PHYS B - HW #6 Spring 4, Solution by David Pace Any referenced equation are from Griffith Problem tatement are paraphraed. Problem. from Griffith Show that the following, A µo ɛ o A V + A ρ ɛ o Eq..4 A V + µ o ɛ o µ oj Eq..5 can be written a, V + L ρ ɛ o 3 where, Solution µ o ɛ o A L µo J 4 L A + µ o ɛ o V It will uffice to write out the new form of thee equation and then verify that they are equivalent to the original form. For, V µ o ɛ o V V µ o ɛ o V V + L + A V + µ o ɛ o 5 ρ ɛ o 6 7 + A + µ o ɛ o V 8 V + A ρ ɛ o 9 For, A L µo J A A µo ɛ o A µo ɛ o A A V + µ o ɛ o A V + µ o ɛ o µ o J
where in the lat line I added parenthee to make thi expreion match the original.. Problem.4 from Griffith Let V and A A o inkx ωt ŷ. Solve for the field in thi cae and how that atify Maxwell vacuum equation. State the relation between ω and k that i required. Solution The field are given by, E V A Eq..3 3 B A Eq.. 4 Since the potential i zero it gradient i alo zero. We need only be concerned with the vector potential in thi problem. Solving for the electric field, E A 5 A o inkx ωt ŷ 6 A o cokx ωt ω ŷ 7 A o ω cokx ωt ŷ 8 Determining the magnetic field i implified by noting only one term in the curl expreion i non-zero. B A y x ẑ 9 x A o inkx ωt ẑ A o cokx ωtk ẑ A o k cokx ωt ẑ At thi point it i worth noting that the electric and magnetic field are perpendicular to each other. The next part of the problem i to how that thee field atify the vacuum Maxwell equation, which would require that they fit the form of electromagnetic wave. Such wave alo feature perpendicular field, o we move on to the next part of the problem with confidence that o far we have the correct anwer.
Maxwell equation in vacuum: E B Thee are given in Griffith a equation 9.4. E B B µo ɛ o E Verify them one at a time. Since the electric field, 8, ha only a y component it divergence i, 3 E E y y 4 y A oω cokx ωt 5 The divergence of the magnetic field follow the ame pattern, 6 B B z z 7 z A ok cokx ωt 8 9 The following write out the curl of the electric field and the negative time derivative of the magnetic field imultaneouly, E B 3 E y x ẑ A ok cokx ωt ẑ 3 x A oω cokx ωt ẑ A o k inkx ωt ω ẑ 3 A o ω inkx ωtk ẑ A o kω inkx ωt ẑ 33 A o kω inkx ωt ẑ A o kω inkx ωt ẑ 34 Verifying the final equation will allow u to how a condition placed between ω and k. 3
Recall that µ o ɛ o /c. B µ o ɛ o E 35 B z x ŷ c A oω cokx ωt ŷ 36 x A ok cokx ωt A oω inkx ωt ω ŷ 37 c A o k inkx ωtk ŷ A oω A o k inkx ωt ŷ A oω c inkx ωt ŷ 38 c inkx ωt ŷ 39 where there i no check mark becaue thi i only atified if k ω /c. We do know, however, that electromagnetic wave in vacuum travel at the peed of light and that thi velocity i related to the angular frequency and wave vector a ω/k c. Thi i the condition impoed on ω and k. 3. Problem.9, Part a Only, from Griffith a Let the current in the wire of example. be It kt for t >. The variable k i a poitive contant i.e. thi current i increaing in time. Find the field generated. Solution The field are found uing 3 and 4. A in example., we hould aume that the wire tay charge neutral and thi lead to V. Before finding the field we mut olve for the vector potential. We ue, A r, t µ o J r, t r R dτ Eq..9 4 Thi i a one-dimenional problem o the integral reduce to; taken directly from the example, A, t µ o It r dz ẑ 4 R noting that we are now in term of retarded time, t r t R/c. Still following the example, at any time t, only the current through that portion of the wire that i within the ditance traveled by light electromagnetic information matter. Any current that i further away would not yet have been felt by the oberver at the given point at the given time. Following the geometry of figure.4 in Griffith the integral become, A, t µ o ct ct kt r dz ẑ 4 + z 4
