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Energy Conservation and Poynting's Theorem in the Homopolar Generator

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Energy Conservation and Poynting's Theorem in the Homopolar Generator Energy conservation and Poynting’s theorem in the homopolar generator Christopher F. Chybaa) Department of Astrophysical Sciences and Woodrow Wilson School of Public and International Affairs, Princeton University, Princeton, New Jersey 08544 Kevin P. Handb) Jet Propulsion Laboratory, California Institute of Technology, Pasadena, California 91109 Paul J. Thomasc) Department of Physics and Astronomy, University of Wisconsin, Eau Claire, Wisconsin 54702 (Received 13 February 2013; accepted 27 August 2014) Most familiar applications of Poynting’s theorem concern stationary circuits or circuit elements. Here, we apply Poynting’s theorem to the homopolar generator, a conductor moving in a background magnetic field. We show that the electrical power produced by the homopolar generator equals the power lost from the deceleration of the rotating Faraday disk due to magnetic braking and review the way that magnetic braking arises within Poynting’s theorem. VC 2015 American Association of Physics Teachers. [http://dx.doi.org/10.1119/1.4895389] I. INTRODUCTION from the deceleration of the rotating Faraday disk due to magnetic braking. We then review how this magnetic brak- 1 Poynting’s theorem for electric field E, magnetic flux ing arises within Poynting’s theorem. density B, and current density J follows from Maxwell’s equations (Faraday’s and Ampe`re’s laws) and the vector identity E ðrBÞ¼B ðrEÞrÁðE  BÞ.2,3 The theorem states that the rate at which work is done on the II. POWER DISSIPATION IN THE HOMOPOLAR electrical charges within a volume is equal to the decrease in GENERATOR energy stored in the electric and magnetic fields, minus the The homopolar generator, or Faraday disk, produces an energy that flowed out through the surface bounding the vol- electromotance (emf) by rotating a conducting disk in a ume. In integral form, it can be written as constant uniform magnetic field.9–11 The disk connects to an ð ð ð ammeter via brushes, one of which makes electrical contact 1 @ 1 1 EÁJdV ¼ E2 þ B2 dV À ðÞEÂB Áda; with the rim of the disk and the other with the conducting V 2@t V l l S axle of the disk. Consider the disk shown in Fig. 1, which (1) has radius b and thickness h, with an axle of radius a aligned along the z-axis. The disk rotates with angular frequency x where S is the surface with area element da bounding the in the laboratory frame, and the material of the disk has volume V, and l and are, respectively, the permeability and permittivity. In Poynting’s words, Eq. (1) means that “we must no lon- ger consider a current as something conveying energy along the conductor. A current in a conductor is rather to be regarded as consisting essentially of a convergence of elec- tric and magnetic energy from the medium upon the conduc- tor and its transformation there into other forms.”1 This can be shown, for example, in the case of a wire segment of length L and radius r, with voltage drop e carrying a steady current I. With E ¼ðe=LÞ^z and B ¼ðlI=2prÞ/^ , integrating the Poynting vector S ¼ E  B/l over the cylindrical area 2prL of the wire gives the power entering the wire’s surface. The result is identical to that obtained from Ohm’s law, with dissipated power P ¼ eI.1,3 In addition to this example, Poynting’s theorem has been applied to a number of station- ary circuits of simple geometry.4–8 Can Poynting’s theorem be applied to circuits in which a conductor is moving in the presence of a constant back- ground magnetic field? The theorem should account for such cases, but elementary electromagnetism texts do not typi- cally consider them. To examine a concrete and well-known system, we apply Poynting’s theorem to the homopolar gen- erator. We first demonstrate that the electrical power pro- Fig. 1. The Faraday disk, or homopolar generator. An ammeter measures duced by the homopolar generator equals the power lost current flowing from point O to point Q. 72 Am. J. Phys. 83 (1), January 2015 http://aapt.org/ajp VC 2015 American Association of Physics Teachers 72 conductivity r. The disk rotates in a constant magnetic field J ¼ rðE þ v  BÞ; (8) of flux density B ¼ Bz^. We first calculate the power dissipation in the disk, work- so the braking force becomes ing in the lab