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MasteringPhysics: Course Home

Chapter 28 Homework AssignmentDue: 11:59pm on Monday, April 23, 2012

Jrint 1 Note: You will receive no credit for late subm iss ions. To learn rrore, read your instructor's Grading Policy[Sw itch to Standard Assignment View 1

Force between Moving Chargesy

:Two point charges, with charges qland 0, are each moving with speed tI'toward the origin. At the instant shownqi is at position (0, d) and > 1 1 2 is at ( d,0). (Note that the signs of the chargesare not given because they are notneeded to determine the magnitude ofthe forces between the charges.)

A1_Il .

o f ' J } ;~~~ - d ~~~-!IiPart AWhat is the magnitude of the electric force between the two charges?Hint A.1 Which law to use

Hint not displayed

Hint A.2 Find the value of r'l

Hint not displayed

: ":

ANSWER: I F = i i : : : Ii ~

Express F in terms of q1' 'f.l, ,d, and fO

Part BWhat is the magnitude of the magnetic force on 0 due to the magnetic field caused by ql?Hint B.1 How to approach the problem

Hint not displayed

I Hint B.2 Magnitude of the magnetic fieldgphysics.com/mycUcourseHome?start=1

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Hint not displayed

Hint B.3 Find the direction of the magnetic fieldHint not displayed

Hint B.4 Computing the forceHint not displayed

Express the magnitude of the magnetic force in terms of q1, .q2, "V , d, and 1 1 0 ". . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . . . .

.

ANSW ER: i : : : : : : : : . (ql~v2. i4l) i " " " " . F = 4''l'f.2.v'2.cPCorrecti ~

Part CAssuming that the charges are moving nonrelativistically (v


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MasteringPhysics: Course HomeIDENTIFY the relevant concepts:The Biot-Savart law allows you to calculate the magnetic field due to a current-carrying wire of anyshape. The idea is to calculate the field due to a representative current element in the wire andthen combine the contributions from all such elements to find the total field.SET UP the problem using the following steps:Make a diagram showing a representative current element and the point P at which the fieldis to be determined (the field point).Draw the current element e n , being careful that it points in the direction of the current.Draw the unit vector f. Note that it is always directed from the current element (the sourcepoint) to the field point P.Identify the target variables. Usually, they will be the magnitude and direction of the magneticfield B .EXECUTE th e s olu tio n as fo l/o VIIS :Use the equations

where r is the distance from the current element to point P and ' is the angle between J iand f, to express the magnetic field at P from the representative current element.Add up all the dB's to find the total field at point P, recalling that d B ' is a vector. To find thetotal jj, you may have to set up a coordinate system and then integrate to find eachcomponent.Sometimes you can use the symmetry of the situation to prove that one component of B 'must vanish. Always be alert for ways to use symmetry to simplify the problem.Look for ways to use the principle of superposition of magnetic fields. If you encounter aconductor of a complex shape that can be represented as a combination of these simpleshapes, you can use superposition to find the field of the complex shape.EVALUATE your answer:Often your answer will be a mathematical expression for jj as a function of the position of the fieldpoint. Check the answer byexamining its behavior at as many limits as you can.

IDENTIFY the relevant conceptsIn this problem, you are asked to find the magnetic field contribution from a given currentelement. This may be calculated straightforwardly using the Biot-Savart law, as outlined inthe strategy above. In other problems that ask for the total magnetic field, you will need tointegrate over all such current elements to find the total field at point P.SET UP the problem using the following steps

Part AIn the figure, draw the vector representing the current element dj and the unit vector f. Thevectors should be in the correct positions relative to the current element and with the correctdirections, but their magnitudes are not important.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,ANSWER: i i

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ViewCorrect

Part BUsing the right-hand rule, determine the direction of the magnetic field contribution d B atpoint P due to the current element at point A. Keep in mind that we are assuming the ycoordinate of point P to be positive (i.e., f always points above the wire).

