Feynman Simplified 2B: Magnetism & Electrodynamics [2 ed.]

Table of contents :
Chapter 14  Magnetostatics
Chapter 15 The Vector Potential
Chapter 16 Electromagnetic Statics Wrap-Up
Chapter 17 Induced Currents
Chapter 18 Exploring Induction
Chapter 19 Maxwell’s Equations
Chapter 20 Wave Solutions in Vacuum
Chapter 21 Solutions with Charges
Chapter 22 Electrical Circuits
Chapter 23 Cavity Resonators
Chapter 24 Review: Feynman Simplified 2B

Citation preview

Feynman Simplified 2B: Magnetism & Electrodynamics Everyone’s Guide to the Feynman Lectures on Physics by Robert L. Piccioni, Ph.D.

Copyright © 2015 by Robert L. Piccioni Published by Real Science Publishing 3949 Freshwind Circle Westlake Village, CA 91361, USA Edited by Joan Piccioni

All rights reserved, including the right of reproduction in whole or in part, in any form. Visit our web site www.guidetothecosmos.com

Everyone’s Guide to the Feynman Lectures on Physics Feynman Simplified gives mere mortals access to the fabled Feynman Lectures on Physics.

This Book Feynman Simplified: 2B covers one quarter of Volume 2 of The Feynman Lectures on Physics. The topics we explore include: Magnetostatics Dynamic Electric & Magnetic Fields Filters & Transmission Lines Electromagnetic Waves in Vacuum Electrical Circuits & Components Circuit & Cavity Resonances To find out about other eBooks in the Feynman Simplified series, click HERE. I welcome your comments and suggestions. Please contact me through my WEBSITE. If you enjoy this eBook please do me the great favor of rating it on Amazon.com or BN.com.

Table of Contents Chapter 14: Magnetostatics Chapter 15: The Vector Potential Chapter 16: Electromagnetic Statics Wrap-Up Chapter 17: Induced Currents Chapter 18: Exploring Induction Chapter 19: Maxwell’s Equations Chapter 20: Wave Solutions in Vacuum Chapter 21: Solutions with Charges Chapter 22: Electrical Circuits Chapter 23: Cavity Resonators Chapter 24: Review: Feynman Simplified 2B

Chapter 14 Magnetostatics In Chapter 4, we learned that in static conditions, when charges and currents are unchanged over time, Maxwell’s four equations reduce to two de-coupled pairs of equations: Electrostatics: Ď•E = ρ/ε

0

Ď×E = 0 Magnetostatics: Ď×B = j/ε

0

Ď•B = 0 In the static case, we can treat electricity and magnetism as if they were unrelated phenomena. They become interconnected only when charges move or currents change. The preceding nine chapters explored electrostatics. We now begin the study of magnetostatics, the physics of static magnetic fields. Magnetic forces are more complex than electric forces, as demonstrated by the Lorentz force that combines both effects. F = q(E + v×B) The electric component of force depends only on the charge q and the electric field E. The magnetic component depends on charge q, magnetic field B, and also on the charge’s velocity v, both on its magnitude and its orientation. The magnetic component of force F is always perpendicular to both v and B. Its polarity is defined by the cross product and the right hand rule. Please ensure you master this sign convention; it would be a shame to solve a complex problem and get the wrong answer because you didn’t orient your hand properly. Different people get the right answer in different ways. The image below shows how I do it: the index finger points along the first vector (v in this case); the middle finger points along the second vector (B); and the thumb points along the cross product (F).

Figure 14-1 shows vectors v and B within the same plane (shown in gray), with θ being the angle between them. You’ll need to twist your wrist to get the right sign here.

Figure 14-1 Magnetic Force: F = v×B

Force F is perpendicular to both vectors and has magnitude: F = v B cosθ Magnetic fields can be measured in many equivalent units. For θ=90º, v=1m/s, and F=1 newton. The strength of the magnetic field is: 1 newton-second / coulomb-meter 1 volt-second / meter 2

1 weber / meter 1 tesla = 10,000 gauss 2

Current & Charge Conservation Electric current is electric charge in motion. A single charge q with velocity v produces a current J given by J=qv. Like fields, currents from multiple charges add vectorially according to the principle of linear superposition. In the same way that we dealt with heat flow in Chapter 2, Feynman defines current density as a vector field j(r). At each point r, j points in the direction of current flow, and the magnitude of j equals the amount of charge passing parallel to j per unit area per unit time. The amount of charge ΔQ passing in time Δt through a small surface of area S, whose unit normal is n, equals: ΔQ / Δt through S = j•n S If S is perpendicular to j, n is parallel to j, and this equation reduces to our definition: the amount of charge ΔQ per unit area per unit time equals the magnitude of j. The prior equation is illustrated in Figure 14-2. The physical situation is three-dimensional; Figure 14-2 shows it in a 2-D cross section. Here, current j is vertical and we represent surface S with a bold diagonal line whose unit normal n is at angle θ to j.

Figure 14-2 Current Flow Through Surface S

If a collection of electric charges has charge density ρ(r) per unit volume, and average velocity v, each charge will move an average distance v•Δt in time Δt. In Figure 14-2, all charges within the gray parallelepiped are below S by a distance of no more than v•Δt; hence that much charge will, on average, move up through S during time Δt. The amount of charge ΔQ contained within the gray parallelepiped is its volume V multiplied by ρ. This is: ΔQ through S = (V) ρ ΔQ through S = (v Δt S cosθ) ρ ΔQ / Δt through S = v•n S ρ ΔQ / Δt through S = j•n S with j = v ρ If the collection of charges consists of N identical particles per unit volume, each with charge q, the last equation becomes: j = v Nq The total electric current J, the total charge per unit time, flowing through any surface S equals the current density integrated over S: J = ∫ j•n dS S

Physicists generally use i or I for current, but that is harder to read in an eBook than j or J (is “l” capital i or lower case L?). Also i may be confused with √–1. For clarity, I will use j and J for electrical current. With Gauss’ law, we can replace the above surface integral with an integral over the volume V contained within a closed surface S: J = ∫ Ď•j dV V

J is then the current flowing out of V. A fundamental principle of physics is that net electric charge is locally conserved. An electron and an antielectron (a “positron”) can annihilate, with both particles disappearing, their mass energies converting into gamma rays with the loss of one negative and one positive charge. An electron / positron pair can also be created, for example in neutral pion decay, with the creation of one negative and one positive charge. Neither reaction changes net charge: the sum of all positive charges minus the sum of all negative charges. Physicists have never observed a reaction that fails to conserve net charge. Local conservation is a stronger requirement than global conservation. If charge were conserved only globally, an electron could disappear in London provided that a new electron appeared in Auckland. Local conservation states that the net charge within any volume V changes only when a current crosses its boundary. If the current carries charge ΔQ into V, the net charge in V increases by ΔQ, which can be positive or negative.

It is the principle of local charge conservation that leads us to focus on how electric currents carry charge from place to place. With local charge conservation we are assured that: dQ/dt within V = ∫ dρ/dt dV = – ∫ Ď•j dV V

V

(Recall that Q within V = ∫ ρ dV). Since the above equation must be true for every volume V, this proves the differential equation: V

dρ/dt = – Ď•j

Magnetic Force on Currents In Chapter 1, we said that magnetic fields exert forces on wires carrying currents. We can now derive an equation for that force. The Lorentz force law provides the force on an individual charge q moving with velocity v in magnetic field B: F = q v×B Let N be the number of charges per unit volume, each with charge q moving with velocity v. We will calculate the force G on a short piece of cylindrical wire of length ΔL and cross sectional area A. As shown in Figure 14-3, the force F on each individual charge q is perpendicular to both B and v, with v parallel to the wire’s axis, the direction of current flow.

Figure 14-3 Magnetic Force F on Charge q

By linear superposition, the total force per unit volume equals NF, (the number of charges per unit volume) multiplied by (the force per charge). The total force G is then NF multiplied by AΔL, the volume of the piece of wire. This means: G = N(qv×B) A ΔL

Recall that j=vNq, and that the total current J equals the integral of j across the surface through which current passes. For a wire with uniform current density: J=jA. Combining all this yields: G = j×B A ΔL G / ΔL = J×B The force on the wire per unit length is G/ΔL that equals J×B. As Feynman stresses in V2p13-3, the force on a wire is proportional to the total current and not on q, the charge of individual particles. We get the same force for positive charges moving right as for negative charges moving left. This is fortunate, since scientists originally incorrectly guessed the carrier polarity.

Magnetic Fields From Moving Charges Also in Chapter 1, we said that current-carrying wires exert forces on magnets. Let’s see how that happens. Newton’s third law of motion, action-begets-reaction, suggests that if a magnet exerts a force on a current-carrying wire, there should be a reaction force exerted on the magnet, the source of the original force. Indeed, from our experience with magnets, we know that they exert non-contact forces upon one another through their magnetic fields. For constant fields and currents, the equations of magnetostatics are: Ď•B = 0 c Ď×B = j / ε 2

0

Feynman pauses here to note that statics — the condition of constant fields and currents — is an idealization, similar to friction-less mechanical motion. We make such approximations to facilitate learning physical principles by minimizing distracting complications. If you are going to design a rocket to go to Mars, you should make more realistic assumptions. If you want to learn physics in order to qualify for that job, read on. Feynman notes that since Ď•Ď×ξ=0 for any vector field ξ, the last equation is valid only if Ď•j=0. From our earlier equation for the local conservation of charge we have dρ/dt=–Ď•j. Thus the equations of magnetostatics require dρ/dt=0. A changing charge density would generate changing electric fields, contrary to our static assumption of constant fields. All this means our assumptions are consistent, even if perhaps highly idealized. A final comment in this regard: Ď•j=0 means that currents must flow in closed loops. In statics, currents cannot start here and end there; they must flow through complete circuits that might include batteries, inductive coils, and similar elements, but they cannot include capacitors. Constant currents

change a capacitor’s charge and therefore its electric field. We will get to electromagnetic dynamics in later chapters. Here we examine the physics of constant fields and currents. The first equation of magnetostatics, Ď•B=0, says there are no magnetic monopoles: no magnetic equivalent of a single electric charge. We said earlier that a magnet’s B field lines start at its north pole and circle around to its south pole. But actually, the field lines continue within the magnet forming a closed loop. Magnetic field lines never start at point X or end at point Y. Feynman says that in some complex situations, the B field lines are not simple loops, but that they nonetheless never stop or start at a point. Everything in this paragraph is true for all of electromagnetism, static or dynamic. The second equation of magnetostatics, Ď×B=j/c ε , enables calculating the B field produced by a current j. Using Stokes’ theorem we can relate the circulation of B to the flux of its curl: 2

0

∫ (Ď×B)•n dS = ∫ B•ds S

Γ

Here, S is a surface enclosed by the curve Γ, n is the unit normal to S at each point, and ds is an infinitesimal tangent vector to Γ at each point. This is illustrated in Figure 14-4. Substituting j/c ε for the curl of B yields: 2

0

∫ (j/c ε )•n dS = ∫ B•ds 2

S

0

Γ

Figure 14-4 Stokes’ Theorem Relating B & j

Thus (the tangential component of B integrated around curve Γ) equals (the total current flux through surface S) divided by (c ε ). The equation, known as Ampere’s law, is: 2

0

∫ B•ds = current J through S / (c ε ) 2

Γ

0

Comparing the equations of electrostatics with those of magnetostatics, we see a reversal of the roles of curl and divergence. For electrostatics, the curl of E is always zero, while the divergence of E is proportional to charge density. For magnetostatics, the divergence of B is always zero, while the curl of B is proportional to current density.

Fields From Wires & Solenoids Our first example of employing Ampere’s law is finding the magnetic field B around an infinitely long, cylindrical wire carrying current J. Define a cylindrical coordinate system with the z-axis along the length of the wire, r being the radial distance from the z-axis, and β being the azimuthal angle around the wire, as shown in Figure 14-5. The dashed circle labeled Γ has radius r, and is centered on and lies in a plane perpendicular to the wire.

Figure 14-5 Magnetic Field Near Wire

In cylindrical coordinates, we can express the most general form of the B field as: B = (R, ß, Z) In general, B‘s components R, ß, and Z may be functions of all three coordinates r, β, and z. But due to the wire’s symmetry, none can be a function of β or z. We can see this by imagining rotating the wire around its axis, or moving the wire along its length. Neither action makes any physical difference due the wire’s infinite length and rotational symmetry. Hence no real physical entity, including the components of the magnetic field, can change as a function of β or z. Each component of B must be a function of r only. In V2p13-5, Feynman says: “We will assume something which may not be at all evident, but which is nevertheless true: that the field lines of B go around the wire in closed circles.” (We prove this in the next chapter.) This means R=Z=0 and B = (0,ß,0). Let’s calculate ß(r), the tangential component of B, on the circle Γ of radius r. Every point along Γ has the same r and therefore the same ß(r). By Ampere’s law: ∫ B•ds = J/c ε 2π r ß(r) = J/c ε ß(r) = J/(2πε c r) 2

Γ

0 2

0

2

0

The tangential component ß is orthogonal to both J and the radial coordinate r. We can express this equation in vector form as: B = J × r/(2πε c r ) 2

2

0

You might check that this cross product employs the right hand rule. Since current-carrying wires produce magnetic fields, and since magnetic fields exert forces on one another (as we know from playing with magnets), it follows that current-carrying wires exert forces on one another. In Chapter 1, we said that parallel wires are pushed together if their currents are in the same direction and pushed apart if their currents are in opposite directions. This is illustrated in Figure 14-6.

Figure 14-6 Forces on Current-Carry ing Wires

Here, current flows toward you in the central three wires, and flows away from you in the outer two wires. This is indicated using a standard convention: the black dots represent the tips of arrows pointing toward you, while the crosses represent the feathers of arrows pointing away from you. Figure 14-6 shows two magnetic field lines (labeled B) produced by the central wire, and the four forces (labeled F) that these fields exert on the other four wires. As the figure demonstrates, the central wire attracts wires with parallel currents and repels wires with antiparallel currents. You might exercise your right hand and verify the polarities of each field and force. A physicist needs to practice this skill until it becomes automatic. Our next example of Ampere’s law is a solenoid: a coiled, current-carrying wire. Figure 14-7 shows current J flowing into a coiled wire at the right and out at the left. The figure also shows two complete magnetic field lines and portions of six others.

Figure 14-7 Solenoid: Image by Hy perPhy sics

Additionally, the figure shows a rectangular closed curve Γ with horizontal length L. One horizontal side of Γ is inside the coil and the opposite side is outside the coil. As the field lines suggest, the magnetic field is much stronger inside the solenoid’s coils than outside. To a first approximation, we can assume B=0 outside the coil. This means Ampere’s law becomes quite simple. The path integral of B around Γ is zero on three sides, and equals BL along Γ’s horizontal side within the coil. The current flux through Γ equals nJL, where n is the number of coils per unit length. The equations are: ∫ B•ds = Current through Γ / c ε B L= n J L/ c ε B=nJ / cε 2

Γ

0

2

0

2

0

One more comment about solenoids. The coils strongly attract one another, because the current flow in each wire is parallel to all others. The wires must be insulated, covered with a non-conductive coating, to prevent current from jumping directly to an adjacent coil rather than flowing along the wire’s length. In high-field solenoids, the coils must also be mechanically restrained. Solenoids vividly demonstrate magnetic fields arising from currents. Indeed all magnetic fields arise from either current flow or changing electric fields. We will discuss the latter when we get to electrodynamics. Another example of the former is an iron bar magnet. We will delve into the complexities of how atoms sustain currents in later chapters. For now, suffice it to say that electrons have magnetic moments due to their intrinsic spin, and that in some materials (such as iron) electrons can align themselves with their individual moments accumulating to macroscopic proportions.

