Phys. 506 Electricity and Magnetism Winter 2004 Prof. G. Raithel Problem Set 1 Total 30 Points. 1. Jackson Points

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Baldric Berry
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1 Phys. 56 Electricity nd Mgnetism Winter 4 Prof. G. Rithel Prolem Set Totl 3 Points. Jckson 8. Points : The electric field is the sme s in the -dimensionl electrosttic prolem of two concentric cylinders, i.e. E(ρ = ˆρE ρ where E denotes the field on the surfce of the inner conductor, which hs rdius. Since for TEM wves propgting in the +z-direction it is H = µcẑ E = Z ẑ E with the plne-wve impednce Z = µ ɛ, it is H(ρ = ˆφE Zρ = ˆφH ρ with H = E Z eing the mgnetic field on the surfce of the inner conductor. The Poynting vector S(ρ = E H = ẑ Z E ρ = ẑz H ρ, nd the power is P = Re [ẑ S] πρdρ = Zπ H ln ( q.e.d. : On the inner surfce, = inner σδ ρ= H(ρ = dl = π σδ H nd on the outer surfce = outer σδ ρ= H(ρ = dl = π σδ H = π σδ H The power loss is the sum of the two, = π σδ H ( + Then, the ttenution constnt γ = P,

2 γ = ( + σδz ln ( q.e.d. c: Voltge: V = E dl = ˆρE ρ ˆρdρ = H Z ln (. Current: For TEM modes, the surfce currents re in z-direction, i.e. on the inner conductor K z = H. The totl current I = πh. Thus, the chrcteristic impednce Z is Z = V I = Z π ln ( q.e.d. Note tht Z is the plne-wve impednce Z = µ ɛ. d: The series resistnce R stisfies RI =, where the fctor complex quntities. Thus, with the ove I nd R = ( πσδ + q.e.d. occurs, s usul, due to the use of The inductnce per length is defined vi the mgnetic-field energy per length, u m = 4 B H d = π 4 B H ρdρ = 4 L I, where the fctor 4 (in plce of the mgneto-sttic vlue is to the use of complex quntities. The given result oviously ccounts for the field energies inside the guide, u m,guide, nd in the inner nd outer skin regions, u m,inner nd u m,outer. In the guide, ( u m,guide = π µ H ln In the inner guide wll it is H = H exp( ξ/δ, where ξ mesures the depth from the guide surfce. Assuming δ, which is good ssumption except in pthetic cses, u m,inner = πµ c 4 where µ c is the permeility of the conductor, nd similrly u m,inner = πµ c 4 Summing over the liner mgnetic-field energy densities, it is H exp( ξ/δdξ = πµ cδ 4 H ( H πµ c δ exp( ξ/δdξ = H 4 [ ( u m = π H µ ln + δµ ( c + ] nd L = 4u m I = µ ( π ln + δµ ( c 4π + q.e.d. Note tht normlly δ < nd therefore the second term is much smller thn the first.

3 . Jckson 8.4 Points : Following the nlysis on pge 369f with the replcement pπ d re k it is seen tht the cutoff frequencies ω M,mn = x mnc R for mode T M mn with m =,,.. nd n =,,.. nd ω E,mn = x mn c R for mode T E mn with m =,,.. nd n =,,.. There, x mn is the n-th zero of the Bessel function J m (x nd x mn is the n-th zero d dx J m(x. The fundmentl mode is T E with ω E, =.84c R =: ω. with c = µɛ. The next four higher modes re: T M with ω M, /ω =.36 T E with ω E, /ω =.659 T E with ω E, /ω =.8 T M with ω M, /ω =.8 : T E : We require ψ to clculte the power P from Eq. 8.5 in Jckson, nd ll mgnetic fields to clculte Here, it is / = σδ H dl. H z = ψ = H J ( x R ρ exp(iφ H t = ik γ tψ = ikh γ exp(iφ [ ˆρ x R J ( x R ρ + ˆφ i ρ J ( x R ρ ] with γ = µɛω E, nd k = µɛ(ω ω E, nd δ = µ c σω. Ug Eq. 8.5, P = π H ω ω E, µ ω E, ɛ ω R ρj ( x R ρdρ where the integrl equls R ( J x (x. Also, on the surfce H z = H J (x nd H t = k H γ R J (x, nd thus = σδ πr( H z + H t = πr H J (x σδ ( + k γ 4 R

4 The ttenution constnt for hollow rss guide, for which µ = µ c = µ nd ɛ = ɛ is then found to e β E, (ω = P = R ɛ σ µ ɛ ω 4 E, R + ω ω E, ω ω ω E, (µ ɛ ω E, R T M : The required fields re, ug γ = x R nd Z = k ɛω, Ug Eq. 8.5, clculting nd evluting β = P E z = ψ = E J ( x R ρ E t = ik γ tψ = ike [ γ ˆρ x R J ( x ] R ρ = ˆρ ike R J x ( x R ρ H t = Z ẑ E t = ˆφ ike R Zx J ( x R ρ = σδ πr H t, for hollow rss guide (µ = µ c = µ nd ɛ = ɛ, it is found β M, (ω = ɛ ω 3 R σ(ω ωm, Expresg the results in normlized frequency, x := ω ω E, nd the dmping constnts in units of R ɛ ω E, σ, ( ɛ ω E, β = β/ R σ, it is β E, (x = β M, (x =.84 + x x x (.84 x 3 x.36

