elmaglightsaslowallanswersnew

8/8/2019 ElMagLightSaslowAllAnswersNew

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Part I

Revised Appendix D - Answers to

Odd-Numbered Problems

1


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R2.1 The ground wire provides an alternate cur-rent path that carries more current.

R2.3 (a) 0 A, 0 W. (b) 2 A, rightward, 20 W.(c) 1 A, rightward, 5 W. (d) 1 A, leftward, 5 W.(e) 2 A, leftward, 20 W.

R2.5 These precautions prevent current from flow-ing through the torso of your body.

R3.1 Computers, VCRs, microwave ovens, etc.

R4.1 (a) 20 . (b) 0.2 W. (c) 960 J. (d) 480 C.

R5.1 (a) No. (b) The monitor cant work withoutpower.

R6.1 Her hair becomes charged and the strandsrepel each other.

R6.3 The source provides a voltage, not a current.

R6.5 You develop a charge by rubbing against theseat as you leave the car. Holding onto the outsidesurface as you exit allows the charge to leave gradu-ally. Touching the outside surface only after you areoutside forces the charge to leave all at once, shockingyou.

R8.1 (a) 3. (b) 3. (c) 3. (d) In the first case thereis a collective effect; in the other cases there is anindividual effect.

R8.3 For an insulator a minus indicates an excesselectron and a plus a deficit of an electron exactly atthat spot. For a conductor they represent an excessor deficit in the average density of the conduction

electrons in one mole.R8.5 (a) Assume that the extra charge carrier isa positive ion. The bulk of the liquid is neutral andall excess charge is due to ions distributed over itssurface. (b) Assume that the extra charge carrier isan electron. The bulk is neutral and all excess charge,due to electrons, is distributed over its the surface.

R8.7 (a) Yes. (b) In pure water electric current isentirely due to H+ and OH ion movement. In saltwater the current is dominated by the movement ofNa+ and Cl ions. In metal wire the current is due tothe movement of electrons. In all cases a small aver-

age velocity is superimposed on the random motionsof charge-carriers.

R9.1 Only (d) would receive full credit.

R9.3 Show that problem.

R9.5 (2, 1, 0) and a 180 rotation about the y-axis

R10.1 (a) Show that problem. (b) Use a = i,b = i , and c = j.

R10.3 (a) Show that problem. (b) The left sideequals 2; the right side equals 0; the two sides areunequal.

R10.5 (a) Show that problem. (b) Show thatproblem. (c) Show that problem.

R10.7 Show that problem.

R10.9 (a) Counterclockwise rotation of 53.1.(b) (3.60, 5.20, 1). (c) (10.40, 2.20, 26).(d) (10.40, 2.20, 26). (e) they are the same. (f)5.385, 6.403, 28.1. (g) 145.45 or 214.55. (h) Yes.R10.11 (10, 5, 3) N-m.R10.13 (4.8

103, 9.6

103, 0) N.

R10.15 Show that problem.R10.17 Show that problem.

R10.19 (0, 0, d/c), (a/e,b/e,c/e), wheree =

a2 + b2 + c2.

R10.21 (a). n (0.254, 0.381, 0.889).(b)

dEdA

E n = 23.0 volt/m.

(c) dE =dEdA

dA = 1.195 105 volt/m.

12.1 See Fig.2.1 and the accompanying discussionof the amber effect.

12.3 They will repel if the positive ends are

brought near each other but attract if the positiveend of one is brought near the neutral end of theother.

13.1 The mechanical motion would be identical,but the induced charge would be opposite the previ-ous induced charge.

13.3 Either end would be attracted to an electri-cally charged object due to electrostatic induction.

14.1 (a) In the 17th century, the most commonway to charge an object was to rub it or to touchit to a previously charged object. Water and ironcannot be charged in this manner. (b) Put object on

insulator, charge by induction and either groundingor sparking or contact.

14.3 Conductors: your body, a penny. Insulators:your clothing, plastic, your comb, a styrofoam cup

14.5 When the person stands on the ground heis subject to the full voltage difference between the


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charge source and ground, which produces a largeenough current to cause the shock.

14.7 Watsons view would predict identical shocksin the two cases. Franklins view (the correct view)would predict a greater shock in the first case.14.9 A is a conductor and B is an insulator.15.1 (a) A Penny Saved Is a Penny Earnedimplies money conservation in a situation where thereis a flow of resource, both in and out. (b) Decreasingtheir energy bill with thermal insulation has the sameeffect as increasing their revenue by the same amount.15.3 After the first connection, A has 4.8 unitsand B has 3.2 units. After the second connection, Ahas 0.96 units and B has 0.64 units.15.5 (a) The jar cannot lose charge to the groundand losing it through the air could take hours or days.(b) As long as the top wire remains charged it willattract charge to the bottom, and thus the LeydenJar remains charged. (c) As long as the bottom wireremains charged it will attract charge to the top, andthus very little charge can be drawn off the top wire.(d) The bottom retains its charge, since it is insu-lated. Once the top wire is connected to ground, thebottom wire will attract charge to the top and theLeyden Jar would regain its strength.16.1 If the tube were left in place, electrostatic in-duction would occur. Then the experiment would be

less reproducible, because the electrostatic inductionwould depend on the placement of the tube.16.3 Rapid discharge of a nearby source causes arapid decrease of the electrostatically induced chargein the human body.16.5 The induced charge on each leaf will be ofthe same sign and hence the leaves will repel.16.7 (a) Rubbing the balloons on clothing chargesthem up by friction. They are attracted to the wallby electrostatic induction and fall as they graduallylose their charge. (b) No. (c) The moisture in the airwill draw off charge. The more moisture, the fasterthe charge will be drawn off.

16.9 When the negatively charged rod is broughtnear the grounded sphere, negative charge driven byelectrostatic induction flows away from the sphere,leaving the sphere positively charged. When theground connection is removed, the sphere remainspositively charged, but also subject to electrostatic

induction from the rod. When the rod is removed,the positive charge on the sphere redistributes.

17.1 +2e.17.3 (a) Allowed. (b) Prohibited. (c) Allowed.17.5 Electrons are transferred from the cloth tothe rod. Protons generally cannot be transferred.17.7 (a) The styrofoam partially discharges, af-ter a diffusion time T across the tube the oven bagdischarges, and after 2T the styrofoam completes itsdischarge. (b) No. (c) The oven bag discharges, thenafter T the styrofoam discharges.18.1 People come in integer values; you are eitheralive or you are dead. There is no such thing as peo-ple conservation, as established by the phenomena ofbirth and death.18.3 6.25 1019 electrons.18.5 3.125 109.19.1 Show that problem. A has units of C/m3.19.3 (a) The charge per unit length for rods 1 and 2are q1/l1 and q2/l2, respectively. (b) (q1+q2)/(l1+l2).(c) (q1 + q2)/l2. (d) (q1 + q2)/2l2.19.5 (a) C has units of C/m3 and B has units ofC/m4. (b) (C+ Br)(4r2dr).(c) (4C)(a3/3) + (4B)(a4/4). (d) C+ (3/4)Ba.19.7 (a) C has units of C/m3 and B has units ofC/m5. (b) (C+ Bz2)a2dz. (c) a2(Cl + Bl3/3).(d) C+ Bl2/3.

19.9 Show that problem.110.1 When the tapes are pulled apart they de-velop a charge. Your finger is neutral. By electro-static induction each will be attracted to your finger.110.3 Use a versorium. It will respond at a greaterseparation to the tape with more charge.110.5 (a) See Fig.4.24. (b) Smaller. (c) The same.111.1 Moist wool is a conductor, and more eas-ily transfer electrons on contacting the (conducting)sphere. The wool and the sphere then would havecharge of the same sign, and would thus repel.111.3 Since silk can be grounded here, it isacting like a conductor in this context. (In other

contexts, silk acts like an insulator; most likely silkhas a very low conductivity, so that for short timesit behaves like an insulator, and for long time it be-haves like a conductor). (1) Unelectrified and insu-lated silk is attracted to (charged) amber by electro-static induction, or polarization, as in Figure 1.1. (2)


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Electrified and insulated silk will repel amber onlyif it has charge of the same type as amber; the silk

must have been electrified by placing it in contactwith amber or something electrified like amber. (3)If the silk is attracted to the amber only when thesilk is grounded, then again the silk is subject to elec-trostatic induction, or polarization. Wheeler clearlymissed the possibility that silk could have resinouselectricity (opposite to that of amber).111.5 (a) Approximating a conducting disk bya uniform charge density and a line charge aroundits perimeter puts greater charge on the perimeter,thus causing greater electrical effects there. (b) Fora charged needle, approximate the charge density asa uniform line density and one point charge at eachend. (c) With an extra charge at the ends, electricaleffects would be greater at the ends, thus providinga qualitative explanation for the power of points.111.7 It suggests there is no electricity on theinner surface of the cup.111.9 (a) The amber is charged and the crumbsare uncharged. (b) Electrical attraction: one pieceis charged and the other is uncharged (the amber ef-fect), or one piece is charged and the other is chargedoppositely. Magnetic atraction: one piece of silvercontains iron (giving it has a permanent magneticmoment), and the other contains iron (without a per-

manent magnetic moment, but a large magnetic po-larizability), so the magnetic analog of the ambereffect occurs. (c) A magnet can attract silver elec-trically if the magnet is charged and the silver is un-charged (the amber effect) or if the magnet is chargedand the silver is charged with electricity of the oppo-site sign. A magnet can attract silver magnetically ifthe silver is not pure, but rather contains iron withinit.111.11 The third of Newtons laws of motion isthat to every force on one object due to a second,there is an equal and opposite force on the seconddue to the first; this is consistent with Fabris ob-

servation that not only does electrified amber attractother (unelectrified) objects, but other (unelectrified)objects attract electrified amber.111.13 (1) Levitation could have been due to re-pulsion between like charges, the feather having beencharged by contact the already-charged globe. (2)

