Phys 6321 Final Exam - Solutions May 3, 2013

Size: px

Start display at page:

Download "Phys 6321 Final Exam - Solutions May 3, 2013"

Zoe Dixon
5 years ago
Views:

1 Phys 6321 Finl Exm - Solutions My 3, 2013 You my NOT use ny book or notes other thn tht supplied with this test. You will hve 3 hours to finish. DO YOUR OWN WORK. Express your nswers clerly nd concisely so tht pproprite credit cn be ssigned for ech problem. There re 6 problems. You must do 5 for full credit. TURN IN ONLY 5 PROBLEMS - I WILL GRADE ONLY THE FIRST 5 PROBLEMS YOU SUBMIT. Full credit for ech problem is 25 points. 1) A squre loop of wire lies in the (x, y) plne (see the figure below). In the rest frme it crries current, I 0. Assume the wire hs squre cross sectionl re with negligible thickness. It is boosted with velocity, V, prllel to the ˆx direction. Find the liner chrge densities in the wire on the top nd bottom of the loop in the moving frme. y Io x V V Figure 1: The geometry of problem 1 Solution 1 In the rest frme, the top nd bottom of the loop hve vnishing liner chrge density since the positive nd negtive chrge on the wire re equl. λ t λ + λ = δq + /δl δq /δl = 0. δq + = δq = δq In the moving frme, the bove eqution equting the chrge is lso true since chrge trnsforms s sclr. However, length is contrcted in the moving frme s compred to the 1

2 rest frme by fctor of γ = 1 1 β 2 with β = V/c. Thus if the negtive chrge moves for provide the current, I, we obtin in the moving frme; For postiive chrge δq/δl mov = Qγ + /δl rest For negtive chrge δq/deltl mov = Q(γ + /γ )/δl Note tht the rtio of the γ fctors bove, (γ /γ u ) trnsforms the length in the frme moving with the current to the system rest frme before the boost, nd then trnsforms this length to the moving frme of the system fter the boost. Then we need vlues for gmm. For the rest frme to the moving frme; γ + = 1 1 β 2. For the trnsformtion of the negtive chrge to the rest frme the velocity using U 1 s the velocity due to the current in the rest frme. U = V ± U 1 ± V U/c 2 γ 1 1 U 2 /c2 Insert the vlue for U pr results in; γ = γ u γ + (1 ± ββ u ) into the bove eqution for γ nd work through the lgebr. This Therefore the liner chrge density fter subtrction of the negtive chrge density form the positive chrge density, is ; λ mov = ββ u γ + λ rest 2) A conducting cube with side lengths,, hs the upper side (z = ) held t potentil, V = V o. All other sides re held t potentil, V = 0. Find; Solution 2 1) The resulting electric field in the interior of the cube; 2) The force on the upper side (z = ) of the cube using the Mxwell stress tensor. Solve Lplce s eqution for the electric potentil using seprtion of vribles in Crtesin 2

3 V = 0 V = 0 V = Vo V = 0 V = 0 Figure 2: The geometry of problem 2 coordintes to get the electric field inside the cube. 2 V = 0 After ppliction of the boundry conditions on ll sides but the top, the solution tkes the form; V = nm A nm cos(nπx/) cos(mπy/) sinh(γz) In the bove, n, m must be odd for the potentil to vnish t x, y = ±/2. Also γ 2 = (nπ/) 2 + (mπ/) 2. Now use orthogonlity of the cosine functions to find A nm so tht V = V 0 when z =. A nm = 4 π 2 nm (V 0/sinh(γ)) The electric field is E = V /2 /2 dx /2 /2 dy cos(nπx/) cos(mπy/) E = nm A nm [(nπ)sin(nπx/) cos(mπy/) sinh(γz)]ˆx + [(mπ)cos(nπx/) sin(mπy/) sinh(γz)]ŷ + [(γπ)cos(nπx/) cos(mπy/) cosh(γz)]ẑ The field tensor is obtined from only the electric field components. By symmetry, only the force in the z direction is non-zero, F z. T zz = (1/2)ǫ 0 [E 2 z E2 x E2 y ] F z = /2 /2 dx /2 /2 dy T zz z= 3

