Physics 15b - Final Exam - Take-home portion

Here is the final version of the takehome exam, along with the old problems which I will followup on the inclass exam on Saturday morning.

You may work on the take-home problems in study groups, as usual, but you should writethem up individually and pay special attention to the coherence and clarity of your solutions. Youshould also list the students with whom you have discussed the problems. The in-class portion ofthe exam will follow up on some of these problems (probably not all of them) so they will alsoserve as a study guide for the exam. We strongly recommend that you think about possiblefollow-up questions these problems. Some of the followups will be conceptual or calculationaltrue-false questions, but one or two may be a more conventional calculation. We hope that many ofthe followups will be obvious if you have understood the problems. Obviously, the more you havethought about these problems, the easier it will be for you to complete the inclass exam in the timeyou are given. We strongly urge you to explore them. Ask yourself whether you know what thefields look like. Are there concepts from later chapters that are relevant to problems from earlierchapters? Think about it.

At the beginning of exam, you will get a packet containing the questions from the take-home,our solutions, and the follow-up questions. You may consult your take-home solutions during thefollow-up process. At the end of class, you will hand in the packet and your problem set togetherwith your take-home inserted into the in-class packet.

Because our solutions will be in the packet, you may not modify your take-home solutionsonce you get to the exam. However, there will be a place in the exam packet where you canmake comments on your take-home solutions if you wish to add anything once you have seen oursolutions.

Because you can consult your take-home solutions during the exam, you may want to includein your solutions any notes that you think may be helpful to you on the exam, even if they are nota part of your solutions to the take-home problems. This is allowed. You can include anythingthat you produce by your own intellectual effort. You may not simply copy something or printa version of someone else’s LATEX file or anything like that. But if you want to write your ownphysics textbook and include it with your solutions, that is fine (though we think that this will notbe an efficient use of your time). You may put your notes in a separate section at the end of yourtake-home, and you must hand them in along with the rest of your take-home at the end of thein-class exam.

Calculators are allowed for the exam, but should not be essential.

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take-home-1. One can make a “2-dimensional dipole” by putting an infinite charged rod withuniform linear charge density q/ε along the z axis and another with linear charge density −q/εparallel to the z axis at y = −ε. As ε → 0 for fixed q, this produces an interesting electric field,independent of z and with no z component.

take-home-1.a. Show explicitly that the resulting electric field has the form

~E(x, y, z) = 2q(2xy, y2 − x2, 0)

(x2 + y2)2(1.1)

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These 2-dimensional dipoles have properties similar to those of 3-dimensional dipoles. Theparticular analogy we will explore in this problem is to the uniformly polarized sphere, which asyou saw in Purcell, has a constant electric field inside and a dipole field outside. An analogousargument yields the result that an infinite cylinder with radius R centered on the z axis with po-larization P0 y in the y direction has an electric field that is uniform and in the −y direction insidethe cylinder for x2 + y2 < R2 and has the form (1.1) for x2 + y2 > R2. You can prove this byanalogy with the argument illustrated in Purcell’s figure 10.22 by considering the uniformly po-larized cylinder to be built out two oppositely uniformly charged cylinders slightly displaced fromone another in the y direction.

take-home-1.b. Assuming this to be correct, find q and the magnitude of the field inside thecylinder in terms of P0.

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You can use the result of the previous parts to solve the following problem. Find the electricfield around an infinite conducting cylinder of radius R, centered on the z axis in an externalelectric field E0 y. The field lines for the resulting field in the x-y plane look something like thoseshown in the figure below.

The thick lines in the figure represent the boundary between the field lines that intersect the surfaceof the cylinder and those that do not.

take-home-1.c. What are the asymptotic values of x of the thick lines as y → ±∞ and how doyou know?

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Comments on your solution to take-home-1:

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take-home-2. Consider the following electric and magnetic fields:

~E(x, y, z, t) = E0y cos[k(x− ct)]− E0y cos[k(x + ct)]

~B(x, y, z, t) = E0z cos[k(x− ct)] + E0z cos[k(x + ct)]for x > 0

~E(x, y, z, t) = ~B(x, y, z, t) = 0 for x < 0

(2.1)

(2.1) describes a plane electromagnetic wave for x > 0 being reflected from a stationary perfectlyconducting plane at x = 0. Describe these fields in terms of the coordinates, (x′, y′, z′, t′), of aframe of reference moving with velocity βc in the +x direction. Choose the origin of the movingframe so that it coincides with origin of the fixed frame at time t′ = t. Find the transformed electricand magnetic fields,

