The homopolar generator: an analytical example

Transcription

1 The homopolar generator: an analytial example Hendrik van Hees August 7, Introdution It is surprising that the homopolar generator, invented in one of Faraday s ingenious experiments in 1831, still seems to reate onfusion in the teahing of lassial eletrodynamis. This is the more surprising as the problem of the eletromagnetism of moving bodies has been solved more than 1 years ago by Einstein in his famous paper, introduing his Speial Theory of relativity (195), and mathematially onsolidated by Minkowski hin his famous talk on spae and time (198). Also one an still find some misleading, if not even wrong, statements on the issue in the more reent literature, and I ould not find any paper using the loal (differential) Maxwell equations and the Lorentz-fore Law whih is always valid, as suggested in the Feynman Letures (vol. II) in onnetion with the use of Faraday s flux law (the integral form of one of the Maxwell equations, E = t B/, see App. A). Here, I try to provide preisely suh a study for the most simple arrangement showing the effets, namely the rotating homogeneously magnetized sphere. 2 The one-piee Faraday generator It is surprising that the so-alled Faraday paradox is still a soure of onfusion although the eletrodynamis of moving bodies is well understood with Einstein s famous speial-relativity paper. Here, I try to give an explanation by avoiding the use of the integral form of Maxwell s equation, whih seems to be the main soure of the onfusion. I ll omment on this in Appendix A. In this Setion, we onsider the most simple possible setup, namely a rotating spherial permanent magnet, assuming onstant polarization, whih we hoose as the diretion of the z axis 1. This magnet is put into rotation around the z axis and one measures the eletromotive fore (emf) between two points on the sphere, using brushed ontats onneted to a high-ohmi volt meter. We assume that the magneti field field is not hanged through this rotation (whih is by far not a trivial assumption and may be heked experimentally). There are two ways to explain the measured emf. The first is to solve the stati Maxwell equations. A permanent magnet s magneti field is due to the alignment of the magneti moments of the atoms, whih an be explained only via quantum theory, but we still an desribe the marosopi features 1 We hoose this somewhat unusual geometry, beause then we an solve for the magneti field in and outside of the magnet in terms of elementary funtions. We shall use this example to study exat analyti solutions of the various homopolar generators that are disussed in the literature (and online), whih should provide the best insight. 1

2 without quantum theory. We simply introdue the magnetization M as the dipole-moment density of the magnetized matter. Due to Gauss s Law for the magneti field, B =, (1) the most onvenient desription of the magneti field is through its vetor potential B = A. (2) Now for a given magneti field, whih is an observable quantity, the vetor potential is determined only up to the gradient of a salar field, beause if A fulfills (2), also A = A+ χ (3) for any salar field, χ, fulfills (2) too. This is part of what is known as the gauge invariane of eletrodynamis. This enables us to introdue an arbitrary onstraint without hanging the physially observable magneti field to fix the gauge. As it will turn out, for our purposes the Coulomb-gauge ondition, A =, (4) is most onvenient. The vetor potential of a magneti point dipole with dipole moment µ, loated at the point x is given by A( x) = µ ( x x ) 4π x x 3, (5) and thus for a ontinuous dipole distribution 2 This an be written as A( x) = 1 4π For M = onst we thus only need the integral Φ( x) = M ( x A( x) = d 3 x ) ( x x ). (6) R 3 4π x x 3 R 3 d 3 x M( x ) 1 x x = S R 3 d 3 x M( x ) 4π x x. (7) d 3 x 1 4π x x. (8) This, however is nothing else than the eletrostati potential of homogeneously harged sphere with harge density 1, and the integral an be easily solved by making use of the symmetry of the problem. Obviously due to spherial symmetry, Φ( x) = Φ(r ), where r = x, and thus we an hoose x = r e z in (8). Introduing standard spherial oordinates for x, osϕ sinϑ x = r sinϕ sinϑ, (9) osϑ 2 I use the Heaviside-Lorentz system of units, i.e., rationalized Gaussian units, beause these are most onvenient for theoretial onsiderations. 2

