Phys. 506 Electricity and Magnetism Winter 2004 Prof. G. Raithel Problem Set 2 Total 40 Points. 1. Problem Points

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1 Phys. 56 Electricity nd Mgnetism Winter 4 Prof. G. ithel Problem Set Totl 4 Points 1. Problem Points : TM mnp : ω mnp = 1 µɛ x mn + p π y with y = L where m, p =, 1,.. nd n = 1,,.. nd x mn is the n-th zero of J m (x. The frequencies of the TM mn do not depend on the cvity length L. TE mnp : ω mnp = 1 µɛ x mn + p π y with y = L where m =, 1,.. nd n, p = 1,,.. nd x mn is the n-th zero of J m (x. The frequencies of ll TE-modes depend on both the cvity length nd rdius. As the figure shows, the ground mode is either TE 1 or TM 1, dependent on L. At y =.67, the fundmentl mode is the TM 1. Its fields re E z = ψ(ρ, φ cos E t = H t = iɛω ˆφ γ E ( pπz L x 1 J By Eq. 8.9 in Jckson, the intrcvity energy is U = E πlɛ ρj The dissiption power due to Ohm-type heting is ( x1 ρ = E J ( x1 ρ iɛω = ˆφ γ E x 1 J 1 ( x1 ρ dρ = E ( x1 ρ πɛl J1 (x P = H d σδ surfce { = E ɛ ω x 1 σδ γ 4 πlj1 (x 1 + π ρj1 } ( x1 ρ dρ

2 Figure 1: Frequencies of the lowest cvity modes = E πɛ ω x [ ( 1 {LJ 1 σδ γ 4 (x 1 + J 1 (x ]} x J1 (x 1 1 where the first term is from the mntle nd the second from the end cps. The Q-fctor then is, using γ = x 1, Q = ω U P = Lσδx 1 { [ ( ]} J 8ɛω L + 1 (x 1 J1 (x x 1 Using the identity xj ν(x + νj ν (x = xj ν 1 (x with ν = 1, it is J 1 (x1 J 1 (x 1 = 1 x 1, nd the result simplifies to Lσδx 1 Q = 4ɛω {L + } = ω σδµ L 4 L + (1

3 where in the second line we hve used tht for the TM 1 -mode ω = 1 µɛ x 1. Since lso the skin depth δ = σωµ c nd c = 1 µɛ nd µ = µ c, this cn be expressed s Q = L σµcx1 + L 1 For numericl result, use µ = µ, x 1 =.45, = cm, L = 3cm, σ = Ωm. Then one obtins quite typicl vlue, Q = 1385

4 . Problem Points TM-modes. We use the convention tht the norml vector ˆn is pointing inwrd. The unperturbed solution, ψ = E z,, is such tht it vnishes on the unperturbed surfce S. The surfce perturbtion function δ(x, y equls the distnce between perturbed nd unperturbed surfce in the direction of ˆn, where ccording to our choice of ˆn n inwrd deformtion counts positive. The perturbtion will not significntly lter the mode function. We my ssume tht the effect of the wll deformtion is tht the mode function ψ shifts together with the wll, thereby mintining the proper boundry condition ψ = on S. The leding dependence of ψ ner the unperturbed surfce S is ψ(ξ = ξ ψ, where ξ mesures the norml distnce from S (ξ > inside the guide. Shifting ψ such tht its zero moves from S to S then is equivlent to ssuming perturbed boundry condition ψ = δ(x, y ψ on S The equtions nd boundry conditions for ψ nd ψ nd the respective eigenvlues γ nd γ re ( t + γ ψ = nd ψ = on S ( t + γ ψ = nd ψ = δ(x, y ψ on S Using Green s nd identity in two dimensions with ˆn pointing inwrd for ψ nd ψ then yields (ψ t ψ ψ t ψd = (ψ ψ S ψ ψ dl Inserting the bove eigenvlue equtions nd boundry vlues on S it is (γ γ ψψd = δ(x, y ψ dl where we hve lso used tht the eigenvlues re rel. Since in the re integrl the difference between ψ nd ψ will only produce higher-order corrections, we my set ψ = ψ in the re integrl nd get (γ γ S δ(x, y ψ dl = ψ d nd for the wvenumbers, k = µɛω γ, (k k S δ(x, y ψ dl = ψ d The sign of the result must be such tht reduction of the guide cross section, corresponding to positive δ, must result in reduction of k (i.e. n increse of the wvelength in the guide. Our result stisfies