Relativity & Electromagnetism In V2p13-6, Feynman explores the intimate relationship between electromagnetism and the special theory of relativity. He first points out that, in stating the Lorentz force F=q(E+v×B), we failed to specify a reference frame — in which frame is v measured? And in which frame do we define E and

B? Feynman says: “It turns out that any inertial frame will do. We will also see that magnetism and electricity are not independent things—that they should always be taken together as one complete electromagnetic field. Although in the static case Maxwell’s equations separate into two distinct pairs, one pair for electricity and one pair for magnetism, with no apparent connection between the two fields, nevertheless, in nature itself there is a very intimate relationship between them that arises from the principle of relativity. Historically, the principle of relativity was discovered after Maxwell’s equations. It was, in fact, the study of electricity and magnetism which led ultimately to Einstein’s discovery of his principle of relativity. But let’s see what our knowledge of relativity would tell us about magnetic forces if we assume that the relativity principle is applicable—as it is—to electromagnetism.” The first question to address is the impact of relativistic velocities on charge. We know that mass, length, and time change at relativistic velocities. So we ask: Do fast electrons and stationary electrons have the same charge? Feynman makes two compelling arguments that electric charge is invariant. Firstly, heating an electrically neutral block of matter, such as copper, does not give it a non-zero charge. At higher temperatures, particles have greater thermal kinetic energy. Since the mass of a single electron is 117,000 times less than the mass of a copper nucleus, the velocities of copper’s negative charges change with temperature 340 times faster than those of its positive charges. If the unit charge changed with velocity, copper’s negative charges would change more than its positive charges, and a block of copper would acquire a net charge. Even a change of one part per billion would produce many tons of electrostatic force between two heated 1 kg copper blocks. No such effect has been observed. Secondly, electron energies and velocities are also changed by various chemical reactions. Feynman says: “a straightforward calculation shows that even a very small dependence of charge on speed would give enormous fields from the simplest chemical reactions. No such effect is observed.” While charge is relativistically invariant, charge density is not. This is because length along the direction of motion is not invariant. The Lorentz contraction (see Feynman Simplified 1C Chapter 26) states that an object that has length L in its own rest frame appears to have a shorter length L* in a frame moving at velocity v in the same direction as L. The equation is: L* = L √(1–v /c ) 2

2

Figure 14-8 illustrates the effect of length contraction on charge density. In the upper image, in the stationary reference frame S, a gray block has length L, cross-sectional area A, and a number of charged particles (black dots) that sum to charge Q. In the lower image, in the frame S* that is moving relative to S at velocity v to the right, the block has length L*, cross-section A, and contains the same particles with the same charge Q.

Figure 14-8 Charge Q in 2 Frames

Relativity says that area A is the same in both frames because it is orthogonal to the direction of motion. The key point is that both volumes contain the same charge Q, because charge is invariant. However, the charge densities observed in S and S* are different because the blocks’ volumes are different. These are: in S: ρ = Q / L A in S*:ρ*= Q / L* A ρ* L* = ρ L = Q / A ρ* = ρ / √(1–v /c ) 2

2

As Feynman notes in V2p13-9, charge density and mass vary with velocity in the same way.

Magnetism as a Relativistic Effect Now, let’s see how relativity and electromagnetism work together. Imagine a charge Q at a distance r from the center of an infinitely long, current-carrying wire. We will take the realistic view that current results from the flow of electrons, while the positive nuclei are stationary: v =0. For negative electrons moving to the right with velocity v , electric current J is p

n

flowing to the left. In a real conductor, not all electrons move; those that remain attached to their atoms do not contribute to current flow. For our purposes, we can ignore stationary electrons and neutral atoms, and consider only the conduction electrons and the positive ions they leave behind. We define the wire’s positive and negative charge densities as ρ and ρ , respectively. p

n

Let’s consider this situation in two reference frames: frame S in which Q is stationary; and frame S* that is moving with the electrons to the right at velocity v relative to S. We define positive velocity to be toward the right. Thus v >0, and velocities are transformed from S to S* by subtracting v . n

n

n

Figure 14-9 shows this situation as viewed in both frames.

Figure 14-9 Wire & Charge Q in 2 Frames

In S, the wire’s magnetic field exerts a vertical force on charge Q that has magnitude F. The force is upward, away from the wire if Q>0, and downward, toward the wire if Q0) with a force given by: F = Q A |ρ | v /(2πε c r) F = Q A ρ v /(2πε c r) n

p

2 n 2 n

2

0

2

0

In the S* frame, charge Q is now stationary: v* =0. Since Q’s velocity is zero, the wire’s magnetic field cannot exert a force on Q. But there must be a force between Q and the wire in every reference frame. Q

To understand why, recall what we learned about special relativity in Feynman Simplified 1C, chapters 25 through 29. The distance between two objects is a relative quantity: its measured value may be different in different inertial reference frames. But how that distance changes is invariant. If the distance increases (or decreases) in one frame, it must increase (or decrease) in all other frames; the amount of change may be different, but the existence of change and its polarity is not. In the limiting case that the objects collide (distance becomes zero), in one frame, they must collide in every frame. Hence, while the wire cannot exert a magnetic force on charge Q in frame S*, where Q is stationary, there must be some other force accelerating Q. The answer to this apparent contradiction lies in the wire. While the wire is neutral in S, we showed

above that it is positively charged in S*. That positive charge exerts a force on Q. The electric field at a distance r from the center of a wire with charge λ per unit length is (see Chapter 5, Figure 5-5): E = λ / (2π ε r) 0

For our wire, (charge per unit length) equals (charge per unit volume) multiplied by (cross-sectional area A). In S*: λ* = (ρ* + ρ* ) A λ* = A ρ {v /c } / √(1–v /c ) p

n

2 n

p

2

2 n

2

The electric force exerted on Q by the charged wire in S* is: F* = Q E* F* = Q A ρ {v /c } / {√(1–v /c ) (2π ε r)} F* = {Q A ρ v / (2πε c r)} / {√(1–v /c ) } 2 n 2 n

p

p

2

2 n

2

2

0

0 2 n

2

The final step is transforming forces from S to S*. The force on charge Q is vertical, orthogonal to the direction of motion. Define the y-axis to be vertical, pointing upward. Any force in the y-direction can be written in terms of the rate of change of y-momentum p as: y

in S: F = Δp / Δt in S*: F* = Δp * / Δt* y

y

y

y

Relativistically, y-momentum is unchanged by a Lorentz transformation along an orthogonal axis, the horizontal axis in this case. But, time does change. For charge Q, S* is its rest frame and Δt* is its proper time. From Q’s frame, clocks seem to run slower in S, which means physical events seem to last longer. (Unstable particles take more clock time to decay in frames in which they are moving.) This means Δt is greater than Δt*, according to: Δt = Δt* / √(1–v /c ) 2 n

2

Combining these equations, we find: Δp = Δt F Δp = Δt {QA ρ v /(2πε c r)} y

y

y

2 n

p

2

0

Δp * = Δt* F* Δp * = Δt* {QA ρ v /(2πε c r)} /√(1–v /c )} Δp * = Δt {QA ρ v /(2πε c r)} y

y

y y

2 n

p

p

2 n

2

0

2 n

2

2

0

We see that Δp has the same value in both S and S* reference frames, as required by special relativity. This means the “magnetic” force in frame S exactly equals the Lorentz transformed electric force in frame S*, the rest frame of charge Q.

One can interpret this to mean that “magnetism” is a relativistic effect, an artifact of not choosing the simplest reference frame: the rest frame of the charge on which forces act. In V2p13-10, Feynman makes this important comment about the concept of fields that you should thoroughly digest: “Electric and magnetic forces are part of one physical phenomenon—the electromagnetic interactions of particles. The separation of this interaction into electric and magnetic parts depends very much on the reference frame chosen for the description. But a complete electromagnetic description is invariant; electricity and magnetism taken together are consistent with Einstein’s relativity. “Since electric and magnetic fields appear in different mixtures if we change our frame of reference, we must be careful about how we look at the fields E and B. For instance, if we think of “lines” of E and B, we must not attach too much reality to them. The lines may disappear if we try to observe them from a different coordinate system. For example, in system S*, there are electric field lines, which we do not find ‘moving past us with velocity v in system S.’ In system S there are no electric field lines at all! Therefore it makes no sense to say something like: When I move a magnet, it takes its field with it, so the lines of B are also moved. There is no way to make sense, in general, out of the idea of ‘the speed of a moving field line.’ The fields are our way of describing what goes on at a point in space. In particular, E and B tell us about the forces that will act on a moving particle. The question ‘What is the force on a charge from a moving magnetic field?’ doesn’t mean anything precise. The force is given by the values of E and B at the charge, and the [Lorentz force] is not to be altered if the source of E and B is moving (it is the values of E and B that will be altered by the motion). Our mathematical description deals only with the fields as a function of x, y, z, and t with respect to some inertial frame. “We will later be speaking of ‘a wave of electric and magnetic fields traveling through space,’ as, for instance, a light wave. But that is like speaking of a wave traveling on a string. We don’t then mean that some part of the string is moving in the direction of the wave, we mean that the displacement of the string appears first at one place and later at another. Similarly, in an electromagnetic wave, the wave travels; but the magnitude of the fields change. So in the future when we—or someone else—speaks of a ‘moving’ field, you should think of it as just a handy, short way of describing a changing field in some circumstances.”

Lorentz Transform of Charge & Current In the last section we made the simplifying assumption that v =v , that the lone charge Q outside the wire had the same velocity as the electrons within the wire. In V2p13-11, Feynman says you might be wondering what happens in a more general case. Does the electromagnetic force exerted on Q by the wire transform properly to any reference frame? n

In this section, we shall prove that it does.

Q

We found that charge density ρ transforms relativistically in the same way that mass does, according to: ρ* = ρ / √(1–v /c ) 2

2

0

Here, ρ is the charge density in its rest frame S, and ρ* is the charge density in frame S* that moves at velocity v relative to S. 0

In S*, this charge is moving with velocity v, producing a current j* given by: j*/c = ρ* (v/c) = ρ (v/c) / √(1–v /c ) 2

2

0

Here and below, I have included extra factors of c to illuminate a point. Compare the last two equations to the transformation equations for energy E and momentum p (see Feynman Simplified 1C Chapter 27). E* = m c / √(1–v /c ) cp* = m c (v/c) / √(1–v /c ) 2

2

2

0

2

2

2

0

We know that the four quantities: (E,cp) = (E, cp , cp , cp ) x

y

z

form a 4-vector that properly transforms relativistically between any two frames. The preceding two pairs of equations show that the four quantities (ρ,j/c) are governed by exactly the same relativistic transformation as (E,cp). Indeed, (ρ,j/c) multiplied by m c /ρ equals the 4-vector (E,cp). This proves that (ρ,j/c) is also a 4-vector governed by the Lorentz transform. For any (ρ,j/c) in frame S, (ρ*,j*/c) in frame S* that is moving in the z-direction with velocity u relative to S is given by: 2

0

0

ρ* = (ρ – j u/c ) / √(1–u /c ) j*=j j*=j j * = (j – ρ u) / √(1–u /c ) 2

2

2

z

x

x

y

y

2

z

2

z

Linear Superposition The equations of magnetostatics are: Ď•B = 0 Ď×B = j / ε c

2

0

Both are clearly linear in both B and j. This means the principle of linear superposition applies to magnetostatics, as it does to all of electromagnetism. The B field produced by two constant currents is simply the vector sum of the fields produced by each current separately.

Right Hand Rule Feynman ends this lecture with a comment about the right hand rule in electromagnetism. There are two ways to define the polarity of a cross product; they correspond to using either one’s right or left hand. The choice is arbitrary in electromagnetism, because nature makes a distinction between left and right only in weak force interactions. (See Feynman Simplified 1D Chapter 49.) Yet, mathematicians and physicists are right-hand bigots, repeatedly instructing students to employ the right hand rule exclusively. (I, of course, am not a bigot; I even have a left-handed friend or two.) Indeed, if we all used the left hand rule exclusively, we would obtain exactly the same final results. Some quantities, such as B, would have the opposite polarity, but all real measurable quantities, such as the direction of particle motion, would be the same. For example, the force F exerted on a charge q by the magnetic field B of current j is: F = q v×B F = q v × (J × r) / (2πε c r ) 2

2

0

A leftie would get the “wrong” sign for each cross product. But, in this case two “wrongs” do make a right: two polarity inversions result in the correct sign of F. The real measurable quantities here are j, v, r, and F, not B. More generally, this issue arises because there are two types of vectors: polar and axial. Polar vectors are the more familiar and include: position, velocity, force, and the electric field E. Axial vectors are mathematical constructs that are not directly observable; these include: angular momentum, torque, and the magnetic field B. All axial vectors are defined by a cross product, and the physical effect due to any axial vector involves another cross product. The polarity of physical observables is defined by nature: two wires pull together if they carry parallel currents and push apart if their currents are antiparallel. We are compelled to get the correct polarity of physical observables. The polarity of mathematical constructs is decided by human convention. The essential point is consistency. If you sometimes use the right hand rule and sometimes use the left hand rule, your answer will be Wrong in a fundamental way, not just contrary to an arbitrary convention. If you always use the left hand rule, you will get the correct answers for physical observables, but your instructor will still mark your answer Wrong.

Chapter 14 Review: Key Ideas • In static conditions, when charges and currents are unchanged over time, Maxwell’s equations

reduce to two de-coupled pairs of equations: Electrostatics: Ď•E = ρ/ε

0

Ď×E = 0 Magnetostatics: Ď×B = j/ε

0

Ď•B = 0 Electricity and magnetism become interconnected only when charges move or currents change.

• A fundamental principle of physics is that net electric charge is locally conserved. Net charge is the sum of all positive charges minus the sum of all negative charges. Local conservation is a stronger requirement than global conservation. Local conservation states that the net charge within any volume V changes only when a current crosses its boundary. If the current carries charge ΔQ into V, the net charge in V increases by ΔQ, which can be positive or negative.

• The force f per unit length on a wire carrying current J in a magnetic field B equals: f = J×B.

• Ampere’s law relates the tangential component of B integrated around curve Γ to the flux of current through a surface S enclosed by Γ, according to: ∫ B•ds = J through S / (c ε ) 2

Γ

0

• The magnetic field from a solenoid carrying current J, with n coils per unit length, is: B=nJ / cε 2

0

• A “magnetic” force on charge Q in one reference frame S corresponds exactly to an electric force in the rest frame of charge Q. One can interpret this to mean that “magnetism” is a relativistic effect, an artifact of not choosing the simplest reference frame: the rest frame of the charge on which forces act.

Chapter 15 The Vector Potential We have solved the equations of magnetostatics Ď•B = 0 Ď×B = j/c ε 2

0

in a number of special, symmetric conditions. We now wish to solve these equations in a more general way. In electrostatics, we found a general analytic strategy: calculate the potential ø for a distribution of charges and employ E=–Ďø. We will now find a corresponding strategy for magnetostatics.

Vector Potential A In electrostatics, the electric field is proportional to the gradient of a scalar field, because the curl of E is always zero: Ď×E=0. Magnetic fields are different: their curls are non-zero, but their divergences are always zero. We learned in our study of vector algebra that any vector field with zero divergence can be represented as the curl of another vector field. We therefore define the vector potential A such that: B = Ď×A In V2p14-1, Feynman writes this out in component notation: B = ∂A /∂ – ∂A /∂ B = ∂A /∂ – ∂A /∂ B = ∂A /∂ – ∂A /∂ x

z

y

y

z

y

x

z

z

x

z

y

x

x

y

Note that: Ď•B = Ď•(Ď×A) = 0 Since the last step is true for any vector field A (see Chapter 2), this representation ensures B always has zero divergence. The equation B = Ď×A defines the curl of A, but does not completely specify A itself. Indeed, for any

scalar field ψ, adding Ďψ to A does not change its curl, as shown below: B = Ď×(A+Ďψ) = Ď×A + Ď×(Ďψ) = Ď×A This is true because Ď×(Ďψ) = 0 for any ψ, as we found in Chapter 2. We have a similar situation in electrostatics: E=–Ďø does not completely define ø. There we usually make ø unique by adding the condition ø(r=∞)=0. In magnetostatics, we will usually make A unique by adding the conditions Ď•A=0 and A(r=∞)=0. These added conditions are arbitrary, and in some situations other choices may be more convenient mathematically. We explore the vector potential beginning with a simple example: a uniform magnetic field B in the +z-direction. 0

B = (0,0,B ) 0

B = 0 = ∂A /∂ – ∂A /∂ B = 0 = ∂A /∂ – ∂A /∂ B = B = ∂A /∂ – ∂A /∂ x

z

y

y

z

y

x

z

z

x

z

0

y

x

x

y

Two possible solutions for A are: A = (–y, 0, 0) B A = (0, +x, 0) B 1

0

2

0

For any arbitrary constant φ, the principle of linear superposition ensures that A φ plus A (1–φ) is also a solution. The endless possibilities include φ=1/2: 1

2

A = (–y, +x, 0) B / 2 0

By defining r = (x, y), we can rewrite the prior equation as: A = B × r / 2 = (0, 0, B ) × (x, y, 0) / 2 0

This means A is perpendicular to both B and r, as shown in Figure 15-1. Here, dots indicate B field lines pointing toward +z (toward you). Two representative examples of r and A are shown as arrows.

Figure 15-1 Vector Potential A for Uniform Field B

The circle of radius r labeled Γ is centered at x=y=0, and lies in a plane orthogonal to B. The area S that Γ encloses is shown in gray. We can also derive a solution for the vector potential A for a uniform B field using Stokes’ theorem. It says the circulation of any vector field A around a closed curve Γ equals the flux of the curl of A through the surface S enclosed by Γ. The equation is: ∫ A•ds = ∫ (Ď×A)•n da ∫ A•ds = ∫ (B)•n da ∫ A•ds = πr B Γ

S

Γ

S

2

Γ

From above, we know A is perpendicular to both B and r, which means it must be parallel to ds. And by symmetry, A•ds must have the same value at every point on the circle. Hence: 2πr A = πr B A= r B / 2 2

This is the same result we obtained above. Feynman notes that in this example we calculated A from B, which is the opposite of the normal process. Just as we can calculate ø from a charge distribution and then calculate E from ø, we will usually calculate A from a current distribution and then calculate B from A, as we will do next.