frame. Divide the disk into concentric cylindri- cal shells of radial thickness dr and circumferential area F ¼ J  B ¼ r½E  B þðv  BÞB: (9) A ¼ 2prh. The Lorentz force generates an emf across a shell ^ given by For v ¼ xr/ ¼ xrðsin /x^ þ cos /y^Þ; B ¼ Bz^, and with no external electric field E, we find de ¼ðv  BÞdr ¼ xrB dr; (2) F ¼rxrB2/^ : (10) where we have used v ¼ xr/^. The emf across the entire disk is then The work done per unit volume is then dW ¼ F Á dl ¼ F Á v ð dt, so the rate at which work is done decelerating the disk is b 1 ðÞ2 2 ð ð ð e ¼ de ¼ xBb À a : (3) b 2p h p a 2 F Á v rdrd/ dz ¼ rB2x2hbðÞ4 À a4 ; (11) a 0 0 2 The emf drives a current that runs from axle to rim, then through the ammeter back to the axle. Since only the disk is which is identical to Eq. (7). Equation (11) makes clear that rotating in the lab frame, the only emf generated in the sys- the power dissipated by the current driven to flow in the tem is within the disk. The resistance of a single cylindrical homopolar disk comes directly from the disk’s kinetic shell is energy of rotation. The magnetic field acts, so to speak, to convey the necessary energy from the rotation of the disk to dR ¼ðrAÞÀ1dr; (4) the electrical circuit, but the magnetic field energy is unchanged. so the resistance of the entire disk is then In Eqs. (9) and (10), we treated B as the constant external ð field, ignoring secondary magnetic fields that must result 1 b dr 1 from the generation of the current J. Lorrain et al.19 have R ¼ ¼ lnðÞb=a : (5) examined the secondary magnetic fields B and B that r a 2prh 2prh disk axle arise as a result of the current flowing in the disk and The Faraday disk is known as an intrinsically high-current axle, respectively. Both are azimuthal, so that 12 low-voltage device, and Eq. (5) can help us understand v  Bdisk ¼ v  Baxle ¼ 0. We note that neither component À1 why. An application of Ohm’s law gives I=e ¼ R contributes to magnetic braking: J  Bdisk is in the axial ¼ 2prh=lnðb=aÞ. Consider a disk made of a typical conduc- direction so it does not slow the rotating disk; and J  Baxle tor, say aluminum, for which r ¼ 3.8  107 S/m.13 If I were 1 is in the radial direction so it does not slow the rotating axle. A, then even if b ¼ 104a, we could only have e ¼ 1 volt if h were 4  10À8 m. Because e in Eq. (3) is fixed for a given B, IV. POYNTING’S THEOREM AND MAGNETIC a, and b, the resistance of the remainder of the circuit only BRAKING decreases I further. High voltages can be achieved by homo- polar generators in astrophysical contexts where the length To apply Poynting’s theorem to the homopolar generator, scales can be 103 km or more. Within our Solar System, it remains to show that the magnetic braking force (per unit however, astrophysical homopolar generators appear to pro- volume) J  B is a consequence of Poynting’s theorem, in vide only minor electrical heating of planetary satellites.14–18 which case the theorem would indeed show that the power production derives from the electromagnetic fields via mag- The total power P dissipated in the disk can be determined 20 by integrating over the increments of power dP dissipated in netic braking. Davidson gives a straightforward demonstra- each concentric shell. Using Eqs. (2) and (4), we have tion. One simply uses Ohm’s law [Eq. (8)] to rewrite the first term in Poynting’s theorem [Eq. (1)]as ðÞde 2 ð ð ð dP ¼ ¼2prB2x2hr3dr; (6) 1 dR E Á J dV ¼ J2 dV þ ðÞJ  B Á v dV; (12) V r V V where the negative sign indicates that power is being lost. Then where we have used –(v  B) Á J ¼ (J  BÐ ) Á v. In our case, E ¼ 0 so in Eq. (12) the Joule heating ðJ2=rÞ dV in the ð V b p disk is equal to the energy lost from the disk by magnetic P ¼ dP ¼ rB2x2hbðÞ4 À a4 : (7) braking. a 2 Other authors obtain magnetic braking from the Poynting theorem via the Lorentz transformations of the electromag- netic fields.21–23 Define two frames: K is the frame in which III. MAGNETIC BRAKING AND ENERGY a conductor is moving at velocity v, and K0 is the frame CONSERVATION moving at v along with the conductor. How does the quantity E Á J transform between the two frames? For v2 c2, where The current density J in the disk interacts with B to decel- c is the speed of light, we have erate the disk by magnetic braking. The magnetic braking force per unit volume is F ¼ J  B.12,19 The disk obeys Ohm’s law for moving conductors E0 ¼ E þ v  B (13) 73 Am.
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