3 the integral (8) beomes Φ(r ) = R 2π dr dϕ π = Θ(R r ) (3R 2 r 2 ) + 6 dϑ r 2 sinϑ 1 4π r 2 + r 2 2r r osϑ Θ(r R)R3. 3r Putting everything together, the vetor potential beomes A = M 3 e z x Θ(R r ) + Θ(r R) R3, (11) and finally the magneti field where B = A = 2 3 r 3 (1) MΘ(R r ) + 3 x( µ x) r 2 µ r 5 Θ(r R), (12) µ = R3 3 M. (13) The magneti field is thus onstant within and a dipole field with the dipole moment (13) outside the sphere. Now it is easy to evaluate the emf. Through the rotation of the magnet on eah ondution eletron ats the Lorentz fore e v B/ whih drives the eletrons against the frition (whih marosopially manifests itself in terms of the resistivity) out of their equilibrium plaes, whih leads to a separation of harges, leading to the buildup of an eletri field. After some short time a new stati equilibrium state has been established, so that the total Lorentz fore F = e E + v B =. (14) Here we assume that the ondution eletrons beome o-moving with the magnet due to frition (or marosopially spoken a finite resistane). As we will now study in detail, the drift of the ondution eletrons into the new equilibrium state leads to a negative harge distribution inside the magnet and to a positive surfae harge on the boundary of the sphere of the same magnitude, establishing the eletri field Sine ω = ω e z we get from (12) E = v B ω x = B for r < R. (15) E = 2M ω 3 x y = ωb x y for r < R. (16) Sine this is a stati eletri field, it must have a potential, whih also follows immediately from the Maxwell equation E = 1 t B =, (17) 3

4 whih is Faraday s Law in loal form, whih is always valid in an unambiguous way. Sine here t B =, the eletri field is url free and thus a potential field. Indeed, the potential for the eletri field inside the magnet is U ( x) = M ω 3 (x2 + y 2 ) for r < R, (18) and thus the voltage measured by a high-ohmi voltmeter is given by the orresponding potential differene between the two points on the sphere. One should note that the measured voltage is independent of the path hosen to evaluate the potential differene, beause the eletrostati field is a potential field! Now we disuss this result from the point of view of the flux rule (39) to show that there is no ontradition with (18). Just using a path going along the wires onneting the voltmeter with the surfae of the magnet, losing it by a time-independent path inside or outside of the magnet simply leads to d x E = 1 Φ B =, (19) f whih is of ourse orret aording to the fat the E is a stati eletri field and thus url free everywhere. Here it is important to remember that v in (39) is the veloity of the boundary of the integation path and not the eletron s veloity. As we see, in this way we have no hane to find the voltage reading of the voltmeter using the flux rule. On the other hand, a quite lever way to obtain the measured voltage via the flux rule is to use a path along the onnetions of the voltmeter ompleted to a losed line with an (arbitrary!) path omoving with the magnet. Then v is the same in (39) and in (14). Thus, the so hosen omoving path within the sphere does not ontribute anything, beause aording to (14) E + v B/ = within the sphere. The part of the path outside of the sphere gives the orret voltage, given by the orresponding potential differene with the potential given by (18) sine E is a onservative field everywhere. For ompleteness we also give the eletri field outside of the magnet, whih must exist, beause the tangent omponents at the boundary must be ontinuous. Using Gauss"s Law, from (15) we find a homogeneous harge density within the magnet ρ = E = 4Mω, (2) 3 whih results from the ondution eletrons that are pulled inwards due to the magneti Lorentz fore. The total harge is Q = 4π 3 R3 ρ = 16πM R3 ω. (21) 9 Sine the total harge is onserved, there must be a surfae-harge distribution with the total opposite harge, Q. From the symmetry properties of the problem and beause the total harge of the matter vanishes, the eletri potential outside of the sphere must be a multipole expansion of the form U ( x) = l =1 A l z r P l +1 l r for r > R, (22) where P l (osϑ) is the Legendre polynomial of degree l. The potential inside of the sphere is desribed by (18). On the other hand it must be the superposition of the potential for the interior of a homogeneously harged sphere with the homogeneous harge density (2), U Q ( x), and a multipole expansion 4

5 with positive powers in r, whih reads z U ( x) = U Q ( x) + B l r l P l r l = for r < R (23) The first piee is found most onveniently by integration of the Legendre equation U Q = ρ, where U Q = U Q (r ). The solution is U Q ( x) = ρr 2 2 2M ωr =. (24) 6 9 Now we have U ( x) U Q ( x) = M ω 9 (x2 + y 2 2z 2 ) = Comparison of both sides of the equation yields z B l r l P l r l = (25) B 2 = 2Mω, B 9 l = for all l 2. (26) Now we look at the ontinuity onditions at the boundary of the magnet. Sine E Q is a radial field, it is perpendiular to the magnet s boundary everywhere. Thus, to fulfill the ontinuity onditions for the tangential omponents of E, the multipole expansions (22) and (25) must oinide for r = R, whih gives A l R l +1 = B l Rl A l = B l R 2l +1. (27) This leads to Finally we find an eletri quadrupole field outside of the sphere E = M R5 ω 3 r 7 A 2 = 2M R5 ω. (28) 9 x(r 2 5z 2 ) y(r 2 5z 2 ) for r > R. (29) z(3r 2 5z 2 ) The surfae-harge distribution on the sphere is given by the disontinuity of the normal omponents, i.e., σ = x r ( E r =a+ + E r =a +) = M Rω [1 5os(2ϑ)], (3) 6 where we have introdued the usual spherial oordinates for the surfae of the sphere x = R(os ϕ sin ϑ, sin ϕ sin ϑ, os ϑ). Integrating this over the surfae, leads to π 2π dϕa 2 sinϑσ = 16πM R3 ω 9 (21) = Q, (31) whih just anels the total harge density, Q, as demanded by the harge neutrality of the magnet. 5