~E ′(x′, y′, z′, t′) and ~B′(x′, y′, z′, t′) (2.2)

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Comments on your solution to take-home-2:

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take-home-3. A magnetic dipole with dipole moment ~m is fixed at the point (0, 0, `) for ` > 0.All the region for z ≤ 0 is filled with the ultimate diamagnetic material with µ = 0. Find themagnetic field for z ≥ 0. Hint: Think about ~B inside the material and the boundary condition onthe surface. Then find an appropriate “image” for z < 0 to give the right field at the boundary.Another Hint: You have actually seen this material in a lecture demonstration, but that was beforewe knew about diamagetism, so we didn’t mention that µ = 0 and we talked about it in a differentway.

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Comments on your solution to take-home-3:

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Below are the problems from previous problem sets that will be followed up on the exam.

1.3.Do Problem 1.33 in Purcell.

Imagine a sphere of radius a filled with negative charge of uniform density, the totalcharge being equivalent to that of two electrons. Imbed in this jelly of negative chargetwo protons and assume that in spite of their presence the negative charge distributionremains uniform. Where must the protons be located so that the force on each of themis zero? (This is a surprisingly realistic caricature of a hydrogen molecule; the magicthat keeps the electron cloud in the molecule from collapsing around the protons isexplained by quantum mechanics!)

Answer: You are asked to find the position of two protons, each with charge e, insidea uniformly charge sphere with radius a and charge−2e, such that the force on each of theprotons vanishes. First note that two protons must lie on a line through the center of thesphere. Otherwise the force on one proton from the electron cloud would not be parallelto the force from the other proton, and the two could not cancel. Also, by symmetry, theyare the same distance from the center. Call it x. Thus the system looks like this:

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The force on the right proton from the other proton is repulsive with magnitude

e2

4x2(3.1)

The force from the electron cloud is (for example from lecture) attractive with magnitude

2e2x

a3(3.2)

Thusx = a/2 (3.3)

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4.3.

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Shown above is a top view of a parallel plate capacitor in the form of a square with side ` and plateseparation s (which you can’s see of course). A solid rectangular conductor with thickness s/3 andwidth w is covered on the top and bottom with a frictionless non-conducting skin (Teflon c© mightwork) of thickness s/3 so that the whole sandwich just fits between the plates, and slides smoothlytouching both, but without making electrical contact. This is shown in the side view below.

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conductor

frictionless nonconductor

frictionless nonconductor

upper plate

lower plate

s

Suppose that the top and bottom capacitor plates have charge ±Q and that the conductor is slid ina distance x as shown. Making appropriate assumptions and approximations (which you shouldexplain in detail) find the force on conductor.Answer: Probably the simplest way to get the answer is to consider the capacitior astwo capacitors connected in parallel, one with area `2− xw in which there is no conductorbetween the plates, and another with are xw in which there is conductor between theplates. The capacitance of the first, in the parallel plate capacitor approximation, is

C1 =`2 − xw

4π s(3.4)

The capacitance of the second isC2 =

xw

4π 2s/3(3.5)

because the conductor does not contribute to the gap between the two plates. There arevarious ways to see this — for example consider this system to be two capacitors eachwith gap s/3 connected in series. The total capacitance of the system is then

C = C1 + C2 =2`2 + xw

8π s(3.6)

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and the energy of the charged capacitor is

U =Q2

2C=

4π s Q2

2`2 + xw(3.7)

The external force required to increase x is then

F =dU

dx= − 4π w s Q2

(2`2 + xw)2(3.8)

Because the external force required to increase x is negative, this means that the con-ductor is being pulled into the capacitor by the electric force. This is similar to the forcebetween the plates of the variable capacitor discussed in the Tuesday lecture. The forcearises from the fringe fields around the edges of the boundary of the conductor.

Meanwhile note that the units of the answer are esu2/cm2, which is right. It seemsa bit odd that the result vanishes as s. But this is also a bit misleading, because thecapacitance blows up in this limit.

8.2. Do problem 6.14 in Purcell.

A coil is wound evenly on a torus of rectangular cross section. There are N turnsof wire in all. Only a few are shown in the figure. With so many turns, we shallassume that the current on the surface of the torus flows exactly radially on the annularend faces, and exactly longitudinally on the inner and outer cylindrical surfaces. Firstconvince yourself that on this assumption symmetry requires that the magnetic fieldeverywhere should point in the “circumferential” directions, that is, that all field linesare circles about the axis of the torus. Second, prove that the field is zero at all pointsoutside the torus, including the interior of the central hole. Third, find the magnitudeof the field inside the torus, as a function of radius.