The charged feather would go to the edges of mate-rials because that is where materials are most polar-

izable (by the power of points). (3) A linen threadattached to the charged globe would be charged likethe globe (it also could be polarized, but other resultsindicated that linen, relative to silk, is a conductor),and thus could attract chaff at the other end by theamber effect. (4) the crackle and glow can be due toatmospheric ions or electrons being attracted to theglobe, and the associated electrons recombining andproviding light, in a manner similar to phosphores-cence.111.15 Dufays suspended, electrified people werecharged up, and would discharge when touchingsomeone on ground, releasing energy in the form of aspark.111.17 According to Le Roy, conical dischargesare associated with positive points. Thus, in Grays1708 conical discharge we conclude that his finger wascharged positively relative to the globe.111.19 (a) Show that problem. (b) Show thatproblem.

22.1 (a) Show that problem. (b) It decreasesby a factor of 1/

2. Yes. (c) 4.07 107 C.

(d) m1 m1 + M/2.23.1 4.77 106 N, attractive23.3 0.626 mC23.5 Show that problem. The maximum forcewill have magnitude kQ2/4r2.23.7 48.5 C.23.9 (a) 3.33 1010 C. (b) 4.8 1010 sC.24.1 = 60, qmax = 2l

mg tan

k ,

qmax = 4.76 107 C.24.3 .24.5 (a) kqQ4R2 sin2(/2) . (b)

kqQ cos(/2)4R2 sin2(/2) .

25.1 (a) 0 N. (b) 28.4 m. (c) 3.59 106 C.25.3 F = 9.53 N, = 19.25.5 Show that problem. Fx = 1.102106 N,Fy = 0.995 10

6

N, F = 1.484 106

N, = 137.9.26.1 Rotating the rod about its perpendicular bi-sector does not change the configuration so it shouldnot change Fy, but this rotation should reverse Fy.We conclude Fy = 0. Yes.


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27.1 (a) Fx =kqQ

a(a+l) , Fy = Fz = 0. (b) Fx kqQa2 .27.3 The y component of the force due to an ar-

bitrary infinitesimal charge dQ between y and y + dyis exactly cancelled by an equal charge dQ betweeny and y dy. Thus Fy = 0.27.5 Fx =

kqQL

1a 1L2+a2

, Fy =

kqQ

a

L2+a2.

27.7 Fx = Fz = 0, Fy = 2kqQa2 .27.9 The force will point directly away from thecenter of the arc, at an angle of /2. The force willhave magnitude F = 2kqQa2 sin(

2 ) .

32.1 (a) Fe = 2.003 1017 N.(b) Fg = 2.935 1025 N. The force of gravity ismuch smaller than the electrostatic force.32.3 (a) 6.12 1011 N/C. (b) 1.76 1013 m/s2.32.5 (a) = tan1

|Q|E

mg

. (b) T = mg sec .

(c) = 4.75 104 degrees, T = 0.412 N.32.7 No.32.9 (a) E = 2500 N/C. (b) |Q| = 0.11 nC.32.11 Scalar fields: pressure, density, temperature.Vector fields: flow velocity.32.13 (a) Gravity is always attractive, so the fielddue to m2 will point toward m2, pulling m1 towardm2. (b) g = i GmiR2i Ri.33.1 (a) 85,000 N/C y. (b) 85,000 N/C y.34.1 Drag-dominated.34.3 (a) The advantage is that a ball of chargecreates field lines inside. (b) The disadvantage is thatthere are too many arrowheads if you want to sketchthe field quickly. (c) The grass seed method has twoproblems: the grass seeds can align with the fieldin either direction and the density of the grass seedsgives only a qualitative representation of magnitude.35.1 (a) 175 N/C . (b) 1100 N/C .(c) 425 N/C .35.3 (a) Up . (b) kQa2 (1 + qQ ) (1 + 2 cos ) .35.5 (a) Ex = 47.0 N/C, Ey = 36.1 N/C.(b) This calculation was relatively simple. Doing the

calculation from scratch would have involved 23 sepa-rate calculations and than the addition of 23 vectors,a very involved calculation.

35.7 (a) E = kQ

1(x+a)2 +

1(xa)2 2x2

x.

(b) E 6kQa2x4 x.

35.9 Show that problem.

36.1 (a) Q = L. (b) E = kLx(x

L)

.

(c) E = k(2xL)x(Lx) .36.3 E = kQ(3

5)

2a210

y.

36.5 (a) Down. (b) E = 4ka y.36.7 (a) 0. (b) p = 2 R

2x. (c) E = k2R x.36.9 E = 2 k (1 cos(2)) z.36.11 (a) E = 4kyy2+a2 y. (b)

E = 4kay2+a2 x.36.13 (a) Place q at (14.7, 0, 0) m. (b) Place inthe xy plane, parallel to the y-axis, and intersectingthe x-axis at x = 288 m. (c) Impossible.36.15 (a) 20 N/C downward. (b) The upper sheet

has charge density 0.442 nC/m2

and the lower sheethas charge density 0.0884 nC/m2.37.1 Along.

37.3 E is zero 2/3 of the way from 2 to .

37.5 Within an insulator we can arrange the chargehowever we please, but within a conductor the likecharges would repel each other toward the surfacedisrupting the uniform volume charge distribution.

38.1 (a) 3.6 1025 J. (b) 90, 1.8 1025 N-m.38.3 (a) 2.5 1025 N-m. (b) 1.17 1025 N-m.(c) 2.207 1025 J.38.5 (a) F = q(2kpx3 ) . (b) F = q( 2kpx3 ) .

(c) p = 7.11 1025

C-m.39.1 (a) F = pAx. (b) 2.37 109 N x.(c) 2.37 109 N x.310.1 (a) 2.59 106 m/s. (b) 4.64 108 s.310.3 (a) 728 N/C (b) 6.44 109 C/m2.(c) 4.3 106 m/s, 21.8.310.5 2.01 105 m.310.7 (a) The velocity v does not depend on theradius r. (b) T = 2r

me2ek .

42.1 (a) The flux will have the opposite sign.(b) 34 N-m2/C. (c) E = 10 Qenc = 4kQenc.

42.3 7.68 106

N-m2

/C.42.5 ER2.

42.7 2kQ

1 d

R2 + d2

.

43.1 (a) 0. (b) 679 N-m2/C. (c) 0.

43.3 (a) 4kQ. (b) (2/3)kQ.


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43.5 (a) (0.8, 0.48, 0.36). (b) 8.06 N/C.(c) 5.64 N/C. (d)

45.6. (e) 1.128

103 N-m2/C.

(f) 9.97 1015 C.44.1 1.77 nC/m2.44.3 3.76 109 C.45.1 (a) 1.94 1012 C. (b) 2.59 105 N/C.45.3 (a) 1.30 1010 C/m2. (b) 2.66 109 N/C.45.5 (a) = (b2a2) . (b) 0,

2k(r2a2)(b2a2)r ,

2kr .

45.7 (a) 1 = 0.722 nC/m, 2 = 0.856 nC/m.(b) 20.1 N/C.

45.9 Take mass density . (a) g =Fgrav

M ,

gr

g ndA = 4GMenc. (b) g = Gr3 r.(c) T = 2

3

4G . (d) 5070 s.

46.1 No.46.3 (a) 25 V. (b) The dipole moment on theconductor will point away from the point charge. Thepoint charge will be attracted to the conductor.46.5 (a) The inner surface has Q and the outersurface has +Q. (b) The inner surface has Q andthe outer surface has zero charge.46.7 Connecting the slats makes them behave al-most like a solid conductor, which can screen out anexternal field.47.1 0.212 nC/m2.47.3 Explain in your own words problem.47.5 (a) Starting to the left and proceeding to

the right, the electric fields are 10k, 18k,2k, 10k. (b) Starting with the left sideof #1, going rightward the charge densities are(5/2) and (9/2), (9/2) and (1/2), (1/2)and (5/2).47.7 (a) 7.96 nC/m2. (b) 4.0 1011 C.(c) E = 0. (d) Er = 400 N/C (radially inward).47.9 (a) 0, kQ/r2, kQ/r2 (radial component offield). (b) Qinnera = 0, Q

outera = Q. (c) Q

innerb = Q,

Qouterb = Q.47.11 (a) 0, 4k/r, 2k/r (radial component offield). (b) innera = 0,

outera = 2. (c)

innerb = 2,

outerb =

.

