Flow of Energy and Momentum in a Coaxial Cable

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1 Flow of Energy nd Momentum in Coxil Cle 1 Prolem Kirk T. McDonld Joseph Henry Lortories, Princeton University, Princeton, NJ (Mrch 31, 007) Discuss the flow of energy nd of momentum in, s well s the electromgnetic forces on, coxil cle tht crries TEM wve. You my ssume the cle to e mde from nerly perfect conductors (with liner medium of dielectric constnt ɛ nd permeility μ etween them), so tht the chrges nd currents re confined to thin lyers t the surfces of the conductors. This prolem is n extension of the cse where the coxil cle crries stedy current [1], for which the discussion concerned the reltion etween electromgnetic field momentum nd hidden mechnicl momentum in the system. Solution.1 Dul Roles of the Poynting Vector nd the Mxwell Stress Tensor This prolem illustrtes the dul roles of the Poynting vector, (in MKSA units) nd the Mxwell stress tensor T, S = E H, (1) T ij = E i D j + B i H j 1 δ ij(e D + B H) =ɛe i E j + μh i H j 1 δ ij(ɛe + μh ), () in liner medium with dielectric constnt ɛ nd permeility μ such tht D = ɛe nd B = μh..1.1 Energy Blnce As introduced y Poynting [], the vector S descries the flow of energy cross unit surfce re in unit time. In more detil, Poynting noted tht the electromgnetic fields do work W on the distriutions ϱ nd J = ϱv of chrge nd current density t the rte per unit volume of 1 dw dt = f EM v =(ϱe + J B) v = ϱv (E + v B) =ϱv E = J E, (3) where f EM = ϱe + J B (4) 1 Eqution (3) contins the insight tht mgnetic fields do no work individul chrges with no intrinsic mgnetic momment. 1

2 is the volume density of the fields on the chrges nd currents. The energy trnsferred to these chrges nd currents ccording to eq. (3) comes from the energy stored in the electromgnetic fields, whose energy density u field ccording to Mxwell is u field = E D + B H = ɛe + μh. (5) The field energy with volume cn lso e chnged y the flow S of energy cross the surfce of tht volume, so tht conservtion of energy cn e expressed s dvol = S dare J E dvol = S dvol J E dvol, (6) ufield which cn e expressed equivlently s the continuity eqution u field + S = J E. (7) Using Mxwell s equtions nd vrious vector clculus identities, Poynting showed tht the energy flow vector S is given y eq. (1)..1. Momentum Blnce Following the spirit of Poynting s rgument, Arhm [9] extended Mxwell s nlysis in terms of stress tensor of electromgnetic forces due to sttic fields to include the cse of time-dependent fields. Recll tht for the cse of sttic fields, Mxwell expressed the electromgnetic force F EM on volume in terms of n integrl of the stress tensor T over the surfce of tht volume, F EM = f EM dvol = T dare. (8) Using Mxwell s (time-dependent) equtions nd vrious vector/tensor clculus identities, Arhm showed tht eq. (8) cn e generlized to the form 3 F EM = f EM dvol = T dare (ɛμs) dvol. (9) If the only forces on the chrges nd currents in the volume re electromgnetic, then Newton s nd lw cn e written pmech F EM = f EM dvol = dvol = dp mech, (10) dt where p mech is the volume density of mechnicl momentum, nd P mech is the totl mechnicl momentum in the volume. Comining eqs. (9) nd (10) we hve ( p mech + (ɛμs) ) dvol = T dare = T dvol. (11) Poynting s derivtion is discussed in most textooks on electromgnetism. See, for exmple, sec. 8.1 of [3], sec..19 of [4], sec of [5], sec. 3.1 nd chp. 7 of [6], sec. 6.7 of [7] nd sec. 8.1 of [8]. 3 For discussions of Arhm s derivtion see, for exmple, sec. 1.1 of [3], secs..6 nd.9 of [4], sec of [5], sec. 3. nd chp. 7 of [6], sec. 6.7 of [7] nd sec. 8. of [8].

