magnetic fields of charged particles in motion

CONCEPTS

28.1 Source of the magnetic field28.2 Current loops and spin

magnetism28.3 Magnetic moment and torque28.4 Ampèrian paths

QUANTITATIVE TOOLS

28.5 Ampère's law28.6 Solenoids and toroids28.7 Magnetic fields due to

currents28.8 Magnetic field of a moving

charged particle

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C H A P T E R 2 8

Magnetic fields of chargedparticles in motion

2 CHAPTER 28 Magnetic fields of charged particles in motion Concepts

I n this chapter we investigate further the relation-ship between the motion of charged particles andthe occurrence of magnetic fields. As we shall see,

allmagnetism is due to charged particles in motion —whether moving along a straight line or spinning aboutan axis. It takes a moving or spinning charged particleto create a magnetic field, and it a takes another moving or spinning charged particle to “feel” that magnetic field. We shall also discuss various methodsfor creating magnetic fields, which have wide-rangingapplications in electromechanical machines and instru-ments.

28.1 Sources of magnetic fieldsAs we saw in Chapter 27, magnetic interactions takeplace between magnets, current-carrying wires, andmoving charged particles. Figure 28.1 summarizes theinteractions we have encountered so far. Figures28.1a–c show the interactions between magnets andcurrent-carrying wires. The sideways interaction between a magnet and a current-carrying wire(Figure28.1b) is unlike any other interaction we have encoun-tered. The forces between the wire and the magnetare not central — they do not point directly from one

object to the other. As we saw in Section 27.7, the magnetic force acting on a current-carrying wire isthe sum of the magnetic forces exerted on many indi-vidual moving charge carriers. Similarly the magneticfield due to a current-carrying wire is the sum of themagnetic fields of many individual moving charge car-riers. Figures 28.1d and 28.1e illustrate the magnetic in-teractions of moving charged particles. Note that fortwo charged particles moving parallel to each other(Figure 28.1e), there is, in addition to an attractive mag-netic force, a (much larger) repulsive electric force.It is important to note that the magnetic interaction

depends on the state of motion of the charged parti-cles. No magnetic interaction occurs between a barmagnet and a stationary charged particle (Figure 28.1f )or between two stationary charged particles (Fig-ure 28.1g). These observations suggest that the motionof charged particles might be the origin of allmagnet-ism. There are two problems with this assumption, how-ever. First, the magnetic field of a wire carrying aconstant current looks very different from that of a barmagnet (Compare Figures 27.13 and 27.19). Second,there is no obvious motion of charged particles in apiece of magnetic material. Figure 28.2a shows the magnetic field line pattern

(a)

1

1 2

2

(b) (c)

(d) (e)

(f) (g)

II I

S N S N S N

F→1B2F

→2B1 F

→1B2F

→2B1

F→1B2F

→2B1

q2q1

q2q1

q

q

F→2E1 F

→2E1

F→b

Bw

F→b

Bq

F→w

Bb

S N

F→q

Bb

S N

F→2E1 F

→2E1

v→ v

→v→

+ + +

+ + +

Figure 28.1 Magnetic interactions between (a)two bar magnets 1 and 2, (b) a bar magnet (m)and a current-carrying wire (w), (c) two current-carrying wires 1 and 2, (d) a bar magnet (m) anda moving positively charged particle (q), (e) twopositively charged particles 1 and 2 moving par-allel to each other. ( f ) There is no magnetic in-teraction between a bar magnet and a stationarycharged particle.(g) Two stationary charged par-ticles 1 and 2 interact only electrostatically.

of a straight wire carrying a constant current. The linesform circles centered on the wire, circles that reflectthe cylindrical symmetry of the wire (Section 24.4). Thehorizontal distance between adjacent circles is smallernear the wire, where the magnitude of the field is larg-er. A single moving charged particle does not have

cylindrical symmetry because, unlike an infinitely longstraight wire, moving the particle up or down along itsline of motion changes the physical situation. Becauseof the circular symmetry of the situation, the field stillforms circles around the line of motion, but the mag-nitude of the field at a fixed distance from the parti-cle’s line of motion decreases if one moves away fromthe particle (Figure 28.2b). It is not at all obvious howthe field pattern shown in Figure 28.2b could give riseto the magnetic field of a bar magnet — there are cer-tainly no poles in the magnetic field of the movingcharged particle.

28.1 Make a sketch showing the directions of themagnetic forces exerted on each other by (a) anelectron moving in the same direction as the cur-rent through a wire, (b) a moving charged parti-cle and a stationary charged particle, (c) twocurrent- carrying wires at right angle to each otheras illustrated in Figure 28.3. (Hint: Determine theforce exerted at points P1 through P5).

28.2 Current loops and spinmagnetismThe circular pattern of magnetic field lines around awire carrying a constant current suggests a method forgenerating a strong magnetic field: if a wire carrying aconstant current is bent into a loop as shown in Figure28.4, all the magnetic field lines inside the loop point inthe same direction, reinforcing one another.*

28.2 Current loops 3

(a) (b)

Iline ofmotion

BB

+

v→

Figure 28.2 Magnetic field patterns of (a) a wire carrying aconstant current and (b) a moving charged particle.

* For now, we’ll ignore how to make charge carriers flow throughsuch a loop. In Section 28.7 we’ll discuss physical arrangementsthat accomplish the situation illustrated in Figure 28.4.

P2

P1

P3

P5

P4

I

I

Figure 28.3 What forces do two crossed current-carryingwires oriented at right angles to each other exert on eachother? (Checkpoint 28.1c).

B→

I

Figure 28.4 When a wire carrying a constant current is bentinto a loop, the magnetic fields from all parts of the loop re-inforce one another in the center of the loop.

What does the magnetic field of such a current-car-rying loop, called a current loop for short, look like?To answer this question, we treat the current loop as acollection of small segments of a current-carrying wireand determine the direction of the magnetic field atvarious points around the loop. As we did in Chapter27, we shall assume all currents to be constant in the re-


B1

B2

B2

B1

B1 = B2

B2

B1

B1B2

B

B

B

B

(a)

12 12

(b)

(c)

12

(d)

(e)

12

12

(f)

12

magneticfield line

of segment 1central

axismagneticfield line

of segment 2A

I

A

A

C

G

D

II

II

I

→

→

→

→ →

→ →

→

→ →

→

→

→

→

Figure 28.5 Mapping the magnetic field of a current loop.The magnetic field contributions from (a) segment 1 and (b)segment 2 at A (c) add up to a vertical field. Magnetic fieldsat (d) point C at the center of the ring, (e) point D below thering, and (f) point G to the right of the ring. Note that in allcases the magnetic field of each segment is perpendicular tothe line connecting that segment to the point at which we aredetermining the field.

mainder of this chapter.We begin by considering the magnetic field due to a

small segment of the current loop at a point on the cen-tral axis that passes perpendicularly through the face ofthe loop (point A in Figure 28.5). Segment 1 carries acurrent that points into the page. The magnetic fieldlines of this segment are concentric circles centered onthe segment. Using the right-hand current rule, we findthat the magnetic field curls clockwise in the plane ofthe drawing, and so the magnetic field due to segment1 at A points up and to the right. Figure 28.5b shows amagnetic field line through A due to segment 2. Be-cause the current through segment 2 points out of thepage, this field line curls counterclockwise, and so themagnetic field due to segment 2 at A points up and tothe left. Figure 28.5c shows the contributions from segments

1 and 2 together; because their horizontal componentscancel, the vector sum of B1 and B2 points verticallyup. The same arguments can be applied to any otherpair of segments lying on opposite sides of the currentloop. Therefore the magnetic field due to the entirecurrent loop points vertically up at A.Now consider point C at the center of the current

loop. As illustrated in Figure 28.5d, the magnetic fieldsdue to segments 1 and 2 point straight up there, too.The same is true for all other segments, and so the fieldat point C also points straight up. We can repeat theprocedure for point D below the current loop, and asshown in Figure 28.5e, the magnetic field there alsopoints up. To find the direction of the magnetic field at a point

outside the current loop, consider point G in Figure 28.5f.Using the right-hand current rule, you can verify thatthe field due to segment 1 points vertically down and thefield due to segment 2 points vertically up. Because G iscloser to 1 than it is to 2, the magnetic field due to 1 isstronger, and so the sum of the two fields points down.

28.2 What is the direction of the magnetic field ata point vertically (a) above and (b) below segment1 in Figure 28.5?

The complete magnetic field pattern of the currentloop, obtained by determining the magnetic field formany more points, is shown in Figure 28.6a. Close to thewire, the field lines are circular, but as you move fartheraway from the wire, the circles get squashed inside theloop and stretched outside due to the contributions ofother parts of the loop to the magnetic field.

As you may have noticed, the field line pattern of acurrent loop resembles that of a bar magnet (Figure28.6b). Indeed, if you shrink the size of both the currentloop and the bar magnet, their magnetic field patternsbecome identical (Figure 28.6c). The magnetic field pat-tern in Figure 28.6c is that of an infinitesimally small mag-netic dipole. Given that a current loop produces a magnetic field

similar to that of a bar magnet, is the magnetic field of abar magnet then perhaps due to tiny current loops in-side the magnet? More precisely, are elementary mag-nets (Section 27.1) simply tiny current loops? The connection between current loops and elemen-

tary magnets becomes clearer once we realize that acurrent loop is nothing but an amount of charge thatrevolves around an axis. Consider, for example, a posi-tively charged ring spinning around a vertical axisthrough its center (Figure 28.7). The spinning charged

particles cause a current moving in a circle exactly likethe current through the circular loop in Figure 28.5. Ifwe let the radius of the ring approach zero, the ring be-comes a spinning charged particle, and its magnetic fieldpattern approaches that of the magnetic dipole illus-trated in Figure 28.6c.This surprising result tells us that

A spinning charged particle has a magnetic fieldidentical to that of an infinitesimally small mag-netic dipole.