48.1 (a) All of the charge resides on the cups outersurface. (b) 3 C. (c) 2 C. (d) 0.4 C.49.1 2

1 s

s2 + d2/4

.

410.1 3 103 C/m2.

411.1 (a) Ex = 0, Ey = 4kbx2+b2 .(b) =

b

(x2

+b2

)

.

411.3 (a) kQ2

8r4 . (b)1b 1a

kQ2

2 .

52.1 94.7 N.53.1 (a) 2 108 J. (b) 2 108 J. In the firstcase we raise the electrical potential energy; in thesecond case we lower the electrical potential energy.53.3 2 V/cm, 2.2 V/cm, 2.4 V/cm.53.5 2.7484 V.54.1 5.69 1014 m.54.3 (a) 35.3 V; the electron goes toward thehigher potential at the end point. (b) 883 N/C, along

y (downward).

54.5 (a) 4.8 105 J. (b) 20000 V, the startingpoint is at a lower potential than the endpoint.(c) 2500 V/cm.54.7 (a) The 6 V plate, 1.78 106 m/s.(b) 1.44 1018 J. (c) 1.44 1018 J. (d) 9 V.54.9 0.058%.54.11 (a) 8.90 104 V. (b) 0.629 107 C.(c) 3.93 105 C/m2.55.1 The field is largest near the two bottom cor-ners and smallest on the line directly between the twoside plates.55.3 The field is largest at the upper right and at

the lower left corners and smallest at the center.55.5 (a) Show that problem. (b) Q.55.7 (a) Spheres centered at the charge. (b) Yes.(c) Yes. (d) No.55.9 The field must be zero at the crossing point.56.1 2.0 V.56.3 (2kq2/a)(2 + 1/

2).

56.5 (a) 480 kV. (b) 1920 kV. (c) 0 J.56.7 (a) Halfway between them. (b) Yes. TakeV = 4kqa .56.9 5.56 1010 C.56.11 442 nC/m2.56.13 12.71 V.57.1 (a) 0. (b) Yes. (c) a2x2 bx.57.3 (a) b2. (b) +b2. (c) 2b2. (d) No.58.1 (a) The potential at the center is the sameas on the surface. (b) The center.58.3 2.02 nC/m.


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58.5 (1/3)Ar3, the equipotential surfaces areconcentric cylinders centered on the z-axis.

58.7 2k (z2 + a2 z).59.1 (a) V(z) = kQ

R2+z2. (b) Ez =

kQz(R2+z2)3/2

.

(c) No. (d) Ex and Ey will in general change underrearrangement.59.3 Qualitatively similar to Figure 5.10(b), butskewed with 2 dominating .

59.5 (a) 0. (b) V(x) = ka

x ln( x+l/2xl/2 ) l

.

(c) Show that problem. (d) Show that problem.59.7 (a) V = kQ/b. (b) V(a) = 0.9kQ/b < 0.(c) For q > 0.1Q.510.1 (a) 24.85 V, 29.25 V. (b)Ex 22 V/m.(c)Ex =

22 V/m.

510.3 (a) V(x) = kQl [ln |x + l| ln |x|].(b) Ex =

kQl

1x

1x+l

. (c) Ex =

kQl

1x

1x+l

.

510.5 Show that problem.510.7 (a) 3.28 V, 7.32 V. (b) 20.2 V/m.(c) 20r3 , 20 V/m.510.9 (2y + 8x)x + (2x + 10y)y.510.11 (a) Ex = 43 Vad ( xd )1/3. (b) In units of V/m:0, 0.84 104, 1.06 104, 1.2 104, 1.33 104.(c) Va9k ( 1xd2 )2/3. Total charge per unit area from 0to d is Ex(d)/4k = 1.18 107 C/m2.511.1 (a) 6300 V, 1125 V. (b) 3.15 105 N/Coutward, 2.81

104 N/C inward. (c) 3

109 C,

6 109 C, 1350 V, 1350 V. (d) 6.75 104 N/Coutward, 3.375 104 N/C outward.511.3 No.511.5 Q10R = 10QR, 10R =

110R, VR = V10R,

E10R =110

ER.

511.7 (a) 9.09 V. (b) 100 V. (c) 100

1 (1011n

V.512.1 1.114 106 N/C.512.3 1.323 1010 N/C.

62.1 (a) 23.6 nF. (b) 2.12 102m.62.3 Excess charge affects the structure of thesolid sphere less than the structure of the shell.

62.5 4.50 108

m.63.1 (a) 0.143 nF. (b) 1.25 nC. Larger sphere haslarger charge.63.3 (a) 5.79 1011 F. (b) 1.736 1012 C.63.5 (a) By how close the plates can be kept with-out touching each other. (b) B how large the plates

can be without making the capacitor unusable.63.7 (a) 1250 V. (b) 0.0905 m2. (c) 6.25

105 N/C.

(d) 5.53 106 C/m2.63.9 This problem involves dielectrics, which arediscussed in section 5. (a) 4.73 cm. (b) 0.775 nF.(c) 5.89 106 C.63.11 (a) 19.24 pF. (b) 4.62 nC. (c) 28.8103 N/C.64.1 (a) Make three parallel arms, each with three2 F capacitors in series. (b) Put two units from part(a) in parallel.64.3 A/4k(d1+d2). C = Q/V has V increaseby the factor (d1 + d2)/d.64.5 (a) 3.428 F, 14 F. (b) 41.14 C, 5.143 V,6.857 V. (c) 12 V, 96 C, 72 C.64.7 154 nF.64.9 4 nF.64.11 7 F.64.13 (a) 9 C on each capacitor, V1 = 3 V,V2 = 1.5 V. (b) Yes. (c) No.65.1 d = 2.67 104 m, A = 242 m2.65.3 3.91.65.5 Above about one volt, electrolysis occurs atthe plates and an ion current would flow. The capac-itor would not be able to hold charge in this circum-stance.65.7 (a) 4.24 cm2. (b) 120 pF, 3.5 kV.65.9 Show that problem.

65.11 Show that problem.65.13 (a) 2.221017 F = 22.2 aF. (b) 0.00721 V.65.15 Ctotal =

1+2

abk(ab) .

66.1 (a) 6.67 108 F, 3.32 106 m.(b) 4.8 104 J. (c) 3.61 107 V/m. (d) 5761 J/m3.66.3 4 106 F.66.5 (a) QA = QB = 160 nC, VA = 4 V,VB = 8 V, UA = 0.32 J, UB = 0.64 J.(b) VA = VB = 5.333 V, QA = 213.3 nC,QB = 106.7 nC, UA = 0.568 J, UB = 0.284 J.(c) 0.108 J. (d) VA = VB = 1.777 V,QA = 71.1 nC, QB = 35.6 nC, UA = 0.063 J,UB = 0.032 J. (e) 0.757 J.

66.7 (a) U = kQ22R

. (b) dU = kQ22R2

dR, Pel = E2

8k.

66.9 In left branch of Figure 6.33, set C1 = 6 F,C2=12 F; in right branch set C3=12 F, C4 = 6 F.(a) Q1= Q2 = Q3 = Q4 =360 C, V1 = V4=60V,V2= V3 =30V. (b)32.4 mJ. (c) Q1 = Q4 =240 C,


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Q2 = Q3 =480 C, V1= V2 = V3= V4 =40 V.(d) 28.8 mJ. (e)

240 C.

66.11 1.128 1012 J.67.1 (a) | Ediel|/| E0| = 1/5. (b) Vdiel/V0 = 1/5,Vdiel = 1.2 V. (c) Qdiel/Q0 = 1. (d) Cdiel/C0 = 5.(e) Udiel/U0 = 1/5.67.3 In the first case, U gives all of the energy.In the second case the capacitor energy must be in-cluded.67.5 Show that problem.67.7 Take C = 12 F initially.(a) Ucap = (1/2)CV

2, Ubatt = CV2,Uheat = (1/2)CV

2; Ucap = CV2,

Whand = (1/2)CV2; U

cap = (1/4)CV2,

U

batt = (1/2)CV2, U

heat = (1/4)CV2.(b) Ucap = (1/4)CV

2, U

batt = (1/2)CV2,

Whand = (1/4)CV2.

68.1 (a) 0.2233 C. (b) 191.4 V.(c) Qa = 0.0106 C, Qb = 0.2127 C.68.3 Show that problem.V1V2 = Q1(p11p21)+ Q2(p12p22)+ Q3(p13p23).The term Q3(p13 p23) is the effect of Q3.68.5 (a) Charge will distribute over the surfaceof a conductor, so that the material is is irrelevantand only the shape of the surface is important. Po-larization of an insulator depends on the dielectricconstant and thus the material. (b) The spheres in-duced dipole moment, due to polarization, is rela-tively small for r a.69.1 (a) Along the length of the molecule.(b) would need to change depending on the direc-

tion of E. In most cases E and p would not evenpoint in the same direction.69.3 Show that problem.610.1 Show that problem.

610.3k2

a.

71.1 (a) No current. (b) No current. (c) 5 mA.