3 Following Arhm, we identify the vector p field = ɛμs = D B (1) s the momentum per unit volume tht is stored in the electromgnetic field, so tht the totl momentum density is p totl = p mech + p field, (13) which oeys the continuity eqution p totl T =0. (14) This leds to second interprettion of the Mxwell stress tensor T, nmelytht T is the flux of momentum in the electromgnetic field. Momentum flux is tensor, eing the vector momentum crossing n (oriented) re element per unit time. Momentum flux hs the dimensions of momentum density times velocity (nd therefore the sme dimensions s energy density nd s pressure). For TEM plne wve, such s in the present prolem, the fields E nd H nd the wve vector k form n orthogonl trid, the wve velocity is v = ˆk/ ɛμ, the fields oey ɛe = μh, the energy density is u = ɛe, the momentum density is p field =(u/v) ˆk nd the Mxwell stress tensor hs the simple form T = u = p field v 0 0 0, (15) if we chose xis 1 long E, xis long H nd xis 3 long k. Interpreting T s the momentum flux, we confirm tht this flux flows only in the k direction (cross plnes perpendiculr to k) nd hs mgnitude equl to the momentum density times the wve velocity. In sttic or qusisttic exmples the Poynting vector, nd hence the momentum density, cn e zero, while the tensor T is nonzero so long s either the electric or mgnetic field is nonzero. In such cses there is formlly momentum flux ut no momentum density. Here it is etter to consider the tensor T to e simply mesure of the stresses cused y the chrges nd currents. In sum, the Poynting vector S = E H hs the interprettion s the flux of energy in electromgnetic field, nd when multiplied y ɛμ s the density of momentum p field = ɛμs = D B. The Mxwell stress tensor T descries the stresses in system due to its chrges nd currents, s well s eing the negtive of the flux of momentum within the system. 4 4 Although we do not need to consider electromgnetic ngulr momentum here, we note tht the vector r p field descries the density of ngulr momentum stored in the electromgnetic fields, nd the tensor ɛ ikl r k T jl descries the flux of ngulr momentum. See, for exmple, pro. 5, chp. 3 of [6]. 3

4 . TEM Wve in Coxil Cle with Perfect Conductors We use cylindricl coordinte system (r, φ, z) withthez xis long the xis of the coxil cle. The nnulr gp etween the conductor extends from r = to, nd this gp is filled with nonconductor with dielectric constnt ɛ nd permeility μ. In the idelized cse of perfect conductors, the electromgnetic fields re nonzero only for <r<. The velocity of the TEM wve is v =1/ ɛμ, nd the wve propgtes in the +z direction...1 E nd H Fields for the TEM Wve The electric field is rdil, nd cn e written E = E cos(kz ωt) ˆr ( <r<), (16) r where E is the field strength t the outer conductor, k =π/λ nd ω = kv. The mgnetic field B hs mgnitude equl to E/v = ɛμe, nd its direction is zimuthl. The mgnetic field H = B/μ cn then e written ɛ ɛ H = μ c ẑ E = μ E r cos(kz ωt) ˆφ = E Z 0 r cos(kz ωt) ˆφ ( <r<), (17) where Z 0 = μ/ɛ 100 Ω is the chrcteristic impednce of the coxil trnsmission line. We recll tht the TEM wve fields (16)-(17) consist of the wve function cos(kz ωt) times fields E sttic nd H sttic tht re possile stedy-stte fields with no z dependence... Energy Density The density (5) of energy stored in the electromgnetic field is u = ɛe + μh = ɛe = ɛe r cos (kz ωt). (18) Note tht ɛe = μh, so the electric nd mgnetic components of the energy density re equl in the TEM wve...3 Poynting Vector The Poynting vector (1) is S = E H = ɛ μ E r cos (kz ωt) ẑ = u ɛμ ẑ = uv ẑ ( <r<). (19) The Poynting vector equls the electromgnetic energy density times the (vector) wve velocity, which confirms the interprettion of S s the flux of electromgnetic energy crried y the wve. 4