Experiments show that most elementary particles,such as electrons and protons, possess an intrinsic angular momentum — as if they permanently spinaround — and such spinning motion indeed would pro-duce a magnetic field of the form shown in Figure 28.6c.The combined intrinsic angular momentum of theseelementary particles inside atoms, then, causes certainatoms to have a magnetic field, causing them to be theelementary magnets we discussed earlier. The reasonwe cannot separate north and south magnetic poles istherefore a direct consequence of the fact that the magnetic field of a particle with intrinsic angular mo-mentum is that of an infinitesimally small magnetic di-pole.

28.3 Suppose a negatively charged ring is placeddirectly above the positively charged ring in Fig-ure 28.7. If both rings spin in the same direction,is the magnetic interaction between them attrac-tive or repulsive?

28.3 Magnetic moment and torqueTo specify the orientation of a magnetic dipole we introduce the magnetic dipole moment. This vector,represented by the Greek letter (mu), is defined to

28.3 Magnetic moment and torque 5

(a) (b)

(c)

S

N

I

Figure 28.6 Magnetic field patterns due to (a) a current loop,(b) a bar magnet, and (c) an infinitesimally small magnetic dipole.

(a) (b)

N

S

m

m

→

→

Comment from Irene (1/5/2010): In reading Summ28 pgh about magnetic dipole moment def, I applied r/h/r to prin �g 28.8a. However, even tho I saw label "electric current" in that �g, I did NOT notice thick black current arrow. Instead, I assumed label "el current" went with all the thinner black curved lines because label is superimposed on a few of these lines. Of course, I soon saw my mistake &, probably more to the point, stu reading Sec 28.3 won't be coming to it totally, totally cold the way I am, not having working on the book in weeks, but still . . . . . wd be good to have more labels. Also, have you and Margot considered di�erent colors for B lines and I lines? Isn't that pretty standard in books these days: black for current but green for mag �eld symbols, including the �eld line & those symbols � and � for �elds into/out of page

I

Figure 28.8 The magnetic dipole moment points (a) alongthe central perpendicular axis of a current loop and (b) alongthe long axis of a bar magnet. In both cases, the direction of

is the same as the direction of the magnetic field. Æ

Æ

spin

Figure 28.7 The magnetic field of a charged spinning ring isidentical to that of a current loop.

point, like a compass needle, along the direction of themagnetic field through the center of the dipole (Fig-ure 28.8). For a bar magnet the magnetic dipole mo-ment points from the south pole to the north pole. Tofind the direction of for a current loop, you can usea right-hand rule: when you curl the fingers of yourright hand along the direction of the current, yourthumb points in the direction of .We now have three right-hand rules in magnetism,

illustrated in Figure 28.9 and summarized in Table 28.1together with the right-hand vector product rule. Whathelps distinguish these right-hand rules from one another is looking at which quantity curls: for a current-carrying wire, it is the magnetic field that curls, where-as for a current loop, it is the current that curls. Theonly additional thing you need to remember is that inthe case of the magnetic force exerted on a current-carrying wire, the fingers are associated with the curl

Æ

Æ

from the direction of the current to the direction of themagnetic field. Note, also, that each rule can be usedin the reverse direction indicated in Figure 28.9. Theright-hand current rule, for example, can be used tofind the direction of the current, given the direction inwhich the magnetic fields lines curl.

28.4 Does the direction of the electric field alongthe axis inside an electric dipole coincide with the direction of the electric dipole moment?


current

B

B B

(a) Right-hand current rule (b) Right-hand force rule (c) Right-hand dipole rule

m

→

→

→

right hand thumbpoints in directionof magnetic force

curl fingers from

current to B→

curl fingers in direction of

current

point fingers ofright hand in

direction of current

1

1

12curl of fingers gives

direction of B→

2

23

right hand thumbpoints in direction

of magnetic dipole moment

point right hand thumb in direction

of current

F→B

w

I

I

I

Figure 28.9 Right-hand rules in magnetism.

F→B

F→B

B→

→

need to add anno to indicate

direction of rotation

m

I

Figure 28.10 Magnetic forces exerted on a current loop withits magnetic dipole moment oriented perpendicular to an ex-ternal magnetic field.

TABLE 28.1Right-hand rules

Right-hand thumb fingersrule points along curl

vector product from to current rule current along B-fieldforce rule magnetic force from current to B-fielddipole rule (parallel to ) along current loop

CÆ

� AÆ

� BÆ

BÆ

Æ

BÆ

AÆ

What happens when a current loop is placed in amagnetic field? To find out, consider a square currentloop of wire placed in a uniform magnetic field withits magnetic dipole moment perpendicular to themagnetic field (Figure 28.10). The current loop expe-riences magnetic forces on the top and bottom sidesbut not on the vertical sides because they are paral-lel to the direction of the magnetic field (Section 28.3).Using the right-hand force rule of Figure 28.9b, wesee that the magnetic forces exerted on the top andbottom sides cause a torque that tends to rotate theloop as indicated in Figure 28.10. When the current loop is oriented with its magnet-

ic dipole moment parallel to the magnetic field, allfour sides experience a magnetic force (Figure 28.11),However, because all the four forces lie in the plane ofthe loop, none of them causes any torque. Because themagnitudes of the four forces are the same, their vec-tor sum is zero and the loop is not accelerated side-ways.

28.5 As the current loop in Figure 28.10 rotatesover the first 90°, do the magnitudes of the (a)magnetic forces exerted on the horizontal sidesand (b) the torque caused by these forces in-crease, decrease, or stay the same? (c) As the looprotates, do the two vertical sides experience anyforce, and, if so, do these forces cause any torque?(d) What happens to the torque as the loop ro-tates beyond 90°?

Summarizing the results of Checkpoint 28.5,

A current loop placed in a magnetic field tends to

rotate such that the magnetic dipole moment ofthe loop becomes aligned with the magnetic field.

This alignment is completely analogous to the align-ment of the electric dipole moment in the direction ofan external electric field, which we studied in Section23.8 (see Figure 23.32).

28.6 Suppose the square current loop in Figure28.10 is replaced by a circular loop with a diameterequal to the width of the square loop and with thesame current. Does the circular loop experience atorque? If not, why not? If so, how does this torquecompare with that on the square loop?

■ Example 28.1: Current loop torque

When placed between the poles of a horseshoe magnetas shown in Figure 28.12, does a rectangular currentloop experience a torque? If so, in which direction doesthe loop rotate?

Getting started: I know that a current loop placedin a magnetic field tends to rotate such that the magnet-ic dipole moment of the loop becomes aligned with themagnetic field. If it tends to rotate, then the loop ex-periences a torque.

Devise plan: The simplest way to answer this ques-tion is to look at the directions of the magnetic fieldand the magnetic dipole moment . By definition,

the magnetic field between the poles of the magnetpoints from the north pole to the south pole. To deter-mine the direction of , I can use the right-hand di-pole rule.

uÆ

uÆ

BÆ

2

1

F→B

F→B

F→B

F→B

B

I

→

m→

Figure 28.11 Magnetic forces exerted on a square currentloop with its magnetic dipole moment oriented parallel toan external magnetic field.

28.3 Magnetic moment and torque 7

NS

I

Figure 28.12 Example 28.1.

Execute plan: I begin by sketching the loop and in-dicating the direction of the magnetic field (Figure28.13a). To find the direction of , I curl the fingers ofmy right hand along the direction of the currentthrough the loop. My thumb shows that pointsstraight up. The current loop rotates in such a way as toalign with . The loop therefore experiences atorque which rotates the loop in the direction shown bythe curved arrow in my sketch (Figure 28.13a). ✔

Evaluate result: I can verify my answer by deter-mining the force exerted by the magnetic field on eachside of the loop and see if these forces cause a torque.The front and rear sides experience no force becausethe current through them is either parallel or antipar-allel to . To find the direction of the force exertedon the left side of the loop, I point my right-hand fin-

Æ

3

BÆ

4

BÆ

Æ

Æ

gers along the direction of the current through thatside and curl them toward the direction of the magnet-ic field. My upward-pointing thumb indicates that the


NS

NS

I I

currentcurl from

current to Bcurl from

current to B

pencil sketch for loops and vectors, but keep rendered hands

m

F→B

F→B

→

B→

B→

(a) (b)

Figure 28.13

The torque caused by the forces exerted on a currentloop in a magnetic field is the basic operating mech-anism of an electric motor. A problem with thearrangement shown in Figure 28.12, however, is thatonce the magnetic dipole moment is aligned with themagnetic field, the torque disappears. Also if the cur-rent loop overshoots this equilibrium position, thetorque reverses direction (Checkpoint 28.5d). Themost common way to overcome this problem is witha commutator— an arrangement of two curved platesthat reverses the current through the current loopeach half turn. The result is that the current loopkeeps rotating. The basic operation is illustrated in Figure 28.14.