72.1 10,800 C, 3.75 1019

electrons/s.72.3 (a) 35,840 C. (b) 2.39 103 s, or about 66.4hrs.72.5 5.09 1011 A.72.7 (a) 133.3 A. (b) 4.76 105 A/m2.(c) Jx = 4.47 105 A/m2, Jy = 1.629 105 A/m2,

Jz = 0. (d) 0.612 A.72.9 No. Opposite directions.

72.11 Jr = 995 A/m2.73.1 (a) 4 , 5 . (b) Non-ohmic.73.3 (a) 2 A. (b) 2 A.73.5 (a) 3.13 105 /m. (b) 0.125 .73.7 8.45 -m.73.9 0.086 mm.73.11 (a) 7 mA. (b) 8.4 nV. (c) 20 n.73.13 (a) 240 , 2.4 . (b) Resistance of the bulbfor home use.73.15 (a) 5 kV. (b) 250 .73.17 (a) 3.6 104 J. (b) 0.833 A. (c) 6000 C.73.19 (a) 28.8 . (b) 28.8 106/m.73.21 (a) 23.9 horsepower. (b) 1000%. (c) No. Itcannot exceed 100%.74.1 (a) 0.28 V/m rightward. (b) 0.532 V. Highervoltage on left, lower voltage on right.74.3 (a) 2.79 108 -m. (b) Aluminum.74.5 (a) For d = 0.04 cm, RCu = 0.01547 ,Rsteel = 8.27 106 . (b) VCu = 0.0309 V,Vsteel = 16.54 106 V.74.7 Charge on the surface of the circuit, includ-ing the wire, makes an electric field that drives thecurrent through the wire.74.9 E points from regions of higher potential toregions of lower potential. The local form of Ohms

Law,J =

E, says that

J is in the same direction asE. Therefore J points from regions of higher poten-

tial to regions of lower potential.75.1 (a) 13 . (b) V2 = 27 V, V1 = 12 V.(c) 39 V. (d) 3 A. (e) 3 A.75.3 (a) 6 . (b) 3.5 A. (c) 35 V. (d) 3.5 A.(e) 10 .75.5 (a) 1.5 . (b) Both are 12 V. (c) 12 V.(d) 11 A. (e) 1.09 .75.7 (a) R1 = 4 , R2 = 2.67 , R3 = 1.2 .(b) I3 = 10 A, I2 = 6 A. (c) 16 V. (d) 28 V. (e) 2.8 .75.9 One has a parallel combination of two resis-tors in series with another parallel combination of two

resistors. The other has a series combination of tworesistors in parallel with another series combinationof two resistors.75.11 96 , 144 , 100 W.75.13 (a) IA = 3 A, PA = 0.9 W. (b) IA = IB =3 A, PA = PB = 0.9 W. (c) IA = IB = 1.5 A,


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8

PA = PB = 0.225 W. (d) IB = 2 A, PB = 0.4 W,IA = IC = 1 A,

PA =

PC = 0.1 W.

75.15 (a) Resistors in parallel add as inverses andcapacitors in parallel add directly. (b) Resistors inseries add directly and capacitors in series add asinverses. (c) The formulas for total resistance andcapacitance take the same form if R 1C and IQ.Then R = VI 1C = VQ .75.17 If a person stands on only one foot, thenmost current would pass through the foot touchingthe ground. If the person were to stand on both feet,there would be a path for current into the torso ofthe body where current is most dangerous.76.1 (a) 1 mV. (b) 20000 /V.

76.3 (a) 1.98 k. (b) 0.5526 . (c) 83.99 mV.76.5 (a) 6.505 V. (b) 0.0769%.77.1 The ion density is high near the plates, butlow farther from the plates.77.3 Starting a car four times a day consumes0.667% of the chemical charge every day: about thesame amount as the non-current-producing sulfationreaction does.77.5 1662 s = 27.7 min.78.1 1.36 V, 0.3 .78.3 (a) The filling, aluminum foil and saliva inthe mouth. (b) The chemical energy of the voltaiccell (the foil and filling are the electrodes and the

saliva is the electrolyte).79.1 (a) 432,000 C. (b) 60 hr. (c) 24 hr.79.3 (a) 2880 C = 0.8 A-hr. (b) 0.1 A. (c) 8 hr.710.1 (a) Voltage gains of 0.6 V and 1.2 V acrosselectrodes, voltage losses Ir = 0.3 V and IR = 1.5 Vacross resistances. (b) 0.6 hr. (c) 7776 J.710.3 (a) 2 A. (b) Voltage loss of 0.3 V and voltagegain of 1.3 V across electrodes; and voltage lossesIr = 0.4 V and IR = 0.6 V across resistances.711.1 (a) F = 1.59 1017 N y,F+ = 1.61 1017 N y.(b) v = 3.18 106 m/s y,v+ = 3.22 10

6

m/s y.711.3 (a) Show that problem.(b) 11.2 106 s/m2. (c) 9.54 107 m.712.1 1.846 106 s.712.3 n = 3.84 1013/m3, vd = 9000 m/s.712.5 (a) dI = rdr. (b) I = a2/2.

712.7 2.415 C/m2.712.9 Show that problem.

713.1 Insulators, semiconductors, conductors.713.3 J = nev. If the critical velocities vc are nottoo different for semiconductors and metals, then thedensities n mostly determine the Jcs.

83.1 (a) E= 24 V; (b) r = 0.024 ;(c) 105.0 A-hr.83.3 (a) 30 days; (b) 22.2 days.83.5 E= 10.72 V; r = 0.0245 .83.7 (a) I = 5.2 A charging the battery;(b) 1.352 W heating;(c) 10.72 W charging; (d) 88.98% efficiency.83.9 (a) R = 0.75 . (b) I = 3 A.84.1 (a) 40 hr. (b) 12 cents.85.1 E= 0.6 V, r = 600 ; (b) 4.02%; (c) 5.14%.85.3 I = 2.25 A.85.5 (a) R = 3.264 ; (b) V = 261 V;(c) M = 1.190 104 kg; (d) 78.6%.85.7 (a) R = 2.4 ; (b) 99.59%; (c) 96%.

86.1 (a) I1 = I + I2; (b) I1 = VR1

, I2 =V

R2,

I =V E

r.

86.3 R = 788 .86.5 (a) 0.05 r/R 0.55;(b) 0.033

r/R

0.367.

86.7 R = 15 , E= 120 V.87.1 (a) I = I1 = I2 = 22.22 A;(b) V1 = V2 = 11.778 V; (c) P1 = 266.7 Wdischarge, P2 = 222.2 W charge. (d) I21r1 = 4.9 W,I22r2 = 39.5 W. To our numerical accuracy, energy isconserved.87.3 (a) Reverse the direction of positive I2 rela-tive to Figure 8.13(b). Then I1 = I+ I2,I1 =

E1Vr1

= 600 100V,I2 =

E2+Vr2

= 500 50V, I = VR = 100V.(b) V = 0.4 V. (c) I = 40 A , I1 = 560 A,I2 = 520 A. (d) PR = 16 W, Pr1 = 3136 W,Pr2 = 5408 W, P1 = 3360 W discharge, P2 = 5200 Wdischarge.87.5 (a) B1 off, B2 off. (b) B1 dim, B2 dim.(c) B1 bright, B2 off. (d) B1 bright, B2 off.87.7 R = (1 + 3)30 = 21.96 .87.9 0.5R.


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9

87.11 R = 7.2 .87.13 Show that problem.

87.15 Show that problem.87.17 (a) All positive currents and V are as in

Fig.8.39, so that I1 =E V

r1, etc. Also, I = V /R

and I = I1 + I2 + I3. (b) V = 4.963 .(c) I = 496.3 A, I1 = 103.7 A, I2 = 251.8 A,I3 = 140.7 A. (d) current is conserved.88.1 (a) Q2 = 0, I1 = 0, I = 6 A. (b) I2 = 0,I = I1 = 1.5 A, Q2 = 40.5 C.88.3 (a) Q = 0, I1 = 4/3 A, I2 = 2/3 A,I3 = 2/3 A, I4 = 4/3 A. I = I1 I2 = 2/3 A goes tothe capacitor. (b) I1 = I3 = 0.8 A, I2 = I4 = 1.0 A,I = 0, Q = 14.4 C.

88.5 (a) IR = 0. (b) IR = I0.89.1 (a) RC = 12 s. (b) I0 = 0.84 A, I = 0.(c) Q0 = 0, Q = 10.08 C. (d) Uheat = 15.88 J.(e) Uheat = 21.17 J.89.3 (a) Show that problem. (b) 96 J.(c) Show that problem.89.5 (a) R = 5.091015 . (b) RC = 2.04107 s.(c) C = 1.965 1013 F.810.1 More turns through small angles require lesssurface charge but are more expensive to make. Two45 might be adequate.810.3 (a) r/R . (b) Let R1eq = r1 + R1.