5 this requirement, but it differs from the result stted in Jckson by minus sign. The result stted in Jckson obviously corresponds to choice of the norml vector ˆn pointing outwrd (which is opposite to the convention used in the corresponding portion of the text in Ch TE-modes. We use the convention tht the norml vector ˆn is pointing inwrd. The unperturbed solution, ψ = H z,, is such tht its norml derivtive ψ vnishes on the unperturbed surfce S. The effect of the wll deformtion is tht the mode function ψ shifts together with the wll, thereby mintining the proper boundry condition ψ = on S. The leding dependence of the norml derivtive of ψ ner the unperturbed surfce S is ψ ξ= (ξ = ξ ψ, where ξ mesures the norml distnce from S (ξ > inside the guide. Shifting ψ such tht the zero of its norml derivtive moves from S to S then is equivlent to ssuming perturbed boundry condition ψ = δ(x, y ψ on S The equtions nd boundry conditions for ψ nd ψ nd the respective eigenvlues γ nd γ re ( t + γ ψ = nd ( t + γ ψ = nd ψ = on S ψ = δ(x, ψ y on S Using Green s nd identity in two dimensions with ˆn pointing inwrd for ψ nd ψ nd inserting the bove eigenvlue equtions then yields (γ γ ψψd = (ψ S ψ ψ ψ dl Setting ψ = ψ in the re integrl nd inserting the boundry vlues on S it is (γ γ S = δ(x, yψ ψ dl ψ d nd for the wvenumbers, k = µɛω γ, (k k S = δ(x, yψ ψ dl ψ d Concerning the sign of the result, note tht δ chnges sign upon reversl on ˆn, while ψ does not. Our result differs from the result stted in Jckson by minus sign, gin reflecting the fct the result stted in Jckson ssumes norml vector ˆn pointing outwrd.

6 b: The depicted deformtion δ(y = δ y b long the verticl sides nd δ = long the horizontl sides. The depicted cse corresponds to positive δ. The line integrl only needs to be evluted long the verticl sides. We cn use unnormlized mode functions. TM. The norml derivtive ψ ψ = E z = sin ( πx ( πy sin b on the verticl sides x = nd x = is ψ = π ( πy sin b The line integrl δ(x, y ψ S dl = π δ b ( y sin πy dy = π δb b The re integrl ψ d = 1 4 b nd γ γ = k k S δ(x, y ψ dl = ψ d = π δ 3 Since k = µɛω π ( b, the perturbed vlue of k is ( k = k π δ = µɛω π 1 + δ π b TE 1. ψ = H z = cos ( πx On the surfces x = nd x =, it is ψ = ±1 nd ψ = π (upper signs for x =, lower for x =. Thus, S δ(x, yψ ψ δ dl = π b b ydy = π bδ

7 The re integrl ψ d = 1 b nd γ γ = k k S = δ(x, yψ ψ dl ψ d = π δ 3 Since k = µɛω π, the perturbed vlue of k is ( k = k π δ = µɛω π 1 + δ

