Energy creation in a moving solenoid? Abstract

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1 Energy cretion in moving solenoid? Nelson R. F. Brg nd Rnieri V. Nery Instituto de Físic, Universidde Federl do Rio de Jneiro, Cix Postl 68528, RJ Brzil Abstrct The electromgnetic energy U em stored in the mgnetic field of n idel solenoid t rest ppers for moving observer s U em, sum of mgnetic nd electric energies stisfying: U em > γu em. We explin this seemingly prdoxicl result clculting the stresses in the solenoid structure nd showing tht the totl energy of the solenoid trnsforms by the expected reltivistic fctor γ. Electronic ddress: brg@if.ufrj.br Electronic ddress: rnieri@ufrj.br 1

2 I. INTRODUCTION We consider n infinitely long idel solenoid with xis z nd rdius, s shown in fig. (1), t rest in n inertil frme S 0. There re n turns of wire per unit length, crrying n electric current i. As is well known (e.g. [1]), the mgnetic field B vnishes outside the solenoid nd hs constnt mgnitude B = µ 0 ni inside the solenoid in the xil direction. We tke the orienttion of the current such tht the field is in the positive z direction. Although this solenoid is infinite, there re neither externl forces nor externl energy fluxes cting on it. So it is n isolted system, for which we expect tht the totl energy should trnsform by fctor of γ = 1/ 1 v 2 /c 2, when going to frme where it is in motion with speed v. Let us consider frme S, where the solenoid moves with constnt velocity v in the negtive direction of the x xis. Considering the reltivistic trnsformtions of the Electromgnetic fields, one finds tht inside the solenoid there is mgnetic field B = γb, in the xil direction, nd n electric field E = vγb, in the negtive y direction. Outside the solenoid the fields remin null. The internl volume of ny finite portion of the solenoid is reduced by fctor of 1/γ becuse of Lorentz contrction in the x direction. In the rest frme S 0, the energy stored in the mgnetic field in volume V = π 2 z of the solenoid is U em = B2 V. (1) In S the sum of the energies of the electric nd mgnetic fields is U em = ( B 2 + ε 0E 2 ) V 2 γ ( ) = γ B2 1 + v2 c 2 V = ( 1 + v2 c 2 ) γ U em. (2) 1 z + FIG. 1: Infinite solenoid 2

3 So, we find for the electromgnetic energy: U em > γ U em. In order to understnd why the electromgnetic energy does not trnsform with the fctor γ, we must complete the nlysis of the trnsformtion of the energy, considering the fct tht the solenoid is continuous system. The energy density for continuous distributions of mtter trnsforms s component of four-tensor tht lso includes the stresses[2], unlike the energy of point prticles tht trnsforms s component of four-vector. The energy density in the frme S includes contribution due to the stresses in the rest frme S 0. This stress show up in the solenoid becuse the wire loops crry n electric current nd re subject to mgnetic field. We know the mgnetic field inside nd outside the wire loops (tht mens, respectively t r < nd r > ). In order to determine the mgnetic force, nd the corresponding stress, on the wire, we must find the field t r = in the frme S 0. With this purpose, we cn remodel the solenoid, tking the wires to hve finite width, mening tht the rdius of the wire loops rnges from r = to r = +. The mgnetic field in point inside the wire t rdius r corresponds to the field produced by the current situted t lrger rdii: r > r. Incresing the rdius by dr we decrese this externl current. Clling the decrese s di, the vrition db of the mgnetic field is db = µ 0 ndi. (3) The force cting on n element of wire with infinitesiml re dr dl is d 2 F = dibdl = BdBdl µ 0 n, (4) nd points in the rdil direction. Considering dl = rdθ nd integrting by prts we hve df = 1 ( ) B(+ ) BrdB dθ = 1 ( ) B 2 (+ ) r µ 0 n B() µ 0 n B 2 () 2 db2 dθ = 1 ( 1 + B 2 ) µ 0 n 2 rb2 (r) r=+ r= 2 dr dθ. (5) Tking now the limit of n idel solenoid with very thin wire: 0, the term + B 2 2 dr becomes negligible. Using the boundry conditions B() = µ 0ni nd B( + ) = 0 the force over the infinitesiml element of the wire tkes the form ( ) µ0 ni df = i dθ. (6) 2 3

4 This is equivlent to the force produced by mgnetic field µ 0 ni/2. So tht we cn tke the mgnitude of the field t r = s one hlf of the mgnitude t r < (interior of the solenoid). II. CALCULATING THE STRESS The mgnetic force pushes ll pieces of the solenoid in the positive rdil direction. Considering tht the solenoid is t equilibrium, this force must be blnced by the internl tensions in the wire. In order to simplify the clcultions, we consider tht our idel wire hs no width but it is surrounded by cylindricl shell of width λ mde of mteril tht holds the stbility of the system s shown in (fig. 2). We clculte the stresses in this shell by considering volume element inside it nd using cylindricl coordintes. The symmetry of the solenoid implies tht only the digonl terms of the stress tensor re non vnishing. The equilibrium of forces leds to the following reltion between the relevnt components: r t rr (r)dzdθ [r + dr] t rr (r + dr)dzdθ + t θθ drdzdθ = 0. (7) This implies: d(t rr r) dr = t θθ. (8) y d F x i 1 λ FIG. 2: Representtion of the shell nd the forces cting on the solenoid 4

5 The energy density for the mtter content of the solenoid in S is given by γ 2 (ρc 2 + t xx (v 2 /c 2 )), with ρ being the mss density nd t xx the stress long x nd on fce orthogonl to this direction, both clculted in S 0. Since the volume of ny prt of the solenoid is reduced by n fctor of 1/γ, the first (mss) term trnsforms s expected, leving the nlysis only to the term involving the stress. The reltion between crtesin nd cylindricl components is: t xx = t rr cos 2 θ+t θθ sin 2 θ. The extr contribution to the energy due to the stress is obtined integrting γ 2 t xx (v 2 /c 2 ) long the cylindricl shell in the S frme. Considering prt of the shell with height z, tht involves the volume V, we cn integrte in S using the cylindric coordintes (r, θ) of S 0. Tking into ccount the contrction in the x direction, the integrl tkes the form U +λ 2π = γ 2v2 c z 2 0 = γ v2 +λ c π z 2 r +λ t xx dθdr = γv2 γ c π z rt 2 rr + rt θθ dr d(r 2 t rr ) dr = γ v2 dr c 2 π2 zt rr () (9) Where we used the fct tht t the outside rdius of the shell there re no externl forces, so t rr ( + λ) = 0. In order to clculte t rr in the inside rdius of the shell we use eq. (6) nd tke surfce element dθ z, contining n z spires to find t rr () = n z df dθ z = nib 2 = B2, (10) so tht U = γ v2 c 2 B 2 V. (11) This is exctly the opposite of the exceeding term tht ppered in the electromgnetic energy in S. So U em + U = γu em (12) Showing tht the whole solenoid indeed behves s n isolted system. Similr problems were discussed for plne nd sphericl cpcitors, respectively in [3] nd [4]. Acknowledgments: The uthors re prtilly supported by CNPq nd FAPERJ. [1] Griffiths, D. J. Introduction to Electrodynmics (Prentice Hll, 1999). 5

6 [2] Rindler, W. Introduction to Specil Reltivity (Clrendon Press, Oxford, 1982). [3] Rindler, W nd Denur, J A simple reltivistic prdox bout electrosttic energy Am. J. Phys , 1988 [4] Brg, N R F nd Sophi, G Reltivistic energy of moving sphericl cpcitor Eur. J. Phys ,

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