Then QCp = (E/R)(r + R)Cp[1 et/ReqC

],IR = QCp/RCp.810.5 (a) 2s. (b) Charge will actually flowfrom one part of the surface through the bulk toanother part of the surface. (c) A combination ofbulk and surface current would provide the least resis-tance, but in practice the resistance to surface currentis very large because of the associated small cross-sectional area.810.7 (a) If 1 < 2, then the electric field islarger in material 1 than in material 2, so a largerfield enters than leaves the surface.(b) S = J(1 2)/4k12.811.1 Many choices are possible. ForR1 = R3 = 0, (9.65

) gives I1 = I3 = 0, whereas(9.65) gives nonzero I1 and I3.812.1 n = 0.862 1024/m3.

92.1 qm1 = 22.56 A-m, qm2 = 15.95 A-m.

92.3 = 1.64 A-m2.

92.5 B = 0.0259 T.

92.7 F = qm B, qm due to one pole of a longmagnet and B = km[ + 3( R)R]/R3 due to ashort magnet. The force on the distant pole of thelong magnet is neglected.

92.9 (a) The magnet is strongly attracted to thesoft iron rod when the soft iron is brought up to themagnets poles, but only weakly attracted when therod brought up to the magnets center. (b) The softiron is strongly attracted to the magnet when themagnet is brought to any part of the soft magnet.


93.1 (a) qm = 5.12 A-m, m = 3.2 105 A/m, = 0.256 A-m

2

. (b) 0.201 T, 1.28 103

T,2.84 105 T.93.3 M = 3.99 105 A/m.93.5 (a) r a, no r-dependence. (b) a r l,| B| r2. (c) l r, | B| r4.94.1 Discontinuity in C is 0( 1) M.95.1 Put a rod of soft magnet in a line betweena pole of a permanent magnet and the region wherethe field is to be intensified.

95.3 When field lines are expelled (concentrated),the object that is their source is repelled (attracted).

95.5 (a) H = 1481 A/m, = 1.688.

(b) Memu = 2.50 mmu/cm3

,Bemu = 50 G, Hemu = 18.6 Oe, emu = 0.1344.

96.1 Mr and Ms are too far in value; the magnetwould not retain its magnetization.

97.1 (a) | H| = 485 A/m, | B| = 0.98 T. (b) small,except perhaps near the poles.

97.3 By keeping in the field lines, the keeper mag-net lets the magnetization of one pole of the magnetmagnetize the other end, and vice-versa.

98.1 In Figure 9.19a there are no poles becauseM is normal to n; hence H 0. In Figure 9.19b the

poles produce a demagnetization field H M thatmakes B = 0( H+ M)

0.

99.1 | B| = 6.63 106 T.99.3 72.8 dip angle.99.5 Show that problem.

910.1 If the Fe and Nd interaction were ferromag-netic, the net magnetic moment would be larger than


12/20

10

when they are antiferromagnetic: 7F e + Nd ratherthan 7F e

Nd .

101.1 (a) B = 0.056x T. (b) B = 0.056x T.101.3 (a) B1x = 0.0371 T. B1y = 0.01486 T.(b) B2x = 0.0742 T. B2y = 0.0297 T.(c) Bx = 0.1113 T. By = 0.01486 T.102.1 (a) South pole up. (b) Uptward force onloop. (c) Downward force on magnet. (d) No torqueon magnet.102.3 (a) Left magnet has moment into page, righthas moment out of page. (b) Attracted to loop onleft.102.5 The wire moves leftward, along x.102.7 (a) = [3.157

i 0.476

j + 6.156

k] A-m

2

.(b) = (0.209 j + 0.0162 k) N-m.102.9 (a) Attractive. (b) Repulsive. (c) Wirescarrying parallel currents will attract.103.1 (a) = 4.52 104z A-m2.(b) = 1.810 106y N-m.(c) = 1.666 106y N-m.103.3 (a) = 1.024 104 A-m2.qm = 8.53 104 A-m, (b) | B| = 94.8 nT.(c) 0.521 nT. (d) | B| = 10.24 nT.103.5 (a) M = 9.27 106 A/m. (b) For a goodmagnet, M 105-106 A/m.104.1 (a)

|F|

= 81.6 N. (b) Reversing the currentor field reverses the force, pumping the blood theopposite way. (c) Reversing both current and fielddoes not change the direction of the force.104.3 Compress.104.5 F = 2Ia| B|y.104.7 (a) F = 2N a2AIz. (b) F = 0.253z N.104.9 (a) Show that problem. (b) F = 0.049y N.104.11 No current was specified. For I = 1 A,we have (a) Fbot = 0.010y N. Ftop = 0.005y N. (b)Fleft = 0.0045x N. Fright = 0.0045x N.(c) Fnet = 0.005y N, compress.

104.13 (a) a = 107 m/s2. (b)|

F|

= 1.4

105 N.(c) I = 2.8 106 A.104.15 F = [2kmqmIa

2/(r2 + a2)3/2]y.105.1 (a) Force is out of the page.

(b) | F| = 2.2061017 N. (c) a = 2.471012g, whereg = 9.8 m/s2.

105.3 | B| = 6.65 1017 T.105.5 F = (

0.399x

0.908y

1.167z)

1014 N,

| F| = 1.532 1014 N.105.7 B = (0.0286j 0.0755k) T.105.9 Show that problem.106.1 (a) Electron moves on a semicircle thatbends left, and comes back out of the field region.(b) Proton moves on a semicircle that bends right,and comes back out of the field region.106.3 (a) It initially deflects along y.(b) v = 1.655 104 m/s.106.5 The period and radius are independent. Themechanics student is wrong.106.7 (a) T = 1.725 108 s. (b) | B| = 7.62 T.(c) 4.17 10

4

V.106.9 Deuteron, 2.55 cm; Triton, 3.12 cm; 3Henucleus, 1.56 cm; 4He nucleus, 1.80 cm.106.11 (a) 26.1 T. (b) 30.1 T.

106.13 (a) F is perpendicular to B.(b) v2 = v

2v2 =

2mEv2, and both Eand v donot change. (c), (d), (e) are show that problems.106.15 (a) = 107.6. (b) R = 1.768 cm.(c) p = 3.53 cm.106.17 (a) circle centered at x = z = mv0/qB,y = 0, with radius R =

2mv0/qB. (b) penetration

is (

2 1)mv0/qB. (c) time in field is (/2)(m/qB).(d) Exits at x0 = 2mv0/qB.

106.19 (a) Show that problem. (b) Show thatproblem. (c) The two formulae coincide.106.21 (a) Show that problem. (b) Show thatproblem.107.1 Because the charge carriers move to the sideof the wire, the force does act on the charge carriersthemselves (i.e. the current). Thus Maxwell was inerror here to err is human.107.3 (a) Positive. (b) vd = 0.882 mm/s.107.5 (a) RH = 7.35 1011 m3/C.(b) Rt = 10.5 109 . (c) l = 1.22 m.107.7 (a) Top is negative. (b) Emot = vBy,Ees = vBy. (c) V = vBl. (d) V = vBlcos.108.1 dWemf = dWpmf = 0.084 J.

112.1 d B = 2.72 1010(2i + k) T.113.1 Show that problem.114.1 (a) 5.29 m. (b) 281 turns.


13/20

11

114.3 2.81 mm.

114.5 (a) d B = (kmIds/a2)

.

(b) B = (kmI/a).114.7 3.67 mA.

114.9 Bx = 6 105 T.115.1 (a) B = 0, B = 4kmKi, B = 12kmKi.

(b) | B| = 0.00314 T.115.3 (a) Two close coils behave like a single coil,with a single maximum at their midpoint. (b) Twodistant coils behave like two independent coils, witha local minimum at their midpoint.(c) By = 2kmIR

2[ 1[(y+ s

2)2+R2]3/2

+ 1[(y s

2)2+R2]3/2

].

(d) s = R.


115.7 Show that problem.115.9 (a) | B| = 168.4z T, into page. (b) | B| =49.1z T, out of page. (c) | B| = 0.678z T, intopage.

115.11 B = 2kmKln( x + w/2x w/2 )y.

115.13 (a) I = Kd, K = M, counterclockwise.

(b) B = kmMdab

z2+(a2+b2)/4

1

z2+a2/4 +1

z2+b2/4

z.

115.15(a)Bx=2kmNI

L

x+L2

(x+L2)2+a2

xL2(xL

2)2+a2

.

(b) For L/a 0, Bx 2kmN Ia2

(x2 + a2)3/2. For L/a ,

Bx 4kmNI/L.115.17 (a) dI/dr = r . (b) I = (a2/2).

(c) Bx = 2km[

x2 + a2 2x + x2

x2 + a2

].

116.1 (a) F = 3.04 1021y N.(b) F = 3.04 1021x N. (c) F = 0.116.3 F =

2evkmI

ri.

116.5 (a) | F| = 2kmI2ln(a/r), to the left, andexpansive. (b) | F| = 9.15 105 N.116.7 Take solenoid current clockwise as viewed

from above. (a) Bof loop =2kmI2

b

.

(b) Fon solenoid =22kmI1I2na

2

b.

(c) B = 2.51 104 T, F = 1.18 105 N.116.9 F/l = kmI

2/a.

117.1 (a) B = 1.721 105 T-m. (b) B = 0.

117.3 (a) Ienc = 19,900 A. (b) B = 0.05 T-m.(c) B = 0 T-m.