Each half of the commutator is connected to one terminal of a battery and is in contact with one end ofthe current loop. In the position illustrated in Figure28.14a, the blue end is in contact with the positive com-mutator (that is, the half of the commutator connect-

ed to the positive battery terminal). Consequently, thecurrent direction is counterclockwise through the cur-rent loop, and the magnetic dipole moment pointsup and to the left. The vertical magnetic field there-fore causes a torque that turns the loop clockwise toalign with . Once the loop has rotated to the position shown in

Figure 28.14b, is aligned with but the contactbetween the ends of the current loop and the com-mutator is broken which means there is no current inthe loop. The loop overshoots this equilibrium posi-tion, but as soon as it does so, the current directionreverses because now the red end of the loop is incontact with the positive commutator (Figure 28.14c).The current direction is now clockwise, and so isreversed. Consequently, the current loop continuesto rotate clockwise, as illustrated in Figures 28.15c–e.After half a revolution, we reach the initial situationagain and the sequence repeats.

BÆ

Æ

Æ

BÆ

Æ

Æ

Electric motors

m

m

m

mno current

add direction of rotation?

commutator

I

II

I+

–

+

–

+

–

+

–

+

–

(a) (b) (c) (d) (e)

B BBBB→ → → → →

→

→

→

→

= 0m→→

Figure 28.14 Operating principle of an electric motor. At instant (b), the current direction through the current loop is reversed.

28.4 Ampèrian path 9

R E • dl = 0

R E • dA =

R B • dl ≠ 0

closedsurface

closedpath 1

closedpath 2

closedsurface

oeqenc R B • dA = 0

closedpath 1

closedpath 2

(a) Surface integral of electric field (Gauss’s Law)

q

q

(b) Surface integral of magnetic field (Gauss’s Law for magnetism)

(c) Line integral of electric field (d) Line integral of magnetic field

B E

E

B

→ → → →

→ →→ →

I

I

Figure 28.15 Surface and line integrals of electric and magnetic fields

force exerted on the left side is upward. Applying theright-hand force rule to the right side of the loop tellsme that the magnetic force exerted on that side is di-rected downward (Figure 28.13b). The magnetic forcesexerted on the left and right sides thus cause a torquethat makes the current loop rotate in the same direc-tion I determined earlier.

The alignment of the magnetic dipole moment of cur-rent loops in magnetic fields is responsible for the op-eration of any device involving an electric motor (seebox on p. ••).

28.7 Describe the motion of the current loop inFigure 28.12 if the magnitude of the magneticfield between the poles of the magnet is larger onthe left than it is on the right.

28.4 Ampèrian pathsIn electrostatics, Gauss’ law provides a powerful toolfor determining the electric field of a charge distribu-tion: the electric flux through a closed surface is determined by the amount of charge enclosed by thatsurface (Section 24.3). The basic reason for Gauss’ lawis illustrated in Figure 28.15a: electric field lines originateor terminate on charged particles, and the number of“field line piercings” through a closed surface is proportional to the amount of charge enclosed by thatsurface. As you saw in Checkpoint 27.6, however,Gauss’ law for magnetism is not as helpful for deter-mining magnetic fields. The reason is illustrated in Fig-ure 28.15b: magnetic field lines form loops and so ifthey exit a closed surface, they must reenter it at someother point. Consequently, the magnetic flux through aclosed surface is always zero, regardless of whether ornot the surface encloses any magnets or current-car-rying wires. In mathematical language, the surface in-tegral of the magnetic field over a closed surface is

always zero.Let us next consider line integrals of electric and

magnetic fields. As we saw in Section 25.5, the line in-tegral of the electrostatic field around a closed path isalways zero. Consider, for example, the line integral ofthe electrostatic field around closed path 1 in Figure28.15c. Because the electrostatic field generated by thecharged particle at the center of path 1, is always perpendicular to the path, is perpendicular to , sothat , and therefore the line integral is alwayszero. Any other path, such as closed path 2 in Figure28.15c, can be broken down into small radial and circu-lar segments. As we saw in Section 25.5, going oncearound a closed path, all the nonzero contributionsalong the radial segments add up to zero. Physicallythis means that as you move a charged particle arounda closed path through an electrostatic field, the workdone by the electrostatic field on the particle is zero.Consider now, however, the line integral around

closed path 1 in Figure 28.15d, which is concentric witha wire carrying a current out of the page. The magneticfield generated by the current through the wire alwayspoints in the same direction as the direction of this path.Therefore is always positive, making the line inte-gral nonzero and positive. Along closed path 2, the com-ponent of the magnetic field tangent to the path is alwaysopposite the direction of the path. Therefore is al-ways negative and the line integral is negative.Let us next compare the line integrals along two

closed circular paths of different radius (Figure 28.16a).The arrowheads in the path indicate the direction alongwhich we carry out the integration. Let the magnitude ofthe magnetic field a distance R1 from the wire at the cen-ter of the paths be B1. Because this magnitude is con-stant along the entire circular path, the line integral

BÆ.dl

Æ

BÆ.dl

Æ

EÆ.dl

Æ� 0

dlÆ

EÆ

along this path is the product of the field magnitude andthe length of the path: = . Alongclosed path 2 we obtain , where B2 is the fieldmagnitude a distance R2 from the wire. As you may sus-pect from the cylindrical symmetry of the wire, the mag-nitude of the magnetic field decreases as 1/r withdistance r from the wire (we’ll confirm this dependencein Example 28.6). This means that as we go from R2 toR1 the field decreases by a factor R2/R1. In other words,B1 = B2(R2/R1) or B1R1= B2R2. Thus we see that the lineintegrals along the two paths are equal: 2pB1R1 =2pB2R2. The same argument can be applied to any otherclosed circular path centered on the wire, from whichwe conclude that the line integral of the magnetic fieldof a straight current-carrying wire over any circular pathcentered on the wire has the same value.

28.8 If the magnitude of the current I through awire is increased, do you expect the line integralof the magnetic field around a closed path aroundthe wire to increase, decrease, or stay the same?

Now consider the noncircular path illustrated in Figure 28.16b. Most of the path lies along a circle ofradius R2, with the exception of one eighth of a revo-lution, which is along two radial segments and an arc ofradius R1. Because the radial segments are perpendic-ular to the magnetic field, they do not contribute to theline integral. How do the line integrals along arcs 1 and2 compare? As we just saw, the line integrals along thetwo closed circular paths in Figure 28.16a are equal, sothe line integrals along one-eighth of each closed pathin Figure 28.16amust also be equal. The same must betrue in Figure 28.16b, which means the line integralsalong arcs 1 and 2 are identical, and so the line inte-gral along the noncircular path in Figure 28.16b is equalto that along any circular path centered on the wire.

2�B2R2

B1(2�R1) 2�B1R1


R1

E E

closedpath 1

arc 1arc 2

closedpath 2

R2 R1R2

(a) (b)

I I

Figure 28.16 (a) Two closed circular paths 1 and 2 concentricwith a current-carrying wire coming out of the plane of thepage. (b) A noncircular path encircling the current- carryingwire. Arc 1 is one-eight of the circumference of the circle ofradius R1. Arc 2 is one-eight of the circumference of the circle of radius R2.

(a) (b)

B→

B→

Figure 28.17 (a) A noncircular closed path encircling a cur-rent-carrying wire. (b) Approximation of the path consistingof small arcs and radial segments.

We can make the deviations from a circular path pro-gressively more complicated, but as illustrated in Figure28.17, any path can always be broken down into smallsegments which are either radial or circular and concen-tric with the wire. The radial segments never contributeto the line integral because the magnetic field is alwaysperpendicular to them, while the circular segments al-ways add up to a single complete revolution. So, in con-clusion,

The value of the line integral of the magnetic fieldalong a closed path encircling a current-carryingwire is independent of the shape of the path.

28.9 What happens to the value of the line inte-gral along the closed path in Figure 28.17a when(a) the direction of the current through the wireis reversed; (b) a second wire carrying an identi-cal current is added parallel to and to the rightof the first one (but still inside the path); (c) thecurrent through the second wire is reversed?

Next let’s examine the line integral along a closedpath near a current-carrying wire lying outside thepath. One such path is shown in Figure 28.18 as twoarcs joined by two radial segments. As before, the ra-dial segments do not contribute to the line integral.The magnitudes of the line integrals along arcs 1 and2 are equal, but because the direction of arc 2 is oppo-site the direction of the magnetic field, the line inte-gral along that arc is negative. Consequently, the lineintegral along the entire path adds up to zero. We canagain extend this statement to a path of a differentform, but as long as the path does not encircle the cur-rent-carrying wire, the line integral is zero.

The line integral of the magnetic field along aclosed path that does not encircle any current- carrying wire is zero.

Putting the above results together, we see that theline integral of the magnetic field along a closed pathtells us something about the amount of current encir-cled by the path:

The line integral of the magnetic field along aclosed path is proportional to the current encir-cled by the path.

We shall put this statement in a more quantitativeform in Section 28.5. As we shall see there, this law,called Ampère’s law, plays a role analogous to Gauss’law: given the amount of current encircled by a closedpath, called an Ampèrian path, we can readily deter-mine the magnetic field due to the current, provided thecurrent distribution exhibits certain simple symmetries.Because the line integral along a closed path dependson the direction of integration, we must always choosea direction along the loop when specifying an Ampèri-an path. Exercise 28.2 illustrates the importance of thedirection of the Ampèrian path.

28.10 Suppose the path in Figure 28.17 were tiltedinstead of being in a plane perpendicular to thecurrent-carrying wire. Would this tilt change thevalue of the line integral of the magnetic fieldaround the path?

