Question 1: The magnetic case

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1 September 6, 018 Corell Uiversity, Departmet of Physics PHYS 337, Advace E&M, HW # 4, due: 9/19/018, 11:15 AM Questio 1: The magetic case I class, we skipped over some details, so here you are asked to show them. 1. Show that choosig A = 0 is equivalet to choosig ψ. Do it by fidig the explicit form of ψ that with A = A + ψ makes A = 0. Assume that A has a localized distributio. The solutio ca be expressed as a uevaluated itegral. Use electrostatics as a aalogy to costruct this itegral. Aswer: What we like to fid is A = A + ψ such that A = 0. Applyig o both sides we obtai The equatio above reduces to Poisso s equatio 0 = A = ψ + A 1 ψ = A where we ca thik of A as a charge desity. We kow the solutio to Poisso s equatio so by aalogy the solutio here is ψ r = A r d 3 r 3 r r so we have the explicit form for ψ.. Show that for a localized curret desity, J = 0 implies J r d 3 r = 0. Do t cofuse with divergece theorem, which says J r d S = 0. Aswer: The textbook gave a somewhat physical proof from Eq..46 to.49. Let us give two other proofs. First ote that sice J = 0, this meas J = K for some vector field K. Moreoever, sice J = 0 far away say at r > L we ca choose Kr > L = 0. We the have J r d 3 r = K r d 3 r 4 Let s look at the x-compoet. We do the itergral i oe of the three dimetio up to a far away poit L. We get [ Kz K ] y d 3 r y z = dx dz [K z x, L, z K z x, L, z ] 1 dx dy [K y x, y, L K y x, y, L], 5

2 ad sice each term vaishes the itegral vaishes. The same is true for the y ad z compoet. Oe other proof is as follows. We defie v = jd 3 r 6 We pick a arbitrary directio, ad show that the compoet of v i this arbitrary directio is 0. Ad sice the directio is chose arbitrarily, the compoet i ay directio is 0, ad hece v must be 0. Without loss of geerality set the z axis to be the chose directio. We cosider a plae parallel to the x y plae iside the regio, ad close it far away that is, we cosider a closed surface where oe face is this plae, ad all other faces are far away, outside of the support of j. Give that it is a closed surface, the Gauss theorm implies that the itegral perpedicular to it vaishes. Give that j = 0 outside of the bouded support, that implies that the itegral must vaish o the plae, sice it vaishes o the rest of the surface. We coclude that the itegral of j vaishes i the z directio, ad thus i ay directio. 3. Cosider J = 1, 0, 0 T ad show that J = 0 ad J r d 3 r 0. Explai how this result is ot i cotradictio to the geeral proof you got i the previous item that stated that J = 0 implies J r d 3 r = 0. Aswer: The curret is ot localized. Questio : More of the square tuel y Φx,L=fx 0,L Φ-L,y=0 -L,0 0,0 L,0 ΦL,y=0 x 0,-L Φx,-L=fX We still did ot cover all that is eeded to solve this questio, but we will do it o moday. Yet, I put the questio up. If, however, because of it you do eed more time, you ca submit the homwork o friday.

3 Cosider a cofiguratio similar to the oe we did i class, see the figure above. Similarly to what we have show i class, the geeral solutio ca be writte as follows: Φx, y = Φ x, y, 7 where Φ x, y = a e βy + b e βy cos β x + c e αy + d e αy si α x, 8 with a positive iteger ad β = 1π, α = π L Cosider a case where fx is eve. Show the that c = d = 0. Aswer: To satisfy the boudary coditio alog the lie y = L, we eed Φx, L = fx 10 Sice fx is eve, this meas that we have Φx, L = Φ x, L, or [ a e βl + b e βl cos β x + c e αl + d e αl si α x ] [ a e βl + b e βl cos β x c e αl + d e αl si α x ] 11 = Matchig coefficiets for the differet siusoidal fuctios, we fid a e βl + b e βl = a e βl + b e βl c e αl + d e αl = c e αl + d e αl 1 a ad b are ucostraied, while we eed c = d = 0.. What ca you say about a, b, c ad d i the case that fx is odd? Aswer: Sice fx is odd, this meas that we have Φx, L = Φ x, L, or [ a e βl + b e βl cos β x + c e αl + d e αl si α x ] [ a e βl + b e βl cos β x c e αl + d e αl si α x ] 13 = Matchig coefficiets for the differet siusoidal fuctios, we fid a e βl + b e βl = a e βl + b e βl 14 c e αl + d e αl = c e αl + d e αl Now c ad d are ucostraied, while we eed a = b = 0. 3