117.5 (a) Field circulates counterclockwise.(b) B/s = 0.04 T. (c) Ienc = 636.6 A. (d) Currentflows out of the page.

118.1 Show that problem. Deformation doesntaffect circulation if no current passes through the de-forming circuit.

118.3 (a) and (b) For both a physical circuit andan Amperian circuit, ds is defined, but only for aphysical circuit does it point along the local currentdirection.

119.1 Take ds to be clockwise, so Ienc > 0 is into thepage. (a) B = 120 T-m. (b) Ienc = 9.55 107 A(out of page). (c) J n = 7.96 10

6

A/m2

.119.3 (a) B = 10ydydx.(b) Ienc = 5ydydx/2km. (c) I/A = 5y/2km.

119.5 (a) B = 5.6 106 T-m.(b) B = 3.56 105 T. (c) Ienc = 4.45 A,Ienc/A = 2266 A/m

2. (d) Into the page.

1110.1 (a) | B| = 9.60 mT. (b) | B| = 15 mT.(c) counterclockwise.

1110.3 (a) r < a and a < r < b counterclockwise,c < r clockwise. (b) concentric circle of radius r > c.

(c) B = 2r| B|. (d) B = 12kmI.(e) | B| = 6kmI/r .1110.5 Take Iinner to be out of the page, and to indicate the counterclockwise tangent.

(a) Jcore =I

a2, Jsheath =

I

(c2 b2) . (b) For r < a,B =

2kmI

a2r; for a < r < b, B =

2kmI

r; for

b < r < c, B =2kmI

r(1 r

2 b2c2 b2 ); for c < r,

B = 0.

1110.7 (a) d B = km(nIz )(ds )/R3.(b) d B =

d B =

2km (nI) (ds )2

.

(c) Show that problem.

1111.1 Use qM

=

l. (a)

|F|

=kmq

2m

4h2, repelled.

(b) h =qm2

kmMg

.

1111.3 (a) Show that problem.

(b) By =kmqm

(y + h)2 for y < 0.


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12

(c) F = qmB =kmq

2m

4h2.

1112.1 (a) Induced surface current K circulatesclockwise as seen from above.

(b) K =qm2

(2 + h2)3/2. (c) F =

kmq2m

(2h)2(z).

1113.1 See Section 11.13.

1114.1 (a) Bgap = 17.2 mT,Bin = Bgap = 17.2 mT. (b) Agap/A = 45.5.

1114.3 (a) Bin = 1.29 T. (b) Bgap = 1.29 T.

For Chapter 12 Food for Thought questions

on pp.512-513, 515, and 517, see last page.

123.1 (a) Calico is an insulator. (b) The greaterthe current, the greater the deflection. (c) Counter-clockwise.

123.3 (a) Counterclockwise. (b) To increase theresponse.

123.5 (a) Smaller. (b) Clockwise.

123.7 (a) Ssource of magnetic field. (b) Counter-clockwise.

124.1 (a) Increases out of the page. (b) Into thepage. (c) Clockwise. (d) Clockwise. (e) Up.(f) Compress. (g) No tendency to rotate.

124.3 No current.

124.5 IfR increases, (a) decrease. (b) to observer.(c) same direction as primary. (d) same direction

as primary. (e) loops attract and expand. (f) If Rdecreases, all answers reverse.

124.7 (a) Increase out of the page. (b) Into thepage. (c) Clockwise. (d) Clockwise. (e) Leftward,compressive. (f) If the field is reversed, then the re-sponses in parts (a-d) reverse, but are the same forpart (e). (g) If the field is tilted, the effects decrease.

124.9 (a) Counterclockwise. (b) Move foil awayand compress the foil.

124.11 See last page of Answers.

124.13 See last page of Answers.

125.1 (a) Eand B are vectors; E, B, and dB/dtare scalars. (b)

E, E, and dB/dt do not change;

Band B reverse. (c) Eand dB/dt.

125.3 (a) B = (16t + 32 103t2) 104 Wb.(b) dB/dt = (16 + 64 103t) 104 Wb/s.(c) E= (16 + 64 103t) 104 V, counterclockwiseas seen from above.

(d) | F| = 2.56(t + 6 103t2 + 8 106t3) 108 N.(e) 3.03 s.

125.5 (a) 2.39 mV. (b) 2.47 mV, a 3.57%change. (c) 0.646 mA.125.7 (a) 0.008 V counterclockwise. (b) 0.32 mAcounterclockwise. (c) 4.43 105 N pushing the loopinto the field region. (d) 1.536 106 N-m, tendingto decrease the angle below 60 degrees.125.9 (a) 0.096 T into the page. (b) 0.00144 Wb/s.(c) 0.036 mA clockwise. (d) 6.912 108 N drawingthe circuit into the solenoid.125.11 (a) 2N BA. (b) 2NBA/R.125.13 (a) E= (2r)(dr/dt)B, I = E/R.(b) dF/ds = (2r)(dr/dt)B2/R. (c) 144 N/m.126.1 (a) 0.2 H. (b) 27 A/s.126.3 (32.4 + 32.4t) mV.126.5 (a)

2kmN1N2ha

. (b)

2kmN1N2ha

.

126.7 (a)22kma

2b2

R3. (b) 1.599 1010 H.

126.9 (a) 42kmNcNsIsa2/l. (b) counterclock-

wise. (c) |E| = dB/dt = 42kmNcNs(dIs/dt)a2/l.(d) 0.1263 mH.127.1 (a) 0.08y V/m. (b) bottom is higher, by18.35 mV.127.3 (a) Vright = 0. (b) Vleft = 0.(c) Vtop = 0.127.5

East wing is higher by 0.135 V.127.7 1.42 V, clockwise viewed from above.127.9 (a) rBb, radially out. (b) b/cd. (c) ForcerB2bcd opposing motion, torque r2B2bcd op-posing motion. (d) Show that problem.128.1 (a) Middle arm. (b) Before, IJ = 0.048 A,Iw = 1.2 A, both right to left, (c) After, Iw isunchanged and IJ = 1.2 A, left to right. (d)Vw = 24 V, VJ = 600 V, both clockwise, andVL = 624 V counterclockwise. (e) See Fig.12.24.129.1 2.703 mH.129.3 61,300 A/s, the sign meaning that I2 andI1 are changing in opposite senses.

129.5 (a) On average, the equivalent ring andsolenoid magnets are opposed. (b) On average, thering and solenoid currents are opposed. (c) An in-crease in Isol is accompanied by a decrease in Iring.This corresponds to out-of-phase magnets, which re-pel.


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13

1210.1 (a) 72 turns. (b) 1.55 mV.1210.3 (a) N2L. (b) 0.1382 mH.

1210.5 L = 2kmN2a ln[(a + b)/b].1211.1 (a) dI/dt = 2.21 105 A/s, VR = 0.966 V,VL = 1.434 V. (b) I = 6.52 A, dI/dt = 1.38105 A/s,VR = 1.5 V.

1211.3 (a) dI/dt = 5.78 105 A/s, 4.8 V.(b) 37.25 ns.

1211.5 (a) 6.73 mH. (b) 0.410 ms. (c) 19.8 mA.

1211.7 Let VL = LdI/dt + IRL. (a) At t = 0+,

VL = LdI/dt = 12 V across the inductance part ofthe inductor. At t = 25 s, VL = IRL = 0.293 Vacross the resistance part of the inductor. (b) Att = 0+, a sketch would show voltage changes of 12 V

across the emf and VL = LdI/dt = 12 V across the in-ductance part of the inductor. At t = 25 s, a sketchwould show voltage changes of 12 V across the emfand voltage drops IRL = 0.293 V and IR = 11.707 V.(c) At t = 0+ the electric field is electromagneticallyinduced. At t = 25 the field is electrostatic.

1212.1 (a) 0.1568 J. (b) 2.306 108 J/m3.(c) 24.1 T.

1212.3 0.416 H/m.

1212.5 (a) 0.008 s. (b) See Figure 12.23.(c) At t = 0, I = 0 and dI/dt = 3 103 A/s. Att = 0.002 s, I = 5.31 A and dI/dt = 2.34 103 A/s.At t =

, I = 24 A and dI/dt = 0. (d) Battery

provides 0, 63.7 W, and 288 W. (e) Resistor uses 0,14.1 W, and 288 W. (f) Inductor builds up energyat rate 0, 49.7 W, and 0. (g) Except for negligibleparasitic capacitance, there is no electrical energy.(h) Pbatt = PL + PR.1213.1 (a) 1.1 N/C. (b) 1.76 N/C.

1213.3 (a) The problem statement should haveasked you to show that |Ez/E| = l/2N a, not2Na/l. (b) 0.00398. (c) |E| = 0.0329 V/m,Ez = 0.1309 mV/m. Note: VL = Ezl = 0.0262 mV.(d) E is electromagnetically induced. (e) Ez is elec-trostatic.

1213.5 Let R2

= RrRl +RrRm +RlRm. IfEr > 0,then Il = ErRm/R2 goes up the left arm andIr = Er(Rm + Rl)/R2 goes down the right arm.(a) IrRr. (b) IlRl. (c) Er = 0.787 mV, soIrRr = 0.562 mV (voltmeter bottom is positive)and IlRl = 0.225 mV (voltmeter top is positive).