5 ..4 Density nd Flux of Momentum The density (1) of momentum stored in the electromgnetic field is p field = ɛμs = ɛμ u ẑ = u ẑ ( <r<), (0) ɛμ v reclling eq. (19). This momentum density flows long with the wve, so the flux of momentum is the momentum density times the wve velocity,..5 Mxwell Stress Tensor momentum flux = p field v = u ẑ ( <r<). (1) The Mxwell stress tensor () hs component in the cylindricl coordinte system, T rr T rφ T rz T = T φr T φφ T φz = 1 ɛe ɛe μh μh 0 T zr T zφ T zz 0 0 ɛe 0 0 μh = ɛe = u ( <r<). () In the interprettion of T s the momentum flux tensor, eq. () implies tht the only nontrivil component of momentum flux is T zz = u =(u/v)v, which corresponds to momentum density of mgnitude u/v flowing with velocity v in the z direction cross surfces perpendiculr to the z xis. This is consistent with the previous result (1) for the momentum flux...6 Forces on the Coxil Cle The stress tensor lso hs the interprettion s eing the electromgnetic force per unit re cross n oriented surfce. The form of eq. () implies tht the only nonzero electromgnetic force in the coxil cle is in the z direction, nd this cts cross surfces perpendiculr to the z direction. In greter detil, oth the electric nd the mgnetic prts of the stress tensor () hve nonzero rdil, zimuthl nd longitudinl components. The rdil electric component, Trr E = ɛe / is positive, nd implies n ttrctive force etween the opposite chrge distriutions on the inner nd outer conductors of the cle, corresponding to Frdy s insight tht there is tension long the (rdil) electric fields lines. The zimuthl nd longitudinl electric components, Tφφ E = T zz E = ɛe /, imply tht there re repulsive forces etween portions of the conductors otin y, sy, slicing them long the plnes x =0,y =0or z = 0. These repulsive forces re qulittively nticipted y Frdy s view tht field lines repel one nother. 5

6 The zimuthl mgnetic component Tφφ H = μh / is positive, nd implies tht there is n ttrctive force etween filments of current on the sme conductor. The rdil nd longitudinl mgnetic components Trr H = T zz H = μh / re negtive nd imply tht there re repulsive rdil mgnetic forces etween the inner nd outer conductor, nd lso etween longitudinl segments of the cle t, sy, positive nd negtive z. These electric nd mgnetic forces exist in the sttic cse where the coxil cle supports DC voltge or DC current (or oth). In the cse of TEM wve, for which ɛe = μh, the rdil nd zimuthl electric nd mgnetic forces cncel one nother, nd only the longitudinl forces remin. This conclusion follows quickly from the form of the Mxwell stress tensor. We now digress to clculte the forces y elementry methods...7 Cnceltion of the Rdil Force The surfce density ς of free chrge on the inside of the outer conductor cn e found from the Mxwell eqution ρ free = D = ɛ E, which implies tht ς = ɛe. (3) The force per unit re on this chrge distriution is rdilly inwrds, with mgnitude Fr E = ςe = ɛe, (4) following the usul rgument tht the electric field flls to zero from E over the smll ut finite rdil thickness of the surfce chrge distriution, so tht the verge field on this chrge distriution is E /. The zimuthl mgnetic field B = ɛμe t the outer conductor cts on the current I in tht conductor to produce n outwrd rdil force. From Ampére s lw we hve tht I =πh =πb /μ. The force on portion of the outer conductor of zimthl extent φ nd length L is (φl/π)ib /, noting tht the mgnetic field on the current vries from B to zero with verge strength B /. The re of the portion of the conductor is φl, sothe outwrd mgnetic force per unit re is F B r = φl πb B π μ φl = B μ = μh = ɛe = F E r. (5) Thus, the repulsive mgnetic force cncels the ttrctive electric force in the rdil direction...8 Cnceltion of the Trnsverse Forces The zimuthl component T φφ of the stress tensor cn e used to clculte the trnsverse force per unit length long the z xis etween the portions of the coxil cle on either side of the plne x = 0. In prticulr, the trnsverse force df x per length dz in the x direction on the portion of the cle t x<0isgiveny df x = T xx da x = dz T xx dy + dz T xx dy =dz T φφ dr (6) 6

