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Unipolar Induction

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Unipolar Inductionvia a Rotating, Conducting, Magnetized CylinderKirk T. McDonaldJoseph Henry Laboratories, Princeton University, Princeton, NJ 08544(August 17, 2008; updated November 13, 2012)1ProblemA conducting cylinder of radius R with permanent magnetization density M0parallel to itsaxis when at rest is rotated about that axis with angular velocity ω = ωˆz with respect tothe lab frame. A voltmeter with very high internal resistance is connected to the rotatingcylinder via wires with sliding contacts, one of which (C1) is on the axis of the cylinder andthe other (C2) is on the circumference, as shown below.Deduce the voltage V observed on the voltmeter by a lab-frame analysis as well as by ananalysis in the rotating frame. You may assume that the velocity ωR is small compared tothe speed of light c. Comment on the electric polarization density P in the cylinder shouldit have (relative) permittivity that differs from unity.This configuration of unipolar induction was first considered by Faraday in 1851 [1], whoalso considered the case of the magnetized cylinder at rest while the voltmeter and contactwires rotated around the axis of the cylinder.12Solution2.1 Analysis Using a Comoving Inertial FrameAs discussed in [2, 3, 4, 5], the best approach to an understanding of lab-frame electrody-namics of a rotating system is via a comoving inertial frame corresponding to some point inthe rotating system.We follow Minkowski [2] in arguing that the local magnetization at a point P in therotating cylinder equals the rest value M0according to an observer in the inertial framethat is instantaneously comoving with point P.ThatisM= M0, where the superscriptindicates quantities observed in the comoving inertial frame.Similarly, we expect that the electric polarization Pnear point P in the comoving inertialframe equals that of the magnetized cylinder in an inertial rest frame, namely P=0.Writing v as the velocity of point P in the lab frame, the field transformations to thecomoving inertial frame are [6] (see also [7]), in Gaussian units and to order v/c where c isthe speed of light in vacuum,B= B −vc× E, D= D +vc× H, E= E +vc× B, H= H −vc× D,M= M +vc× P, P= P −vc× M, (1)and the inverse transformations areB = B+vc× E, D = D−vc× H, E = E−vc× B, H = H+vc× D,M = M−vc× P, P = P+vc× M. (2)We now find the lab-frame polarization and magnetization densities inside the rotating cylin-der to beP =vc× M0, M = M0. (3)As a next step we deduce the magnetic fields B and H in the lab frame from the mag-netization density M = M0along the axis of the cylinder. Formally, the magnetic field Bcan be deduced from a vector potential A,B = ∇ × A, (4)where in the absense of free currents the vector potential is related byA =1cc∇ × MRdVol +1c surfacecM × dAreaR, (5)while the H field can be deduced (in the absense of free currents) from a magnetic scalarpotential φMaccording toH = −∇φM, (6)andφM=−∇ · MRdVol + surfac eM · dAreaR. (7)2In the present example with uniform magnetization only the surface integrals contributeto eqs. (5) and (7), where the source is the magnetization surface current density on thecylindrical surface in eq. (5), and a uniform surface density of magnetic poles on the flat endsurfaces in eq. (7). For a long cylinder the interior B field is essentially uniform,Bin≈ 4πM0, and Hin= B − 4πM0≈ 0 (long cylinder), (8)while for a thick disk the interior H field is essentially uniform,Hin≈−4πM0, and Bin= H +4πM ≈ 0 (thin disk). (9)In all cases the conduction electrons must be at rest with respect to the rotating cylinder,which implies that the interior electric field vanishes in the comoving frame, 0 = Ein=Ein+ v/c × Bin, recalling eq. (2). Thus,Ein= −vc× Bin. (10)A unipolar generator requires a nonzero internal electric field, so that there is a nonzeroelectric potential difference between the sliding contacts on the rotating disk. Hence, a thindisk does not make a good unipolar generator (although this is often used in ”cartoons” ofthese devices).We restrict our discussion in the rest of this note to the case (8) of a long cylinder.Then,Ein= −vc× 4πM0= −ω × rc× 4πM0= −4πωM0cr⊥, (11)and the electric potential inside the rotating cylinder can be written,Vin= −2πωM0cr2⊥, (12)The potential difference between a point on the cylindrical surface, at radius R,andoneonthe axis isΔV =2πRωM0c. (13)For completeness, we note that the polarization density inside the cylinder is, recallingeq. (3),Pin=vc× M0=ωM0cr⊥, (14)so the bound volume and surface charge densities (on the cylindrical surface) areρbound= −∇ · P = −2ωM0c, and σbound= P(R−) ·ˆr⊥=ωRM0c(15)with no surface charge on the flat ends of the cylinder. The total bound charge is, of course,zero. Finally, the electric displacement isDin= Ein+4πPin=0. (16)This is not a trivial result, and in the related case of a rotating, conducting magnetizedsphere the electric displacement is nonzero inside the sphere [8].32.2 Analysis in the Lab FrameThe analysis in the lab frame follows Chap. E III of [9], where it is naively assumed thatthe magnetization of the rotating cylinder in the lab frame is M0. See also [11]. Strictlyspeaking, the analysis of sec. 2.2 does not hold without having first made the arguments ofsec. 2.1.2.2.1 Analysis via Faraday’s LawThe current in the circuit ABCDA in the figure below is negligible because of the highresistance of the voltmeter, so the resistive voltage drop in the circuit can be ignored.Then, the reading V on the voltmeter equals the electromotive force around the circuit.According to Faraday’s law, the electromotive force in the lab frame is given byV = E · ds = −1cdΦBdt, (17)in Gaussian units, where ΦB=lo opB·dArea is the magnetic flux linked by the loop ABCDA.Since the magnetic field B = H+4πM =4πM0ˆz inside the (long) cylinder has no azimuthalcomponent in this example, it might seem that ΦB= 0 and hence, V = 0. However, becausethe cylinder is rotating, we can argue that the portion BCD of the loop deforms into BB’C’Dduring time dt where the arc BB’ has length vdt= ωR dt. Hence, the flux through the loopincreases by amount ωR2Bdt/2 during time dt, and the voltage according to eq. (1) isV = −ωR2B2c= −2πωR2M0c. (18)The negative sign of the voltage (18) means that it is higher at points B


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