1213.7 (a) 4 A clockwise. (b) 8 V across left, 0 Vacross top, 4 V across right, 0 V across bottom.

(c) VA = 5 V, VB = 2 V, VC = 3 V, VD = 0 V.(d) 5 V, 3 V, 1 V, 3 V.1214.1 (a) I1 = I2 = 6 A, right to left.(b) I = 1.714 A in both arms, circulating clockwise;I = 0.968 A in both arms, circulating clockwise.1214.3 (a) I1 = I2 = 0. (b) I1 = I2 = 0.(c) I1 = 0.545 A, I2 = 0.15 A. (d) I = 0.346 A,circulating counterclockwise. (e) 68.6 s, 0.193 A.

131.1 Verify that problem.132.1 From best to worst: iron rods, iron, plastic(either rods or solid). Iron rods have a large magne-

tization, but a relatively large resistance.133.1 The motor is an external source of mechani-cal power, and the generator uses the electrical power.134.1 (a) 10 . (b) 100 V.134.3 (a) 92.31 A. (b) 5.94 N, 10.24 m/s2.(c) 0.01481 V. (d) 92.08 A. (e) 0.224 s.135.1 (a) 429 s. (b) 7.33 A, 0.0703 N.135.3 (a) 228 s. (b) 2.89 104 m/s. (c) 771 A.136.1 0.0403 s.136.3 88.7%137.1 (a) 80 s. (b) 0.08 N, 1.667 m/s2.(c) 69.1 m/s, 1.97 m/s.138.1 2.69

106 m/s, 2.69

105 m/s,

2.69 102 m/s, 2.69 101 m/s,138.3 Show that problem.138.5 (a) opposite source current. (b) same assource current.

141.1 Vrms = 1/

3, V = 0.141.3 (a) 0.0127 s per radian and 0.08 s per period.(b) 12.5 s1 and 78.5 rad/s. (c) 12.5 s1 is f and78.5 rad/s is .141.5 Vrms = Vm/

3; V = Vm/2.

141.7 Vrms = Vm/

3; V = 0.142.1 Decrease either L or C by a factor of 4.

142.3 (a) 2.093 109

F to 0.264 109

F.(b) 5.23 1012 J and 0.660 1012 J.142.5 16.24 pF.143.1 (a) Current starts at zero but with finiteslope, going through some damped oscillations untilit is zero at long times. (b) Charge starts at zero and


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14

with zero slope, going through some damped oscilla-tions until it is finite at long times.

143.3 Show that problem.143.5 (a) 0 = 2.49 104 rad/s,f0 = 3.97 103 Hz, Rc = 1197 . (b) For R = 5 ,we have R < Rc; very underdamped. A sketch wouldshow that the voltage initially rises quadratically intime, and then goes to a slowly damped oscillationaround the applied emf.144.1 (a) 29.4 cm2. (b) 800 turns.145.1 (a) 5.86 , 56.1. (b) energy is absorbed.14-5.3 (a) B provides power, A receives power.(b) 55.145.5 (a) 500 , 6.37 F, 90. (b) 510 mJ.145.7 (a) Scalar. (b) The y-component of therotating vector, or phasor, gives a scalar that equalsthe voltage.145.9 (a) At low f capacitors serve as open cir-cuits (finite voltage), at high f capacitors serve asclosed (or short) circuits (zero voltage). (a) At lowf inductors serve as closed (or short) circuits (zerovoltage), at high f inductors serve as open circuits(finite voltage).146.1 5 ms was a misprint. The intended time lagwas 0.5 ms. This gives Z = 40 , = 0.1885 rad =10.80, R = 39.3 , XL = 7.50 , L = 0.0199 H.146.3 XL = 1.57 , Z = 20.06 ,

= 0.0784 rad = 4.49, tlead = 0.250 ms,Im = 4.23 A.146.5 XC = 3.02 104 , R = 1.433 105 ,Z = 1.464 105 , = 11.86.146.7 Z = 160 , XL = 30.14 , R = 157.1 , = 0.1894 rad = 10.85, tlag = 6.85 105 s.146.9 = 0.1520 rad, XL = 1.839 ,L = 0.665 mH, Z = 12.14 .146.11 (a) Irms = 4.87 A for all circuit elements.(b) VR,rms = 116.8 V, VL,rms = 27.5 V.(c) PR,rms = 569 W, PL,rms = 0. (d) = 0.231 rad,tlag = 0.614 ms.146.13 (a) Inductor, with XL = 39.25 ,

L = 0.1041 H. (b) P= 148.4 W.147.1 (a) f = 62.5 kHz. (b) ZR = R = 200 ,ZL = XL = 3571 . (c) power surges have high fre-quencies, for which inductors have a high impedance.147.3 Units misprint: capacitance is in F, notH. (a) Capacitor. (b) Resistor. (c) VR,m = 2.40 V,

VC,m = 24.9 V. (d) VR,m = 17.35 V, VC,m = 18.00 V.147.5 Show that problem.

148.1 (a) Yes. (b) No. (c) Yes.148.3 (a) XL = 6.283 , XC = 0.796 ,Z = 6.79 , = 53.9. (b) Im = 3.53 A,VL,m = 22.2 , VR,m = 14.14 , VC,m = 2.81 .(c) Voltage leads current by 0.374 103 s.(d) 14.41 V, voltage lags current by 0.781 104 s.(e) 26.3 V, voltage leads current by 3.99 104 s.(f) 19.39 V, voltage leads current by 6.25 104 s.148.5 (a) 1340 Hz, Irms = 0.05 A.(b) VL,rms = 6.316 V, VR,rms = 1.20 V,VC,rms = 6.316 V. (c) Z = 191 , = 82.8,Irms = 0.00628 A. (d) VL,rms = 1.587 V,VR,rms = 0.1507 V, VC,rms = 0.397 V.(e) Rc = 252.6 . Definitely a resonance, but ratherbroad, since Q = 5.625 is fewer than the 6.26 radiansthat correspond to a full period.148.7 Show that problem.149.1 For water, a valve permits a small force tocontrol a large flow of water; for electricity, a valvepermits a small voltage to control a large flow of elec-tricity.149.3 See Figure 14.14b. The grid voltage de-termines whether or not electrons are drawn off thecathode; once off the cathode, they go to the anode.1410.1 Neither of them can dissipate energy, and

on average neither of them can store energy.1410.3 (a) P> 0 when refinery is buying power.(b) P< 0 when refinery is selling power.1410.5 (a) cos0 = 0.423. (b) Add capacitor withC = 1.957 F. (c) Z = 1215 .1411.1 (a) 6.25 turns. (b) 62.5 turns.1411.3 100 V.1411.5 (a) Ns/Np = 1/10. (b) Lower voltage sidehas higher current; thicker wire decreases the Jouleheating rate.1411.7 (a) Pgen = 4800 W.(b) P= Pgen = 4800 W. (c) Pwires = 2.56 W.(d) Pload = 4797.44 W.1411.9 See Section 14.11.3.1411.11 (a) R = 1 . (b) Rc/R = 50.(c) L = 13.37 mH, C = 21.39 F. (d) Ns/Np = 100.1412.1 Toaster.1412.3 (a) Em = 0.400 V. (b) Z = 20.8 , =54.8. (c) Im = 8.16 A. (d) F = 0.1011 N.


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1412.5 Use the geometry of Figure 14.18a, withan inductor and capacitor in series with the circuit.

(a) IR + LdI/dt + Q/C = Emsint,Em = Bmlx. (b) I = (Em/Z)sint, Z and as in(14.71) and (14.72). (c) F = (EmbBm/2Z)sin .(d) If > 0, L dominates; if < 0, C dominates.1413.1 (a) 23.9 m. (b) d = 0.001 inch is 25.4 m,so that the field will penetrate, but down by a factorof ed/ = 0.346.1413.3 Show that problem.

152.1 (a) 2011 m/s. (b) 2 min. (c) 70 min.(d) 60 flagmen.153.1 2.83 1016 N/C-s.154.1 0.01096/m.154.3 (a) dy/dx = x/(a y),d2y/dx2 = a2/(a y)3. (b) 1/a. (c) d2y/dx2 = 1/R,where R = a is the radius of curvature.155.1 120 m/s.155.3 3 N.155.5 (a) Energy flowing along the string is moreconcentrated. (b) Yes. (c) 3361 N. (d) It correspondsto hanging a mass of 342.6 kg; unlikely.155.7 Show that problem.155.9 (a) 2.048 m/s. (b) 0.36 m. (c) 0.653 m/s.(d) 0.626 m/s2.155.11 The large phases require high accuracy.

(a) 251.3274 m1

. (b) 2010.6193 s1

. (c) 320 Hz.(d)0.003125 s.(e) 8048.23 rad 0.527rad.(f) In cm,y(x, t) = 0.03 sin[251.3274x + 2010.6193t + 0.527].(g) 0.0200 cm.155.13 (a) 3.0 kg/m. (b) 0.72 N.155.15 (a) 147.3 m/s. (b) 170 cm and 85 cm.(c) 86.6 Hz and 173.3 Hz.155.17 (a) 7.48 m. (b) 18.70 m. (c) 0.01282 s.(d) 78.0 Hz. (e) 0.0358 m, 2.65 m/s, 8600 m/s2.155.19 (a) d

2ydx2 = v

2( d2y

dt2 ) andd2zdx2 = v

2( d2z

dt2 ).