Finding A from Current J The principle equation of magnetostatics is: Ď×B = j/c ε 2

0

Inserting the definition of A yields: Ď×(Ď×A) = j/c ε Ď(Ď•A) – Ď A = j/c ε 2

0

2

2

0

Since we conventionally set the divergence of A equal to zero, this becomes: Ď A = – j/c ε 2

2

0

In component notation this is: Ď A = – j /c ε Ď A = – j /c ε Ď A = – j /c ε 2

2

x

x

0

2

2

y

y

0

2

2

z

z

0

These equations are identical to the electrostatic equation: Ď ø = – ρ/ε 2

0

As Feynman says in V2p14-4: “All we have learned about solving for potentials when ρ is known can be used for solving for each component of A when j is known!” Adapting the general equation for ø found in Chapters 4 and 13, we obtain: A(r) = (1/4πε c ) ∫ j(σ) dV / |r–σ| 2

0

V

Here, the integral runs over all points σ within volume V. To avoid mathematical pitfalls, this equation should be solved in rectilinear coordinates (x,y,z), rather than polar coordinates, or other systems in which the coordinate axes change direction. For magnetostatics, where Ď•j=0, this equation ensures that Ď•A=0. The same techniques used to solve for electric potential ø are applicable to solving for vector potential A, from which we obtain B. As Feynman says: “It’s a little more complicated than electrostatics, but the same idea.” Actually, it can be significantly more complicated mathematically. Let’s see how some simple physical situations can be addressed by analogy to electrostatics.

A Current-Carrying Wire First, we reexamine the field from a current-carrying wire. We solved this problem before with Stokes’ theorem; now, we will employ the equation for A and the electrostatic analog. Let’s define cylindrical coordinates with current J flowing in the +z-direction, r being the distance from the wire’s axis, and β being the azimuthal angle around the wire, as shown in Figure 15-2. As stated in the prior chapter, the magnetic field B is circumferential and counterclockwise as viewed from +z.

Figure 15-2 Vector Potential A of Current J

In Chapter 5, we proved that the electric field from a wire with charge λ per unit length is: E(r) = λ / (2πε r) 0

This means the electric potential ø(r) is: ø(r) = – ∫ E(u) du Here, for a wire of radius w, we integrate outside the wire, from u=w to u=r. ø(r) = – (λ/2πε ) ∫ du / u ø(r) = – (λ/2πε ) { ln(u) | } ø(r) = – (λ/2πε ) {ln(r) – ln(w)} ø(r) = + (λ/2πε ) ln(w/r) 0

r

0

w

0

0

In V2p14-5, Feynman uses the mathematically improper equation: ø(r) = – (λ/2πε ) ln(r) 0

This has the correct r-dependence, but logarithms are only defined for pure numbers that have no units. The ln(1 meter) and ln(100 cm) are not defined quantities. In both his and my equations, ø is not defined according to the normal convention: ø(∞)=0. That isn’t possible with a logarithmic function. The problem is not the math, but our invalid assumption of an infinitely long wire. A real wire cannot be infinitely long. At distances from the wire much farther than its length, both E and ø must approach zero. Fortunately, we can still use this equation near the wire, because only changes in ø have physical

significance. Since ø and A are governed by the same equation with different constants, the solution in magnetostatics is the same as the solution in electrostatics with charge density per unit volume ρ replaced by current density per unit area j divided by c . 2

For a wire with cross-sectional area πw : 2

λ = πw ρ πw j = J 2

2

z

Making these substitutions, we obtain: A (r) = (πw [j /c ] / 2πε ) ln(w/r) A (r) = (J / 2πε c ) ln(w/r) 2

2

z

z

0

2

z

0

We get B by taking the curl of A (r). Let’s first calculate the partial derivative of the logarithm. z

∂ ln(w/r) /∂x = ∂/∂x {ln(w) – ln(r)} ∂ ln(w/r) /∂x = – ∂ ln(r) / ∂x ∂ ln(w/r) /∂x = – (1/r) ∂r/∂x ∂ ln(w/r) /∂x = – (1/r) ∂/∂x (x +y ) ∂ ln(w/r) /∂x = – (1/r) (1/2) (2x) (1/r) ∂ ln(w/r) /∂x = – x/r 2

2 1/2

2

Now take the curl of A. B = ∂A /∂ – ∂A /∂ B = ∂A /∂ – ∂A /∂ B = ∂A /∂ – ∂A /∂ x

z

y

y

z

y

x

z

z

x

z

y

x

x

y

B = (J / 2πε c ) (–y/r ) B = (J / 2πε c ) (+x/r ) B =0 x

0

y

0

2

2

2

2

z

In vector notation, this is: B = (1 / 2πε c r ) J×r 2 2

0

This confirms the assertion of the prior chapter than B is entirely circumferential, and the final equation that we derived therein.

A Solenoid We next reexamine the field from a solenoid that we previously derived with Stokes’ theorem.

Consider an infinitely long solenoid of radius w, carrying current J, with n coils per unit length. As before, we assume the coils are parallel and infinitesimally thin, and that each coil lies in a plane perpendicular to the z-axis that runs through the solenoid’s center. (We are neglecting the coils' slight winding pitch.) The x- and y- axes are indicated in the cross-sectional image in Figure 15-3.

Figure 15-3 Fields From A Solenoid

Here the interior of the solenoid is shown in light gray. Per the right hand rule, B points toward +z (out of the screen towards you) for a counterclockwise j, as shown. We define j to be the current density per unit area. Vector field j is circumferential and non-zero only on the solenoid’s surface. The magnitude of j equals the product of (coils per unit length) multiplied by (current per coil): j=nJ. We see from the figure that the components of j are: j (x,y,z) = – j sinθ j (x,y,z) = + j cosθ j (x,y,z) = 0 x y z

Our task is solving for vector potential A, and from A, solving for B. Let’s first solve for A outside the solenoid (r>w). x

Using the electrostatic analogy, a solution for ø from charge density ρ is also a solution for A from current density j , with ρ=j /c . In this case, we have surface charges and currents, so we will use the symbol σ for the surface charge density, with: x

2

x

σ = – σ sinθ σ = j/c 0 2

0

x

Feynman has a trick to solve this problem. In V2p14-5, he says our σ is the same as the net charge density from two solid, uniformly charged, opposite polarity, cylinders that are slightly displaced in the y-direction. Since the equations are the same, the solutions are the same, even if the physical situations are not. The proof of this claim is in the next section. As we proved earlier, for a cylindrical wire of radius w, with charge λ per unit length and with r=√(x +y ), the electric potential is: 2

2

ø(x,y) = + (λ/2πε ) ln(w/r) 0

Here, the wire’s axis is in the z-direction and its center is at x=y=0. For one wire centered at (x,y) = (0,0) with charge density +λ and a second wire centered at (0,+Δy) with charge density –λ, the total potential is: ø* = ø(x,y) – ø(x,y–Δy) Note that moving the second wire up is the same as moving the y-axis down. We’ve done this before. ø* = ∂ø/∂y Δy ø* = (λ/2πε ) {–∂[ln(r)]/∂y} Δy ø* = – (λ/2πε ) (1/r) (1/2r) (2y) Δy ø* = – (λ/2πε ) Δy y/r 0

0

2

0

We see that the sign is correct: the negative cylinder at y=+Δy leads to a negative potential for y>0. To reduce clutter, define K=(λ/2πε c )Δy. By analogy: 2

0

A = – K y/r

2

x

By symmetry: A = + K x/r

2

y

The magnitude of A outside the solenoid (r>w) is: A = √(A + A ) = (K/r ) √(y +x ) = K/r 2

2

x

2

2

2

y

In the prior chapter, we said B is zero outside an infinitely long solenoid. Let’s check that. B = Ď×A B = ∂A /∂ – ∂A /∂ = 0 B = ∂A /∂ – ∂A /∂ = 0 B = ∂A /∂ – ∂A /∂ B /K = ∂(x/r )/∂x + ∂(y/r )/∂y B /K = (1/r )+x(–1/r )(2x) + (1/r )+y(–1/r )(2y) x

z

y

y

z

y

x

z

z

x

z

y

x

x

y

2

2

z

2

z

4

2

4

B /K = (2/r ) – 2(x +y )/r = 2/r – 2r /r B =0 2

2

2

4

2

2

4

z z

This confirms the claim that B is zero outside an infinitely long solenoid, even though A is not zero. We can check that our result is consistent with Stokes’ theorem. B equals the curl of A, which is circumferential. This means the path integral of B around a circle of radius r (r>w) centered on the solenoid equals 2πrA. By Stokes’ theorem this equals the flux of B through that circle, which is πw B , where B is the magnetic field inside the solenoid. The result is: 2

0

0

2πrA = πw B 2πr (K/r) = πw nJ / c ε K = w nJ / 2c ε 2

0

2

2

0

2

2

0

and: A = w nJ / 2c ε r 2

2

0

Note that the circular path integral of A equals 2πK for all circles larger than the solenoid. Because the flux of B is entirely within the solenoid, all circles with larger radii enclose the same flux.

Potential of Offset Cylinders Here is the proof that the net charge distribution of two displaced, oppositely charged, cylinders is sinusoidal. Those wishing to avoid the math can skip to the next section. Those wishing to learn how to solve such problems should read this carefully. Figure 15-4 shows two superposed, solid cylinders each of radius r. One has uniform charge density –ρ per unit volume, and is centered at (x,y) = (0,+Δy). The other has uniform charge density +ρ, and is centered at (0,0). Let L be the length (the direction perpendicular to the screen) of each cylinder. Also, let θ be the azimuthal angle, with θ=0 being the +x-direction.

Figure 15-4 Two Oppositely Charged Cy linders

The figure shows a horizontal band with infinitesimal azimuthal width dθ and height dy. The key geometric relationships are: r = √(x +y ) x = r cosθ y = r sinθ dy = d(r sinθ) = r cosθ dθ = x dθ 2

2

Before we move away from Figure 15-4, note that a thin cylindrical shell of radius r and length L has a surface area S contained within the horizontal band equal to: S = 2 L r dθ For the positive solid cylinder, the charge Q contained within the horizontal band is: +

Q = L 2x dy ρ Q = 2L x dθ ρ Q = 2L [r –y ] dθ ρ +

2

+

2

2

+

For the negative cylinder, the corresponding charge Q is: –

Q = – 2L [r –(y–Δy) ] dθ ρ 2

2

–

The sum of charges from both cylinders in the horizontal band is: Q=Q +Q +

–

Q = 2L dθ ρ {[r –y ] – [r –(y–Δy) ]} Q = 2L dθ ρ {–2yΔy+Δy } Q = – 4L dθ ρ y Δy 2

2

2

2

2

For the cylindrical shell to have the same charge Q within its surface S contained in the horizontal band, its charge density per unit area σ must be: shell

σ σ σ

shell shell shell

S = – 4L dθ ρ y Δy L 2r dθ = – 4L dθ ρ y Δy = – 2ρ sinθ Δy

This is the result we set out to prove. To match this with the solenoid’s x-component of surface charge density, we require: – j/c sinθ = – 2ρ sinθ Δy Δy = j / (2ρc ) 2

2

The solenoid’s y-component of surface charge is: j = + j cosθ y

This corresponds to the net charge density of two oppositely charged cylinders displaced in the xdirection. The positive cylinder is centered at (+Δx,0), and the negative cylinder is centered at (0,0). For a vertical band of width dx and dθ, the key results are: Q = 2L (r –x ) dθ ρ Q + Q = + 4L dθ ρ x Δx σ = + 2ρ cosθ Δx Δx = j / (2ρc ) 2

2

+ +

–

shell

2

QED

A Rotating Cylindrical Shell In V2p14-6, Feynman points out that a charged cylindrical shell spinning about its axis produces the same fields as a solenoid. Imagine an infinitesimally thin shell, the surface of a solid cylinder with radius w and surface charge density σ that is spinning with angular velocity Ω, as shown in Figure 15-5. The speed of each charge in the shell is wΩ. The surface current density j and internal magnetic field B are: j=σwΩ B=σw Ω/ cε 2

0

Figure 15-5 Rotating Shell With Wire

Now imagine that a tiny wire (shown in light gray) is mounted perpendicular to the shells’ inner surface. Let the wire be attached in a way that does not allow charge to flow between the wire and the shell. As the shell rotates and the wire moves through the magnetic field, its charges experience a Lorentz force: F=qv×B. Right-hand-ruler’s will confirm that the wire’s positive charges are pushed outward while its negative charges are pushed inward. This creates an electric field within the wire that seeks to pull the opposite charges together. At equilibrium, the electric force exactly balances the magnetic force. Feynman says: “By measuring the charge on the end of the wire, we could measure the speed of rotation of the system. We would have an ‘angular-velocity meter’! But are you wondering: ‘What if I put myself in the frame of reference of the rotating cylinder? Then there is just a charged cylinder at rest, and I know that the electrostatic equations say there will be no electric fields inside, so there will be no force pushing charges to the center. So something must be wrong.’ But there is nothing wrong. There is no ‘relativity of rotation.’ A rotating system is not an inertial frame, and the laws of physics are different. We must be sure to use equations of electromagnetism only with respect to inertial coordinate systems.”

Field From Magnetic Dipole We next consider a rectangular loop carrying current J in a counterclockwise direction. The loop has width a in the x-direction and depth b in the y-direction, as shown in the upper half of Figure 15-6. We wish to find the magnetic field at a point P located at position r, with x=y=z=0 at the center of the loop.

Figure 15-6 Magnetic Dipole & Ax/ø Equivalent

Consider the x-component of vector potential A that arises from currents flowing in the top and bottom sides of the rectangular loop — the two sides of length a that are parallel to the x-axis. Again, the equation for A in terms of J/c is the same as the electrostatic equation for ø in terms of ρ. Since J flows in opposite directions along the two sides, the electrostatic equivalent is an electric dipole, whose positive and negative charges are separated by distance b, as shown in lower left of Figure 15-6. 2

x

If point P is far away (if r=|r| is much greater than both a and b), the potential ø from an electric dipole moment µ is: ø(r) = (1/4πε ) µ•r / r

3

0

With λ being the charge per unit length, the magnitude of dipole moment µ equals λa, the charge on one side, multiplied by b, the distance between opposite charges. µ=–λabn Here n is the unit normal to the loop’s surface, which is the +y-direction in this case. As the minus sign indicates, µ points in the opposite direction, from the negative charge to the positive charge. µ•r = – λ a b n•r µ•r = – λ a b y

Here, we used r = (x,y,z). The electric potential and A are therefore: x

ø(r) = – λ ab y / (4πε r ) A = – J ab y / (4πε c r ) 3 0 2 3

x

0

The A component is calculated similarly, but with its dipole pointing in the +x-direction, because the positive polarity current, the one flowing toward +y, is on the +x side of the loop, as shown in the lower right of Figure 15-6. We therefore have: y

A = + J ab x / (4πε c r ) 2 3

y

0

The magnitude of A is proportional to Jab: (current J) × (loop area ab). We define this to be the magnetic dipole moment µ of the current loop. µ = J × (area within loop) We derived this for a rectangular current loop, but it is also true for loops of any shape. Feynman suggests you try to prove that. Guidance is provided at the end of this chapter. Since there are no magnetic monopoles, µ is often simply called the magnetic moment. The magnetic moment is a vector whose direction is normal to the loop area, with its polarity determined by the right hand rule applied to the direction of current flow. In vector form, the potential A at position r far from magnetic moment µ is: A(r) = µ×r / (4πε c r ) 2 3

0

As Feynman says in V2p14-8, since A is proportional to –y and A is proportional to +x, A circles the z-axis, as shown in Figure 15-7. x

y

Figure 15-7 Vector Potential From Dipole

As indicated in the figure, current J flows counterclockwise within the xy-plane. The final step is calculating B from A. To reduce clutter, let K=µ/(4πε c ). Let’s first calculate the partial derivative of 1/r with respect to x; the y and z partial derivatives will then be evident. 2