5 TM TE Figure : Dmping constnts in scled units

6 3. Jckson 8.5 Points : The prolem hs discrete π/ rottion symmetry. Thus, ny mode T X mn is degenerte with T X nm, where X = E or X = M. For m n, we cn use the superposition principle to form liner superpositions of degenerte modes tht stisfy the pplicle oundry conditions on the digonl (in ddition to the sides tht form right ngle. TM-modes: E z (x, y = E [ ( mπx ( nπy ( nπx ( mπy ] for n, m =,, 3.. nd n m. It is E z (x, y = for x = y. The corresponding cutoff frequencies re ω M,mn = π m + n µɛ The lowest TM-mode hs m = nd n =. TE-modes: ( mπx H z (x, y = H [cos cos ( nπy + cos ( nπx cos ( mπy ] for n, m =,,,.. ut not n = m =. As required, the norml derivtive on the digonl, n H z = ( x H z y vnishes for x = y. The corresponding cutoff frequencies re lso ω E,mn = π m + n µɛ The lowest TE-mode hs m = nd n =. Note. Since ech pir of degenerte modes of the squre guide gives only one mode of the tringulr guide, there re out hlf s mny modes in the tringulr guides s there re in the squre guide. This oservtion mkes sense, ecuse generlly the mode density (=numer of modes per frequency intervl t lrge frequencies is pproximtely proportionl to the guide cross section. : T E : We need to clculte the power nd the mgnetic field on the surfces. With n (unnormlized ψ = H z = cos ( πy + cos

7 nd, y symmetry, tringle ψψ dxdy = squre ψψ dxdy = squre Mgnetic-field mplitudes on the sides: Use H z = ψ nd [ ( cos πx H t = ik γ th z = ikπ [ ˆx γ ( + cos πy + cos + ŷ ( πy ] cos ( πy ] dxdy = to see tht on the side x = it is H z (y = + cos ( πy nd Ht (y = kπ γ ( πy. Then, long tht side it is y= x=,y= [ H 3 dy = + k π ] γ 4 The y = -side yields the sme result. On the digonl x = y it is H z (y = cos ( πy nd Ht (y = kπ γ ( πy. Then, long the digonl it is y= x=y,y= H dl = y= H dy = [ + k π ] x=y,y= γ 4 nd the sum over ll three sides, H dl = (3 + + k π γ 4 ( + = (3 + + k π γ 4 ( + Use k = µɛ(ω ω E, nd γ E, = ω E, µɛ = π to get H dl = ( + + ω ωe, ( + With Eq. 8.5 nd = σδ β E, = σδ E, H dl nd β = P ɛ µ ω ω E, ω E, ω [ ( + ω E, ω + ( + ] where δ E, = µ c σω E,. The result hs een written in form nlogous with Eq in Jckson. T M : We use which yields ψ = ( πy ( πx ( πy

8 tringle Also, H z =, nd the mgnitude of the mgnetic field is ψψ dxdy = 4 On the side x = it is H = H t = Z E t = ɛω γ tψ = ɛω [ π ( πy γ ˆx cos [ π ( πy ŷ cos π ( πx cos π ( πx cos ( πy ( πy ] ] y= x=,y= H dy = 5 ɛ ω π γ 4 The sme pplies on the side y =. On the digonl x = y, it is nd H t = ɛω γ y= π cos x=y,y= ( πx π ( πx cos Ht dl = y= Ht dx = 5 ɛ ω π x=y,y= γ 4 The line integrl over ll three sides in H dl = ɛ ω π γ 4 Use γ M, = ω M, µɛ = 5 π nd Eq. 8.5 nd β M, = σδ M, ɛ µ = σδ ω 4 ω M, H dl nd β = P ω E, ω to find For the corresponding modes in squre guide, doule the re integrls ψψ dxdy, nd for the line integrls doule the results over the verticl nd horizontl sides of the tringulr guide nd leve out the digonls. T M E, : The re integrl doules nd the line integrl ecomes H dl = 4 + ω ω E,

9 The resultnt dmping constnt is β E,,squre = β E,,tringle + 4 ω E, ω ( + + ( + ω E, ω T M : The re integrl doules nd the line integrl remins unchnged. Thus, β M,,squre = β M,,tringle

Homework Assignment #1 Solutions

Homework Assignment #1 Solutions Physics 56 Winter 8 Textook prolems: h. 8: 8., 8.4 Homework Assignment # Solutions 8. A trnsmission line consisting of two concentric circulr cylinders of metl with conductivity σ nd skin depth δ, s shown,