■ Exercise 28.2: Crossed wires

Consider the Ampèrian path going through the collec-tion of current-carrying wires in Figure 28.19. If themagnitude of the current is the same in all the wires, isthe line integral of the magnetic field along the Ampèrian path positive, negative, or zero?

28.4 Ampèrian path 11

R1

B

R2

arc 1

arc 2

Figure 28.18 A noncircular closed path not encircling a current-carrying wire.

2I

I I

II

3

4

1

5

Ampèrian path

Figure 28.19 Exercise 28.2.

Solution: For each wire, I must determine whether ornot the loop encircles the wire and, if it does, whetherthe component of that wire’s magnetic field tangent tothe loop points in the same direction as the loop. I seethat, wires 1 and 3 go through the loop but the otherthree wires lie either on top of the loop or beneath it. The direction of current through wire 1 is forward

out of the plane of the page, and so the magnetic fieldlines around this wire curl counterclockwise — oppo-site the direction of the Ampèrian path — giving a neg-ative contribution to the line integral. The direction ofcurrent through wire 3, is down into the plane of thepage, and so it yields a positive contribution to the lineintegral. Because the two currents are equal in magni-tude, the contributions to the line integral add up to zero.

28.11 How do the following changes affect theanswer to Exercise 28.2: (a) reversing the currentthrough wire 1, (b) reversing the current throughwire 2, (c) reversing the direction of the Ampèri-an path?


[(H1F)] 13

[(H1F)] 14

SE L F - Q U I Z

14

1. Determine the direction of the magnetic force exerted at the center of the wire or on the particles in Figure 28.20.

2. Determine the direction of the magnetic field at P due to (a) the current loop in Figure 28.21a and (b) seg-ments A and C of the current loop in Figure 28.21b:

3. Determine in which direction the current loop rotates (a) in Figure 28.21c and (b) in Figure 28.21d.4. (a) Determine the current encircled by the five Ampèrian paths in Figure 28.21e: (b) Rank the loops accord-

ing to the magnitude of the line integral of the magnetic field along each loop, greatest first.

ANSWERS:1. (a) No magnetic force exerted by the magnet on the wire because the magnetic field at the

location of the wire and the current are antiparallel. (b) The magnetic force acting on theparticle is upward. To see this, consider the moving positively charged particle to be currentin the direction of the velocity of the particle and then use the right-hand force rule, whichmakes your thumb point upward. (c) The magnetic dipole moments of the two spinning par-ticles both point up and so the particles attract each other, just like two bar magnets orient-ed the way the spinning particles are (Figure 28.22a). (d) The magnetic dipole moment ofthe negative particle points down, that of the positive particle points up, and so the two par-ticles repel each other. The bar magnet comparable orientation is shown in Figure 28.22b.

2. (a) The right-hand dipole rule tells you that the magnetic dipole of the loop and the magnetic field produced by the looppoints to the right at P. (b) Segments A and B both contribute a magnetic field that points out of the page at P and so themagnetic field due to both segments also points out of the page at P.

3. (a) for the current loop points to the right and so the current loop rotates counter-clockwise about an axis perpendic-ular to the page and through the center of the loop. (b) for the current loop points out of the page, and so the loop ro-tates about an axis aligned with the vertical sides of the loop. The right side of the loop moves up out of the page, and theleft side moves down into the page.

4. (a) Loop 1 encircles two currents I in the same direction as the magnetic field: +2I. Loop 2 encircles the same two currentsin the opposite direction: –2I. Loop 3 encircles 2I in the direction opposite the magnetic field direction: –2I. Loop 4 en-circles all three currents, which add up to zero. Loop 5 encircles I in the direction opposite the magnetic field direction and2I in the same direction as the magnetic field: +I. (b) Each line integral is proportional to the current enclosed, makingthe ranking 1 = 2 = 3, 5, 4.

Æ

Æ

(a) (b) (c) (d)

S

N +e

+e

+e

S N

vÆ

+e

–e

I

(a) (b) (c) (d) (e)

A

C

PP

B

B2

1

4

3

5

2II→

→

I

II

II

Figure 28.20

Figure 28.21

Figure 28.22(a) (b)

NS

NS

NS

NS

μ→

μ→

μ→

μ→

28.5 Ampère’s law 15

28.5 Ampère’s lawIn the preceding section, we saw that the line integral of the magnetic field arounda closed path, called an Ampèrian path, is proportional to the current encircled bythe path, . This can be expressed mathematically as

(constant currents), (28.1)

where is an infinitesimal segment of the path and the proportionality constantis called the permeability constant. To define the ampere (Section 27.5), its

value is set to be exactly

.

Equation 28.1 is called Ampère’s law, after the French physicist André Marie Ampère (1775–1836). Let us illustrate how Eq. 28.1 is used by applying it to the closed path in Figure

28.23. From the figure we see that the path encircles wires 1 and 2 but not 3, whichlies behind the path. To calculate the right side of Eq. 28.1, we must assign an alge-braic sign to the contributions of currents I1 and I2 to Ienc. As we saw in Section 28.4,we can do so using the right-hand current rule. Putting the thumb of our right handin the direction of I1, we see that the magnetic field of I1 curls in the same directionas the direction of the integration path (the Ampèrian path) indicated in the diagram.Applying the same rule to I2, tells us that the magnetic field of this current curls inthe opposite direction. Thus I1 yields a positive contribution and I2 yields a negativecontribution to the line integral. The right side of Eq. 28.1 thus becomes.To calculate the line integral on the left side of Eq. 28.1, we first divide the closed

path into infinitesimally small segments , one of which is shown in Figure 28.23.The segments are directed tangentially along the loop in the direction of integra-tion. For each segment , we take the scalar product of with the magneticfield at the location of that segment, , and then we add up the scalar prod-ucts for all segments of the closed path. In the limit that the segment lengths ap-proach zero, this summation becomes the line integral on the left in Eq. 28.1. In thesituation illustrated in Figure 28.23, we cannot easily carry out this integration be-cause of the irregular shape of the path. Just like Gauss’ law in electrostatics, how-ever, Ampère’s law allows us to easily determine the magnetic field for highlysymmetric current configurations. The general procedure is outlined in the proce-dure box on page ••. The next two examples illustrate how this procedure can beapplied to simplify calculating the line integral in Ampère’s law.

■ Example 28.3: Magnetic field generated by a long straight current-carryingwire

A long straight wire carries a current of magnitude I. What is the magnitude ofthe magnetic field a distance d from the wire?

Getting started: I begin by making a sketch of the wire. I know that the mag-netic field is circular, so I draw one circular field line around the wire. Using theright-hand current rule, I determine the direction in which the magnetic field pointsalong the circle and indicate that with an arrowhead in my drawing (Figure 28.24).

1

BÆ

BÆ.dl

ÆdlÆ

dlÆ

o(I1 � I2)

dlÆ

o � 4� � 10�7 T.mA

o

�O BÆ.dlÆ

� oIenc

Ienc

dlÆ

I2

I1

I3

3

1

2

B

dl

θ

→

→

Figure 28.23 A closed path en-circling two of three straightcurrent-carrying wires.

SE L F - Q U I Z

Devise plan: To determine the magnitude of the magnetic field, I can use Am-père’s law (Eq. 28.1). If I let the field line in my sketch be the Ampèrian path, themagnetic field is constant in magnitude and tangential all along the circular loop,simplifying the integral on the left in Eq. 28.1. The direction of the Ampèrian pathis the same as that of the magnetic field; I denote the radius of the circular loop byd (Figure 28.24).

Execute plan: With my choice of Ampèrian path, and are always point-ing in the same direction, and so . Because the magnitude B of themagnetic field is the same all around the loop, I can write for the left side of Eq.28.1

.

The line integral on the right is the sum of the lengths of all segments aroundthe Ampèrian path; that is to say, it is equal to the circumference of the circle.Therefore I have

.

Because the Ampèrian path encircles the current I in the same direction as themagnetic field generated by the current, the right side of Eq. 28.1 yields

2

�OBÆ.dl

Æ� B �Odl � B(2�d)

dlÆ

�O BÆ.dlÆ � �OB dl � B�Odl

BÆ.dl

Æ� B dl

BÆ

dlÆ

3

16 CHAPTER 28 Magnetic fields of charged particles in motion Quantitative Tools

4. Use the right-hand current rule to determine the di-rection of the magnetic field of each current encir-cled by the loop. If this magnetic field and theAmpèrian path have the same direction, the contri-bution of the current to Ienc is positive. If they haveopposite directions, the contribution is negative.

5. For each Ampèrian path calculate the line inte-gral of the magnetic field along the loop. Expressyour result in terms of the unknown magnitudeof the magnetic field B along the Ampèrian path.

6. Use Ampère’s law (Eq. 28.1) to relate Ienc and theline integral of the magnetic field and solve for B. (Ifyour calculation yields a negative value for B, thenthe magnetic field points in the opposite directionyou assumed in step 1.)

You can use the same general approach to determinethe current given the magnetic field of a current dis-tribution. Follow the same procedure, but in Steps4–6, express Ienc in terms of the unknown current Iand solve for I.

For magnetic fields with straight or circular fieldlines, Ampère’s law allows you to calculate the mag-nitude of the magnetic field without having to carryout any integrations.

1. Sketch the current distribution and the magneticfield by drawing one or more field lines using theright-hand current rule. A two-dimensional draw-ing should suffice.

2. If the field lines form circles, the Ampèrian pathshould be a circle. If the field lines are straight, itshould be rectangular.