0

3

∂(x +y +z ) /∂x = (–3/2) r (2x) = –3x/r 2

2

2 –3/2

–5

5

We found above that for µ = µ e : z

A(r) = (–Ky/r , +Kx/r , 0) 3

3

B = ∂A /∂ – ∂A /∂ B = 0 –∂(+Kx/r )/∂ B = –Kx (–3z/r ) x

z

y

y 3

z

x

z

5

x

B = ∂A /∂ – ∂A /∂ B = +∂(–Ky/r )/∂ – 0 B = –Ky (–3z/r ) y

x

z

z

x

3

y

z

5

y

B B B B B B

z z z z z z

= ∂A /∂ – ∂A /∂ = +∂(+Kx/r )/∂ – ∂(–Ky/r )/∂ = +K(1/r ) +Kx(–3x/r ) +K(1/r ) +Ky(–3y/r ) = +K(2/r ) –3K(x +y )/r = +K{ (2r ) –3(r –z ) }/r = +K{ –r +3z ) }/r y

x

x

y

3

3

x

y

3

5

3

2

2

2

2

2

2

2

5

3

5

5

5

B = +3µxz / (4πε c r ) B = +3µyz / (4πε c r ) B = –µ (r –3z ) / (4πε c r ) x

0

y

0

2

2

5

2

5

2

2

z

5

0

In V2p14-8, Feynman says these magnetic field components are the same as the electric field components from an electric dipole. He adds that “magnetic dipole” might be a misleading term: a magnetic dipole is produced by a current loop, not by two oppositely charged “poles”. Feynman says: “It is curious, though, that starting with completely different laws, Ď•E=ρ/ε and Ď×B=j/c ε , we can end up with the same kind of a field. Why should that be? It is because the dipole fields appear only when we are far away from all charges or currents. So through most of the relevant space the equations for E and B are identical: both have zero divergence and zero curl. So they give the same solutions. However, the sources whose configuration we summarize by the dipole moments are physically quite different.” 2

0

0

The Biot-Savart Equation With the normal convention Ď•A=0, the general equation for vector potential A from current distribution j is: A(r) = (1/4πε c ) ∫ j(σ) dV / |r–σ| 2

0

V

For the case of current flowing in a very thin wire, this equation can be slightly simplified. The volume element dV becomes the product of S, the wire’s cross-sectional area, and dσ, an infinitesimal displacement along the wire’s axis. For a thin wire and a constant current, we can assume j is constant throughout each cross-section S and is parallel to dσ. In this case: j(σ) dV = j S dσ = J dσ Here, J, the total current flowing through the wire equals its current density j multiplied by area S. The equation for the vector potential becomes: A(r) = (1/4πε c ) ∫ J dσ / |r–σ| 2

0

σ

Here, the integral is over σ, the coordinate along the wire’s axis. The usual rule of superposition applies: the potential from several current-carrying wires is the vector sum of their individual potentials. Next, recall that in electrostatics, ø is governed by a nearly identical equation, and E=–Ďø. There, we can take the gradient of the integral, or move the gradient inside the integral sign. The latter results in taking the integral of a gradient. Mathematically this is: ø(r) = (1/4πε ) ∫ ρ(σ) dV / |r–σ| 0

V

E(r) = –Ďø = –(1/4πε ) ∫ ρ(σ) dV Ď(1/|r–σ|) 0

V

It is essential to remember here that we are taking the gradient only with respect to r not σ. Similarly, B(r) = Ď×A(r) B(r) = (1/4πε c ) ∫ Ď×{j(σ) dV / |r–σ|} 2

0

V

Again, the curl operates only on r not σ. The x-component of the integrand is: j ∂(1/|r–σ|)/∂y – j ∂(1/|r–σ|)/∂z = j (r –σ )/|r–σ| – j (r –σ )/|r–σ| z

y

3

z

y

3

y

y

z

z

This is the x-component of the vector expression: j × (r–σ) / |r–σ|

3

The y- and z-components are similar, with the result that: B(r) = (1/4πε c ) ∫ j × (r–σ) / |r–σ| 2

0

3

V

In the case that j flows through a thin wire, this equation becomes: B(r) = (1/4πε c ) ∫ J dσ × (r–σ) / |r–σ| 2

0

3

σ

This is called the Biot-Savart law.

Philosophy of Fields In V2p14-10, Feynman says: “You may wonder: ‘What is the advantage of the vector potential if we can find B directly with a vector integral? After all, A also involves three integrals!’ Because of the cross product, the integrals for B are usually more complicated... Also, since the integrals for A are like those of electrostatics, we may already know them. Finally, we will see that in more advanced theoretical matters (in relativity, in advanced formulations of the laws of mechanics, like the principle of least action to be discussed later, and in quantum mechanics) the vector potential plays an important role.” In V2p15-7, Feynman explores the question of whether or not A is a real field. He says: “First we should say that the phrase ‘a real field’ is not very meaningful. For one thing, … the whole idea of a field is a rather abstract thing. You cannot put out your hand and feel the magnetic field. Furthermore, the value of the magnetic field is not very definite; by choosing a suitable moving coordinate system, for instance, you can make a magnetic field at a given point disappear.

“[For our purposes] a real field is a mathematical function we use for avoiding the idea of action at a distance. If we have a charged particle at the position P, it is affected by other charges [even those far way]. One way to describe the interaction is to say that the other charges make some ‘condition’—whatever it may be—in the environment at P. If we know that condition, which we describe by giving the electric and magnetic fields, then we can determine completely the behavior of the particle—with no further reference to how those conditions came about. “… A ‘real’ field is then a set of numbers we specify in such a way that what happens at a point depends only on the numbers at that point. We do not need to know any more about what’s going on at other places.” One could easily spend the rest of one’s life debating: “What is Real?” If we were standing next to one another, how would you know if I am a real person, an advanced robot, a remarkable hologram, or merely sensory stimuli in your brain? You would have to cut me open to be sure, but I’d really prefer you didn’t. Most often, we make the simplest assumptions and move forward to more productive pursuits. If a concept is useful, that is sufficient reason to employ it and teach it to others. Physicists cannot prove that electrons are Real, but assuming they are real has immeasurably enriched mankind.

Dipole Moment & Loop Shape We showed above that the magnetic dipole moment of a rectangular current loop is proportional to the area of the loop. Here we show that the moment is proportional to the loop area for any loop shape. Following the logic of Chapter 3, a loop of any shape can be subdivided into a vast number of infinitesimal rectangular loops without any overlaps or gaps. Every point in the large loop must be inside one and only one small 4-sided loop. Let each small loop carry the same current in the same orientation as the large loop. Every small loop side that is interior to the large loop is a boundary between two small loops, with currents flowing in opposite directions; these sides make zero contribution to the magnetic field. Only the exterior sides of the small loops contribute. The sum of all such sides is identical to the boundary of the large loop.

Chapter 15 Review: Key Ideas • The vector potential A is defined by: B = Ď×A Additionally, we normally set the divergence of A equal to zero and require A(r=∞)=0. Thus: Ď×Ď×A = Ď×B = j/c ε 2

0

Ď A = – j/c ε 2

2

0

which matches the electrostatic equation: Ď ø = – ρ/ε 2

0

Since the same equations have the same solutions, we have: A(r) = (1/4πε c ) ∫ j(σ) dV / |r–σ| 2

0

V

Here, the integral runs over all points σ in all space V. This equation should be solved in rectilinear coordinates (x,y,z), in which the coordinate axes do not change direction.

• For a wire with radius w carrying current J in the +z direction, and for r being the distance from the wire’s axis: A (r) = (J / 2πε c ) ln(w/r) 2

z

0

B(r) = (1 / 2πε c r ) J×r 2 2

0

• For a long solenoid with n coils per unit length, each carrying current J, and for r being the distance from the solenoid’s axis, the internal magnetic field B and the vector potential are: 0

B = nJ / c ε 2

0

0

A = w nJ / 2c ε r 2

2

0

The magnetic dipole moment, simply called the magnetic moment, of a loop of area S carrying current J is: µ=JS

• The magnetic moment is a vector whose direction is normal to S, with its polarity determined by the right hand rule applied to the direction of current flow. In vector form, the potential A at position r from magnetic moment µ is: A(r) = µ×r / (4πε c r ) 2 3

0

Chapter 16 Electromagnetic Statics Wrap-Up This chapter concludes the study of electrostatics and magnetostatics. A comprehensive comparison of the equations of static and dynamic electromagnetism is presented in the Review section at the end of this chapter.

Forces on Magnetic Dipoles As developed in the prior chapter, a current loop produces a magnetic dipole moment µ given by: µ=JSn Here, J is the current circulating around a loop enclosing an area S, whose unit normal is n. The direction of n is determined by the right hand rule: if you position your right hand as if you were signaling a “thumbs-up”, and then turn your hand so that your right thumb points along n, your fingers will point in the direction of current flow. Magnetic dipoles experience forces and have potential energy in magnetic fields, just as electric dipoles do in electric fields. Consider a magnetic dipole with moment µ, carrying current J around a rectangular loop of size a×b, as shown in Figure 16-1.

Figure 16-1 Forces on Magnetic Dipole

Here, a uniform external magnetic field B points in the +z-direction. The plane of the dipole loop is rotated about the y-axis by angle θ relative to the z-axis. The angle between B and µ is therefore θ. The current flow in magnetic field B exerts forces on each side of the dipole loop force, labeled F through F in the figure.

1

4

In a uniform external field, the forces are equal and opposite in pairs: F =–F , and F =–F . F and F are also collinear, so they cancel one another completely. However, F and F are offset, hence they exert a torque τ on the loop. 1

1

2

3

4

3

4

2

The currents that produce forces F and F are both parallel to the xy-plane and orthogonal to the zaxis and to B. To calculate these forces, let the conduction charges have velocity u and charge density ρ per unit volume. Let S be the cross-sectional area of the loop’s conductor, and let b be the loop’s length in the y-direction. The total charge on that side is: Q=ρSb. The current density (the charge per unit time per unit area) is j=ρu, and the loop’s total current (charge per unit time) is J=jS. The magnitudes of these forces are: 1

F F F F

1 1 1 1

=F =F =F =F

2 2 2 2

= (Q) u B = (ρ Sb) u B = j Sb B =JbB

Both forces have a lever arm equal to: (a/2) sinθ Hence the total torque is:

2

|τ| = 2 (JbB) (a/2) sinθ = µB sinθ τ=µ×B In V2p15-1, Feynman notes that although we have proven this only for a rectangular loop, it is true for any loop shape. See the last section of the prior chapter for an explanation of this generalization. The prior equation is identical to that of an electric dipole µ in an electric field E: τ=µ×E

Energy of Magnetic Dipoles Next, we consider the energy of a magnetic dipole due to its orientation in an external magnetic field. Feynman says that since there is a torque, there must be energy associated with an orientation angle. This is because the principle of virtual work says that forces (or torques) equal minus the derivative of distance (or angle). The equations are: F = – ∂U/∂x τ = – ∂U/∂θ x

θ

For the magnetic dipole in Figure 16-1, we have: dU/dθ = – µB sinθ U = – µB ∫ sinθ dθ U = – µB cosθ + constant mech

The minus sign confirms that energy is minimized when µ is parallel to B, the most stable condition. Conversely, energy is maximized when µ is antiparallel to B, the least stable condition. Feynman adds the subscript “mech” to stress that U is the potential energy associated with the dipole’s mechanical orientation. This is only one part of its total energy; we will discuss its electrical energy shortly. And, since only potential energy differences have physical consequences, we can set the arbitrary integration constant to zero. mech

By now, I’m sure you realize that we almost never need to include every type of energy in every analysis. We routinely ignore all types of energy that do not change in the process being examined. These often include gravitational potential energy, mass energy, thermal energy, chemical energy, and others. We focus on what matters: energy changes. In vector form the prior equation can be written: U

mech

= – µ•B

The corresponding electric dipole potential energy is:

U = – µ•E elec

In V2p15-2, Feynman says U is “the true energy”, but U is “not the real energy.” This seems confusing. I believe the distinction Feynman intends is that electric dipoles can often persist without an external energy source, whereas magnetic dipoles often cannot. For example, the electric dipole of a water molecule will exist forever without consuming energy, so long as the molecule exists. But, current in a normal wire quickly stops unless driven by an external energy source, such as a battery. elec

mech

There are however, many counterexamples on both sides: electrons have permanent magnetic dipole moments with no external energy source, while a charged capacitor in a changing external electric field requires an energy source to maintain its electric dipole moment. A better statement is: U and U are both forms of potential energy that may or may not persist without a sustaining energy source. In either case, if an energy source is required, we must account for that energy, as we shall see shortly. mech

elec

Feynman next shows that U is also the work done in bringing the dipole into a magnetic field B, assuming µ and B are both constant over time. We said earlier that in a uniform field, the net force on the current loop is zero. (In Figure 16-1, F and F may exert a torque because these forces are not collinear, but their components along each axis do cancel one another, yielding zero net force.) However, in any real situation, B cannot be uniform everywhere; it must become zero at some great distance from its source. mech

1

2

Imagine that B is zero at x=–∞. As we move the dipole’s center from x=–∞ to x=0, forces F and F make no contribution to the work done, because they are in the y-direction, orthogonal to the motion in x. We consider therefore only the work done against forces F and F . When the dipole’s center is at x=0, its left side is at x=–β, where β=(a/2) cosθ, and its right side is at x=+β. I copied Figure 16-1 below for your convenience. 3

1

2

4

Define W and W to be the work done against forces F and F , respectively. With β=(a/2)cosθ, the work equations are: 1

2

1

2

W = – ∫ F dx, from x=–∞ to x=–β W = – ∫ F dx, from x=–∞ to x=+β 1

1

2

2

W = + Jb ∫ B dx, from x=–∞ to x=–β W = – Jb ∫ B dx, from x=–∞ to x=+β 1 2

Here, the signs correspond to F being in the –x-direction and F being in the +x-direction. We now separate the W integral into two parts. 1

2

2

W = – Jb ∫ B dx, from x=–∞ to x=–β – Jb ∫ B dx, from x=–β to x=+β 2

The upper line here exactly cancels W , making the sum of both contributions: 1

W + W = –Jb ∫ B dx, from x=–β to x=+β 1

2

Assuming B is uniform near the origin: W = W + W = –Jb B 2β W = –Jb a cosθ B W = – µ•B 1

2

This exactly equals U . mech

In V2p15-3, Feynman highlights a mathematically equivalent procedure. Define B to be the field on side 1 (left side) and B the field on side 2 (right side). The total force is then: 1

2

F = F + F = Jb (B – B ) 1

2

2

1

F = Jb (a cosθ dB/dx) Here, a•cosθ = 2β is the x-distance from side 1 to side 2, and we assume B changes slowly with x. We can then calculate work W in one integral, with x now being the coordinate of the loop’s center. W = – ∫ F dx, from x=–∞ to x=0 W = – Jb a cosθ ∫ dB/dx dx, from x=–∞ to x=0 W = – Jb a cosθ B W = – µ•B Just as above.

Electrical Energy of Dipole Now let’s examine the electrical energy associated with current flowing through a loop. The analysis of the prior section assumed an unchanging dipole moment µ. In some special circumstances, such as elementary particles and superconductivity, magnetic moments persist without consuming energy. But normally, an external energy source is required to maintain a constant current flowing through a loop. As we found above, moving a current-carrying loop through a non-uniform magnetic field results in work. Positive work must be supplied to move a dipole into a stronger field if µ•B0. This means current J flows to the left in the suspended wire and counterclockwise around the loop. (Actually, current is carried by electrons flowing clockwise.) The galvanometer’s arrow indicates the magnitude of current flow. In V2p16-2 Feynman says: “The force pushes the electrons along the wire. But why does this move the galvanometer, which is so far from the force? Because when the electrons which feel the magnetic force try to move, they push—by electric repulsion—the electrons a little farther down the wire; they, in turn, repel the electrons a little farther on, and so on for a long distance. An amazing thing.” Now consider the opposite motion: move a magnet past a stationary wire. As Faraday discovered, a magnet moving with velocity –v produces the same current as a wire moving with velocity +v. Of course, this would not surprise Einstein: relativity says absolute velocity is meaningless, only relative velocities have physical significance. But, in Faraday’s time, this was a great surprise. In either case — moving wire or moving magnet — the induced current peaks when the wire and the magnet are closest, and drops to zero at very large separations. Faraday also discovered a related phenomenon: changing magnetic fields produces currents. One way to expose a stationary wire to a changing magnetic field is illustrated in Figure 17-4. Here, a loop of wire is near a coil whose current can be switched on or off.

Figure 17-4 Induced Current From Changing Field

With the switch open, no current flows through the coil; the coil’s field B and the wire’s current J are both zero. When the switch closes, current flows through the coil, B increases, current J starts flowing through the wire, and the galvanometer needle moves, as indicated by the dashed arrows. When B stops increasing, J drops to zero. If the switch is then opened again, B decreases and current –J flows through the wire. Finally, when B returns to zero, so does J. These new changing-magnetic-field effects cannot be explained solely with the Lorentz force, since the wire is stationary (v=0). We need new physics! Feynman notes that in a changing magnetic field, charges may experience different forces in different parts of a circuit. The net current is determined by the net electromotive force (“emf”). Emf is defined to be the integral along the entire circuit of the tangential electromagnetic force per unit charge. In V2p16-2, Feynman says that Faraday discovered: “emfs can be generated in a wire in three different ways: by moving the wire, by moving a magnet near the wire, or by changing a current in a nearby wire.” Feynman then states the flux rule for this new physics: A circuit’s emf equals the rate of change of magnetic flux through that circuit. We will be mathematically precise in later chapters. For now, let’s just explore the resulting effects. Consider the two devices shown in Figure 17-5. Each device is identical to the motor in Figure 17-1. First, let’s imagine that the devices are back-to-back and that both coils are mounted on one common axle (dashed central line), as shown.