(b) v =

F . (c) f1 =

v2L .

156.1 12000 x V/m.156.3 (a) (D/c) sin(qxt). (b) (D/c) sin(qyt).156.5 (a) 0.009375 m. (b) 670.2 m1.(c) 2.01 1010 s1. (d) 0.15 sin(qy t) mT.157.1 (a) 303 m. (b) 3.367 m.158.1 (a) r1. (b) r1/2.158.3 1.584 105 V/m; 0.528 mT.

159.1 (a) 4.17 108 m/s2. (b) 19,200 s.(c) 8

104 m/s. (d) 7.68 m.

159.3 (a) uE = uB = 8.95sin2(qy t) mJ/m3.(b) S = 5.37 106 sin2(qy t) y W/m2.(c) P= 0.0179 sin2(qy t) N/m2.159.5 (a) S = 2kmn2rIdIdt r.(b) P/l = 42kmn2r2I| dIdt |, flowing radially inward.(c) dP/l = 82kmn2rdrI|dIdt |.(d) d(dUB/dt)/l = 8

2kmn2rdrI|dIdt |. This equals

P/l, so dP= d(dUB/dt).1510.1 (a) 3.93 m, (b) 1.132, (c) 1.282.1510.3 (a) 5.07 1014 Hz. (b) 5.07 1014 Hz.(c) 443.7 nm. (d) refl = 23

, refr = 17.04.1510.5 (a) crownglassc = 41.14

,

diamondc = 24.41. (b) Crown glass. (c) Diamond.1511.1 2.3 W/m2, polarized vertically.1511.3 (a) 0. (b) 0.134 W/m2, at 35 to the xaxis.1511.5 2.58 W/m2, polarized vertically.1511.7 (a) 53.1. (b) 36.9.1512.1 (a) and (b) See Section 15.12.1512.3 2881 x A/m.1513.1 Show that problem.1514.1 (a) 2.4 m. (b) 2.4 m.1514.3 R/Rc 0.25. This corresponds toQ = Rc/2R 2.0 radians, or 0.32 of a full oscillation.It damps out very quickly.1515.1 (a) L = (A/2L)2L. (b) Show thatproblem.1515.3 (a) 5.09 103 m/s. (b) 1.964 103 s.1515.5 24.9 m.1515.7 Using (15.82), so vwater = 1.432103 m/s,gives 114.8 Hz. (Using vwater 4.3vair, as in exam-ple 15.7, so vwater 1.475103 m/s, gives 118.3 Hz.)

162.1 (a) 76.6 mm. (b) 287 Hz.162.3 4(d2m22)y24m22D2=m22(d2m22).162.5 (a) 0.212 m. (b) 1604 Hz.162.7 (a) one line of maxima, no lines of minima.

(b) one line of maxima, two lines of minima. (c) fivelines of maxima, six lines of minima.162.9 (a) See Fig.16.6a. (b) See Fig.16.7a. (c) SeeFig.16.8a.162.11 Use tan sin . (a) 4.08 m. (b) 12.53.163.1 Show that problem.


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16

163.3 (a) = 2 + 2 6. (b) sin2 = 1(n/3)21(1/3)2 .(c) For n = 4/3,

180 = 51.0.

164.1 See Section 16.4.164.3 See Section 16.4.164.5 At 90 (top of circle) get maximum increasein period; At 270 (bottom of circle) get maximumdecrease in period;164.7 Grimaldis experiment showing light withinthe geometrical shadow, and bright and dark fringes

just outside the geometrical shadow.164.9 Birefringent crystals, unless oriented prop-erly, will give two images.165.1 (a) Show that problem. (b) vs. (c) vp.165.3 For small amplitude disturbances the

equations are linear, which leads to an amplitude-independent sound velocity. Hence the relativeamounts of reflection and refraction are amplitude-independent.166.1 (a) 109.87 Hz. Out of tune by 0.12%. (b)Tone at (fA fE )1.(c) 7.5 s.166.3 / = v/c. Since the experiments indicatethat in the film is less than in air, we deducethat v < c.166.5 (a) The sources have a coherence time muchshorter than the measuring time of the eye. (b) Thesources have a coherence time longer than the mea-

suring time of the ear.166.7 (a) No. (b) No.166.9 (a) 489.44 nm. (b) 611.8 nm and 407.9 nm.166.11 (a) 691.3 nm, 518.5 nm, and 414.8 nm.(b) 592.6 nm and 460.9 nm.166.13 (a) 912 nm. (b) 606.5 nm, 485.2 nm, and404.3 nm.166.15 113.1 nm.166.17 Correct for the misprint of 1575, ratherthan 1515, lines at atmospheric pressure.(a) n1 = 1.000293. (b) vac = 690.2 nm.(c) d = 0.4998 m.

166.19 (a) N = 2D/. (b) 600 nm.166.21 (a) 409 nm. (b) 0.422 degrees, 8.1 mm.(c) 611.166.23 (a) 858 nm. (b) 613 nm.167.1 Show that problem.167.3 (a) 1.536 mm. (b) 0.541 mm.

167.5 Break up the illuminated part of the planeinto tiny strip-shaped sources parallel to the finite

opaque strip (or into tiny annulus-shaped sourcesconcentric with the finite opaque disk). At the cen-ter of the geometrical shadow, wavelets from all strips(or annuli) add in phase, giving an interference max-imum.167.7 366.5 nm.167.9 (a) 162.5 m. (b) 0.220, 0.660.167.11 (a) 0.0766. (b) 0.0321 cm.167.13 For m = 1 minima: (a) 184.7 m.(b) 0.428, 0.486 cm.167.15 197.1 m.167.17 (a) Show that problem. (b) Show that

problem.167.19 (a) 3.93 mm. (b) 1.5 108 radians.(c) 150 radians unresolvable.167.21 (a) 2.161012 m. (b) 3100 times the radiusof the sun. (c) 6.79 1014 m.167.23 (a) 0.671 mm. (b) 0.0419 mm. (c) 3.5 m.(d) 5.75 m.168.1 (a) 2d = (m + 12 )/nO,E. (b) O-beam has520 nm (m = 3) and 404.4 nm (m = 4); E-beam has557.1 nm (m = 3) and 433.3 nm (m = 4).168.3 (a) 63 nm. (b) 0.00451.169.1 (a) maxima out to m = 6. (b) m = 0 at0; m =

1 at

9.04; m =

2 at

18.32; m =

3

at 28.13; m = 4 at 38.94; m = 5 at 51.79;m = 6 at 70.54. (c) maxima out to m = 4.(d) 0.0212 rad = 1.213. (e) 43.9 nm.169.3 Interpret third maximum to be third order(m = 3). (a) 513.4 nm. (b) mmax = 10.(c) 6 = 36.3

.169.5 400 nm 595 nm.169.7 4030 lines.169.9 1216 slits.169.11 (a) Show that problem. (b) Show thatproblem. (c) Multiples of three.1610.1 [scattering angle, angle of incidence rela-

tive to normal]: (a) [20, 80]. (b) [160, 10].1610.3 0.16 nm.1610.5 m = 4 gives = 0.01139 nm, m = 5 gives = 0.00911 nm.1610.7 (a) 0.205 nm, 1.4671018 Hz. (b) 0.270 nm,1.111 1018 Hz.


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1610.9 2dsin = m and nV = nA/d = constant re-lates the volume and area densities and the plane sep-

aration. Higher nA means smaller d and thus larger and larger scattering angle 2.


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Chapter 12 Food for Thought questions on pp.512-513, 515, and 517

pp.512-513 See Figure 12.6.

1. Primary moves leftward, with IP counterclockwise as seen from the right.Then Bext is along , d Bext/dt is along , and Bind is along .

2. Primary moves rightward, with IP clockwise as seen from the right.Then Bext is along , d Bext/dt is along , and Bind is along .

3. Primary moves leftward, with IP clockwise as seen from the right.Then Bext is along , d Bext/dt is along , and Bind is along .

4. There is no general correlation between Bext and Bind;There is a general correlation whereby Bind opposes d Bext/dt.

p.515 See Figure 12.8(a). In all cases, the observer is to the right.

Situationd Bext

dtBind Eind, Iind Fnet Compress or Expand

pull magnet from loop Counterclockwise Expandpush loop to magnet Clockwise Compress

pull loop from magnet Counterclockwise Expandpush reversed magnet to loop Counterclockwise Compress

p.517 See Figure 12.9(a). The observer is to the right.

Situationd Bext


reverse primary Counterclockwise Compress

Table problems 12-4.11 and 12-4.13

124.11

Observerd Bext

dtBind Eind, Iind Compress or Expand

reader Counterclockwise opposite Expand

124.13 See Figure 12.32.

Observerd Bext


from above Counterclockwise Compress

elmaglightsaslowallanswersnew

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