3. Position the Ampèrian path in your drawing suchthat the magnetic field is either perpendicular ortangent to the loop and constant in magnitude.Choose the direction of the Ampèrian path suchthat it coincides with the direction of the magnet-ic field. If the current distribution divides spaceinto distinct regions, draw an Ampèrian path ineach region where you wish to calculate the mag-netic field.

PROCEDURE: Calculating the magnetic field using Ampère’s law

I

Bd

dl→

→

Figure 28.24

28.5 Ampère’s law 17

Btot

B1

P

d

2d

l

AC

D G

(b)

P

d

K

B

(a)

1 2

(c)

(d)

B2

P

d

Ampèrianpath

sheetlength sheet

width

dl→→

→

→

→

Figure 28.25

,

substituting the right sides of this equation and the preceding one into Eq. 28.1, Iget or

✔

Evaluate result: My result shows that the magnitude of the magnetic field is pro-portional to I, as I would expect (doubling the current should double the magnet-ic field), and inversely proportional to the perpendicular distance d from the wire.I know from Chapter 24 that the electric field is also inversely proportional to din cases exhibiting cylindrical symmetry, so my result makes sense.

28.12 Suppose the wire in Example 28.3 has a radius R and the current is uni-formly distributed throughout the volume of the wire. Follow the procedureof Example 28.3 to calculate the magnitude of the magnetic field inside (r <R) and outside (r > R) the wire.

■ Example 28.4 Magnetic field generated by a large current-carrying sheet

A large flat metal sheet carries a current. The magnitude of the current per unit ofwidth is K. What is the magnitude of the magnetic field a distance d above thesheet?

Getting started: I begin by drawing the sheet and indicating a point P a distanced above the sheet where I am to determine the magnetic field (Figure 28.25a). Idraw a lengthwise arrow to show the current through the sheet. If the magnitudeof the current per unit of width is K, I know that a strip of width w of the sheet car-ries a current of magnitude Kw.

Devise plan: I’ll solve this problem using Ampère’s law (Eq. 28.1). I first needto determine the direction of the magnetic field on either side of the sheet. To thisend, I first divide the sheet into thin parallel strips the length of which run in thedirection of the current. I then treat each strip as a current-carrying wire Figure28.25b is a cross-sectional view of the sheet after I have divided it into strips. Theperspective here is looking into the sheet in the direction opposite the direction ofthe current. Using the right-hand current rule, I find that the strip right under-neath point P contributes a magnetic field at P that points parallel to the sheetand to the left. Next I look at the contributions from the two strips labeled 1 and2 in Figure 28.25c, equidistant on either side of P.The two large gray circles showthe magnetic field lines from these two strips that go through P. Because the mag-nitudes of the contributions from the two strips are equal, the vector sum of theircontributions to the magnetic field also points parallel to the sheet and to theleft.The same argument can be applied to any other pair of strips that are equidis-tant on either side of P. Thus, at P the magnetic field due to the entire sheet mustbe parallel to the sheet and to the left. As similar reasoning shows that the mag-netic field below is also parallel to the sheet and points to the right.

B(2�d) � oI

oIenc � � oI

B � oI2�d

.

2

4

1

To exploit what I know about the magnetic field, I choose the rectangular loopABDG in Figure 28.25d as my Ampèrian path. I let the direction of the loop be thesame as that of the magnetic field, as shown in Figure 28.5d. The width of this loopis w and its height is 2d.

Execute plan: I can write the line integral around the loop as the sum of fourline integrals, one over each side of the loop:

.

Along the two vertical sides CD and GA the magnetic field is perpendicular tothe path and so is zero; along each of the two horizontal sides andpoint in the same direction, and so .Symmetry requires that the magnitude of the magnetic field a distance d below

the sheet be the same as the magnitude a distance d above it (because flipping thesheet upside down should not alter the physical situation). Therefore I can take thismagnitude B out of the integral:

.

These two line integrals , yield the lengths of the two horizontal sides and sothe left side of Ampère’s law (Eg. 28.1) becomes

. (1)

To calculate the right side of Ampère’s law, I must first determine the amountof current encircled by the Ampèrian path. If the magnitude of the current perunit of width through the sheet is K, the magnitude of the current through theclosed path of width w is Kw. The right side of Eq. 28.1 thus becomes

. (2)

Substituting the right sides of Eqs. 1 and 2 into Ampère’s law, I getor

. ✔

Evaluate result: Because the sheet has planar symmetry (Section 24.4), I expectthat, in analogy to the electric field of a flat charged sheet, the magnetic field oneither side of the sheet is uniform. That is, the magnitude and direction of the mag-netic field do not depend on distance from the sheet. Indeed, my result does notdepend on the distance d to the sheet. It also makes sense that the magnitude ofthe magnetic field is proportional to the current per unit width K through thesheet.

BÆ

dlÆ

3

�O BÆ.dlÆ � �C

ABÆ.dl

Æ� �D

CBÆ.dl

Æ� �G

DBÆ.dl

Æ� �A

GBÆ.dl

Æ

4

B � 12 oK

2Bw � oKw

oIenc � oKw

�OBÆ.dl

Æ� B(2w) � 2Bw

�dl


AB dl � �G

DB dl � B�C

Adl � B�G

Ddl

BÆ.dl

Æ� B dl

BÆ.dl

Æ


K

K

K

K

(a)

(b)

Figure 28.26 A “Magnetic capacitor” (Checkpoint 28.13).

28.6 Solenoids 19

28.13 (a) What are the direction and magnitude of the magnetic field be-tween the parallel current-carrying sheets of Figure 28.26a. What is the direc-tion of outside these sheets? (b) Repeat for the sheets of Figure 28.26b.

28.6 Solenoids and toroidsA solenoid is a long, tightly wound helical coil of wire (Figure 28.27a). In general,the diameter of the coil is much smaller than the length of the coil. When a cur-rent enters a solenoid at one end and exits at the other end, the solenoid generatesa strong magnetic field. If a magnetic core is placed in the solenoid, the solenoidexerts a strong magnetic force on the core, turning electrical energy into motion.Solenoids are therefore often used in electrical valves and actuators. Because so-lenoids are generally very tightly wound, we can treat the windings of a solenoidas a stack of closely spaced current loops (Figure 28.27b).Figure 28.28a shows that the magnetic field inside a long solenoid must be di-

rected along the axis of the solenoid. Consider, for example, point P inside the so-lenoid. The figure shows the field lines of two of the loops, one on either side of Pand equidistant from that point. The magnetic field contributions of the two loopsgive rise to a magnetic field that points along the axis of the solenoid. The same ar-gument can be applied to any other pair of loops and to any other point inside thesolenoid. Therefore the magnetic field everywhere inside a long solenoid must bedirected parallel to the solenoid axis. Because magnetic field lines form loops, all the lines that go through the sole-

noid must loop back from one end of the solenoid to the other. Because there ismuch more space available outside the solenoid, the density of the field lines as theyloop back is much smaller than the field line density inside the solenoid. The longerthe solenoid, the lower the field line density in the immediate vicinity outside thesolenoid. In the limit of an infinitely long solenoid, we can expect the magneticfield outside the solenoid to approach zero. We can now use Ampère’s law to determine the magnitude of the magnetic

field when a current of magnitude I flows through the solenoid. Exploiting whatwe know about the direction of the magnetic field, we choose the rectangular pathACDG in Figure 28.28b as the Ampèrian path. Along AC, the magnetic field isparallel to the path, and so . Along CD and GA, the magnetic field isperpendicular to the path, which means that these segments do not contribute:

. Because the magnetic field is zero outside the solenoid, the segmentDG also does not contribute. The line integral on the left side of Ampère’s law(Eq. 28.1) thus becomes

. (28.2)

The cylindrical symmetry of the solenoid requires the magnitude of the magneticfield to be constant along AC, and so we can pull B out of the integral:

, (28.3)

BÆ


AB dl

BÆ.dl

Æ� 0

BÆ.dl

Æ� B dl


AB dl � B�C

Adl � Bl

I

I

I

(b)

(a)

axis

windings

current loops

Figure 28.27 (a) A solenoid is atightly wound helical coil ofwire. (b) A solenoid can be ap-proximated as a stack of manyparallel current loops.

where l is the length of side AC.What is the current encircled by the Ampèrian path? Each winding carries a cur-

rent of magnitude I, but the loop encircles more than one winding. If there aren windings per unit length, then nl windings are encircled by the Ampèrian path,and the encircled current is I times the number of windings:

. (28.4)

Substituting Eqs. 28.3 and 28.4 into Ampère’s law (Eq. 28.1), we find

, (28.5)

(infinitely long solenoid). (28.6)

This result shows that the magnetic field inside the solenoid depends on thecurrent through the windings and on the number of windings per unit length. Thefield within the solenoid is uniform — it does not depend on position inside the so-lenoid. Although Eq. 28.6 holds for an infinitely long solenoid, the result is prettyaccurate even for a solenoid of finite length. For a solenoid that is at least fourtimes as long as it is wide, the magnetic field is very weak outside the solenoid andapproximately uniform and equal to the value given in Eq. 28.6 inside. An exam-ple of the magnetic field of a finite solenoid is shown in Figure 28.29. Note how themagnetic field pattern resembles that of the magnetic field around a bar magnet.If a solenoid is bent into a circle so that its two ends are connected (Figure

28.30a), we obtain a toroid. The magnetic field lines in the interior of a toroid (thatis, the donut-shaped cavity enclosed by the coiled wire) close on themselves, thusthey do not need to reconnect outside the toroid, as in the case of a solenoid. Theentire magnetic field is contained inside the cavity. Symmetry requires the fieldlines to form circles inside the cavity; the field lines run in the direction of your rightthumb when you curl the fingers of your right hand along the direction of the cur-rent through the windings. Figure 28.30b shows a few representative magnetic fieldlines in a cross section of the toroid where the current goes into the page on the

Ienc � nlI

B � onI

Bl � onlI


l

AC

D G

I into page

Ampèrianpath

I out of page

(b)

the hanging “l” should really be accompanied by a dimension arrow

(width of the loop)Annotations too visible; should

all be “annotation color”

(a)

PB

tot

B1

1 2

1 2B2

field lineof loop 2

field lineof loop 1

B dl→→

→

→

(b)(a)

Photograph of magnetic

field of finite coil

(see cover of Sep 09 Phys Today)

Figure 28.29 (a) Magnetic field line pattern in solenoid of finite length. (b) Magnetic field line pattern of a finite solenoid,made visible via iron filings.