Figure 17-5 Motor & Generator

As before, driving current J through the lower coil forces the axle to rotate. That turns the coil in the upper device, rotating it through a magnetic field and changing the magnetic flux through that coil. This creates an emf that drives current Ĵ in the upper device. In this mode, the lower device is an electric motor and the upper one is an electric generator. Alternatively, the devices could have two separate axles, and have their coils connected so that the same current flows through each coil (J=Ĵ). If we now mechanically rotate the lower device’s axle, the changing flux creates an emf that generates current J in both coils. That current in the upper device forcefully rotates its axle. In this mode, the lower device is a generator and the upper one is a motor. The devices are identical; their names change depending on how they are operated. In the first mode, the motor-generator driven by current J produces current Ĵ. In the second mode, the generator-motor driven by a rotating axle rotates a second axle. Even though the output is the same type of energy as the input, motor-generator sets can be quite useful. A motor-generator isolates one current from another. The input might be 50 Hz, 60 Hz, or 0 Hz (DC current), while the output could have any frequency and any voltage we wish. In the generator-motor mode, a large source of mechanical power can generate electricity that drives a compact motor at a remote location. The generator can be powered by burning any type of fuel, or by the motion of wind or water. This is how Hoover Dam keeps food cold in my refrigerator.

Another bidirectional device that employs emfs to operate between mechanical and electrical realms is a sound transmitter/receiver. The original telephone of Alexander Graham Bell was comprised of two “earphones” connected by a long loop of wire; one such “earphone” is sketched in Figure 17-6.

Figure 17-6 Bell’s Earphone

Here, a permanent magnet (dark gray) supports two posts, called yokes, that are made of soft iron (described below). Just above the yokes is a thin iron diaphragm that vibrates when exposed to sound waves, which are represented by dotted horizontal lines. A long loop of wire is coiled around one yoke. When sound waves vibrate the diaphragm, the changing yoke-diaphragm gap changes the magnetic fields in the yokes. The changing magnetic flux through the coil produces an emf that drives a current in the wire. When that current reaches the coil in the opposite “earphone”, that yoke’s magnetic field changes, causing its diaphragm to vibrate and produce sound waves. Sound waves are converted into electrical signals and then back into sound waves. Magnetic materials, such as iron, are described as being either hard or soft, depending on whether or not they can be permanently magnetized. In an external magnetic field, atomic magnetic moments become aligned. In hard materials, that alignment remains when the external field is removed, thus making a permanent magnet. In soft materials, atomic moments randomize when the external field is removed, and the material is no longer magnetic on a macroscopic scale. Soft iron in electromagnets make them easy to switch on, off, or reverse.

Transformers Our new principle is emf = ∂/∂t(flux): the emf in a circuit equals the rate of change of magnetic flux through that circuit. A less mathematical statement of this principle is: a changing magnetic field produces an electric field that induces a current in a conductor. Perhaps the most important practical device employing this effect is the transformer, an example of which is shown in Figure 17-7. Here two nearby coils on a common axis influence one another through their magnetic fields.

Figure 17-7 Two Coils & Bulb

The left coil is powered by an alternating voltage (“VAC”) source that alternates between positive and negative voltages. That source creates an alternating magnetic field represented by the array of horizontal lines encircled by the coils. That alternating field changes the flux through the right coil, creating an emf that drives an alternating electric current that lights a bulb. There is no direct mechanical or electrical contact between the left and right coils. Nonetheless, power is transmitted between them. Note that the number of turns in the two coils is different: the left has 8 turns whereas the right has 4 turns. If the coils are properly positioned, the same magnetic flux passes through each turn in both coils, This means the total emf in the left coil is twice that in the right coil — the right coil produces half the voltage applied to the left coil. By adjusting the number of turns in each coil, we can make the left/right voltage ratio anything we wish. This is the essence of a transformer. In a step-up transformer, the output coil has more turns, making the output voltage greater than the input voltage. In a step-down transformer, the output coil has fewer turns, making the output voltage less than the input voltage. In V2p16-4, Feynman introduces the concept of self-inductance. The changing flux in the left coil of Figure 17-7 affects both coils, not just the right coil. The emf in the left coil due to the changing field in the left coil is said to be self-induced. The self-induced emf acts in opposition to the driving AC voltage, resisting change. It is in some sense akin to mechanical inertia. Each time the AC voltage ramps up, energy is required to build up the magnetic field. The self-induced emf in effect absorbs the energy needed to build the field. When we stated our new principle, emf = ∂/∂t(flux), we weren’t specific about the emf polarity. This polarity is defined by Lenz’s rule: the emf always opposes the flux change. If flux changes because current J increases, the induced emf drives a current J* that is opposite to J. If flux changes because current J decreases, the induced emf drives a current J* in the same direction as J. In all cases, the emf acts to minimize change of both currents and fields. The electromagnet shown in Figure 17-8 provides an example of self-inductance. Here, an electromagnet, the big dark gray “C”, is activated by a coil whose current can be switched on and off.

Figure 17-8 Circuit with Electromagnet

If the switch has been closed (”on”) for some time, the coil current and magnetic field will be stable at their maximum values. If the switch is then opened, the coil current and magnetic field drop rapidly. The flux change in the coil creates an emf that strives to maintain a constant coil current and magnetic field. Large electromagnets have enormous self-inductances that create enormous emfs. Without a bulb or similar device to close the circuit, such powerful emfs create their own return paths, often causing dangerous and damaging arcs across an open switch or between coil windings. Feynman says: “The high voltage that appears might also damage … you, if you are the person who opens the switch!” The bulb provides a safe path for the emf-driven current as it gradually decreases to zero. When the switch is opened, the bulb shines brilliantly, and then gradually fades.

Magnetic Fun & Practicality We will next examine several amusing magnetic phenomena, all of which have important practical applications that enrich our lives. This discussion will be particularly enjoyable since we won’t tackle any equations at this time. Lenz’s rule is dramatically demonstrated by an aluminum ring placed atop an electromagnet, as in Figure 17-9.

Figure 17-9 Ring Repelled by Electromagnet

When the magnet is switched on, the changing flux through the aluminum ring produces an emf that drives a current around the ring, creating its own magnetic field. The ring’s field opposes the electromagnet’s field, per Lenz’s rule — when the electromagnet’s top becomes a north pole, so does the ring’s bottom. The opposing magnetic fields repel one another, driving the ring forcefully upward, as the arrows indicate. If a small slice of the ring is removed, so that current can no longer flow around its perimeter, the changing flux through the ring no longer creates an opposing magnetic field, and the upward force disappears. In V2p16-6, Feynman says that if the ring is replaced by a solid aluminum disk we get the same result. The changing flux still produces an emf that drives circulating currents that create an opposing magnetic field that forces the disk upward. A similar effect is shown in Figure 17-10. Here a dark gray magnet is placed above a flat, light gray, superconducting plane. Superconductors have zero resistance: a current flowing through a superconductor has zero voltage drop — not a very small drop, but rather absolutely zero voltage drop.

Figure 17-10 Levitating Electromagnet

The magnet’s field induces eddy currents in the superconductor. These currents create a magnetic field that exactly cancels the magnet’s field within the body of the superconductor. With their opposing magnetic fields, the magnet and plane repel one another, lifting the magnet above the plane. Since the repulsive force decreases with separation, there is some elevation at which the downward force of gravity exactly balances the upward repulsive magnetic force. For an ordinary, “non-super” conductor, one with non-zero resistance, the energy in eddy currents dissipates into heat. The eddy currents wane as the rate of change of flux and its induced emf diminish. The magnet gradually settles down onto the plane. Moving a magnet horizontally across an ordinary conductor changes the magnetic flux in various regions within the conductor. This creates new emfs, new induced currents, and new opposing magnetic fields. As Feynman notes, the opposing field strength is proportional to the rate of flux change, which is proportional to the magnet’s horizontal velocity. This resistive force is analogous to the viscous drag on an object moving through a dense fluid. The phenomenon of eddy currents is dramatically demonstrated by the abrupt stop of a pendulum swinging into a magnetic field, as illustrated in Figure 17-11, from a side view in the left image and from a front view in the right image.

Figure 17-11 Two Views of Pendulum & Magnet

Here, a light gray, conducting, rectangular plate swings from a support rod and passes between the poles of a dark gray electromagnet. In the left image, the B field is represented by arrows pointing to the right. In the right image, the B field points out of the screen towards you and is represented by arrows’ tips (black dots). If the electromagnet is off, the pendulum swings freely between the poles. When the electromagnet is powered, the conducting plate experiences a rapid flux change as it enters the strong B field region. That rapid flux change creates a large emf and strong eddy currents within the plate. These currents may absorb all of the plate’s kinetic energy, bringing the plate to a dead stop at the magnet’s center. These eddy currents are shown on magnified scale in two images in Figure 17-12. In both images, the B field points out of the screen, and the plate is shown partially in the strong B field region.

Figure 17-12 Eddy Currents in Two Plates

The left image shows a normal conducting plate with strong eddy currents (black circles) driven by a rapidly changing high level of flux. Such strong eddy currents can bring the swinging plate to a

complete stop. The right image shows the same scenario, but with a slotted plate. Here, the flux is divided into several smaller separated sections. In each section, the emf and the eddy currents are much less than those in the unslotted plate. The drag force is much weaker, and the pendulum may swing back and forth through the pole gap many times before finally coming to rest. In V2p16-6, Feynman says the eddy current drag force can also be demonstrated by detaching the conducting plate from its support, holding it between the magnet poles, and then releasing it. Rather than falling with the normal acceleration of gravity, the plate will very slowly move downward. This is because the plate’s motion generates strong eddy currents that resist that motion. Our next trick is spinning a suspended, conducting ring about its vertical axis with a rotating magnetic field. Rather than making a rotating field by simply turning a magnet, we examine a more exotic scheme. Imagine a torus (the shape of a donut) made of soft iron with six coils wrapped around its perimeter, as shown in Figure 17-13. Powering the left and right coils in opposite directions (tiny arrows) creates a magnetic field (large arrow) pointing up, as shown in the upper left image of Figure 17-13.

Figure 17-13 Six Coils Select Six B Directions

Powering other pairs of opposite coils orients the B field in five other directions, as the figure shows. Cycling through the six combinations steps the field through six orientations, effectively rotating the field direction. If this torus is now laid flat underneath a suspended ring, as shown in Figure 17-14, the ring will rotate about the vertical axis, tracking the torus’ magnetic field. The changing field flux creates emfs that drive eddy currents in the ring, which exert torques that turn the ring.

Figure 17-14 Rotating B Field Spins Ring

Our last example is a shaded pole motor, sketched in Figure 17-15. An electromagnet, powered by an AC generator, lies under one side of a conducting disk that is free to rotate about a vertical axle. A rectangular conducting plate covers, or shades, half of the magnet’s upper pole.

Figure 17-15 Shaded Pole Motor

With just the disk and magnet, Feynman says in V2p16-7: “we do not yet have a motor. There are eddy currents in the disc, but they are symmetric and there is no torque. If we now cover only one-half of the magnet pole with an aluminum plate [as shown in the figure], the disc begins to rotate….” Timing is the key issue here. Changing magnetic flux from the electromagnet produces eddy currents in the rectangular plate that create a secondary magnetic field that opposes the magnet’s field. Feynman says the plate’s field oscillates with the magnet’s field, but with a delay. Two magnetic fields affect the disk in its shaded region: the magnet’s field and the delayed plate field. The disk’s

unshaded region is affected only by the magnet’s field. Feynman says the net effect is equivalent to a half-sized magnet moving from the unshaded region toward the shaded one. He says interactions between fields and eddy currents exert torques forcing the disk to rotate. In V2p16-8, Feynman concludes this lecture with a long and charming description of a remarkable engineering accomplishment. In what follows Feynman refers to Hoover Dam by another name; I replaced that with its proper name for clarity. “This problem of … making [something] work in the most practical way is engineering. It requires serious study of design problems, although there are no new basic principles [of physics]. But there is a long way to go from the basic principles to a practical and economic design. Yet it is just such careful engineering design that has made possible such a tremendous thing as Hoover Dam and all that goes with it. “What is Hoover Dam? A huge river is stopped by a concrete wall. But what a wall it is! Shaped with a perfect curve that is very carefully worked out so that the least possible amount of concrete will hold back a whole river. It thickens at the bottom in that wonderful shape that the artists like but that the engineers can appreciate because they know that such thickening is related to the increase of pressure with the depth of the water. … “Then the water of the river is diverted into a huge pipe. That’s a nice engineering accomplishment in itself. The pipe feeds the water into a “waterwheel”—a huge turbine—and makes wheels turn. (Another engineering feat.) But why turn wheels? They are coupled to an exquisitely intricate mess of copper and iron, all twisted and interwoven. With two parts—one that turns and one that doesn’t. All a complex intermixture of a few materials, mostly iron and copper but also some paper and shellac for insulation. A revolving monster thing. A generator. “Somewhere out of the mess of copper and iron come a few special pieces of copper. The dam, the turbine, the iron, the copper, all put there to make something special happen to a few bars of copper—an emf. Then the copper bars go a little way and circle for several times around another piece of iron in a transformer; then their job is done. “But around that same piece of iron curls another cable of copper which has no direct connection whatsoever to the bars from the generator; they have just been influenced because they passed near it—to get their emf. The transformer converts the power from the relatively low voltages required for the efficient design of the generator to the very high voltages that are best for efficient transmission of electrical energy over long cables. “And everything must be enormously efficient—there can be no waste, no loss. Why? The power for a metropolis is going through. … If one percent of the power were left in the transformer, that energy would need to be taken out somehow. If it appeared as heat, it would quickly melt the whole thing. There is, of course, some small inefficiency, but all that is required are a few pumps which circulate some oil through a radiator to keep the transformer from heating up. “Out of the Hoover Dam come a few dozen rods of copper—long, long, long rods of copper perhaps the thickness of your wrist that go for hundreds of miles in all directions. Small rods of copper carrying the power of a giant river. Then the rods are split to make more rods … then to

more transformers … sometimes to great generators which recreate the current in another form … sometimes to engines turning for big industrial purposes … to more transformers … then more splitting and spreading … until finally the river is spread throughout the whole city— turning motors, making heat, making light, working gadgetry. The miracle of hot lights from cold water over 600 miles [1000 km]—all done with specially arranged pieces of copper and iron. Large motors for rolling steel, or tiny motors for a dentist’s drill. Thousands of little wheels, turning in response to the turning of the big wheel at Hoover Dam. Stop the big wheel, and all the wheels stop; the lights go out. They really are connected. “Yet there is more. The same phenomena that take the tremendous power of the river and spread it through the countryside, until a few drops of the river are running the dentist’s drill, come again into the building of extremely fine instruments … for the detection of incredibly small amounts of current … for the transmission of voices, music, and pictures … for computers … for automatic machines of fantastic precision. “All this is possible because of carefully designed arrangements of copper and iron—efficiently created magnetic fields … blocks of rotating iron six feet [1.8 m] in diameter whirling with clearances of 1/16th inch [1.5 mm] … careful proportions of copper for the optimum efficiency … strange shapes all serving a purpose, like the curve of the dam. “If some future archaeologist uncovers Hoover Dam, we may guess that he would admire the beauty of its curves. But also the explorers from some great future civilizations will look at the generators and transformers and say: ‘Notice that every iron piece has a beautifully efficient shape. Think of the thought that has gone into every piece of copper!’ “This is the power of engineering and the careful design of our electrical technology. There has been created in the generator something which exists nowhere else in nature. It is true that there are forces of induction in other places. Certainly in some places around the sun and stars there are effects of electromagnetic induction. Perhaps also (though it’s not certain) the magnetic field of the earth is maintained by an analog of an electric generator that operates on circulating currents in the interior of the earth. But nowhere have there been pieces put together with moving parts to generate electrical power as is done in the generator—with great efficiency and regularity. “You may think that designing electric generators is no longer an interesting subject, that it is a dead subject because they are all designed. Almost perfect generators or motors can be taken from a shelf. Even if this were true, we can admire the wonderful accomplishment of a problem solved to near perfection. But there remain as many unfinished problems. Even generators and transformers are returning as problems. It is likely that the whole field of low temperatures and superconductors will soon be applied to the problem of electric power distribution. With a radically new factor in the problem, new optimum designs will have to be created. Power networks of the future may have little resemblance to those of today. [We are still waiting for that 50 years later.] “You can see that there is an endless number of applications and problems that one could take up while studying the laws of induction. The study of the design of electrical machinery is a life

work in itself. We cannot go very far in that direction, but we should be aware of the fact that when we have discovered the law of induction, we have suddenly connected our theory to an enormous practical development. … [Physics] supplies the base—the basic principles that apply, no matter what. “Modern electrical technology began with Faraday’s discoveries. The useless baby developed into a prodigy and changed the face of the earth in ways its proud father could never have imagined.”

