 2 t 2 28) whee the value of 2 is witten because that is what appeas in the elevant euation. Now the suae oot tem in 22) can be simplified. I ll just take the inside of the suae oot hee fo claity. t v) 2 + v 2 ) 2 t 2 ) t v) 2 + v 2 )R 2 + 2t R v + v 2 t 2 t 2 ) 29) 3

4 c 4 t 2 2 t v + v) 2 + v 2 )R 2 + 2t R v + v 2 t 2 t 2 ) 30) c 4 t 2 2 t R v + v 2 t) + R v + v 2 t) 2 + R 2 + 2t R v + v 2 t 2 t 2 ) v 2 R 2 + 2t R v + v 2 t 2 t 2 ) 3) c 4 t 2 2 tr v 2 v 2 t 2 + R v) 2 + 2v 2 tr v + v 4 t 2 + R tr v + v 2 t 2 c 4 t 2 v 2 R 2 2tv 2 R v v 4 t 2 + t 2 v 2 32) R v) 2 + R 2 v 2 R 2 33) R 2 v 2 cos 2 θ + R 2 v 2 R 2 34) ) v R 2 2 cos2 θ + v2 v R 2 2 c 2 sin2 θ) + v2 ) R 2 v2 c v2 2 sin2 θ + v2 ) R 2 v2 sin2 θ ) 35) 36) 37) 38) Plugging this back into 22) gives, V, t) c R 2 v2 sin 2 θ ) 39) R v2 sin 2 θ 40) 3. Poblem 0.7 fom Giffiths Deive the following esult, A by fist showing, Solution c Rc R v) 3 Rc R v) v + R a c ) + R c c2 v 2 + R a) v Rc R u E ) E ) 4

5 Fom Giffiths, wt ) ct t ) E ) Using the above as the stating point, wite R 2, R wt ) E ) R 2 R R t t ) 2 45) Take the non-etaded) time deivative of R 2, ) ) R R) R R + R R 2R R The time deivative of the ight hand side of 45) is, c2 t t ) 2 2 t t ) ) 46) 47) Euating 46) and 47), 2 R R R R 2 t t ) ) cr ) whee the last step makes use of R ct t ), which follows fom the initial euations of the solution. Witing out the left side of 49) gives, R R 48) 49) R wt )) 50) R wt ) ) 5) R wt ) ) 52) R v ) 53) Putting 53) back into 49) poduces, R v ) R v ) cr ) cr cr 54) 55) cr R v) cr 56) 5

6 cr cr R v 57) and it emains to show that the denominato above is eual to R u whee u c ˆR v E. 0.64). R u R c ˆR v) 58) cr R v 59) cr R u 60) Now fo the deivation of euation Begin with the euation fo the vecto potential as witten in 2). Taking the time deivative of this gives, A v v V V + v V + v V 6) 62) Replace the scala potential with its geneal fom, ), A v c R u) + v c R u) 63) whee the denominato in the V tem has been eplaced accoding to 59) fo simplicity. It will be easie to cay out the algeba in this fom and then eplace the R u tem in the final expession. A c v c c a cr a R u c R u) 2 R u + v R u R u + v ) R u) 2 R u) ) R u v R u) 2 cr a v R u) R u) 64) 65) 66) 67) 6

7 At this point it is wothwhile to wite out the time deivative in 67). R u) cr R v) 68) c R R v R v 69) The fist tem is, R ct t ) c ) The second tem is solved in 53), and the thid tem is solved in euations 6) though 65). The esult is, R u) ) v ) v R a ) 7) 70) + v2 R a + v 2 R ) a + v 2 R ) cr a R u ) 72) 73) 74) Plug the esult of 74) back into 67). A c R u) 2 cr a v + v 2 R ) cr a R u )) 75) ) cr R c R R u) 2 R u u) a v u cr c2 + v 2 R a) ) 76) R R u) a R u) c v R v2 v + v + R a) v 77) R u) 3 R R u) 3 R R u) 3 c R c R u) 3 R u) a c v R ) + c2 v 2 + R a) v ) R u) R a c v) + R c c2 v 2 + R a) v R u) R a c v) + R c c2 v 2 + R a) v 78) 79) 80) Finally, eplacing the R u tems and eaanging gives, A c Rc R v) 3 which matches 4) as intended. Rc R v) v + R a c ) + R c c2 v 2 + R a) v 8) 7