Figure 28.28 (a) Cross sectionof a solenoid showing the con-tributions of two loops to themagnetic field at a point P in-side the solenoid. (b) Ampèrianpath ACDG for calculating themagnetic field inside a solenoid.

28.7 Magnetic fields due to currents 21

outside rim of the toroid (as at point P, for example) and out of the page on theinside rim (as at point Q).To find the magnitude of the magnetic field, we apply Ampère’s law (Eq. 28.1)

to a circular path of radius r that coincides with a magnetic field line (Figure28.30b). Because the field is tangential to the integration path, we have

. Furthermore symmetry requires the magnitude of the magnetic fieldto be the same all along the field line, and so we can pull B out of the integration:

. (28.7)

The Ampèrian path encircles one side of all of the windings, and so if there are Nwindings, the enclosed current is

. (28.8)

Substituting these last two equations into Ampère’s law (Eq. 28.1), we find

(toroid). (28.9)

This result tells us that in contrast to a solenoid, the magnitude of the magnetic fieldin a toroid is not constant — it depends on the distance r to the axis through thecenter of the toroid.

■ Example 28.5: Square toroid

A toroid has 1000 square windings carrying a current of 1.5 mA (Figure 28.31). Thelength of each side of the squares is 10 mm, and the toroid’s inner radius is 10 mm.What is the magnitude of the magnetic field at the center of the square windings?

Getting started: The fact that the loops are square does not change the mag-netic field pattern. The magnetic field is still circular as in Figure 28.30b and anyof the arguments given in the derivation of Eq. 28.9 still apply.

Devise plan: I can use Eq. 28.9 to determine the magnitude of the magneticfield.

Execute plan: The distance from the center of the toroid to the center of thesquare loops is 10 mm + 5 mm = 15 mm, and so

✔

Evaluate result: The magnetic field I obtain is rather small — comparable to theEarth magnetic field, but the current through the toroid is very small, so my answeris not unreasonable.

BÆ.dl

Æ� B dl

B � (4� � 10�7 T.mA)(1000) (1.5 � 10�3 A)

2�(15 mm)� 20 � 10�6 T

Ienc � NI

�O BÆ.dlÆ � �OB dl � B�Odl � B(2�r)

4

3

2

1

B � oNI2�r

AmpèrianloopI

I outof page

I intopage

magneticfield line

r

(b)

(a)

Q

P

Q

Pdl

→

B→

show cut-away? See IMN 22-2

Figure 28.30 (a) A toroid is asolenoid bent into a ring. (b)Cross sectional view of thetoroid, showing the circularfield lines inside the toroid andchoice of Ampèrian path.

10 mm 10 mm

1000 windings1.5 mA each

show cross section? square loops? See IMN 23-1need to adjust sizes



28.14 Use Ampère’s law to find the magnetic field outside a toroid at a dis-tance r from the center of the toroid (a) when r is larger than the toroid’souter radius and (b) when r is smaller than the toroid’s inner radius.

28.7 Magnetic fields due to currentsAmpère’s law allows us to determine the magnetic field in only a few symmetri-cal situations involving current-carrying conductors. For any other situation, suchas the one illustrated in Figure 28.32a, Ampère’s law is of little help. Suppose, forexample, that we are interested in determining the magnetic field at point P dueto the current of magnitude I through this conductor.To calculate this field, we de-velop a procedure that parallels the procedure we developed in Section 23.7 tocalculate the electric field of continuous charge distributions. We begin by treating the wire as a current path and dividing it into small seg-

ments each of length dl. For each segment we can define a vector that haslength dl and points in the direction of the current. One such vector segment isshown in Figure 28.32b. Let the magnetic field at P due to this segment be . Ifwe can find an expression for the magnetic field due to the segment at an ar-bitrary position, we can determine the contributions of all the segments making upthe wire to the magnetic field at P and sum these to find . In the limit of infini-tesimally small segments, this summation becomes an integral:

, (28.10)

where the integration is to be taken along the path followed by the current — thatis a path in the shape of the wire and in the direction of the current.Before we can carry out the integration, we need to find an expression for the mag-

netic field of a current-carrying segment . Because the magnetic field causedby one small segment is too feeble to measure, we cannot determine this field experi-mentally. From the expression for the magnetic field of a long straight current-carryingwire, however, it is possible to deduce what the field of a segment such as should be.Let us first examine the direction of the magnetic field contribution due to

the segment that is located such that is perpendicular to the vectorpointing from to P (q = 90°, see Figure 28.32c). The field lines of this segmentare circles centered on the line of motion of the charge carriers causing the cur-rent through (see Figures 28.2a and 28.32c). Therefore the magnetic field at Pdue to is tangent to the circular field line through P. The direction of canbe determined by associating the direction of with a vector product. To thisend, we denote the unit vector pointing from the segment to P by , as shownin Figure 28.32c. The direction of the cross-product is that of — ifyou curl the fingers of your right hand from to in Figure 28.32c, your thumbpoints in the direction of .We expect the magnitude dBs to be proportional to the current I through the

segment and to the length dl of the segment — the larger the current or the longerthe segment, the stronger the magnetic field. We also expect dBs to depend on the

dlÆ

rÆ

sPdlÆ

dBÆ

s

dBÆ

s

dBÆ

s

dBÆ

s

dBÆ

s

dlÆ

dlÆ

dlÆ

dlÆ

BÆ

� �currentpath

dBÆ

s

BÆ

dlÆ

dBÆ

s

dlÆ

dlÆ

dlÆ

� r̂sP

dlÆ

dBÆ

s

r̂sP

r̂sP

I

(b)

(a)

I

dl→

(c)

current path

I

P

P

dBs

→

P

= 90°u

dBsdl

→→

r→s

O

r→P

r→sP

r̂sP

Figure 28.32 (a) The magneticfield at point P due to an arbi-trarily shaped current-carryingwire cannot be obtained fromAmpère’s law. (b) A small seg-ment dl of the wire contributesa magnetic field . (c) Thedirection of can be foundby taking the vector product of

, which has length dl and di-rection given by the currentthrough the wire, and ,which points from the segmentto point P: .dl

Æ� r̂sP

r̂ sP

dlÆ

dBÆ

s

dBÆ

s


distance rsP to the segment. The field line picture for magnetic fields suggests thatthe magnetic field should decrease with distance rsP as , just like the electricfield. Finally, for any other segment than the one shown in Figure 28.32cwe expectthe magnetic field to depend on the angle q between and : for q = 0 (whichis along the direction of the current) the magnetic field is zero and for q = 90° themagnetic field is maximum. The trigonometric function that fits this behavior issin q. Putting all this information in mathematical form, we obtain

(0 � � � �). (28.11)

The proportionality factor is obtained by deriving this expression from Am-père’s law, but the mathematics is beyond the scope of this book. Instead, we usethe reverse approach and show in Example 28.6 that Eq. 28.11 yields the correct re-sult for a long straight current-carrying wire.Incorporating what we know about the direction of , we can also write Eq.

28.11 in vector form:

(constant current), (28.12)

where is the unit vector pointing along . Equation 28.12 is known as theBiot-Savart law. By substituting Eq. 28.12 into Eq. 28.10, we have a prescription forcalculating the magnetic field produced by any constant current.

■ Example 28.6: Another look at the magnetic field generated by a longstraight current-carrying wire

A long straight wire carries a current of magnitude I. Use the Biot-Savart law toderive an expression for the magnetic field produced at point P a distanced from the current-carrying wire.

Getting started: I begin by making a sketch of the wire (Figure 28.33). I choosethe x axis along the direction of the wire. Because the magnetic field produced bythe wire has cylindrical symmetry, I can set the origin anywhere along the axiswithout loss of generality. For simplicity, I let the origin be at the height of pointP. I assume the wire is of infinite length.