Chapter 17 Review: Key Ideas Motors and sensitive electrical instruments often employ an electric current in a magnetic field: to deliver mechanical power, or to precisely measure current flow. The electromotive force (“emf”) in a circuit is defined as the integral along the entire circuit of the tangential electromagnetic force per unit charge. Feynman says: “emfs can be generated in a wire in three different ways: by moving the wire, by moving a magnet near the wire, or by a changing current in a nearby wire.” A circuit’s emf equals the rate of change of magnetic flux through that circuit. Emfs always oppose changing currents or fields.

Chapter 18 Exploring Induction In V2p17-1, Feynman says: “In the last chapter we described many phenomena which show that the effects of induction are quite complicated and interesting. Now we want to discuss the fundamental principles which govern these effects.” All induction phenomena are governed by two fundamental equations that are always valid: The Lorentz force: F = q(E + v×B) Maxwell’s equation #2: Ď×E = – ∂B/∂t In the prior chapter, we introduced emf and the associated flux rule. These are shortcuts that can simplify analysis, but which involve assumptions that are not valid in all cases. We explore here when and how to properly use these shortcuts. The definition of emf and the flux rule for emfs in electrical circuits are: Definition: emf = + ∫ ds•F/q Γ

Flux Rule: emf = – d/dt ∫ B•n da S

The first equation says: the emf in circuit Γ (a closed loop) equals the integral around Γ of the tangential component of the electromagnetic force per unit charge (F/q=E+v×B). The second equation says: the emf in circuit Γ equals minus the rate of change of the magnetic flux through the area S enclosed by Γ. Here, da is the infinitesimal area being integrated over, and n is the unit normal to da. You will often see n da replaced by da; the latter is a vector with magnitude da and direction parallel to n. Feynman isn’t explicit or entirely consistent about emf polarities, only saying that emfs always oppose flux changes. I shall strive to provide the most commonly used polarity conventions where possible. Note that magnetic flux may change due to changes within a circuit, motion of the circuit as a whole, changes in the magnetic field, or any combination thereof. Let’s first consider a simple case: a circuit of a changing size situated in a constant magnetic field. Figure 18-1 shows a rectangular circuit formed by a stationary U-shaped wire and a movable

crossbar (thick black line). The crossbar moves upward at velocity v, while remaining perfectly horizontal. The circuit formed by the crossbar and three sides of the U is rectangular with width w and length L.

Figure 18-1 A Changing Rectangular Circuit

The black dots in the figure indicate the constant magnetic field coming out of the screen. Let’s calculate the circuit’s emf from the integral of F/q=v×B. The charges in the crossbar move upward at velocity v, while all other charges are stationary. The integral thus reduces to the sum along the crossbar. Vectors v and B are orthogonal and v×B points to the right, which is anti-parallel to ds. (The right hand rule specifies counterclockwise line integrals.) The integral is then: emf = ∫ ds•F/q = – w v B Γ

Note that we consider only the forces exerted on free charges, the current carriers in conductors. In a normal conductor, some electrons are not attached to specific atoms but move freely carrying electric current. Conductors also contain ions, with the opposite charge that experience the opposite Lorentz force, but which cannot move because they are secured in place by neighboring atoms. The securing forces cancel the Lorentz forces on the ions. Now let’s compare the above result with the emf calculated from a changing flux. At each moment, the flux through the circuit equals BwL, field B multiplied by the enclosed area wL. Since L changes as the crossbar moves upward, the flux changes, resulting in an emf equal to: emf = – d/dt (flux) = – d/dt (wLB) emf = – w B dL/dt = – w B v This is the same result obtained from integrating the electromagnetic force per unit charge – the flux rule and the emf definition agree.

This emf drives a current j that flows clockwise around the rectangle, as indicated by the arrows in Figure 18-1. Feynman simplifies the problem by assuming the circuit is made of high-resistance wire, making j small enough to neglect its effects, including the magnetic field it produces. Since j flows clockwise in this case, its polarity is negative according to the right hand rule, and it generates a magnetic field into the screen that opposes the external field B. Our result can be generalized to circuits of any shape. As we have seen several times before, any closed loop can be subdivided into numerous small rectangles. Our above proof applies separately to each small rectangle. By the principle of linear superposition, when we sum over all small rectangles, we find that the total emf and total flux change are governed by these same linear equations. We next consider another source of flux change: a stationary loop in a changing magnetic field. Again, let the loop be rectangular, with width w and length L. The emf from the flux rule is: emf = – d/dt (flux) = – d/dt (wLB) emf = – w L dB/dt We wish to compare this with the emf from the integral of electromagnetic force – but where is the force? For a stationary circuit, the velocities of all charges are zero and the force equation F/q=E+v×B reduces to F/q=E. It seems that the magnetic field has no effect. But it actually does. One of the seminal moments in the understanding of electromagnetism was Faraday’s experimental discovery that changing magnetic fields produce electric fields. Maxwell developed the equation for this effect, which we list as his second law: Ď × E = – ∂B/∂t To honor his profound discovery, this is called Faraday’s law. Let’s recast this law. Recall Stokes’ theorem for any vector field β: ∫ β•ds = ∫ (Ď×β)•n da Γ

S

As usual, Γ is a closed loop enclosing surface S, and n is the unit normal to infinitesimal area da within S. With Stokes’ theorem, Faraday’s law becomes: ∫ E•ds = ∫ (–∂B/∂t)•n da ∫ E•ds = – ∂/∂t (flux of B through S) Γ

S

Γ

When v=0 for all charges, the Lorentz force becomes F/q=E, yielding: ∫ ds•F/q = – ∂/∂t (flux of B through S) emf = – ∂/∂t (flux) Γ

We again recover the flux rule. Thus emfs are created by both moving charges and changing fields. Feynman notes that two quite distinct phenomena yield the same outcome, and that the flux rule applies equally to both. In V2p17-2, Feynman says: “We know of no other place in physics where such a simple and accurate general principle requires for its real understanding an analysis in terms of two different phenomena. Usually such a beautiful generalization is found to stem from a single deep underlying principle. Nevertheless, in this case there does not appear to be any such profound implication. We have to understand the ‘rule’ as the combined effects of two quite separate phenomena. “We must look at the ‘flux rule’ in the following way. In general, the force per unit charge is F/q=E+v×B. In moving wires there is the force from the second term. Also, there is an E field if there is somewhere a changing magnetic field. They are independent effects, but the emf around the loop of wire is always equal to the rate of change of magnetic flux through it.” Feynman stresses an important difference between the two emf-producing phenomena. The v×B term of emf requires matter: particles that have both q and v. Most commonly, that matter will be a physical wire, but a beam of charged particles also suffices. But the E term of emf can exist without matter. Wherever B changes, even in completely empty space, it creates an E field. Even an imaginary loop in empty space that encloses a changing magnetic flux will have a non-zero curl of E, a non-zero integral of E•ds. This is quite different from the E fields from static electric charges that always have zero curl, Ď×E=0.

Flux Rule Cautions We now examine two cases in which the flux rule cannot be directly applied. This will highlight the assumptions implicit in the flux rule that require caution in its use. In the first case, an emf exists but there is no flux change. Figure 18-2 shows a disk spinning on an axle, and exposed on one side to a constant magnetic field. Let the disk and its axle be made of a nonmagnetic, electrical conductor, such as copper.

Figure 18-2 Magnet & Spinning Disk

One conducting brush rubs against the disk and another against the axle, forming a closed electrical circuit, passing through a galvanometer that measures current flow. The circuit runs from the galvanometer to the brush contact point on the axle, up to the disk’s center, to the right along a horizontal radial line to the brush contact point at the disk’s edge, and back to the galvanometer. The circuit does not move, and the magnetic field does not change; hence the flux does not change. Yet, there is an emf. Free charges in the conducting disk move through the magnet’s field and therefore experience a Lorentz force: F=qv×B. In the second case, the flux change is large but the emf is much smaller. Figure 18-3 shows two images, each with a galvanometer connected to two long rectangular conductors with a small conducting bar (black square) that electrically connects the rectangles at one spot.

Figure 18-3 Changing Circuit

A uniform, constant magnetic field B fills the space containing the two rectangles and the gap between them. B points into the screen, as indicated by the crossed circles that represent the tails of arrows. The black bar is at the left in the upper image, and at the right in the lower image. The dashed lines show the path of current flow through the rectangles. The complete circuit encloses a much larger area in the lower image than in the upper image. Hence, much more flux passes through the lower circuit than the upper one. As shown in the figure, w is the total width of the two rectangles and the gap between them, w is the gap width, and L is the distance from left edge to the black bar. 1

2

With w >>w , the circuit’s area within the magnetic field is approximately Lw /2, and the flux through that area equals BLw /2. As the black bar slides to the right at velocity v, the derivative of flux is: 1

2

1

1

d/dt (flux) = d/dt (BLw /2) = B v w /2 1

1

Let’s now compare this to the emf from the integral of F/q. The emf in a circuit never comes from charges flowing in that current. The velocity u of current-carrying charges is always parallel to the direction of the circuit path ds, therefore u×B is orthogonal to ds and (u×B)•ds is zero. The emf in this circuit comes entirely from free charges in the sliding bar. emf = – ∫ (v×B)•ds = – v B ∫ds = – v B w

2

By our construction w >>w , so the flux change has a much greater magnitude than does the emf. 1

2

This is the key point: the flux rule works when there is no change in the material of which the circuit is comprised – the actual atoms through which current flows must remain the same. The flux rule may not work if the circuit changes by redirecting current through different atoms. In the spinning disk in a static magnetic field (see Figure 18-2), current flows through the disk at the same coordinate positions but through a continually changing group of atoms. This is why there is an emf despite no flux change, contrary to the flux rule. As the bar slides in a static magnetic field (see Figure 18-3), current flows through different atoms in the rectangular conductors. This is why there is a large flux change with a much smaller emf, again contrary to flux rule. We say the flux may not work if the circuit material changes, but it might work in special circumstances. If we change the sliding bar example so that w =w , we return to the situation of Figure 18-1, the U-shaped wire with a sliding crossbar, which we found does obey the flux rule. 1

2

When in doubt, go back to Maxwell’s equations and the Lorentz force, which are always valid.

Particle Accelerators In V2p17-3, Feynman explains the operation of a betatron, one of the first particle accelerators. Betatrons are able to accelerate electrons up to energies of 300 million-electron-volts (300 MeV). (The name arises because electrons were first identified as beta rays.) After discussing betatrons, we will briefly describe modern particle accelerators that have accelerated electrons up to 50 billion eV (50 GeV) and protons up to 6.5 trillion eV (6.5 TeV). Betatrons accelerate electrons by employing varying magnetic fields and the electric fields that they induce. Figure 18-4 shows two views of a betatron B field: above is a side view, and below is the top view. The magnetic field is strongest at the center, and points upward everywhere.

Figure 18-4 Betatron Magnetic Field

Everything in a betatron is symmetric about the central vertical axis. Figure 18-5 shows the top view of an electron’s circular orbit in a betatron, indicating its velocity v and the qv×B part of the Lorentz force that holds the electron in orbit.

Figure 18-5 Electron Orbit in Betatron

Here, the B field points out of the screen everywhere (upward in the betatron). Also shown is the induced, circumferential E field caused by a constantly increasing B field. Since dB/dt is positive, Ď×E=–dB/dt is negative, yielding a clockwise circulation according to the right hand rule.

Since q is negative, the qE part of the Lorentz force is everywhere antiparallel to E, and therefore everywhere parallel to v. This means the induced electric field accelerates electrons orbiting in the magnetic field. The key to operating a betatron is balancing the electric field that accelerates electrons with the magnetic field that maintains their orbits. For a stable circular orbit of radius r, the flux enclosed within the orbit is πr B , where B is the average value of B within that radius. The circulation of the induced electric field is 2πrE, the orbital circumference times the field strength everywhere along that orbit. Faraday’s law says: 2

av

av

Ď×E = –∂B/∂t 2πrE = –∂/∂t (πr B ) E = –(r/2) ∂B /∂t 2

av

av

The minus sign confirms that E points clockwise, corresponding to a negative curl according to the right hand rule. Electrons are accelerated by the force F=qE. Relativistically, force is the time derivative of momentum p, according to: F = qE = dp/dt We have the vectors in the right directions (E is accelerating the electrons), so we need only deal with the magnitudes. Hence we have: E = (1/q) dp/dt = (r/2) ∂B /∂t dp/dt = (qr/2) ∂B /∂t av

av

Integrating with respect to time yields: p = p + (qr/2) ΔB 0

av

Here, p is the electron’s momentum at an initial time t=0, and ΔB is the change from time t=0 in the average magnetic field within radius r. 0

av

We have one more relationship to impose: the magnetic field at orbital radius r must provide the correct force to maintain that orbit. We analyzed orbital mechanics in Feynman Simplified 1A Chapter 8. For a stable circular orbit: centripetal acceleration = v / r centripetal force = pω = p v/r 2

In our case, the centripetal force is qvB , with B being the magnetic field strength at radius r. We therefore have: r

qvB = p v/r r

r

Assuming that the initial values of electron momentum and magnetic fields are much less than their final values, we can combine the above equations to yield: qvB = (qr/2) B (v/r) 2B = B r

r

av

av

Thus, the betatron field must increase so that the average field within radius r is twice the field at r. This ensures that the electron’s momentum always matches the field needed to maintain its orbit. Betatrons were first developed in the 1940s, and have been used to produce high-energy x-rays and gamma rays for diverse scientific, industrial, and medical applications. The maximum practical electron energy in betatrons is limited by the ability to produce adequate fields and by energy loss to intense radiation emitted by circulating high-energy electrons. Higher energy electron accelerators must employ other techniques. The highest energy electron accelerator ever built is the Stanford Linear Accelerator (SLAC) in Palo Alto, California. It now achieves electron energies of up to 50 GeV, corresponding to electron velocities of 0.999,999,999,95 times the speed of light, the fastest man-made charged particles ever produced. Electrons are accelerated by klystrons, high-power radio-frequency (RF) amplifiers, through a straight 2-mile-long tube. SLAC claims to be the world’s straightest object. Remarkably, despite the need for straightness, SLAC is adjacent to the San Andreas Fault and passes under Interstate Highway 280. The highest-energy particle accelerator ever built is the Large Hadronic Collider (LHC) near Geneva, Switzerland. The LHC features two colliding proton beams, each with a maximum energy (so far) of 6.5 TeV. The LHC is a circular accelerator with a circumference of 27 km (17 miles), and claims to be the largest and most complex scientific facility, and the largest machine of any type ever built. LHC employs over 1600 superconducting magnets, weighing up to 27 tons each. About 75% of these magnets bend protons through the circular beam tubes, and 25% focus the beams. RF cavities provide particle acceleration, as at SLAC.

An Apparent Paradox In V2p17-5, Feynman presents an apparent paradox, saying: “A paradox is a situation which gives one answer when analyzed one way, and a different answer when analyzed another way, so that we are left in somewhat of a quandary as to actually what should happen. Of course, in physics there are never any real paradoxes because there is only one correct answer; at least we believe that nature will act in only one way (and that is the right way, naturally). So in physics a paradox is only a confusion in our own understanding.” The apparent paradox is illustrated in Figure 18-6. The axle of a non-conducting disk pivots on frictionless bearings (not shown). Attached to the disk are six separated conducting balls, and a loop of superconducting wire carrying current j.

Figure 18-6 Disk, Balls & Current Loop

The disk is initially stationary. As long as the superconducting loop (central black circle) remains cold enough, it has zero electrical resistance, and the current j will flow inside it forever. The circulating current creates a magnetic field pointing upward. But as the superconductor warms, its temperature eventually exceeds the limit for superconductivity. The current j then drops rapidly to zero, and the magnetic flux drops to zero as well. The flux change induces a circumferential electric field with E everywhere parallel to the original current j, because the induced emf seeks to maintain the prior B field by opposing its reduction. Our first analysis is: the induced emf applies an azimuthal force to the conducting charges in each of the six balls. These forces exert six torques in the same direction that spin the disk counterclockwise, the same direction as the original current j. Our second analysis is: by the principle of conservation of angular momentum, the total angular momentum must be the same before and after current j stops. Since the disk and all its components were stationary initially, they cannot be spinning after the current stops. Our two analyses are antithetical – one must be wrong - hence the apparent paradox. Feynman leaves this as a puzzle for you to solve. He says the solution “does not depend on any nonessential feature… the solution is not easy, nor is it a trick …When you figure it out, you will have discovered an important principle of electromagnetism.” You can check your answer in the section at the end of this chapter. The remainder of this chapter provides detailed analyses of phenomena discussed qualitatively in the prior chapter.

AC Current Generator We begin with an alternating-current generator. Figure 18-7 shows a conducting loop mounted on a conducting axle, with a uniform, constant magnetic field B that points out of the screen (as represented by the dots).