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Answer: Let us begin by understanding the direction of the magnetic field. The systemlooks like this

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where the current runs up along the inner cylinder, radially out along the top, down alongthe outer cylinder, and radially in along the bottom. The system is symmetrical with re-spect to reflections in any plane through the axis. Under this transformation, points onthe plane go into themselves. The magnetic field at a point on the plane has a compo-nent perpendicular to the plane that goes into itself. But its components parallel to theplane change sign and therefore they must vanish. This shows that the magnetic field isazimuthal anywhere that it is non-zero. More formally, if we take the axis of the torus tobe the z axis, the magnetic field at a point ~r can be written as

~B(r) = z × r⊥ b(r⊥) (3.9)

where r⊥ is the distance from the z axis to ~r, and r⊥ is a unit vector from the z axis towards~r:

~r⊥ ≡ ~r − z(z · ~r) , r⊥ ≡ |~r⊥| , r⊥ = ~r⊥/r⊥ (3.10)

The rotation symmetry about the axis guarantees that the magnitude is constant oncircles centered on the axis in perpendicular planes. The magnetic field lines are thereforecircles, as shown:

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We can use these field lines as the closed curve for the line integral in Ampere’s law. Theline integral of the magnetic field around the curve is

2π r⊥ b(r⊥) (3.11)

If the field line is inside the torus, the current flowing through the surface bounded by thefield line is NI where I is the current in the wire. Thus

2π r⊥ b(r⊥) = 4π NI/c (3.12)

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and the magnetic field inside the torus is

~B(r) = z × r⊥2NI

c r⊥(3.13)

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9.1. Do problem 7.14 in Purcell.

A metal crossbar of mass m slides without friction on two long parallel conductingrails a distance b apart. A resistor R is connected across the rails at one end; comparedwith R, the resistance of bar and rails is negligible. There is a uniform field ~B per-pendicular to the plane of the figure. At time t = 0 the crossbar is given a velocity v0

toward the right. What happens then?

(a) Does the rod ever stop moving? If so when?

(b) How far does it go?

(c) How about conservation of energy?

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−→

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Answer: Call v the velocity of the bar at some time. The rate of change of flux is thenBvb, and the EMF is therefore Bvb/c. The current is then

I =Bvb

cR

This current produces a force on the bar

F = −IbB

c= −B2vb2

c2R

Lenz’s law assures us that this is a frictional force - opposing the change in the flux - andthus opposing the motion of the bar into the magnetic field! The rest is mechanics.

F = −B2b2

c2Rv = ma = m

dv

dt

v(t) = v0e−t/T

where

T =mc2R

B2b2

If the rod starts at x = 0, then

x(t) = v0T(1− e−t/T

)

So

(a) The rod never stops moving.

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(b) But it goes a finite distance v0T in an infinite time.

The power that goes into slowing down the bar is Fv

Fv = −B2b2

c2Rv2

The current flowing through the circuit causes an I2R power loss in the reisitor

Power lost in the resistor = I2R =

(Bvb

cR

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R

Thus

(c) Conservation of energy is satisfied because

Fv + I2R = 0

The kinetic energy that is lost by the bar goes into heat in the resistor.

In addition, answer the following question.

9.1-d. Work is being done to slow down the bar. What force is producing this work? Explainyour answer in detail.Answer: The force slowing down the bar is the electrical force of the current-carryingelectrons on the rigid lattice of the bar. This is equal and opposite by Newton’s third lawto the electrical force of the positive lattice on the electrons which is the force responsiblefor keeping the moving electrons in the bar. The magnetic force cannot produce anywork on the moving charged particles, either the positive lattice of the bar or negativeelectrons, because it acts perpendicular to the direction of motion. But what it can do isto change the direction of the electron’s motion. If the ~B field in the figure is up out of thepage, a current is produced downwards in the bar, which means that the current-carryingelectrons are moving up through the bar as it moves to the right. But the magnetic fieldthen curves the paths of the electrons to the left. Because of this curving of their motion,the electrons fall behind the positive lattice and pile up on the left side of the bar. Theelectrons then pull the bar to left, slowing it down. The positive lattice of the bar is losingenergy because it must do work on the electrons to keep them in the bar. The magneticfield does no work, but what it does do is to maintain the peculiar charge distribution withinthe bar that allows the electrical forces to do work.

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