8 3. Problem Points : Dmping-induced mixing of TM-modes.. Idel degenerte guide modes without dmping nd common eigenvlue γ: ( t + γ ψ (i = nd ψ (i = on S with i = 1,,..N. For smll dmping, the mixed modes re well-defined liner superpositions of the degenerte idel modes, ψ = N i=1 i ψ (i The objective is to find the coefficients i. Since there re s mny mixed modes s there re unperturbed ones, there will be N independent mixed modes (chrcterized by independent sets of i. As shown in Sec. 8.6 of Jckson, the mixed modes re slightly ltered due to the surfce conductivity such tht they stisfy n eqution with perturbed boundry condition ( t + γ ψ = nd ψ = f ψ = f i ψ (i i on S where f = (1 + i µ cδ ω µ ω with the unperturbed cutoff frequency ω = γ ɛµ nd the skin depth δ = µ c σω. Using Green s nd identity in two dimensions with ˆn pointing inwrd for ψ nd ψ (j then yields (ψ t ψ (j ψ (j t ψd = (ψ (j ψ S ψ ψ(j dl Inserting the bove eigenvlue equtions, the sum expression for ψ, the boundry vlues on S, nd using the relity of the eigenvlues, it is N (γ γ i i=1 ψ (i ψ(j d = f N i=1 (i ψ i ψ (j dl Since by ssumption n orthogonlity condition ψ (i ψ(j d = N i δ ji with norms N i pplies, the eqution cn be resorted into

9 N [ Ni (γ γδ ] ji + ji i = i=1 with ji = f S ψ (i ψ (j dl. Q.e.d. Mixing of TM-modes due to surfce deformtion. The equtions nd perturbed boundry conditions for the mixed modes ψ re, s seen in Problem 8.1, ( t + γ ψ = nd ψ = δ(x, y ψ = δ(x, y i ψ (i i on S (ˆn pointing inwrd. Using Green s nd identity in two dimensions with ˆn pointing inwrd for ψ nd ψ (j then yields N [ Ni (γ γδ ] ji + ji i = i=1 with ji = S δ(x, y ψ(i ψ (j dl. Q.e.d. Note. Hving ˆn point outwrd reverses the sign of δ; this produces the result given in the textbook. Mixing of TE-modes due to surfce deformtion. The equtions nd perturbed boundry conditions for the mixed modes ψ re, s seen in Problem 8.1, ( t + γ ψ = nd ψ = δ(x, ψ y = δ(x, y i ψ (i i on S (ˆn pointing inwrd. Using Green s nd identity in two dimensions with ˆn pointing inwrd for ψ nd ψ (j then yields N [ Ni (γ γδ ] ji + ji i = i=1 with ji = + S δ(x, y ψ (i ψ (j dl. Q.e.d. Note. Hving ˆn point outwrd reverses the sign of δ; this produces the result given in the textbook.

10 b: The given mode functions re orthogonl. We cn set B = 1. The norm vlues re then both equl to N : = N + = N = π = π ρ J 1 ( x ρ = π J 1 (x ( x ρ dρ + ρ 1 1 ρ x ρj1 ρ= ( 1 1 x J 1 ( x ρ The surfce deformtion cn be written s δ(φ = cos(φ. It is then seen tht ++ = =. Also, + = ( cos(φ ψ S ψ (+ dφ where ψ ( = ( x J 1 ( x ρ ρ= exp( iφ = ( x J 1 (x exp( iφ. Also, + = +. Thus, : = + = + = = π x J 1 (x J = π x J 1 (x = π x J 1 (x 1 (x [ J 1 (x ( x J 1 (x J 1 (x exp( iφ cos(φdφ S ( 1 x 1 ( 1 x 1 J 1(x 1 ] x The new eigenvlues γ re found by setting the determinnt x=x ( N(γ γ N(γ γ = yielding γ = γ ± N = γ ± ( x = γ (1 ± Therefore, the prmeter λ sked for in the problem is 1. Eigenfunctions From ( N(γ γ N(γ γ ( + =