Figure 18-7 AC Generator & Load

Wires connected to opposite sides of the coil are attached to a load, the resistor R. Let θ be the coil rotation angle. The flux through the coil at angle θ is given by: flux = B S cosθ Here, S is the surface area enclosed by the coil, and θ=0 is the orientation illustrated in the figure, which is the orientation of maximum flux. If the coil rotates at a constant angular velocity ω (θ=ωt), the emf in the coil is: emf = – ∂(flux)/∂t = + B S ω sin(ωt) Feynman says we can equate this emf with an electric potential, voltage V, provided that we do so in a region where ∂B/∂t=0 and therefore Ď×E=0. Our current equation for the electric potential is not well defined when the curl of E is non-zero. Since that is not an issue in this case, we can define V and V by: 0

V = B S ω sin(ωt) = V sin(ωt) 0

The alternating voltage V is what most of us have in our home wiring. The current j flowing through the load resistor R is:

j = V / R = V sin(ωt) / R 0

This current j is called an alternating current (AC), since it alternates between positive and negative values. In V2p17-7, Feynman adds this comment: “Since there is an electric field between the wires [coming from the coil], they must be electrically charged. It is clear that the emf of the generator [the spinning coil] has pushed some excess charges out to the wire until the electric field from them is strong enough to exactly counterbalance the induction force. Seen from outside the generator, the two wires appear as though they had been … charged to the potential difference V, as though the charge was being changed with time to give an alternating potential difference. There is also another difference from an electrostatic situation. If we connect the generator to an external circuit that permits passage of a current, we find that the emf does not permit the wires to be discharged but continues to provide charge to the wires as current is drawn from them, attempting to keep the wires always at the same potential difference.” Now let’s examine the energy delivered by the generator and the energy required to spin the coil. Let n be the number of unit charge carriers per unit length of wire, and let v be the velocity and F be the force on each carrier. The power required to drive this current in an infinitesimal length of wire ds is: power = n ds F•v This is: (the number of unit charge carriers in ds) multiplied by (the power exerted per unit charge). Recall that work equals F•x, the dot product of force and distance moved, and that power is the time derivative of work (F•dx/dt=F•v). In any wire, v must be along the wire parallel to ds, hence we can rewrite the prior equation as: power = nv F•ds Current j is simply nv, the number of unit charges multiplied by their velocity. Also, the integral throughout the circuit of F•ds equals the emf. We therefore have: electrical power = j • emf This result is evident even without all that math: electrical power is always the driving voltage (emf) multiplied by the current (j). Feynman just wanted to provide us some mental exercise. Now let’s calculate the mechanical power required to rotate the coil, which is the source of the electrical power delivered to the circuit. In Chapter 15, we found that the magnetic moment µ of a coil equals the current j flowing through it multiplied by the area S that it encloses. The mechanical potential energy U of magnetic dipole moment µ in field B is:

U = – µ•B = – (jS) B cosθ The torque τ required to rotate the coil is the derivative of U with respect to angle θ is: τ = ∂U/∂θ. This yields: τ = j S B sinθ The mechanical power required to drive torque τ at angular velocity ω equals τω, hence: mechanical power = ω j S B sinθ Recall that: emf = ω S B sinθ electrical power = j • emf We see that the electrical power output exactly equals the mechanical power input (energy is conserved once again).

“Viscous” Forces We stated earlier that the polarity of an induced emf always opposes flux changes. Recall the first example at the start of this chapter: a U-shaped wire with a sliding crossbar in a static magnetic field pointing out of the screen. This time, we place resistor R along the bottom of the U, as shown in Figure 18-8.

Figure 18-8 Changing Circuit with Resistor

The flux change and emf are unchanged by the addition of the resistor; we still have: emf = – v B w The resistor does change current j, which becomes: j = emf / R = – v B w / R The minus sign indicates the current flows clockwise. Current j leads to a Lorentz force on the crossbar. Let n be the number of unit charge carriers per unit length of crossbar, and let u be the velocity and F be the force on each carrier. The total Lorentz force on the crossbar is: F = nw u×B Here nw is the total number of unit charge carriers in the crossbar, and the direction of u×B is downward, per the right hand rule, antiparallel to the crossbar’s velocity v. Since u and B are orthogonal, and nu=j, we have: F=wBj Substituting the above equation for j yields: F = wB (–vBw/R) = – B w v / R 2

2

The Lorentz force is antiparallel to and proportional to v. Note that the mechanical power required to move the crossbar, is: mechanical power = – F•v = + B w v / R 2

2

2

Note the minus sign; work is done in moving against a force. The electrical power output is: j • emf = (–vBw / R) • (–vBw) electrical power = + B w v / R 2

2

2

Again, all mechanical power input is converted into electrical power output. In V2p17-8, Feynman comments on the proportionality of the induced Lorentz force: “Such a ‘velocity-proportional’ force, which is like the force of viscosity, is found whenever induced currents are produced by moving conductors in a magnetic field. The examples of eddy currents we gave in the last chapter also produced forces on the conductors proportional to the velocity of the conductor... “It is often convenient in the design of mechanical systems to have damping forces which are

proportional to the velocity. Eddy-current forces provide one of the most convenient ways of getting such a velocity-dependent force.” Feynman describes one application of this effect that is near and dear to all our pocketbooks: the kilowatt-hour meter that determines your electric bill. Traditionally, these devices contain an aluminum disk rotating in a magnetic field. An electric motor, whose torque is proportional to the power being consumed, spins the disk. Eddy currents in the disk provide a velocity-proportional damping force, making the spin rate proportional to power consumption. The meter simply counts disk rotations. Some modern meters, like mine, are entirely electronic.

Mutual Inductance We turn now to the mutual inductance between two coils of wire. The effect is best demonstrated between two solenoids. Let solenoid #1 have length L and n turns of wire each carrying current j , and let solenoid #2 have length L and n turns carrying current j . Let the solenoids be adjacent, so that the entire magnetic field (horizontal lines within solenoids) produced by either solenoid enters the other, as shown in Figure 18-9. 1

2

2

1

1

2

Figure 18-9 Mutual Inductance: 2 Solenoids

For clarity, we show the solenoid windings as being separated. But in transformers, the most common application of mutual inductance, the windings are on top of one another. Feynman cautions that we have so far only considered the production of magnetic fields by constant currents. More interesting effects are discussed in later chapters. Here, we assume all currents change slowly enough that we can approximate their fields by constant-current equations. The current j produces a magnetic field B given by: 1

B = n j / (c ε L ) 2

1

1

0

1

Let S be the cross-sectional area of each solenoid. The flux passing through each turn of each solenoid due to j is: 1

flux per turn = B S

If j varies slowly, the emf in solenoid #2 is: 1

emf = – d(flux)/dt = – n S dB/dt emf = – n S n (dj /dt) / (c ε L ) emf = – M (dj /dt) with M = n n S / (c ε L ) 2

2

2

2

2

2

1

21

1

0

1

1

2

21

1

2

0

1

Feynman says M is a geometric property of the circuit-pair called its mutual inductance. 21

Any pair of current loops has similar behavior: the induced current in one loop is proportional to the current in the other loop, and the proportionality constant, their mutual inductance, is determined only by the geometry of the loops. Even the bizarre loops shown in Figure 18-10 have a mutual inductance that we can calculate.

Figure 18-10 Mutual Inductance: 2 Loops

Assume initially that we drive a changing current in loop #2 and do nothing to loop #1. The emf induced in loop #1 by a changing magnetic field B is. emf = – ∂/∂t { ∫ B•n da } 1

S1

Here, the integral is over S , the surface area enclosed by loop #1. Recall from Chapter 15 the relation between a surface integral of magnetic field B and the line integral of vector potential A: 1

∫ A•ds = ∫ B•n da Γ

S

Here, Γ is the curve enclosing surface S. Using this equation we obtain: emf = – ∂/∂t { ∫ A•ds } 1

Γ1

1

The vector potential A produced by current j in coil #2 is given by the general equation: 2

A(r ) = (1/4πε c ) ∫ j (r ) ds / |r – r | 2

1

0

Γ2

2

2

2

1

2

This equation equates the vector potential A at point r with an integral over all of loop #2. The integrand is current j at point r divided by the distance between r and r . Lastly, we divide by the 1

2

2

1

2

scaling factor 4πε c . 2

0

Note that when we integrate along a loop, such as a wire, j (r ) ds can be rewritten as: 2

j ds 2

2

2

2

This is because j has the same value everywhere within the loop and the direction of the current must be parallel to the direction of the wire. The equation for A becomes: 2

A(r ) = j (1/4πε c ) ∫ ds / |r – r | 2

1

2

0

Γ2

2

1

2

Note that the integral contains only constant geometric factors. With this equation for A, we obtain a double integral for the emf in loop #1: emf = – M ∂j /∂t 1

12

2

where M , the mutual inductance of the two loops, is defined as: 12

M = (1/4πε c ) ∫ ∫ ds •ds / |r – r | 2

12

0

Γ1

Γ2

2

1

1

2

Note that the mutual inductance is independent of time and current flow. It is evident that M must equal M for any pair of loops. The subscripts on M are usually deleted if only two loops are being considered. 21

12

The normal polarity convention is that M is positive if both circuit loops wind in the same direction.

Self-Inductance We assumed above that loop #2 had a current driven by an external voltage source and loop #1 had an induced current driven by the magnetic field from loop #2. We did not address what the field from the current in loop #1 does to loop #2. Clearly, there can be currents in both loops. The flux linking the two loops would then be the sum of each flux, since magnetic fields are governed by the principle of linear superposition. In this general case, the emfs in the two loops can be written: emf = – M ∂j /∂t – M ∂j /∂t emf = – M ∂j /∂t – M ∂j /∂t 1

11

1

12

2

2

21

1

22

2

We see that each emf is the sum of two terms, each proportional to the current in one loop. The quantities M and M are called self-inductances and are commonly written: 11

22

L =M L =M 1

11

2

22

In the most common convention, these are positive quantities. Feynman uses a different convention in

V2p17-11. All current loops have self-inductance, even those completely isolated from other currents. For a single circuit around a closed loop, the emf due to a change in its own current j is: emf = – L dj/dt The emf opposes the change of current, and is often called a back emf. This is akin to mechanical inertia. In a circuit with self-inductance, energy is required to change its current, the rate of charge flow, just as energy is required to change a massive object’s velocity. The upper portion of Figure 18-11 shows a circuit carrying current j with a voltage source V and a coil with self-inductance L. The lower portion shows a mechanical analog: force F pushing mass m that has velocity v.

Figure 18-11 Electrical & Mechanical Inertia

The electrical equation relating V, L, and j is: V = L dj/dt The mechanical equation relating F, m, and v is: F = m dv/dt

The key quantities in electrical and mechanical phenomena have the following correspondence:

Inductance & Magnetic Energy In the above table, we see the correspondence between mechanical momentum mv and the electrical quantity Lj. Lj satisfies equations of the same form as equations satisfied by mechanical momentum mv. But, Feynman cautions that Lj is not the momentum on an electrical circuit. That circuit might be stationary and have no momentum at all. We also see the correspondence of kinetic energy mv /2 and the electrical quantity Lj /2. Here not only are the equations similar, but so are the physical meanings: Lj /2 is in fact the energy U of a circuit with current j flowing through an inductor, with inductance L. 2

2

2

To demonstrate that Lj /2 really is electrical energy, consider the rate of change of work (the derivative of energy U): 2

Mech: dU/dt = Fv Elec: dU/dt = Vj Integrating with respect to time yields: Mech: U = ∫ Fv dt Mech: U = ∫ (m dv/dt) v dt Mech: U = m ∫ v dv = m v /2 2

Elec: U = ∫ Vj dt Elec: U = ∫ (L dj/dt) j dt

Elec: U = L ∫ j dj = L j /2 2

What we have calculated in each case is the energy stored in the moving mass or the current-carrying inductor. In both derivations above, we assumed ideal conditions with no energy lost to friction or heat. For a two-loop system, such as the two solenoids in Figure 18-9, the total stored energy is: U = L j /2 + L j /2 + M j j 2

1

2

1

2

2

1

2

We can show that this equation is correct by imagining two loops with no initial currents. We then turn on current j , and subsequently turn on current j . Feynman says the work done in turning on j is just L j /2. When we turn on j , energy L j /2 is required to overcome the self-inductance of loop #2, but additional energy is required to hold j constant by overcoming the emf (M dj /dt) in loop #1 from loop #2. The latter energy requirement is given by the integral of j •emf: 1

2

2

1

1

2

1

2

2

2

1

2

1

∫ j (emf in #1 from #2) dt = ∫ j (M dj /dt) dt = M j ∫ dj =Mj j 1

1

2

1

2

1

2

Next, consider the force between two such current loops. Let x be the distance between coil #1 and coil #2. The principle of virtual work in one dimension is: F = –dU/dx. In the equation for U, the only variable that seems to change with x is the mutual inductance M, leading us to: F = – dM/dx j j , but this is wrong! 1

2

We encountered the same issue in calculating the force between capacitor plates in Chapter 8. To get the right answer, we must include energy changes in the devices that maintain constant currents j and j while the coil separation x changes. 1

2

In V2p17-13, Feynman reminds us that in Chapter 8 we found that, for capacitors, the total energy equals –U , the mechanical energy. He says that is true here as well, so that: mech

F = – dU/dt = +d(U )/dt F = + dM/dx j j mech

1

2

Now let’s go back to the equation for the stored energy in two coils. U = L j /2 + L j /2 + M j j 2

1

1

2

2

2

1

2

Feynman suggests a trick to express M in terms of L and L . Rewrite this equation as: 1

U = L (j + M j /L ) /2 + (L – M /L ) j /2 2

1

1

2

1

2

2

1

2 2

2

You might wish to verify this, just for fun. Since stored energy can never be negative, U must be >= zero for any combination of currents and inductances. In the special case of j =–Mj /L , the first term is zero, which means the second term must be >= zero. Thus the following must always hold: 1

2

1

(L – M /L ) j /2 >= 0 L >= M /L L L >= M 2

2

2

1

2

2

2

1

2

1

2

This equation is usually written: | M | = k √(L L ) 1

2

Here, k, the coupling coefficient of the two coils, is a number between 0 and 1. Two coaxial coils in immediate proximity that completely share their magnetic flux have a k value very close to 1, while two coils very far apart have a k near 0. Note that M may be positive or negative depending on how the coils are wound. Let’s now explore the energy of the magnetic field from current loops. Recall the equation for the mutual inductance of two coils: M = (1/4πε c ) ∫ ∫ ds •ds / |r – r | 2

12

0

Γ1

Γ2

2

1

1

2

Feynman says we might try using this equation to compute the self-inductance of one coil by setting Γ =Γ . That attempt would fail due to a zero denominator whenever r =r . The failure of this equation to provide self-inductance is due to an implicit approximation made in its derivation. We assumed the cross-section of the wires in each coil were infinitesimal, much smaller than the separation between coils. That assumption is invalid if the coil separation is zero, which occurs when coil #1 and coil #2 are the same. To do this analysis properly, we would have to consider current densities within wires of non-zero width. 1

2

1

2

An easier approach, Feynman says in V2p17-14, is to first find the magnetic energy of a single coil. From Chapter 16, the general equation for magnetic energy is: U = ∫ j•A dV / 2 To evaluate j•A, let’s first employ Maxwell’s fourth equation (assuming ∂E/∂t=0). j = ε c Ď×B 2

0

2U = ∫ j•A dV = ∫ ε c Ď×B•A dV 2

0

In V2p17-15, Feynman shows that ∫Ď×B•A can be replaced by ∫Ď×A•B, provided that A and/or B is zero at the limits of the integration volume. Indeed, this replacement is valid for any two vector fields

meeting that condition. If you wish to skip the proof, go forward to the “QED” line. ∫ Ď×B•A dx dy dz = ∫ A (∂B /∂y – ∂B /∂z) dx dy dz + ∫ A (∂B /∂z – ∂B /∂x) dx dy dz + ∫ A (∂B /∂x – ∂B /∂y) dx dy dz x

z

y

y

x

z

z

y

x

Now recall the technique known as integration by parts: ∫ u dv = u v – ∫ v du You may confirm this by taking the derivative of both sides. Applying integration by parts to the first of the above six terms yields (here we suppress ∫dxdz for brevity): ∫

A ∂B /∂y dy = A B | – ∫ B ∂A /∂y dy

+∞ –∞

x

z +∞ z –∞

x

+∞ –∞

z

x

The first term above must be evaluated at the integration limits y=±∞. Since any physical system, such as a coil, has finite size, A (±∞) = B (±∞) = 0. This leaves only the second term. x

∫

+∞ –∞

A ∂B /∂y dy = – ∫ x

+∞ –∞

z

z

B ∂A /∂y dy z

x

Doing the same to all terms and regrouping yields: ∫ Ď×B•A dx dy dz = ∫ (–B ∂A /∂y + B ∂A /∂z) dx dy dz + ∫ (–B ∂A /∂z + B ∂A /∂x) dx dy dz + ∫ (–B ∂A /∂x + B ∂A /∂y) dx dy dz z

x

y

x

x

y

z

y

y

z

x

z

∫ Ď×B•A dx dy dz = ∫ (B ∂A /∂y – B ∂A /∂z) dx dy dz + ∫ (B ∂A /∂z – B ∂A /∂x) dx dy dz + ∫ (B ∂A /∂x – B ∂A /∂y) dx dy dz x

z

x

y

y

x

y

z

z

y

z

x

∫ Ď×B•A dx dy dz = + ∫ Ď×A•B dx dy dz QED We therefore have: 2U = ∫ ε c Ď×B•A dV = ∫ ε c Ď×A•B dV 2

0

But, since B=Ď×A: U = ∫ ε c B•B dV /2 2

0

2

0

This equation has the same form as the equation we found for the energy of an electric field: U = ∫ ε E•E dV /2 0

Feynman emphasizes that these energy equations are valid for any E and B fields, static or dynamic. Now, we can address self-inductance. Earlier, we found the equation for the energy of two coils. By setting the current in coil #2 to zero, that equation becomes: U = L j /2 2

1

Combining this with the magnetic energy equation yields: L = ∫ ε c B•B dV / j 2

2

0

Since B is proportional to the magnitude of the current j, we find that self-inductance L is a function only of the geometry of the circuit, not the current through it, as we would expect. For a solenoid of radius r, length Z, current j, and with n windings per unit length, the magnetic field is nearly zero outside the solenoid, and has a uniform magnitude inside, with: B= nj / εc

2

0

The inside volume of this solenoid is πr Z, making the above integral equal to: 2

L = ε c (nj/ε c ) πr Z / j L = π n r Z/ε c 2

0

2 2

2

2

0

2 2

2

0

Does the Disk Spin? Recall the apparent paradox that Feynman suggested you solve. The disk in Figure 18-6 (repeated below) is initially stationary, with current j flowing through the black central circular conductor. The circulating current creates a magnetic field B parallel to the disk’s axis. The disk is not conductive, but contains six isolated conducting balls, shown as black dots.