Hint B.1 The right-hand rule for the magnetic field due to a current elementHint not displayed

ANSWER: o d B is parallel to d i .o d B points into the screen.o d B is opposite to J i .@ d B points out of the screen.Correct

For any source point in the wire, the magnetic field contribution will always point out ofthe screen at any point in the plane of the screen that is above the wire (i.e., at anypoint for which f points above the wire). The situation is reversed if you consider pointsbelow the wire, that is, if y is negative. In that case, the direction of the magnetic fieldcontribution is always into the screen.Since you now know the direction of d iJ at any point Pwith positive ycoordinate, the onlything left to find is the magnitude of d B ' , which you will find in the next part.

i EXECUTE the solution as followsPart CWhat is the magnitude dB of the magnetic field contribution at point P, with coordinates(z, y ) , due to the current element J l at point A, with coordinates (a,O)?

gphysics.com/mycUcourseHome?start=1I I

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MasteringPhysics: Course HomeHint C.1 How to approach the problem

Hint not displayed

Hint C.2 Find m n ,Hint not displayed

Hint C.3 Find the distance between the two pointsHint not displayed

Express your answer in terms of some or all of the varlables je, y, 1(1, 1, d l (themagnitude of d h , and 1 1 0 ' .ANSWER:

i EVALUATE your answerPart D

In Part C, you found that the magnetic field contribution at point P, with coordinates (z~ ' I I ) ' hasmagnitude

What is the behavior of dlJ in the limiting cases?.'];"s held constant, while 1 ,1 goes to infinity.1 ,1 is held constant, while.'];"goes to infinity.ANSWER: The magnitude of the field contribution, dB, goes to zero in@ both cases.

The magnitude of the field contribution, dB, goes to infinity ino case 1 and zero in case 2.The magnitude of the field contribution, dB, goes to zero ino case 1 and infinity in case 2.The magnitude of the field contribution, dB, goes to infinity in

o both cases.Correct



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MasteringPhysics: Course HomeNotice that in case 1, as y becomes arbitrarily large, the denominator (z - a f J . +,"2becomes more and more like y ' , 2 , because the ( : 7 ) - a ) : : l term becomes negligible incomparison. Thus, as y goes to infinity, the fraction Y / ( ( S i - ( 1 , ) ' " +y2 ) s< ,n behaves likeY / f : u ' 2 ) ' J J 2 = ]/,"2, and so it goes to zero. Similarly, in case 2, as r n ; " becomes arbitrarilylarge, the denominator acts like (z - a ) : 2 :R:i z~, because the y and a values becomenegligible in comparison. Thus, as : K : " goes to infinity, the fraction y/((x _ ( 1 , ) 2 + y 2 ) S / : : lbehaves like Y / ' . l P and so goes to zero.This is what you should expect on physical grounds. The magnetic field contribution from aninfinitesimal current element should go to zero as you move away from the current element,regardless of what direction you move in. Ifyou had found an answer in Part C that did not goto zero for large r n ; " or y , then you would know, based on your physical intuition, that you hadmade a mistake somewhere.

Magnetic Field due to Semicircular WiresAloop of wire is in the shape of two concentric semicircles as shown.The inner circle has radius a; the outercircle has radius b.Acurrent 1 flowsclockwise through the outer wire andcounterclockwise through the inner wire.

Part AWhat is the magnitude, B, of the magnetic field at the center of the semicircles?Hint A.1 What physical principle to use

Hint not displayed

Hint A.2 Compute the field due to the inner semicircleHint not displayed

Hint A.3 Direction of the field due to the inner semicircleHint not displayed



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MasteringPhysics: Course HomeHint A.4 Compute the field due to the straight wire segments

Hint not displayed

Express B in terms of any or all of the following: I. Q;. b. and /lo,'

ANSWER: r : - - ~ - - ~ 5 ~ = ~ ) - - - - - - - - 1i J

To see whether .B "'" l/a and .B "'" ljb makes sense, think of the scaling of differentquantities. The size of the current element scales as the radius, whereas the power ofr in the denominator is 2 (and equals the radius also, in this case). So over all, youwould expect the magnetic field to scale as 1/radius. Note that such an argumentworks only because the field due to each point is in the same direction, so you aredoing a much simpler integral.

Part BWhat is the direction of the magnetic field at the center of the semicircles?ANSWER: a into the screen

@ out of the screenCorrect

Conceptual Question 28.01

Part AA vertlca: wire carries a current straight down. To the east of this wire, the magnetic fieldpointsANSWER: a toward the east.a downward.a toward the west.a toward the north.

@ toward the south.Correct

Current SheetConsider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current1uns in



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MasteringPhysics: Course Homethe -y direction through each wire. There are N/a wires per unit length in the x direction.

Part A

Write an expression for . 8 ' (d), the magnetic field a distance d above the xy plane of the sheet.Use P o ' for the permeability of free space.

Hint A.1 How to approach the problemHint not displayed

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A N S W E R : i igphysics.com/mycUcourseHome?start=1

Hint A.2 Find l~.