11 nd (γ γ = γ it follows = γ N + With N = γ we then find = ± + where the upper sign corresponds to the lrger perturbed vlue of γ. The solutions stisfying the boundry conditions on the deformed guide wlls re: ( x ψ 1 = H J 1 ρ cos φ with γ = γ(1 + nd ( x ψ = H J 1 ρ sin φ with γ = γ(1 Interprettion. The electric field is trnsverse nd found to be E 1,t = iµωh γ [ ˆφγ J 1 (γ ρ cos φ + ˆρ 1 ] ρ J 1 (γ ρ sin φ nd E,t = iµωh γ [ ˆφγ J 1 (γ ρ sin φ ˆρ 1 ] ρ J 1 (γ ρ cos φ Noting tht γ γ, ner the xis, where ρ, the electric field reduces to E 1,t = ŷ iµωh γ E,t = ˆx iµωh γ Thus, the mixed modes hve trnsverse electric (nd mgnetic fields tht re ligned with the mjor nd minor xes of the ellipse present fter deformtion. Note (unwrrnted. For the dopted inwrd direction of the norml vector ˆn nd for > the deformtion is such tht the guide becomes stretched in y- nd squished in x-direction. Thus, the mode the eigenvlue γ of which shifts upwrd (corresponding to downshifting k nd upshifting guide wvelength hs n electric field polrized prllel to the long xis of the deformtion ellipse. The mode the eigenvlue

12 γ of which shifts downwrd hs n electric field polrized trnsverse to the long xis of the deformtion ellipse. Note. You recll tht the direction of the norml vector ˆn plyed role in the nlysis. In the present cse, reversing the choice of the direction of ˆn reverses the sign of. As result, the mixed modes 1 nd nd the ssocited eigenvlues become swpped, ensuring tht the physicl results remin the sme. The number vlues for the perturbed eigenvlues must be independent of the direction of ˆn, nd the ssocition between the polriztions of the up- nd downshifting modes nd the orienttion of the deformtion ellipse must be independent of the direction of ˆn.

13 4. Problem 8. 1 Points : TM-modes. The origin is chosen in the lower left corner, nd we use the normlized ode functions Eq nd For the current element, we use ( J(xd 3 sin φ x = I dl = I cos φ dφ nd x = ( cos φ h + sin φ There, is the loop rdius, h the height of the loop center, nd π/ < φ < π/. We ssume, without loss of generlity, z =. Then, the normlized-mode mplitudes re A ± i = Z i J(x E i d3 x which for the TM-modes of the given rectngulr guide nd current yields A ± mn = Z T M πi γ mn b π/ π/ + n b cos φ sin ( mπ cos φ { ( m mπ cos φ ( sin φ cos ( nπ(h + sin φ cos b sin } dφ ( nπ(h + sin φ b Noting tht the integrnds re totl differentils, A ± mn = Z T M πi π/ {[ ( ] ( d mπ cos φ nπ(h + sin φ γ mn b π/ dφ sin sin b [ ( ] ( } d nπ(h + sin φ mπ cos φ 1 + dφ sin sin b π dφ = Z T M πi π/ [ ( ( ] d mπ cos φ nπ(h + sin φ 1 sin sin γ mn b π/ dφ b π dφ = Z [ ( ( ] π/ T M πi 1 mπ cos φ nπ(h + sin φ sin sin γ mn b π b π/ = Interprettion. The wire loop couples to the longitudinl mgnetic field, which is bsent for TM-modes. Thus, for the given geometry the TM excittion mplitudes vnish. Note. The integrnd cn be expnded for the cse, b. For TM-modes, the result lso is A ± mn =.

14 b: TE 1 -mode. From Eq in Jckson nd the subsequent comment it follows tht the normlized fields re E z = E x = π ( πx E y = γ 1 b sin from which Use Z T E = µω/k nd k = A ± E1 = Z T E πi γ 1 b π/ π/ π/ π/ ( π cos φ sin cos φdφ ω c γ nd γ 1 = π/ nd π/ π/ sin(x cos φ cos φdφ = πj 1(x to see ( A ± E1 = µωi π π J 1 ω c π b c: Power in TE 1 -mode. It is, ccording to Eq nd the subsequent comment, Use Z T E = µω/k nd k = ψ = H z = A ± E1 H z,e1 = µωi π J 1 ω c π b ( π iγ1 cos kz T E b ( πx ω c γ nd γ 1 = π/, nd Eq. 8.51, P = const ψ d, to find the power P = I Zπ J1 4b ( π For smll rgument J 1 (x = x/. Thus, for it is P I Z 16 b ( 4 π Note. The result lso follows from Eq , P = 1 Z i A i