Figure 18-6 Disk, Balls & Current Loop

The question is: When j stops flowing, does the disk spin? My answer is: Yes. Initially, the disk and balls are stationary, but the charged particles in current j are flowing around the axis, and thus have angular momentum. As j and B drop to zero, ∂B/∂t creates a circumferential electric field (Ď×E=–∂B/∂t) that exerts a torque on the charge carriers in each of the six balls, thereby spinning the disk. Note that the electric field is transitory; it dissipates as it pushes the balls. Hence, there is no real paradox. The essential realization is that the electric field must also have angular momentum. The angular momentum (always up in the figure) is transferred from the current to the electric field, and then through the six balls to the disk.

Chapter 18 Review: Key Ideas • The flux rule and induced emf for a circuit loop Γ (a coil) that encloses surface area S are: Flux Rule: emf = – d/dt ∫ B•n da Definition: emf = + ∫ ds•F/q where the Lorentz force is: F=q(E+v×B) S

Γ

Emf polarities are such that they always oppose flux changes. Magnetic flux may change due to changes within a circuit, motion of the circuit as a whole, changes in the magnetic field, or any combination thereof. The flux rule works when there is no change in the material of which the circuit is comprised – the actual atoms through which current flows must remain the same. The flux rule may not work if the circuit changes by redirecting current through different atoms. When in doubt, go back to Maxwell’s

equations and the Lorentz force, which are always valid.

• The self-inductance of a single circuit loop is given by: L = ∫ ε c B•B dV / j 2

2

0

The mutual inductance of coils #1 and #2 is given by: M = (1/4πε c ) ∫ ∫ ds •ds / |r – r | 2

12

0

Γ1

Γ2

2

1

1

2

| M | = k √(L L ) 12

1

2

Here, k is the coupling coefficient of the two coils, which is between 0 and 1. The stored energy U of a two-loop system with currents j and j is: 1

U = L j /2 + L j /2 + M j j 2

1

2

2

1

2

2

12

1

2

The force between two such coils is: F = + dM /dx j j 12

1

2

• The inductance of a solenoid of radius r, length Z, current j, and n windings per unit length is: L = π n r Z/ ε c 2 2

0

2

Chapter 19 Maxwell’s Equations In V2p18-1, Feynman says: “In this chapter we come back to the complete set of the four Maxwell equations that we took as our starting point in Chapter 1. Until now, we have been studying Maxwell’s equations in bits and pieces; it is time to add one final piece, and to put them all together. We will then have the complete and correct story for electromagnetic fields that may be changing with time in any way. Anything said in this chapter that contradicts something said earlier is true and what was said earlier is false—because what was said earlier applied to such special situations as, for instance, steady currents or fixed charges. Although we have been very careful to point out the restrictions whenever we wrote an equation, it is easy to forget all of the qualifications and to learn too well the wrong equations. Now we are ready to give the whole truth, with no qualifications (or almost none).”

Primary Equations of Classical Physics Feynman lists here the primary equations of classical physics, physics before relativity and quantum mechanics. These equations would be “the whole truth” if the speed of light were infinite and Planck’s constant were zero. In later chapters, we will explore relativistic effects, including time delays in the propagation of electromagnetic fields due to the finite speed of light. But, first we will complete our exploration of classical electromagnetism.

• Below are Maxwell’s four equations of electromagnetism. Within classical caveat, these equations are valid in all situations, static or dynamic, in matter or in empty space. In what follows, Γ is any closed loop, S is any closed surface, and V is the volume within S. Recall that Ď = (∂/∂x, ∂/dy, ∂/∂z) Gauss’ law: the flux of E through S equals (the charge contained in V)/ε . 0

Ď•E = ρ / ε

0

Faraday’s law: the curl of E equals minus the time derivative of B. Ď×E = – ∂B/∂t

No magnetic monopoles: the flux of B through S equals zero. Ď•B = 0 The curl of B is proportional to current j plus the time derivative of E. c Ď×B = j/ε + ∂E/∂t 2

0

We have already explored the first three equations and half of the fourth. We will shortly examine the effects of the ∂E/∂t term in the fourth equation.

• Three additional principle equations are also essential to electromagnetism. The Lorentz force on a particle from electromagnetic fields arises from both its charge q and its velocity v. F = q (E + v×B) Force equals the time derivative of momentum. This equation remains valid in relativity and in quantum mechanics. F = dp/dt, with p = mv/√(1–v /c ) 2

2

Charge is conserved: the amount of charge flowing outward through S equals the decrease in the amount of charge within S. Ď•j = –∂ρ/∂t

• Completing the primary equations of classical physics is Newton’s law of gravitation, which is an excellent approximation when masses and velocities are not too great. Newton’s gravitational force between two bodies of mass m and M is proportional to the product of their mass and inversely proportional to the square of the distance between them. Einstein’s theory of gravity supersedes Newton’s. General relativity states that gravity is not a force but rather the effect of motion through a non-Euclidean, 4-dimensional spacetime, which is “curved” by all forms of energy, including mass. F = – G mMr / r

• Having listed these equations, Feynman says:

3

“[Here] we have all that was known of fundamental classical physics, that is, the physics that was known by 1905. Here it all is, in one table. With these equations we can understand the complete realm of classical physics. “First we have the Maxwell equations—written in both the expanded form and the short mathematical form. Then there is the conservation of charge… Next, we have written the force law, because having all the electric and magnetic fields doesn’t tell us anything until we know what they do to charges. Knowing E and B, however, we can find the force on an object with the charge q moving with velocity v. Finally, having the force doesn’t tell us anything until we know what happens when a force pushes on something; we need the law of motion, which is that the force is equal to the rate of change of [relativistic] momentum. “If we really want to be complete, we should add one more law—Newton’s law of gravitation —so we put that at the end. “Therefore in one small table we have all the fundamental laws of classical physics. This is a great moment. We have climbed a great peak. We are on the top of K2—we are nearly ready for Mount Everest, which is quantum mechanics. We have climbed the peak of a ‘Great Divide,’ and now we can go down the other side… “Now that we have the whole thing put together, we are going to study what the equations mean —what new things they say that we haven’t already seen. We’ve been working hard to get up to this point. It has been a great effort, but now we are going to have nice coasting downhill as we see all the consequences of our accomplishment.” Don’t take “coasting” too seriously. There are plenty of challenges ahead. We first focus on the new term, ∂E/∂t, in the fourth Maxwell equation.

History of Maxwell’s 4th Equation Faraday’s experiments linked electric currents to magnetic fields. That established half of the fourth equation of electromagnetism, which in our notation is: c Ď×B = j/ε 2

0

In V2p18-1, Feynman says: “Maxwell began by considering these known laws and expressing them as differential equations, as we have done here… He then noticed that there was something strange about [the above equation]. If one takes the divergence of this equation, the left-hand side will be zero, because the divergence of a curl is always zero. So this equation requires that the divergence of j also be zero.” However the divergence of j cannot always be zero. When charge flows out of region X and into

neighboring region Y, Ď•j is negative in X and positive in Y. This is precisely the meaning of the charge conservation equation: Ď•j = –∂ρ/∂t. Maxwell solved this dilemma by adding a final term, resulting in the complete fourth equation: c Ď×B = j/ε + ∂E/∂t 2

0

In V2p18-2, Feynman concludes this story with: “It was not yet customary in Maxwell’s time to think in terms of abstract fields. Maxwell discussed his ideas in terms of a model in which the vacuum was like an elastic solid… Today, we understand better that what counts are the equations themselves and not the model used to get them. We may only question whether the equations are true or false. This is answered by doing experiments, and untold numbers of experiments have confirmed Maxwell’s equations. If we take away the scaffolding he used to build it, we find that Maxwell’s beautiful edifice stands on its own. He brought together all of the laws of electricity and magnetism and made one complete and beautiful theory.” I agree with much of that, but I don’t believe models are unimportant. In 1889, physicists derived the correct equation for length contraction, but they believed it was because objects moving through the luminiferous ether were squeezed along the direction of motion. The equation was right, but the model was wrong. Progress was made only when Einstein proposed his model: the relativity of space, time, energy, and momentum. The ubiquitous belief in the ether model also retarded progress, until it was replaced by the concept of fields, as Feynman stresses elsewhere. Let’s explore the implications of the ∂E/∂t term. Taking the divergence of the fourth equation yields: c Ď•(Ď×B) = Ď•j/ε + Ď•∂E/∂t 0 = Ď•j/ε + ∂/∂t Ď•E 2

0

0

In the last step, we reversed the order of the time and spatial derivatives, because partial derivatives commute. Now, using Maxwell’s first equation for Ď•E: 0 = Ď•j/ε + ∂/∂t ρ/ε Ď•j = – ∂ρ/∂t 0

0

This is the charge conservation equation. Thus Maxwell’s complete equations prove that charge is conserved. But, we have not proven the converse: that charge conservation proves Maxwell’s fourth equation. If we add the curl of any vector field to the fourth equation, the result would still be consistent with charge conservation, because the divergence of any curl is always zero. The proof of Maxwell’s fourth equation is the only proof that matters in science: experimental validation. Now let’s examine situations impacted by the ∂E/∂t term.

The New Term In Action Feynman’s first example is a spherically symmetric current emanating radially from a very small source at the origin of our coordinate system. Feynman imaginatively suggests this source might be a block of Jello injected with excess protons at its center. These protons repel one another and gradually permeate outward through the Jello. The charge distribution peaks at the center creating a radial electric field. Figure 19-1 shows representative radial current vectors j as dashed lines, and radial electric field vectors E as solid lines. We assume everything here has spherical symmetry.

Figure 19-1 Purported B Field Around Current j

What is the magnetic field of this current distribution? Figure 19-1 shows one possible answer: a circulating B field around each current vector j. If there were only one current vector, that B field would be valid, but it cannot be valid with equal currents flowing in all directions. In fact, spherical symmetry requires B=0 everywhere. Symmetry is a powerful tool in physics; let’s carefully examine why it requires B=0. Consider a single point P on a sphere of radius r, centered at the origin O of our coordinate system. Spherical symmetry means that no real measurable quantity can change if we rotate everything by any

angle about any axis that passes through O. Choose the axis that passes through both O and P. The only vectors at P that are not changed by an arbitrary rotation about this axis are vectors that are entirely radial, those pointing toward or away from O. Any vector tangent to the sphere at P is perpendicular to the rotation axis. That vector will spin around and point in a different direction as everything rotates about this axis, which is exactly what spherical symmetry precludes. But if B were radial at P, B would have the same radial component at every point on the sphere, by symmetry. B would then have a divergence, which Maxwell’s third equation forbids. Hence B=0 everywhere in any spherical symmetric situation. Now let’s calculate E(r), the magnitude of the electric field at r. For spherical symmetry, Maxwell’s first equation, Gauss’ law, says: 4πr E(r) = ρ(r) / ε 2

0

Here ρ(r) is the total charge within radius r. Charge conservation says: Ď•j = – ∂ρ/∂t Again by symmetry, the divergence of j equals (the area of a sphere of radius r) multiplied by (the value of j at r). The equation is: 4πr j(r) = – ∂/∂t ρ(r) 2

Substituting ρ(r) using Gauss’ law, we obtain: 4πr j(r) = – ∂/∂t { 4πr E(r) ε } j(r)/ε = – ∂/∂t { E(r) } 2

2

0

0

This matches exactly with Maxwell’s fourth equation for the case of B=0 everywhere: 0 = j/ε + ∂E/∂t 0

We see that the new term is essential to a consistent theory. The next example involves charging a capacitor. Figure 19-2 shows current j flowing into the upper plate of a capacitor, making it positively charged; j also flows out of the lower plate, making it negatively charged. Loop Γ encircles the wire leading to the upper plate, while loop Γ is midway between the plates, completely enclosing the electric field (arrows) between capacitor plates. The two loops are entirely horizontal; they are shown slightly tilted for clarity. 1

2

Figure 19-2 Charging a Capacitor

Let S be the surface enclosed by Γ , and S the surface enclosed by Γ . Both are shown in light gray. 1

1

2

2

Assume for simplicity that the capacitor plates are disks (axially symmetric) and that charge Q is changing slowly enough for us to ignore radiation, field propagation delays, and other “higher-order” effects. Current j and charge Q are related by: j = dQ/dt Current j creates a magnetic field that circulates around the upper wire, going clockwise as viewed from above, per the right hand rule. Let’s calculate the magnetic field at loop Γ and elsewhere. 1

Maxwell’s fourth equation is: c Ď×B = j/ε + ∂E/∂t 2

0

For a highly conductive wire, the electric field near the upper wire is negligible. By Stokes’ theorem: ∫ B•ds = ∫ Ď×B•da = ∫ j•da / ε c Γ1

S1

S1

2

0

If the radius of Γ is r, we have: 1

– 2πr B(r) = j/ε c

2

0

The minus sign arises because B circulates clockwise, and line integrals are defined with counterclockwise circulation. Feynman notes that if we lower Γ until it touches the upper plate, j abruptly drops to zero, and so it seems must B. He correctly notes that an instantaneous drop seems strange: nature abhors infinite 1

derivatives. Now consider Γ . We learned from Gauss’ law that the electrostatic field between oppositely charged plates is Q/ε . Since there is no current flow between plates, Maxwell’s fourth equation applied to Γ yields: 2

0

2

c Ď×B = ∂E/∂t = dQ/dt / ε – c 2πr B(r) = dQ/dt / ε – 2πr B(r) = j/ε c 2

0

2

0

2

0

Thus we find that, in Figure 19-2, the magnetic field does not vary vertically: B(r) has the same value regardless of vertical position.

Traveling Fields We next consider more interesting situations that involve all of Maxwell’s equations together. We begin as simply as possible, examining a one-dimensional problem: a moving sheet of charge. Imagine an infinitesimally thin sheet of matter positioned at x=0 and extending in the yz-plane to infinity. Let the sheet be electrically neutral, with equal numbers of uniformly distributed positive and negative charges. Initially all the charges are stationary, and E and B are zero everywhere. At t=0, let all the positive charges rapidly accelerate to velocity u in the +y-direction and maintain that velocity indefinitely. While the sheet remains electrically neutral, there is an upward current j=σu everywhere within x=0 plane. As the positive charges rapidly accelerate, j rapidly increases as well. We can calculate the magnetic field near the sheet by examining a rectangular loop Γ that encloses part of the sheet, as seen in the top view shown in Figure 19-3. Γ has width w in the z-direction and width Δx in the xdirection. Current j points out of the screen as indicated by the black dots.

Figure 19-3 Loop Near Current Sheet

By making Δx infinitesimal, we can ignore any B component normal to the sheet and also any flux of ∂E/∂t through loop Γ. The flux of j through the enclosed surface creates a magnetic field circulating counterclockwise, as seen in the top view. By Stokes’ theorem and Maxwell’s fourth equation, the line integral of B•ds counterclockwise around Γ is proportional to jw, the current flux. We therefore have: 2w B = jw / ε c B = j / 2ε c

2

0

2

0

After t=0, as j rapidly increases from 0 to σu, the magnetic field follows, increasing from 0 to j/2ε c . Since it circulates counterclockwise, B points toward –z for x>0 and toward +z for x0, and minus that value at all x