Energy conservation and Poynting's theorem in the homopolar generatorChristopher F. Chyba, Kevin P. Hand, and Paul J. Thomas Citation: American Journal of Physics 83, 72 (2015); doi: 10.1119/1.4895389 View online: http://dx.doi.org/10.1119/1.4895389 View Table of Contents: http://scitation.aip.org/content/aapt/journal/ajp/83/1?ver=pdfcov Published by the American Association of Physics Teachers Articles you may be interested in Energy Blocks — A Physical Model for Teaching Energy Concepts Phys. Teach. 54, 31 (2016); 10.1119/1.4937969 Interference and the Law of Energy Conservation Phys. Teach. 52, 428 (2014); 10.1119/1.4895362 • Energy Conservation, Energy Literacy and the DoE eere.energy.gov/education/energy_literacy.html Phys. Teach. 51, 446 (2013); 10.1119/1.4820873 Conserving energy in physics and society: Creating an integrated model of energy and the second law ofthermodynamics AIP Conf. Proc. 1513, 114 (2013); 10.1063/1.4789665 Energy conservation Phys. Teach. 49, 520 (2011); 10.1119/1.3651743

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Energy conservation and Poynting’s theorem in the homopolargenerator

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

Poynting’s theorem1 for electric field E, magnetic fluxdensity B, and current density J follows from Maxwell’sequations (Faraday’s and Ampere’s laws) and the vectoridentity E � ðr � BÞ ¼ B � ðr � EÞ � r � ðE� BÞ.2,3 Thetheorem states that the rate at which work is done on theelectrical charges within a volume is equal to the decrease inenergy stored in the electric and magnetic fields, minus theenergy that flowed out through the surface bounding the vol-ume. In integral form, it can be written as

ðV

E �JdV¼�1

2

@

@t

ðV

�E2þ1

lB2

� �dV�1

l

ðS

E�Bð Þ�da;

(1)

where S is the surface with area element da bounding thevolume V, and l and � are, respectively, the permeability andpermittivity.

In Poynting’s words, Eq. (1) means that “we must no lon-ger consider a current as something conveying energy alongthe conductor. A current in a conductor is rather to beregarded 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 canbe shown, for example, in the case of a wire segment oflength L and radius r, with voltage drop e carrying a steadycurrent I. With E ¼ ðe=LÞz and B ¼ ðlI=2prÞ/, integratingthe Poynting vector S¼E�B/l over the cylindrical area2prL of the wire gives the power entering the wire’s surface.The result is identical to that obtained from Ohm’s law, withdissipated power P¼ 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 aconductor is moving in the presence of a constant back-ground magnetic field? The theorem should account for suchcases, but elementary electromagnetism texts do not typi-cally consider them. To examine a concrete and well-knownsystem, we apply Poynting’s theorem to the homopolar gen-erator. We first demonstrate that the electrical power pro-duced by the homopolar generator equals the power lost

from the deceleration of the rotating Faraday disk due tomagnetic braking. We then review how this magnetic brak-ing arises within Poynting’s theorem.

II. POWER DISSIPATION IN THE HOMOPOLAR

GENERATOR

The homopolar generator, or Faraday disk, produces anelectromotance (emf) by rotating a conducting disk in aconstant uniform magnetic field.9–11 The disk connects to anammeter via brushes, one of which makes electrical contactwith the rim of the disk and the other with the conductingaxle of the disk. Consider the disk shown in Fig. 1, whichhas radius b and thickness h, with an axle of radius a alignedalong the z-axis. The disk rotates with angular frequency xin the laboratory frame, and the material of the disk has

Fig. 1. The Faraday disk, or homopolar generator. An ammeter measures

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

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conductivity r. The disk rotates in a constant magnetic fieldof flux density B ¼ Bz.

We first calculate the power dissipation in the disk, work-ing in the lab frame. Divide the disk into concentric cylindri-cal shells of radial thickness dr and circumferential areaA¼ 2prh. The Lorentz force generates an emf across a shellgiven by

de ¼ ðv� BÞ � dr ¼ xrB dr; (2)

where we have used v ¼ xr/. The emf across the entiredisk is then

e ¼ðb

a

de ¼ 1

2xB b2 � a2ð Þ: (3)

The emf drives a current that runs from axle to rim, thenthrough the ammeter back to the axle. Since only the disk isrotating in the lab frame, the only emf generated in the sys-tem is within the disk. The resistance of a single cylindricalshell is

dR ¼ ðrAÞ�1dr; (4)

so the resistance of the entire disk is then

R ¼ 1

r

ðb

a

dr

2prh¼ 1

2prhln b=að Þ: (5)

The Faraday disk is known as an intrinsically high-currentlow-voltage device,12 and Eq. (5) can help us understandwhy. An application of Ohm’s law gives I=e ¼ R�1

¼ 2prh=lnðb=aÞ. Consider a disk made of a typical conduc-tor, say aluminum, for which r¼ 3.8� 107 S/m.13 If I were 1A, then even if b¼ 104a, we could only have e¼ 1 volt if hwere 4� 10�8 m. Because e in Eq. (3) is fixed for a given B,a, and b, the resistance of the remainder of the circuit onlydecreases I further. High voltages can be achieved by homo-polar generators in astrophysical contexts where the lengthscales can be � 103 km or more. Within our Solar System,however, astrophysical homopolar generators appear to pro-vide only minor electrical heating of planetary satellites.14–18

The total power P dissipated in the disk can be determinedby integrating over the increments of power dP dissipated ineach concentric shell. Using Eqs. (2) and (4), we have

dP ¼ � deð Þ2

dR¼ �2prB2x2hr3dr; (6)

where the negative sign indicates that power is being lost.Then

P ¼ðb

a

dP ¼ �p2

rB2x2h b4 � a4ð Þ: (7)