4 3. Cosider the case where fx = Φ b 1 cos such that Φ b 1 ad Φ b are give. coeffiets, write them dow, ad fid their values. Aswer: + Φ b si, 15 L Explai why there is a fiite umber of o-zero To satisfy the boudary coditio alog the lie y = L, we eed [ a e βl + b e βl cos β x + c e αl + d e αl si α x ] = Φ b 1 cos 16 + Φ b si L Matchig coefficiets for the differet siusoidal fuctios, we fid a 1 e β 1L + b 1 e β 1L = Φ b 1, c 1 e α 1L + d 1 e α 1L = Φ b 17 To satisfy the boudary coditio alog the lie y = L, we eed [ a e βl + b e βl cos β x + c e αl + d e αl si α x ] = Φ b 1 cos 18 + Φ b si L Matchig coefficiets for the differet siusoidal fuctios, we fid a 1 e β 1L + b 1 e β 1L = Φ b 1, c 1 e α 1L + d 1 e α 1L = Φ b 19 Solvig the simultaeous equatios for a 1, b 1, c ad d, we fid that a 1 = b 1 = c = d = All other a, b c ad d are zero. Φ b 1 e β 1L + e β 1L = Φb 1 cosh π Φ b e α 1L + e αql = Φb cosh π 0 4. Next fid Φx, y for the followig case fx = Φ 0 x L. 1 Use the fact that fx is eve to simplify both the Fourier series ad the matchig of Φx, y to the boudary coditios. You ca use the followig itegral 1 1π x 1 cos x dx = π

5 Aswer: As discussed i sectio, we ca decompose a fuctio fx defied over D/ x D/ satisfyig f D/ = fd/ = 0 as fx = A = D B = D 1π A cos D D/ D/ D/ D/ π x + B si D x 1π fx cos x dx D π fx si D x dx As a remider, the trick is to first traslate the fuctio so that the iterval becomes 0 x D, repeat the fuctio i the appropriate maer so that we get a odd periodic fuctio, the decompose the fuctio as a Fourier sie series. Traslate the sie fuctios back to the iterval D/ x D/ to get the above mixture of sies ad cosies. Back to the questio. I our case, D =, ad the fuctio is eve, so we oly expect the cosie Fourier compoets to survive, i.e. B = 0 for all, while A = L 1π Φ 0 x L cos L 1 1π = Φ 0 L z 1 cos 1 = Φ 0 L π 3 x dx z dz To summarise, we foud that fx ca be decomposed ito fx = chage variables x = Lz 3 4 A cos β x 5 where A is give by the above expressio. Now we wat to fid Φx, y. First we ote that fx is eve, so c = d = 0 for all. To satisfy the boudary coditio alog the lie y = L, we eed a e βl + b e βl cos β x = Matchig coefficiets for the differet cosie fuctios, we fid A cos β x 6 a e βl + b e βl = A 7 5

6 Doig the same alog the lie y = L, we fid Solvig the simultaeous equatios for a ad b, we get a e βl + b e βl = A 8 a = b = A e βl + e βl = 1 16Φ 0 L 1 3 π 3 cosh 1π 9 5. We ow cosider the case where istead of Φx, L ad Φx, L the perpedicular electric field is give. I particular we cosider the case where E y x, L = E 0 cos, E y x, L = E 0 cos, 30 Fid Φx, y. Aswer: This is the case of a Neuma BC, sice we are basically give y Φx, y y=±l. The trick is as follows. First, we fid the geeral expressio for y Φ: y Φ x, y = a β e βy b β e βy cos β x + c α e αy d α e αy si α x. 31 To satisfy the boudary coditio alog the lie y = L, we eed a β e βl b β e βl cos β x + c α e αl d α e αl si α x = E 0 cos 3 Sice β 1 = π, by orthogoality of the trigoometric fuctios, we require that a1 β 1 e β1l b 1 β 1 e β 1L = E 0 a β e βl b β e βl = 0 for 1 c α e αl d α e αl = 0 33 To satisfy the boudary coditio alog the lie y = L, we eed a β e βl b β e βl cos β x + c α e αl d α e αl si α x = E 0 cos 34 So we also require that Solvig the simultaeous equatios for the coefficiets, we fid that all coefficiets are zero, except for a 1 = b 1 = E 0 β 1 e β1l e β 1L 1 = E 0L π 6 1 sihπ/ 35 36

7 So the full solutio is Φx, y = E 0L π 1 sihπ/ e πy + e πy cos = E 0L π cosh πy πx sihπ/ cos 37 7