Hint not displayed

Hint A.3 Determine the direction of the magnetic fieldHint not displayed

Hint A.4 Magnitude of the magnetic fieldHint not displayed

Hint A.5 Evaluate f B . a tHint not displayed

Express the magnetic field as a vector in terms of any or all of the following: d. 1.N. 6. P o , . and the unit vectors,i. y. and/or , .

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MasteringPhysics: Course Home= IN- Prj" -. _-~2a

Correcti ~

This equation is analogous to E = (1'/ 'J6 0 ' on either side of a infinitely charged sheet. Thecorrespondence seems more obvious if you set the current per unit length .NI!a =1.Then the magnetic field you just calculated is

U IJ, .\ 8-_F"'._",- 2 .

The electric field, though, points along the perpendicular to the surface.

Do you see why you had to pick the rean loop you used? That is, whywould any other loopnot have worked.Did you notice that byusing Ampere's law you could find the field byusing a much simplerintegral than Biot-Savart's law? The drawback is that you may not always be able to find aconvenient loop in situations where the current distribution is more complicated.

Exercise 28.4An alpha particle (charge +2e) and an electron move in opposite directions from the samepoint, each with the speed of 2.50x 105 m / ; a (the figure ).

E J .. I..ectron 'Alpha particle

Part AFind the magnitude of the total magnetic field these charges produce at point P, which is1.85 om from each of them.

: ':

ANSWER: I~____. ; ~ : ~ _ _ _ _ _ _ _ _ jPart BFind the direction of this magnetic field.



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MasteringPhysics: Course HomeANSWER: o into the page

@ out of the page

Correct

Exercise 28.10A short current element , r t f " " " " {I).SOOmm).i carries a current of 9.00 A in the same directionas oJ. Point P is located at r= (-n.,130m) i + (ll,390m)k'.Part AFind the magnetic field at p produced by this current element.

Enter the ~, '1), and z components of the magnetic field separated by commas .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ,

ANSWE R: !::::::.. !::::::,'. , l B , , ...IID".,llD! = 0.310,0,0.580 'nT,tji 'm ' '~I:I ' ,tjiLj.l!: Correcti ~

Exercise 28.26A rectangular loop with dimensions 4.20 em by 9.50 em carries current 1. The current in theloop produces a magnetic field at the center of the loop that has magnitude 4.30 x 1 0-5 T anddirection away from you as you view the plane of the loop.Part AWhat is the magnitude of the current in the loop?Express your answer with the appropriate units.

: -:

ANSWER: ! ~ _ ~_: ; ; ,Part BWhat is the direction of the current in the loop?ANSWER: @ clockwiseo counterclockwise

Correct

Conceptual Question 28.05gphysics.com/mycUcourseHome?start=1 10/17


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Part AThe figure shows two long wires carrying equal currents 11 and 1 2 flowing in oppositedirections. Which of the arrows labeled A through D correctly represents the direction of themagnetic field due to the wires at a point located at an equal distance d from each wire?

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ANSWER: OA@ BOco Do The magnetic field is zero at that point.Correct

Force between an Infinitely Long Wire and a Square LoopA square loop of wire with side length Q; carries a current h. The center of the loop is locateda distance d from an infinite wire carrying a current ' ' 2 . The infinite wire and loop are in thesame plane; two sides of the square loop are parallel to the wire and two are perpendicular asshown.



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wire loop

Part AWhat is the magnitude, F, of the net force on the loop?Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Determine the direction of force

Hint not displayed

Hint A.3 Determine the magnitude of forceHint not displayed

Hint A.4 Find the force on the section of the loop closest to the wire

Hint not displayed

Hint A.5 Find the magnetic field due to the wire

Hint not displayed

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ANSWER: ! ~ _ __ _~ ~ ~ ~ ~ ~ ~ ; ~ ~ ~ ~ ~ _ ~ iExpress the force in terms of 1111'2'a;. ,d. and P c , .Part BThe magnetic moment 'n 1 of a current loop is defined as the vector whose magnitude equals



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MasteringPhysics: Course Homethe area of the loop times the magnitude of the current flowing in it ( \ I ' J ~ = IA), and whosedirection is perpendicular to the plane in which the current flows. Find the magnitude, F, ofthe force on the loop from Part A in terms of the magnitude of its magnetic moment.Express F in terms of fi:t, 1'2' a;, > I i , and 1 1 - 0 ' .

: ":

ANSWER: ! F = 2 ~ m ~ ) !i JThe direction of the net force would be reversed if the direction of the current in eitherthe wire or the loop were reversed. The general result is that "like currents" (i.e.,currents in the same direction) attract each other (or, more correctly, cause the wiresto attract each other), whereas oppositely directed currents repel. Here, since the likecurrents were closer to each other than the unlike ones, the net force was attractive.The corresponding situation for an electric dipole is shown in the figure below.