7 since the re element on the plne x =0isdA x = dy dz = dr dz, nd on the plne x =0 we hve tht T xx = T φφ. The electric prt of the trnsverse force per unit length is df E x dz = Tφφ E dr = ɛ E dr = ɛ E r dr = ( ) ɛe. (7) The negtive sign mens tht the force is in the x direction, s expected due to the repulsion etween the like chrges on the two hlves of the conductors. Similrly, the mgnetic prt of the trnsverse force is df H x dz = Tφφ H dr =μ H dr =μ H = ɛe ( ) r dr =μh ( ) = df E x dz. (8) The totl trnsverse force vnishes ecuse T H φφ = T E φφ. We now ttempt to verify the trnsverse forces (7) nd (8) y elementry methods. First, note tht the (trnsverse) electric force per unit length df E /dz on wire of liner chrge density λ tht is prllel to the z xis nd psses through point r =(r, φ, 0) due to prllel wire of chrge density λ tht psses through point r =(r,φ, 0) is df E dz = λλ r r πɛ r r. (9) We sudivide the inner nd outer conductors into wires of zimuthl exeunt dφ, such tht their liner chrge densities re dλ 0 = ɛe dφ on the wire segments of the outer conductor, nd dλ I = ɛe dφ = ɛe dφ= dλ O on the inner conductor. The portion of the cle t x<0corresponds to π/ <φ<3π/, nd tht t x>0to π/ <φ<π/. The x-component of the electric force on the portion of the cle t x<0 is then, F E x = x<0 + + dλ O x<0 x<0 dλ O dλ I = ( + ) ɛe 4π dλ O x>0 ( + ) ɛe π dλ I x>0 dλ O x>0 3π/ π/ 3π/ cos φ cos φ 4πɛ[1 cos(φ φ )] x<0 + dλ I cos φ cos φ πɛ[ + cos(φ φ )] cos φ cos φ πɛ[ + cos(φ φ )] π/ dφ dφ cos φ cos φ π/ 1 cos(φ φ ) π/ dφ π/ π/ dλ I x>0 cos φ cos φ 4πɛ[1 cos(φ φ )] dφ cos φ cos φ + cos(φ φ ). (30) The result of eq. (30) could well e the sme s eq. (7). Certinly it is simpler to use the Mxwell stress tensor to otin the force thn it is to use elementry methods. The mgnetic force per unit length on the portion of the cle t x<0 cn in principle e clculted y n ppliction of the Biot-Svrt force lw. We sudivide the currents on the conductors into filments sutending ngle dφ, leding to integrls very similr to those in eq. (30). 7

8 ..9 Significnce of the Longitudinl Force on the Cle The nonzero vlue of component T zz = (ɛe + μh )/ of the Mxwell stress tensor () implies tht there re longitudinl forces on the coxil cle. These forces exist in the sttic limit s well. The electric prt of the longitudinl force etween portions of the cle t, sy, z<0 nd z>0 is redily scried to the repulsion etween the like chrges in these two regions. However, it is hrder to identify the source of the longitudinl mgnetic force, since the currents flow only longitudinlly (in the sttic limit, nd lso for TEM wves on cle mde of idel conductors). Note tht the totl longitudinl force on ny finite portion of the cle, sy z 1 <z<z. vnishes in the sttic limit, ecuse the longitudinl force on the two ends of this portion is equl nd opposite. A nonzero totl longitudinl force is otined for the intervl z 1 <z<z if one end of the cle lies within this intervl, so tht T zz = 0 t either z 1 or z. In this cse, there must e rdil currents t the termintion of the cle, nd these rdil currents interct with the zimuthl mgnetic field to produce the postulted longitudinl force. DC Current For exmple, consider coxil cle tht extends only for z<z 0 nd which crries DC current I in the +z direction on its inner conductor nd current I on its outer conductor. The termintion t z = z 0 is vi uniform resistive disk of thickness d (tht extends from z = z 0 to z 0 +d), so tht the rdil current density J in the terminting resistor for <r< is The zimuthl mgnetic field t z = z 0 is J = I ˆr. (31) πrd B = μi πr ˆφ, = μh r ˆφ, (3) nd flls to zero t z = z 0 d. Tht is, the verge mgnetic field is 1/ tht of eq. (3), B( <r<,z 0 <z<z 0 + d) = μh r ˆφ. (33) The mgnetic force on the terminting resistor is π z0 +d F = J B dvol = dr rdφ dz H 0 z 0 rd μh r ẑ = μh π ln ẑ. (34) We compre this with clcultion using the Mxwell stress tensor for cylinder of rdius R>nd longitudinl extent z 1 < z < z,wherez 1 <z 0 nd z >z 0 + d so tht the terminting resistor lies within this intervl. Then the component T zz is μh /= μh /r t z 1 ut vnishes t z. All components of the tensor T vnish on the cylindricl surfce t r = R since this is outside the cle. The (outwrd pointing) surfce re element t z = z 1 is dare z (z 1 )= πr dr, so the totl force on the cle within this intervl is F = T dare = T zz (z 1 ) dare z (z 1 ) ẑ = ( 8 μh r ) ( πr dr) ẑ = μh π ln ẑ, (35)