8 4. Poblem 0.9 fom Giffiths a) Use E, t) v 2 / ˆR v2 sin 2 θ/) E ) 3/2 R 2 to solve fo the electic field a distance d away fom an infinite wie that is caying a unifom line chage λ that moves down this wie with constant speed v. b) Use B c v E) E ) 2 to solve fo the magnetic field due to this wie. Solution Let the wie lie along the x axis. This poblem uickly educes to a fom simila to chapte 2 in Giffiths. Notice that fo the linea chage density the in the expession fo the electic field may be taken in infinitesimal pieces, d λ dx. The electic field is, E, t) v2 / λ dx ˆR v2 sin 2 θ/) 84) 3/2 R 2 whee the λ is also a constant and may be taken outside the integal. It emains to set the limits of this integal and pefom the mathematical steps, all of which have been employed peviously. Fom the geomety of the poblem we know that the x component of the electic field will be zeo. Choosing any location along the wie, the symmety will cause the x components of the electic field due to that pat of the wie to the left to cancel those x components due to that pat on the ight. This geomety is simila to figue 0.9 in Giffiths, but with the distance fom the line to the obsevation point set at a length d. The y component will be all that emains of the electic field and this is ecoded though ˆR sin θ ŷ. Futhemoe, sin θ d/r so we can make the eplacement /R 2 sin 2 θ/d 2. The integal is now, E, t) λ v2 / ) dx v2 sin 2 θ/) sin θ 3/2 θŷ)sin2 85) d 2 It appeas as though conveting the integal in tems of the angle θ is the best method. This is once again accomplished fom geometical concens, tan θ d x 86) x d tan θ d cot θ 87) dx d cs θ) dθ 88) d sin 2 θ dθ 89) 8

9 The integal becomes, E, t) λ v2 / ) v2 sin 2 θ/) sin θ d 3/2 θŷ)sin2 d 2 sin 2 θ dθ 90) λ v2 / ) π 0 sin θ d v 2 sin 2 θ/) 3/2 dθ ŷ 9) whee the limits follow fom the natue of the infinite wie. When the chage density is all the way to the left, the angle is zeo. As the chage density gets all the way to the ight this angle goes to π. This integal may be solved with the following substitution, Making these substitutions leads to, Let z cos θ dz sin θ dθ sin 2 θ z 2 92) E, t) λ v2 / ) dz d v 2 z ) ) ŷ 93) 3/2 2 whee the limits have changed accoding to the substitution. This poblem is being solved in Catesian coodinates though it clealy has cylindical symmety. We use this to ou advantage by claiming that the vecto natue of ou solution is simply known fom geomety. I keep the ŷ diection hee because that will facilitate the coss poduct necessay fo finding the magnetic field. Since the linea chage density is unifom it is known that the electic field lines will be diected along the adial vecto. Hee is an integal solution, This bings the solution to, E, t) λ v2 / ) d E, t) λ v2 / ) d λ d λ d v 2 dz dx x a + cx 2 ) 3/2 a a + cx 2 v 2 v2 v 2 z v 2 + v2 z 2) /2 + v2 ) /2 + v2 z 2) 3/2 ŷ 94) z + v2 z 2) /2 95) ŷ 96) ŷ 97) v 2 + v2 ) /2 ŷ 98) λ 2) ŷ 99) d λ 2πɛ o d ŷ 00) 9