++++

'++'++++-::.....

,---_ ,-. . . . . . . . . _ _

Exercise 28.35The magnetic field around the head has been measured to be approximately 3.00x10-8 gau.ss.Although the currents that cause this field are quite complicated, we can get a rough estimateof their size by modeling them as a single circular current loop 16.0 em (the width of a typicalhead) in diameter.Part AWhat is the current needed to produce such a field at the center of the loop?

: ":

ANSWER : I ~_~_~:;!~~;~~.~.~: .~: :~~~~~.~ ,Exercise 28.50

A toroidal solenoid has an inner radius of 12.0 am and an outer radius of 15.0 em- It carries agphysics.com/mycUcourseHome?start=1 13/17


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MasteringPhysics: Course Homecurrent of 1.50 A.Part AHow many equally spaced turns must it have so that it will produce a magnetic field of 3.75mT at points within the coils 14.0 em from its center?

Magnetic Field of a Current-Carrying Wire

j

Find the magnetic field a distance r from the center of a long wire that has radius a andcarries a uniform current per unit area j in the positive z direction.

Part AFirst find the magnetic field, . B O O ' ~ ( r : l , outside the wire (i.e., when the distance r is greaterthan a).

Hint A.1 Ampere's law with current densityHint not displayed

Hint A.2 Find the direction of the fieldHint not displayed

Hint A.3 Find the left-hand side of Ampere's lawHint not displayed

Hint A.4 Find the right-hand side of Ampere's lawHint not displayed



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Express , B O O ' ~ ( r ) in terms of the given parameters, the permeability constant 1 1 0 " thevariables Q;, j(the magnitude of 1 ) , ' 1 " ' , (J , and z, and the corresponding unit vectorso f , 0 , and k '. You may not need all these in your answer.ANSWER: l 4 J J o ; ' 2 , : " ,

. . . . = ' =f}B O O ' t ( r : l = 2 , 3 ' "Correct

Part BNowfind the magnetic field , B ; I I ( f ! ' J inside the wire (i.e., whenthe distance r is less than a:) .

-Hint B.1 Establish the relationiship to Part A

Hint not displayed

Hint B.2 Integrate over the Amperean loop

Hint not displayed

Express , B I I I ( r ) in terms of the given parameters, the permeability constant / 1 0 " thedistance ' I ' " from the center of the wire, and the unit vectors;', 0 , and k '. You maynot need all these in your answer.

: ":

ANSWER: i B~(~ = ~~t ii ~

Problem 28.60gphysics.com/mycUcourseHome?start=1 15/17


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MasteringPhysics: Course HomeAt a particular instant, charge >til = 4.40x 10-6 C is at the point (0, 0.250 01, 0) and haswlocity ih """(9 . . .20 X l,(fi m/a. ) E . Charge > 1 1 2 = -2.20x10-6 C is at the point (0.150 01, 0, 0)and has wlocity ~ =-5.30 x loS mJa) 3 .Part AAt this instant, what is the magnetic force that qi exerts on . q " 2 ?Enter the x, y, and z components of the force separated by commas.

: ":

ANSWER: I ~ : : _ ~ : : _ ~ . : ~ ~ ~ ; t ~ _ _ _ _ _ _ _ _ jConceptual Question 28.08

Part ATwowry long parallel wires are a distance d apart and carry equal currents in oppositedirections. The locations where the net magnetic field due to these currents is equal to zeroare . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .ANSWER: a distance dl.,,/2 to the left of the left wire and also a distance

0 clJV2 to the right of the right wire.

0 a distance d to the left of the left wire and also a distance d tothe right of the right wire.0 midway between the wires.0 a distance dl2 to the left of the left wire and also a distance d12 to the right of the right wire.@ The net field is not zero anywhere.

Correct. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Problem 28.70A pair of long, rigid metal rods, each of length L, lie parallel to each other on a perfectlysmooth table. Their ends are connected by identical, wry light conducting springs of forceconstant keach(the figure) and negligible unstretched length. Ifa current 1runs through thiscircuit, the springs will stretch.



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I>

IPart AAt what separation will the rods remain at rest? Assume that k' is large enough so that theseparation of the rods will be much less than L.Express your answer in terms of the variables 1, L, ,""and appropriate constants (1 1 0 " and 'lr).ANSWER:

= " [ii;;L.[~ V 4iiiiCorrect

Score Summary:Your score on this assignment is 87.4%.You received 41.1 out of a possible total of 47 points.

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