Devise plan: To use the Biot-Savart law I need to divide the wire into smallsegments, determine the magnetic field due to the segments, and then take theirsum. In the limit of infinitesimally small segments, this sum becomes the integralgiven by Eq. 28.10 where the wire is the current path along which the integrationis carried out.I indicate one such segment in my sketch; the magnetic field at P due to a

segment of the wire is given by Eq. 28.12. The unit vector in this expres-sion points from to P. The direction is given by the right-hand currentrule: I point my right thumb along the direction of the current and curl my fingersaround the wire to find the direction of (into the page at P in my sketch). Al-ternatively, I can use the vector product to find the direction of : Iline up the fingers of my right hand along in Figure 28.33 and curl them to-ward , which points from the segment to P. When I do this, my thumb points

1rsP2

dlÆ

dBÆ

s

dBÆ

s

dBs � o

4�

I dl sin�

rsP2

dxÆ

2

1

BÆ

r̂sP rÆ

sP

dBÆ

s � o

4�

I dlÆ

� r̂ sPrsP2

o4�

dBÆ

sdxÆ

r̂sP

dBÆ

s

dxÆ

� r̂ sP dBÆ

s

r̂sP

dxÆ

dxÆ

r̂sP

I

d

x

x

(into page)

O P

u

dBs

dx→

rsP

→

→

r̂sP

Figure 28.33

in the direction of the magnetic field. Both methods yield the same result:points into the page.Note that all the segments along the wire produce a magnetic field in the

same direction. This means I can take the algebraic sum of the magnitudes ofto determine the magnitude of the magnetic field at P, and so I can use Eq. 28.11to express dBs in terms of dx and integrate the resulting expression fromto to determine the magnitude of the magnetic field at P.

Execute plan: Because rsP and q in Eq. 28.11 both depend on x, I need to ex-press them in terms of x before I can carry out the integration. By the Pythagore-an theorem,

,

and remembering that , I write

.

Substituting these last two results into Eq. 28.11, I get

,

and integrating this result over the length of the wire, I have

,

, ✔

Evaluate result: This is identical to the result I obtained using Ampère’s law inExample 28.3.

28.15 Imagine a long straight wire of semi-infinite length, extending from x= 0 to x = +, carrying a current of constant magnitude I. What is the mag-nitude of the magnetic field at a point P located a perpendicular distanced from the end of the wire that is at x = 0?

The expression obtained in Example 28.6 for the magnitude of the magnetic fieldgenerated by a long straight wire, can lead us to an expression for the forces ex-erted by two current-carrying wires on each other. Consider the situation illustrat-ed in Figure 28.34: two parallel wires, of length l and separated by a distance d,carry currents of magnitudes I1 and I2. To determine the magnetic force exerted by

x � �

x � �

dxÆ

dBÆ

s

3

rsP2 � x2 � d2

4

B � oI2�d

B � oId4� ��

�

dx[x2�d2]32 �

oId4� � 1d2

x[x2�d2]12 �

x��

x��

dBs � oI4�

d

x2�d21

x2 � d2dx �

oId4�

dx[x2�d2]32

sin � �drsP

�d

x2�d2

sin� � sin(180°��)

dBÆ

s



wire 1 on wire 2, we must first determine the magnitude and direction of the mag-netic field generated by wire 1 at the location of wire 2 and then substitute this in-formation into Eq. 27.8, which gives the magnetic force exerted on a straightcurrent-carrying wire of length l in a magnetic field :

, (28.13)

where is a vector whose magnitude is given by the length l of the wire and whosedirection is given by that of the current through the wire. Using the right-handcurrent rule, we find that the magnetic field generated by wire 1 at the locationof wire 2 points into the page (Figure 28.34). The result I obtained in Example 28.6gives the magnitude of this field:

. (28.14)

Because the magnetic field is perpendicular to , Eq. 28.13 yields for the magni-tude of the magnetic force acting on wire 2

, (28.15)

or, substituting Eq. 28.14,

(parallel straight wires carrying constant currents). (28.16)

The direction of this force follows from the vector product in Eq. 28.13: using theright-hand force rule (place the figures of your right hand along in Figure 28.34,which points in the direction of I2, and bend them toward ), you can verify thatthe force points toward wire 1.

■ Example 28.7: The magnetic field of a circular arc

A circular arc of radius R subtending an angle f carries a current of magnitude I(Figure 28.35). Use the Biot-Savart law to derive an expression for the magneticfield produced at the point P at the center of the arc.

Getting started: I begin by evaluating what a small segment of length dl alongthe arc contributes to the magnetic field. I therefore make a sketch showing onesegment and the vector pointing from the segment to P (Figure 28.36).Using the right-hand vector product rule, I see that the direction of (andtherefore the magnetic field ) is into the page at P.

Devise plan: Because all segments contribute a magnetic field in the same di-rection, I can simply integrate Eq. 28.11 over the arc to obtain the magnitude of themagnetic field at P.

Execute plan: Because and are always perpendicular to each other (Fig-ure 28.36), I can write for Eq. 28.11

lÆ

FÆ

B � IlÆ

� BÆ

BÆ

BÆ

1

rÆ

sP

B1 � oI12�d

dBÆ

s

dlÆ

� rÆ

sP

rÆ

sPdlÆ

dlÆ

3

2

1

BÆ

BÆ

1

lÆ

F12B �

olI1I22�d

F12B � I2lB1

lÆ

I

R

P

f


I1

d

d

B1

I2

1 2

F1B2

l→

→

Figure 28.34 Calculating themagnetic force exerted by one current-carrying wire on another.

P

(into page)

dBs

dl→

→

rsP

→

Figure 28.36


, (1)

where I have substituted the radius R for the magnitude of . To change the integration variable from dl to the angular variable df, I substi-

tute dl = R df into Eq. 1:

.

,

✔

Evaluate result: My result shows that the magnetic field at P is proportional tothe current through the arc, as I would expect. It is also proportional to the angle� of the arc. This is what I would expect, given that two such arcs should yieldtwice the magnetic field. Finally, my result is inversely proportional to the radiusR of the arc. That dependence on the radius makes sense, as increasing the radiusincreases the distance between R and the arc, diminishing the magnetic field.

28.16 What is the magnitude of the magnetic field (a) at the center of a cir-cular current loop of radius R and (b) at point P near the current loop inFigure 28.37? Both loops carry a current of constant magnitude I.

28.8 Magnetic field of a moving charged particleLet us now use the Biot-Savart law to find an expression for the magnetic fieldcaused by charged particles moving at constant velocity.* Consider first a straightwire carrying a current of magnitude I and aligned with the x axis, as in Figure28.38a. The magnetic field generated at point P by a small segment is given byEq. 28.12:

, (28.17)

where is a unit vector pointing from the segment to the point at which wewish to determine the magnetic field, and rsP is the distance between the segmentand the point P. Suppose the segment contains an amount of charge dq. Let thecharge carriers responsible for the current take a time interval dt to have displace-ment (Figure 28.38b). According to the definition of current (Eq. 27.2), we

rÆ

sP

dBs � oI4�R

d�

dxÆ

dxÆ

dxÆ

B � �arcdBs �

oI4�R ��

0d�

B � oI�4�R

4

r̂ sP

dBÆ

s � o

4�

I dxÆ

� r̂ sP

rsP2

dBs � o

4�

I dl sin 90°R2 �

o

4�

I dlR2

* In the derivation that follows, we assume and ignore any relativistic effects as describedin Sections 27.4 and 27.8.

v �� co

dq I

t + dt

t

dq I

I

(b)

(c)

(a)

P

rsP

→

dx→

dx→

t + dt

t

q

q v

v

dx→

→

→

Figure 28.38 (a) A small seg-ment of a straight current-car-rying wire causes a magneticfield at point P. In a time inter-val dt, (b) an amount of chargedq and (c) a particle carryingcharge q have a displacement

. This displacement causesthe current I through the wire.dxÆ

I

2R

R

P

Figure 28.37 A circular currentloop carrying a current of con-stant magnitude I (Checkpoint28.16).

have

, (28.18)

and so , (28.19)

where is the velocity at which the charge carriers move down the wire. In thelimiting case where the segment contains just a single charge carrier carryinga charge q (Figure 28.38c), dq becomes q and

. (28.20)

Substituting this result into Eq. 28.17, we obtain an expression for the magnetic fieldof a single moving charged particle:

(single particle), (28.21)

where rpP is the distance between the particle and P, and is the unit vectorpointing from the particle to P.

■ Example 28.8: Magnetic field of a moving electron

An electron carrying a charge e = –1.60 ¥ 10–19 C moves in a straight line at a speedv = 3.0 ¥ 107 m/s. What are the magnitude and direction of the magnetic field causedby the electron at a point P 10 mm ahead of the electron and 20 mm away fromits line of motion?

Getting started: I begin by drawing the moving electron and the point P atwhich I am to determine the magnetic field (Figure 28.39).

Devise plan: The magnetic field of a moving charged particle is given by Eq.28.21. The unit vector points from the electron to P, and so points intothe page. Because the charge of the electron is negative, the magnetic field pointsin the opposite direction, out of the page.

Execute plan: From the figure I see that the magnitude of is rpP =22 mm, and sin q = (20 mm)/(22 mm) = 0.89. Substi-

tuting these values into Eq. 28.11 thus yields for the magnitude of the magnetic field

✔

Evaluate result: The magnitude of the magnetic field I obtained is much toosmall to be detected, but that’s what I would expect for a single electron. I can ver-ify the direction of the magnetic field by applying the right-hand current rule to thecurrent caused by the electron. Because the electron carries a negative charge, itsmotion to the right causes a current to left. Point my right-hand thumb to the left,

I �dqdt

dxÆ

vÆ

IdxÆ

�dqdt

dxÆ

� dqdxÆ

dt� dq v

Æ

r̂ sP

rÆ

pP

r̂pP vÆ

� r̂ pP

IdxÆ

� qvÆ

B�(4��10�7 T.mA)

4�

(1.60�10�19C) (3.0�107 ms) (0.89)

(22 � 10�3 m)2�8.6�10�16 T

(10 mm)2 � (20 mm)2 �

BÆ

� o

4�

qvÆ

� r̂ pP

rpP2

4

3

2

1

[(H1F)] 27

–e

p

v

r

10 mm

20 m

m

r̂pP

θ

→

→pP

Figure 28.39

I find that my right-hand fingers curl out of the page at P, in agreement with whatI determined earlier.