The contribution are the ame for the region of negative and poitive z, o the limit are implified and a factor of pulled out. A, t µ ct o Continue with the olution for the vector potential, A µ ct o π µ ct o π µ ok π µ ok π µ ok π ct kt r dz ẑ 43 + z kt r dz ẑ 44 + z k t R c dz ẑ + z 45 R t + z c + z dz ẑ 46 ct t + z + z c dz ẑ 47 + z ct t + z dz ẑ 48 c The econd integral i eay, the firt one i given by, dx ln a + x + x 49 a + x where the extra arbitrary contant i neglected. A µ ok π µ ok π µ ok π t ln z + + z ct ct c t ln ct + ct + ln ct c ct + ct ct t ln c ẑ 5 ẑ 5 ẑ 5 The electric field i the negative time derivative of A. Taking thi derivative i tediou becaue of the non-trivial time dependence, but it i not difficult. E A µ o k π ct t ln + ct ct c 53 ẑ 54 5
µ ok ct ln + ct c t + t π ct + ct ct + c c t ẑ 55 c ct µ ok π ct ln + ct ct + ct + ct ct + ct ct ct ct ẑ 56 µ ok π µ ok π ln ct ln + ct ct + ct ct ẑ 57 ct ct + ct ẑ 58 where thi i valid for time when the electromagnetic effect ha traveled to the location in quetion. For time given by t < /c the electric field i zero. We have to wait until the electromagnetic information reache u before we can detect it. Solve for the magnetic field by keeping the only part of the cylindrical curl that i not zero, µ ok π B A z ˆφ 59 µ ok ct t ln + ct ct ˆφ 6 π c t ct + ct c ct + ct + ct ct ˆφ 6 µ ok π t ct + ct ct ct + ct + c ct Thi i the olution, although further algebra can produce, ˆφ 6 B µ o k ct πc ˆφ 63 6
4. Problem. from Griffith Let the current denity be contant in time. Problem 7.55 then how that ρ r, t ρ r, + ρ r, t. Show that in thi cae Coulomb law hold with the charge denity evaluated at the non-retarded time, i.e. E r, t ρ r, t ˆR dτ R 64 Solution Uing retarded potential the electric field i, E r, t ρ r, t r ρ r ˆR, t r + R cr ˆR J r, t r c R dτ Eq..9 65 Since we are told to et the current denity contant we can immediately ay J. The charge denity i given in a generic form that can be applied to both normal time and the retarded time. ρ r, t r ρ r, + ρ r, t r 66 where, again, the retarded time i t r t R/c. The time derivative of the charge denity become, ρ r, t r r ρ r, + ρ r, t r ρ r, 67 Making thee replacement in 65, ρ r E r, t, + ρ r, t r ρ r ˆR, + R cr ˆR dτ 68 ρ r, ρ r ˆR, t R + c ρ r ˆR, + R R cr ˆR dτ 69 ρ r, ρ r ˆR, t + R R ρ r ˆR, R c ρ r ˆR, + R cr ˆR dτ 7 ρ r, + ρ r, t ρ r ˆR, R c ρ r ˆR, + R R cr ˆR dτ 7 ρ r, t ρ r ˆR, R cr ˆR + ρ r, cr ˆR dτ 7 ρ r, t R ˆR dτ 73 7
5. Problem. from Griffith Let the current denity change lowly enough that it may be approximated in an expanion by keeping term only through the firt derivative. Jt r Jt + t r t Jt 74 The patial dependence doe not need to be explicitly written here becaue it will not be relevant to the problem. Show that the Biot-Savart law hold with the current denity evaluated at the non-retarded time, i.e., B r, t µ o J r, t ˆR R dτ 75 Griffith Note: Thi problem illutrate that for lowly changing current the magnetotatic approximation i quite good. The error in uch an approximation include neglecting retardation and dropping the time derivative factor in equation.3 from the text, but thee mitake cancel each other out. Solution Conidering retardation effect, the magnetic field i given by, B r, t µ o J r, t r + R J r, t r cr ˆR dτ Eq..3 76 Since we are told to keep only the firt derivative for the current denity we have, Jt r Jt 77 becaue applying the derivative to the ret of the Taylor expanion would generate a econd order term. Making thee replacement in 76, B r, t µ o Jt + tr t Jt + R µ o Jt + t R t c Jt + R Jt cr ˆR dτ 78 Jt cr ˆR dτ 79 µ o Jt Jt Jt R cr + cr ˆR dτ 8 µ o J r, t ˆR R dτ 8 8