10 This solution is time-independent. It also matches the solution we would have found fo a unifomly chaged wie in the electostatic case hence the claim that this poblem follows those of chapte 2 in the text). b) The magnetic field is found using this solution fo the electic field. Since the velocity is also a constant we have, B c vˆx λ 2 2πɛ o d ŷ λv 2πɛ o dc ẑ 0) 2 This can be witten in a moe familia fom noting /µ o ɛ o. which is again the common static esult. B µ oλv 2πd ẑ 02) The magnetic field still foms loops aound the wie. The z diection is not uniue in this case and as we otate aound the wie the diection of the magnetic field will otate accodingly. 5. Poblem 0.20 fom Giffiths Using the setup of poblem 0.3 find the fields at the cente. Using you solution fo this magnetic field, deive the field at the cente of a loop caying a steady cuent I. Compae this solution to that of example 5.6 in the text. Solution We solved fo the potentials in the setup of poblem 0.3 at the beginning of this poblem set. It is tempting to simply apply the established elations between the fields and these potentials to uickly get an answe fo this poblem. It is not this simple, howeve, as section in the text shows. The diffeentiation due to the etaded tems is non-tivial and the expessions we must use fo the fields ae, E, t) R R v 2 ) u + R u a) u) 3 E ) B, t) c ˆR E, t) E ) We can still use some of the esults fom poblem 0.3, and these esults ae simple now because we ae concened with the cente of the loop i.e. z 0). The plan is to detemine the electic field fist. Relevant tems needed fo 03) that can be taken fom 0.3 at the cente of the loop ae, R wt ) acosωt ) ˆx + sinωt ) ŷ Fom 5) 05) R a Fom 8) 06) 0

11 a vt ) velocity given in ) 07) aω 2 cosωt ) ˆx + sinωt ) ŷ) 08) ω 2 wt ) 09) whee we must be caeful not to confuse a the loop adius) with a the acceleation at the etaded time). We also need to solve fo u c ˆR v. The ˆR tem is, ˆR R R cosωt ) ˆx + sinωt ) ŷ) 0) Leading to, u ccosωt ) ˆx + sinωt ) ŷ) aω sinωt ) ˆx + cosωt ) ŷ ) ˆx c cosωt ) + aω sinωt )) + ŷ c sinωt ) aω cosωt )) 2) R u acosωt )ˆx+sinωt )ŷ) ˆx c cosωt )+aω sinωt ))+ŷ c sinωt ) aω cosωt ))) ac cos 2 ωt ) a 2 ω cosωt ) sinωt )) + ac sin 2 ωt ) + aω cosωt ) sinωt )) 3) accos 2 ωt ) + sin 2 ωt )) + a 2 ωcosωt ) sinωt ) cosωt ) sinωt )) 4) ac 5) Fo the tiple poduct use the following vecto identity the geneal fom is in the font cove of Giffiths), R u a) R a) u R u) a 6) The last tem to solve fo is, R a acosωt ) ˆx + sinωt ) ŷ) ω 2 wt )) 7) wt ) ω 2 wt )) 8) ω 2 w 2 9) ω 2 R 2 20)

12 Theefoe, R u a) ω 2 R 2 u ac a 2) All of the tems ae known so we etun to the geneal expession fo the electic field. E, t) a ac) 3 v 2 )ˆx c cosωt ) + aω sinωt )) +ŷ c sinωt ) aω cosωt ))) + ω 2 R 2 u ac a 22) The final esult is, E a 2 a 2 ω 2 ) cosωt ) + acω sinωt ) ) ˆx The magnetic field is found fom, + a 2 ω 2 ) sinωt ) acω cosωt ) ) ŷ 23) B c ˆR x E y ˆR y E x )ẑ 24) Algeba and substitutions... 25) a 2 c 3 acω cos 2 ωt ) acω sin 2 ωt ) ) ẑ 26) ω a ẑ 27) To convet 27) to an expession fo the field due to a loop of steady cuent we need to define I λv. In this case λ is the chage density and must be λ /2πa by definition since the chage is entiely contained within this loop. The velocity is known fom the adius of the loop and the angula velocity at which it is spinning, v aω. λ2πa) 28) λ I v 29) I aω 2πI ω 30) 3) Replacing this into the pevious solution fo the magnetic field gives, ) ω 2πI B ẑ 32) a ω I 2ɛ o a ẑ 33) 2

13 Once again employing the elation /µ o ɛ o, B µ oi 2a ẑ 34) and this is euivalent to the solution povided by euation 5.38 in Giffiths example