III. MAGNETIC BRAKING AND ENERGY

CONSERVATION

The current density J in the disk interacts with B to decel-erate the disk by magnetic braking. The magnetic brakingforce per unit volume is F¼ J�B.12,19 The disk obeysOhm’s law for moving conductors

J ¼ rðEþ v� BÞ; (8)

so the braking force becomes

F ¼ J� B ¼ r½E� Bþ ðv� BÞ � B�: (9)

For v ¼ xr/ ¼ xrð�sin /x þ cos /yÞ; B ¼ Bz, and withno external electric field E, we find

F ¼ �rxrB2/: (10)

The work done per unit volume is then dW¼F � dl¼F � vdt, so the rate at which work is done decelerating the disk is

ðb

a

ð2p

0

ðh

0

F � v r dr d/ dz ¼ �p2

rB2x2h b4 � a4ð Þ; (11)

which is identical to Eq. (7). Equation (11) makes clear thatthe power dissipated by the current driven to flow in thehomopolar disk comes directly from the disk’s kineticenergy of rotation. The magnetic field acts, so to speak, toconvey the necessary energy from the rotation of the disk tothe electrical circuit, but the magnetic field energy isunchanged.

In Eqs. (9) and (10), we treated B as the constant externalfield, ignoring secondary magnetic fields that must resultfrom the generation of the current J. Lorrain et al.19 haveexamined the secondary magnetic fields Bdisk and Baxle thatarise as a result of the current flowing in the disk andaxle, respectively. Both are azimuthal, so thatv�Bdisk¼ v�Baxle¼ 0. We note that neither componentcontributes to magnetic braking: J�Bdisk is in the axialdirection so it does not slow the rotating disk; and J�Baxle

is in the radial direction so it does not slow the rotating axle.

IV. POYNTING’S THEOREM AND MAGNETIC

BRAKING

To apply Poynting’s theorem to the homopolar generator,it remains to show that the magnetic braking force (per unitvolume) J�B is a consequence of Poynting’s theorem, inwhich case the theorem would indeed show that the powerproduction derives from the electromagnetic fields via mag-netic braking. Davidson20 gives a straightforward demonstra-tion. One simply uses Ohm’s law [Eq. (8)] to rewrite the firstterm in Poynting’s theorem [Eq. (1)] as

ðV

E � J dV ¼ 1

r

ðV

J2 dV þð

V

J� Bð Þ � v dV; (12)

where we have used –(v�B) � J¼ (J�B) � v. In our case,E¼ 0 so in Eq. (12) the Joule heating

ÐVðJ2=rÞ dV in the

disk is equal to the energy lost from the disk by magneticbraking.

Other authors obtain magnetic braking from the Poyntingtheorem via the Lorentz transformations of the electromag-netic fields.21–23 Define two frames: K is the frame in whicha conductor is moving at velocity v, and K0 is the framemoving at v along with the conductor. How does the quantityE � J transform between the two frames? For v2� c2, wherec is the speed of light, we have

E0 ¼ Eþ v� B (13)

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and

J0 ¼ J; (14)

so

E0 � J0 ¼ E � J� v � ðJ� BÞ: (15)

Equation (14) is more complicated if J has a component par-allel to B, but for the Faraday disk that is not the case. Ofcourse, the frame of a rotating disk is an accelerating frameand Maxwell’s equations must be modified as a result, butthese modifications are of order (v/c)2 so we may ignorethem here.24 In K0 Ohm’s law is just J0 ¼ rE0, so E0 � J0¼ J02=r ¼ J2=r and, because E¼ 0 in our case, Eq. (15)gives J2/r¼ –(J�B) � v. Integration over the relevant vol-ume once again shows that the Joule heating equals theenergy lost from the disk by magnetic braking.

V. LINEAR ANALOG

For completeness, we now show that the same approachyields consistent results for the linear analog to the homopo-lar generator. Consider an infinitely long (along the x-axis)conducting rectangular bar moving with velocity v ¼ vxthrough a field B perpendicular to the bar and direction ofmotion (Fig. 2).25 The bar has finite width l and height h inthe y and z directions, respectively. If the bar were instead fi-nite along the x-axis and moving along connected stationaryrails, it would be a so-called rail gun.26–29

The emf across an increment dl of the bar in the y-direc-tion is de¼ vBdl. We calculate power dissipated per volumedV¼ hwdl, where w is some specified distance in the x-direc-tion. Equation (4) gives dR¼ dl/rwh, so by Eq. (6), thepower dissipated per volume is

� deð Þ2

dR dV¼ �rv2B2: (16)

This is to be compared to the dissipated power calculatedfrom magnetic braking. Using Ohm’s law and following Eq.(9), the corresponding magnetic braking force per volume is

F ¼ �rvB2x; (17)

so the work done per unit volume is

F � v ¼ �rv2B2; (18)

which is identical to Eq. (16).

VI. CONCLUSION

Poynting’s theorem has been successfully applied to thecase of the homopolar generator (as well as its linear analog),in which a current flows in a circuit due to the motion of aconductor through a uniform background magnetic field. Thepower generated derives from the magnetic braking deceler-ation of the rotating Faraday disk. This magnetic braking is anatural consequence of Poynting’s theorem.

ACKNOWLEDGMENTS

P.J.T. thanks the Office of Research and SponsoredPrograms at the University of Wisconsin-Eau Claire forsabbatical support. K.P.H. acknowledges support from theJet Propulsion Laboratory, California Institute ofTechnology, under a contract with the National Aeronauticsand Space Administration (NASA), and from the NASAExobiology Program.

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