6. Problem. from Griffith Begin with a platic ring of radiu a and glue charge to it. Now the charge denity may be decribed a λ λ o inθ/. Spin thi loop about it axi with angular velocity ω. Solve for the exact calar and vector potential at the center of the ring. Griffith Hint: Solution A µ oλ o ωa 3π in ωt ωa c ˆx co ωt ωa c The charge denity change in time due to the pinning of the ring. Thi i accounted for in the decription of the charge denity by letting θ φ ωt. The charge denity i, φ ωt ρ r, t λ o in 83 Thi problem doe not allow u to neglect the calar potential becaue the ring i not charge neutral. The exact calar potential i given by, ŷ 8 V r, t ρ r, t r R dτ Eq..9 84 where expreing the charge denity in term of the retarded time require only exchanging t with t r. Thi problem feature charge ditributed along only the φ coordinate uing a cylindrical ytem, o the integral i taken in one dimenion. In the cylindrical coordinate ytem we have that the differential length along the phi direction i dl dφ, furthermore, in thi problem the radiu of the ring i fixed o R a. V r, t π π λ o in φ ωtr R Rdφ 85 φ ωtr λ o in dφ 86 Ue the following ubtitution to olve the integral; conidering thi at any fixed value of t r, θ φ ωt r dθ dφ 87 V r, t π π φ ωtr λ o in dφ 88 θ λ o in dθ 89 where the limit are unchanged becaue at any one moment of time we want to conider the entire ring. Through thee limit the value of inθ/ i alway greater than zero, o the abolute value ign are no longer an extra difficulty. 9
V r, t λ π o θ co 9 λ o πɛ o 9 We could have olved thi problem without any concern for the time dependence. The calar potential depend only on how far away the charge are from the obervation point. Since thee charge are moving a the ring pin and we want the potential they create at the center of that ring, the ditance between each charge and the obervation point i contant. We can ue the charge ditribution at time t and treat it a though thi i fixed ince even a the charge move they will alway make the ame contribution to the potential at the center. V r, t π λ o in φ a a dφ 9 λ o πɛ o 93 Granted, after the charge are initially placed on the ring it will take a time t a/c for an oberver at the center to notice the calar potential. After thi time ha paed, however, the calar potential will be oberved to be contant. The vector potential depend on the current, which in turn depend on the amount of charge flowing pat a certain point. Thi mean that where the charge are affect the olution and we mut carefully incorporate their time dependence. The expreion for the exact vector potential i given in 4. The implified form for thi ring geometry i, A µ π o I ˆφ R dφ 94 R Recall that for current in a wire I λv, where v i the velocity of the charge flow. For a pinning ring thi i given by v ωr ωa. Therefore, A µ π o λ o ωa in φ ωt a dφ a ˆφ 95 µ oλ o ωa π in θ dθ ˆφ 96 taking advantage of all the tep ued in the calculation of the calar potential. At thi point we reach the difficulty in taking integral over an angular vector. The ˆφ vector change direction through the integration and i dealt with here by witching to Carteian coordinate. Alo, becaue of the change of variable ued, φ θ + ωt r. ˆφ in φ ˆx + co φ ŷ 97 inθ + ωt r ˆx + coθ + ωt r ŷ 98
A µ oλ o ωa π θ in inθ + ωt r ˆx + coθ + ωt r ŷ dθ 99 Taking the integral in 99 one at a time, Firt Integral π θ in π co in inθ + ωt r ˆx dθ θ + ωt r θ + ωt r 3θ co + ωt r π 3θ 3 in + ωt r dθ ˆx inπ + ωt r inωt r 3 in3π + ωt r + 3 inωt r inωt r + 3 inωt r ˆx ˆx 3 ˆx 4 Thi ued the trigonometric identitie, 4 3 inωt r ˆx 5 inu inv cou v cou + v 6 cou co u 7 inu + v inu cov + cou inv 8 Second Integral π π θ in co in coθ + ωt r ŷ dθ 9 θ 3θ + ωt r + in θ + ωt r + ωt r π 3θ 3 co + ωt r dθ ŷ ŷ coπ + ωt r coωt r 3 co3π + ωt r + 3 coωt r coωt r + 3 coωt r ŷ ŷ 3 4 3 coωt r ŷ 4
The following identitie were ued above, inu cov inu + v + inu v 5 inu in u 6 Inert the integral reult into 99, cou + v cou cov inu inv 7 A µ oλ o ωa µ oλ o ωa 3π 4 3 inωt r ˆx 4 3 coωt r ŷ 8 inωt r ˆx coωt r ŷ 9 Finally, expreing the retarded time in term of the non-retarded time will put 9 in the form of Griffith given olution, 8.