6 A Faraday s Law The original Faraday Law has been formulated in integral form, and it is often presented as suh in introdutory letures on eletromagnetism. Of ourse, nowadays, as we know eletromagnetism from a quite more fundamental point of view, there are in fat no ambiguities in the use of the law also in integral form. As we shall see, there is also no paradox onerning the homopolar generators in terms of an apparent ontradition with Faraday s Law in integral form. The fundamental laws of eletromagnetism are nowadays understood as originating from a loal relativisti field theory, and its quantized version (quantum eletrodynamis) an be onsidered as the most suessful physial theory disovered by men! So the unambiguous starting point for all onsiderations onerning eletromagnetism are the loal Maxwell equations. One of these is Faraday s Law in loal form, E = 1 t B. (32) It should be noted that the notation in this way, sometimes already leads to misleading ideas! Sometimes this law is read as if a time-varying magneti field is the soure of a solenoidal eletri field. As the study of the omplete Maxwell equations shows, this is not orret in the literal sense, beause the solutions of the Maxwell equations show that the ausal soures of the eletromagneti field, onsisting of eletri and magneti omponents ( E and B, respetively) are the harge and urrent densities, upon whih the eletromagneti field depends in the sense of retarded integrals, showing learly the ausal onnetion between the harge-urrent distribution and the fields. One annot make suh a statement from a loal law like (32), and indeed the attempt to express E (or parts of it) in terms of B leads to quite ompliated non-loal expressions. After this side remark, we derive Faraday s Law from its loal form (32) by integrating over an arbitrary surfae. In the following, the surfae and/or its boundary urve an be moving, i.e., time dependent, but the integral has to be taken at a fixed time, t. The surfae-normal vetors d 2 f and the tangent vetors of its boundary f are relatively oriented in the usual sense of the right-hand rule. Then we an use Stokes s integral theorem on the left-hand side, leading to d 2 f ( E) = d x E = 1 d 2 f t B. (33) Of ourse, this is always valid as is the loal form (32). Now usually, Faraday s Law is formulated in terms of the magneti flux, Φ B (t) = d 2 f B. (34) Of ourse to get an expression, appearing on the right-hand side (33), we have to take the time derivative of the magneti flux. However, this does not simply lead to the desired integral, beause the surfae may be time dependent. To get the orret expression, we evaluate the flux at an infinitesimally later time t +dt up to order dt: Φ B (t + dt) = d 2 f B(t + dt, x) = d 2 f dt t B(t, x) + d f B(t, x) + (dt 2 ). (35) 6

7 d f t+dt d f = d x vdt dt v(t, x) dv d f t Figure 1: The ylinder-like infinitesimal volume to evaluate (35). In the first integral we ould write instead of f (t + dt), beause the integrand is already of order dt and thus the orretion from integrating over instead over f (t + dt) would beome of order (dt 2 ). Now we use a trik to evaluate the seond integral in (35) further: We apply Gauss s integral theorem to evaluate the integral of B (forgetting for the moment that this expression vanishes due to Gauss s Law for the magneti field) over the infinitesimal volume depited in Fig. 1. This volume is swept out by the surfae f during the infinitesimal time interval (t, t + dt). It is like a ylinder of infinitesimal hight, i.e., it onsists of the two surfaes f (t + dt) (in Gauss s integral theorem with the surfae-element normal vetor d 2 f t+dt ) and (surfae-normal vetor d 2 f t as well as on the infinitesimal mantle (with the surfae-normal vetor given by d 2 f = dtd x v. The volume element itself an be written as dtd 2 f v. Thus, Gauss s integral theorem gives to order (dt 2 ) d 3 x B = dt d 2 f v( B) dv = d 2 f B = V d 2 f B d 2 f B + dt (d x v) B. Thus we get d 2 f B d 2 f B = dt d 2 f v( B) dt (d x v) B + (dt 2 ). (37) Plugging this into (35) and dividing by dt leads to the desired time derivative of the flux Φ B = d 2 f [ t B + v( B)] d x ( v B). (38) In the last integral we have used (d x v) B = d x ( v B). Now using B = and omparing with (33) leads, after some simple algebra, to the orret Faraday Law of Indution, also known as the flux law: d x E + v B = 1 Φ B (t). (39) (36) 7