9 s found in eq. (34). TEM Wve We return to the cse of TEM wve on the coxil cle, nd clculte the electromgnetic forces on portion of the cle tht does not include the terminting resistor. For this, we pply eq. (9) to cylinder of rdius R>nd longitudinl extent 0 <z<z 1, for which the relevnt (outwrd pointing) re elements re dare z (0) = πr dr nd dare z (z 1 )=πr dr. Then, reclling eqs. (0)-(), the force on this portion of the cle is F = T dare d p field dvol dt z1 (ɛμs) = T zz (0) dare z (0) ẑ + T zz (z 1 ) dare z (z 1 ) ẑ πr dr dz 0 ɛe = [cos ωt cos (kz r 1 ωt)] πr dr ẑ ɛ ɛμ μ E r πr dr z1 ω cos(kz ωt)sin(kz ωt) dz ẑ 0 = ɛe π ln [cos ωt cos (kz 1 ωt)] ẑ = 0, +ɛ ɛμ ω k E π ln [cos (kz 1 ωt) cos ωt] ẑ noting tht ω/k = v =1/ ɛμ. Thus, lthough the forces ssocited with the Mxwell stress tensor on portion of the coxil cle re nonzero, they ct to chnge the electromgnetic field momentum in tht portion of the coxil cle, rther thn producing mechnicl force on the conductors of the cle. The force on portion of the cle tht includes terminting resistor t z = z 0 cn e otined from the nlysis contined in eq. (36) y replcing z 1 with z 0 nd omitting the contriution from the stress tensor t z = z 1, (36) F = ɛe π ln cos (kz 0 ωt) ẑ =μh π ln cos (kz 0 ωt) ẑ. (37) Another view is tht since the terminting resistor sors the momentum flowing long the cle, it experiences force equl to the momentum flux into the resistor, nmely References F = T zz (z 0 )πr dr = ɛe π ln cos (kz 0 ωt) ẑ. (38) [1] K.T. McDonld, Hidden Momentum in Coxil Cle (Mr. 8, 00), [] J.H. Poynting, On the Trnsfer of Energy in the Electromgnetic Field, Phil. Trns. Roy. Soc. London 175, 343 (1884), 9

10 [3] M. Arhm nd R. Becker, The Clssicl Theory of Electricity nd Mgnetism (Hfner, New York, 193). [4] J.A. Strtton, Electromgnetic Theory (McGrw-Hill, New York, 1941). [5] W.K.H. Pnofsky nd M. Phillips, Clssicl Electricity nd Mgnetism, nd ed. (Addison-Wesley, Reding, MA, 196). [6] J. Schwinger, L.L. DeRd, Jr, K.A. Milton nd W.-Y. Tsi, Clssicl Electrodynmics (Perseus Books, Reding, MA, 1998). [7] J.D. Jckson, Clssicl Electrodynmics, 3rd ed. (Wiley, New York, 1999). [8] D.J. Griffiths, Introduction to Electrodynmics, 3rd ed. (Prentice Hll, Upper Sddle River, New Jersey, 1999). [9] M. Arhm, Prinzipien der Dynmik des Elektrons, Ann. Phys. 10, 105 (1903), 10

This final is a three hour open book, open notes exam. Do all four problems.

This final is a three hour open book, open notes exam. Do all four problems. Physics 55 Fll 27 Finl Exm Solutions This finl is three hour open book, open notes exm. Do ll four problems. [25 pts] 1. A point electric dipole with dipole moment p is locted in vcuum pointing wy from