4 3) A cylindricl wve guide is constructed of perfect conductors in coxil geometry. Although the TEM is the lowest mode, the geometry lso supports both TE nd TM modes. Find n expression for the lowest frequency of the TM mode. The inner conductor hs rdius,, nd the outer conductor rdius, b Inner rdius Outer rdius b Figure 3: The geometry of problem 3 Solution 3 The eqution for wve trveling in the z direction in the wve guide is; [ 2 + µ 0 ǫ 0 ω 2 k 2 ]E z = For the TM mode the mgnetic field in the z direction vnishes. Thus we solve for E z nd pply the boundry condition tht E z = 0 for ρ =, b using cylindricl coordintes. Seprtion of vribles gives; E z = ν A ν D o (γ ν ρ) e ikz In the bove, we choose the zeroth order cylindricl Bessel function, D 0 (γρ) to give the lowest mode nd the solution in independent of the zimuthl ngle. To mtch the boundry conditions t ρ =, b the Bessel function tkes the form; D 0 (γρ) = J 0(γρ) J 0 (γ) N 0(γρ) N 0 (γ) Here, J 0 nd N 0 re the cylindricl Bessel nd Neumnn functions, respectively. Then α ν re the zeros of D 0 (α nu ) = 0. Thus, γ nu b = α ν nd the dispersion reltion is; (α ν /b) 2 = ν 0 ǫ 0 ω 2 k 2 z From this choose the lowest zero, α 0, to get the lowest frequency. 4

5 4) Show tht the eqution of continuity (chrge conservtion) results directly from Mxwell s equtions. Solution 4 Mxwell s equtions re; E = ρ/ǫ B = 0 E = B t B = µ J + 1/c 2 E t Then consider; 5) ( B) = 0 = µ J + (1/c 2 ) E t J + ǫ t E J + ρ t A chrge flls from rest under the influence of grvity. Using n pproximtion s guided below, find the pproximte time it tkes the chrge to fll distnce, d time. 1) Write the eqution for the system energy including rdition s function of 2) Write n eqution for the energy blnce t the time when the chrge reches the distnce, d. 3) Assume the chrge flls without rdition, nd write the eqution in (2) bove using the time to rech d without rdition. Solution 5 4) Solve the eqution in (3) for the time. 5

6 The rditive power loss is given by the Lrmor eqution (non-reltivistic). The ccelertion is v = g. P = = E dt = (2/3)(q2 /c 3 ) 2 The energy loss due to the rdition is obtined by integrtion. E totl = (2/3)(q 2 /c 3 ) T 0 The energy blnce t position, d, is; 0 2 dt mgd = (1/2)mv E t In the first pproximtion, = g, T 0 = 2d/g, nd v 0 = gt 0. Substitution for T 0,, nd v 0 gives; Solving for T 0. (gt 0 ) 2 + αg 2 T 0 2gd = 0 T 0 = (1/2)[ α ± α 2 + 8d/g] Expnsion yields for smll α yields; T 0 2d/g α/2 + α2 g 4d 2dg In the bove, neglect the term in α 2, s in generl there re dditionl terms of this order which re not included in this expression. 6) Two equl chrges, ech Q/2, re plced 180 prt, nd lie in the (x, y) plne. The chrges spin with ngulr velocity, ω, bout the ẑ xis keeping their rdil distnce,, from the origin constnt. Find the power rdited in the lowest multipoles for both the electric nd mgnetic rdition fields. (Note tht you needt to write the chrge motion in terms of e iωt in order to use the expressions for the rdition source components in the notes) Solution 6 The chrge density is; 6

7 Q z ω y x Q Figure 4: The geometry of problem 6 ρ = Q/2 (δ(r )/ 2 ) δ(cos(θ)) [δ(φ φ 0 ) + δ(φ (φ + π))] Integrtion over the sphericl volume gives the totl chrge, Q, s it should. We let φ 0 = ωt below. Now to write the chrge density in form with time dependence e iωt, pply Fourier time de-composition. ρ = n ρ n cos(nωt) = Re n ρ n e inωt ρ n = (ω/2π) 2π/ω dt ρ e inωt Substitute in to the bove eqution the expression for ρ nd integrte over time. ρ n = (Qω/2π)(δ(r )/ 2 ) δ(cos(θ))e inωt [1 + ( 1) n ] Thus n must be even or 0, however 0 hs no time dependence nd the lowest possible vlue would be n = 2. Subsitute into the source term for the electric multipole. There is no mgnetiztion term, M = 0. Q m l = d 3 xr l Y m l Note tht Y 2 2 = (1/4) 15/2π sin 2 e i2φ with φ = ωt. Thus; Q 2 2 = Qω 8π sqrt15/32π d 3 xr 2 sin 2 (θ) δ(r )/ 2 δ(cos(θ))e i2φ e i2φ Q 2 2 = Qω 8π 15/2π To obtin the mgnetic rdition source component the current density is; J = ρ n V ˆφ 7

8 Then r J = ρ 2 ωẑ. The divergence of this vnishes fter converting to cylindricl coordintes or converting ẑ to sphericl coordintes. Thus there is no mgnetic component. The electric rdition component is; E = ck4 i(5!!) 3/2Q 2 2 The rdited power is; P = Z 0 2k 2 E 2 Z 0 = µ 0 /ǫ 0 8

Phys 4321 Final Exam December 14, 2009

Phys 4321 Final Exam December 14, 2009 Phys 4321 Finl Exm December 14, 2009 You my NOT use the text book or notes to complete this exm. You nd my not receive ny id from nyone other tht the instructor. You will hve 3 hours to finish. DO YOUR