We can now combine the expression for the magnetic field caused by a movingcharged particle (Eq. 28.21) with that for the magnetic force exerted on anothermoving charged particle (Eq. 27.19 ) to find the magnetic interactionbetween two moving charged particles. Consider the situation illustrated in Figure28.40. The magnetic field caused by particle 1 at the location of particle 2 is given byEq. 28.21:

, (28.22)

where r12 is the magnitude of the vector pointing from particle 1 to particle 2and is a unit vector pointing along . Substituting this expression into Eq.27.19, we obtain for the magnetic force exerted by particle 1 on particle 2

. (28.23)

Notice the appearance of the double vector product: we must first take the vectorproduct of and and then take the vector product of with the result ofthe first vector product. Equation 28.23 also shows that we need a moving charged particle in order to gen-

erate a magnetic field ( ) and another moving charged particle to “feel” thatmagnetic field ( ), in agreement with our discussion in the first part of this chap-ter.In contrast to the electric force, the magnetic force does not satisfy Eq. 8.15

( ). To see this, consider the two positively charged moving particles shownin Figure 28.41. Particle 1, carrying a charge q1 travels in the positive x direction, whileparticle 2, carrying charge q2, travels in the positive y direction. The force exerted by1 on 2 is given by Eq. 28.23. Applying the right-hand vector product rule, we find thatthe vector product on the right side of Eq. 28.23 points out of the page. Ap-plying the right-hand vector product rule again to the vector product of and

, we see that points in the positive x direction.*The force exerted by 2 on1 is obtained by switching the subscripts 1 and 2 in Equation 28.23:

, (28.24)

where we have used that q1q2 = q2q1 and . Applying the right-hand vectorproduct rule twice we find that points in the positive y direction, and so the mag-netic forces that the two particles exert on each other, while equal in magnitude do notpoint in opposite directions, as we would expect from Eq. 8.15.

FÆ

pB � qv

Æ� B

Æ

vÆ

1 � r̂12

vÆ

1 � r̂12

BÆ

1(rÆ

2) � o

4�

q1vÆ

1 � r̂ 12

r122

rÆ

12

r̂12 rÆ

12

FÆB12 � q2v

Æ

2 � BÆ

1(rÆ

2) � o

4�

q1q2r122 v

Æ

2 � (vÆ

1 � r̂12)

r212 � r221

FÆB21 �

o

4�

q1q2r122 v

Æ

1 � (vÆ

2 � r̂21)

FÆB12

vÆ

2

FÆ

12 � �FÆ

21

vÆ

2 � 0vÆ

1 � 0

vÆ

2vÆ

1 r̂ 12

FÆB21


q1

q2

O

v1

v2

r1 r12

r2

r̂12

→

→→

→

→

Figure 28.40 Magnetic interac-tion of two moving chargedparticles.

r12 →

q2

q1

v2

v1

→

→

B-�eld due to 1 out of page here

B-�eld due to 2 into page here

→F21

B

→F12

B

y

x

Figure 28.41 While the forcesthat moving charged particlesexert on each other are equal inmagnitude, they do not neces-sarily point in opposite direc-tions.

* Here is another way you can determine the direction of the magnetic force exerted by particle 1 onparticle 2. The motion of particle 1 corresponds to a current directed in the positive x direction. Usingthe right-hand current rule, it follows that the magnetic field of 1 points out of the xy plane at anylocation above the x axis. The magnetic field due to 1 at the location of 2 thus points out of theplane. The motion of particle 2 corresponds to a current directed in the positive y direction. To findthe direction of the magnetic force exerted on this “current” we can use the right-hand force rule.When you curl the fingers of your right hand from the positive y axis to the direction of at par-ticle 2 (out of the page), your thumb points in the positive x direction, as we found before.

BÆ

1

We derived Eq. 8.15 from the fact that the momentum of an isolated system of twoparticles is constant, which in turn follows from conservation of momentum, one of themost fundamental laws of physics. The electric and magnetic fields of isolated movingcharged particles, however, are not constant and as we shall see in Chapter 30, we canassociate a flow of both momentum and energy with changing electric and magneticfields. Therefore we need to account not only for the momenta of the two chargedparticles in Figure 28.41, but also for the momentum carried by their fields — whichgoes beyond the scope of this book. Even though Eq. 8.15 is not satisfied by the mag-netic force, the momentum of the system comprising the two particles and their fieldsis still constant.Substituting Coulomb’s law and Eq. 28.23 into Eq. 27.20, we obtain an expression

for the electromagnetic force that two moving charged particles exert on each other.

. (28.25)

By comparing the result obtained in Example 28.6 with Eq. 27.38 and substituting(Eq. 24.7), we see that or , where co is

the speed of light. Using this information, we can write Eq. 28.25 as

( ). (28.26)

This equation is the most general expression for the electromagnetic interactionbetween moving charged particles. Because in most applications the speeds ofcharged particles are significantly smaller than the speed of light, the second termin the square brackets in Eq. 28.26 is much smaller than the first term, which rep-resents the electric contribution to the electromagnetic force from Coulomb’s law.

28.17 Consider two protons 1 and 2, each carrying a charge +e = 1.6 ¥ 10–19

C, separated by 1.0 mm moving at 3 ¥ 105 m/s parallel to each other and per-pendicular to their separation. (a) What is the direction of the magnetic forcethat each proton exerts on the other? (b) Determine the ratio of the magni-tudes of the magnetic and electric forces that the two exert on each other.

v �� co

k � 1(4��o) o�o � 1c2o o � 1(�oc2o)

FÆEB12 �

14��o

q1q2r122 [r̂12 � o�ov

Æ

2 � (vÆ

1 � r̂12) ]

FÆEB12 �

14��o

q1q2r122 � r̂ 12 �

vÆ

2 � (vÆ

1 � r̂ 12)c2o �

[(H1F)] 29

28.8 Magnetic fields of moving charged particles 31

Chapter glossary(SI units of physical quantities are given in parentheses)

Ampère’s law: The line integral of the magnetic fieldalong a closed path (called an Ampèrian path) is pro-portional to the current enclosed by the path, :

(28.1)

In analogy to Gauss’ law in electrostatics, Ampère’slaw allows us to determine the magnetic field for cur-rent distributions that exhibit planar, cylindrical, ortoroidal symmetry.

Biot-Savart law: Expression giving the magnetic field ata point P due to a small segment of a wire carryinga current I:

, (28.12)

where is a unit vector pointing from the segmentto the point at which the magnetic field is evaluated.The Biot-Savart law can be used to calculate the mag-netic field of a current-carrying conductor of arbitraryshape by integration:

(28.10)BÆ

� �currentpath

dBÆ.

�O BÆ.dlÆ

� oIenc.

r̂ sP

dBÆ

s � o

4�

I dlÆ

� r̂ sPrsP2

dlÆ

Ienc

Ampèrian path: Closed path along which the magnet-ic field is integrated in Ampère’s law.

Current loop: A current-carrying conductor in theshape of a loop. The magnetic field pattern of a currentloop is similar to that of a magnetic dipole.

Magnetic dipole moment (Am2): Vector pointingfrom the S-pole to the N-pole for a bar magnet or alongthe axis of a planar current loop in the direction givenby the right-hand thumb when the fingers of that handare curled along the direction of the current throughthe loop. In an external magnetic field, the magnetic di-pole moment tends to align in the direction of the ex-ternal magnetic field.

Permeability constant (Tm/A): Constant relatingthe current enclosed by an Ampèrian path and the lineintegral of the magnetic field along that loop. In vacu-um:

.

Solenoid: Long, tightly wound helical coil of wire. Themagnetic field of a current-carrying solenoid is similarto that of a bar magnet.

Toroid: Solenoid bent into a circle. The magnetic fieldof a toroid is completely contained within the wind-ings of the toroid.

o

Æ

o � 4� � 10�7 T.mA

32 CHAPTER 28 Magnetic fields of charged particles in motion Solutions

Solutions

28.1 (a) See Figure S28.1a. (b) The two particles exertno magnetic forces on each other because it takes amoving charged particle to detect the magnetic fieldof another moving charged particle. (c) See FigureS28.1b. The forces exerted on the horizontal wire causea torque that tends to align that wire with the verticalone. Conversely, the forces exerted on the vertical wirecause a torque in the opposite direction, tending toalign that wire with the horizontal one.

28.2 See Figure S28.2.

28.3 Comparing Figures 28.6b and 28.7; you see thatthe positively charged ring has a magnetic field similarto the field of a bar magnet, with its north pole up. Thenegatively charged ring thus has its north pole down,and this north pole is directly above the north pole of

the positive ring. Because two north poles repel eachother, the interaction is repulsive.

28.4 No. By definition the electric field points frompositive charge carriers to negative charge carriers(Section 23.2). The electric field along the axis passingthrough the poles of an electric dipole therefore pointsfrom the positive end to the negative end. The electricdipole moment, however, by definition points in theopposite direction (Section 23.8).

28.5 (a) Because the magnetic field remains perpen-dicular to the horizontal sides of the loop, you knowfrom Eq. 27.4 that the magnitude of the magnetic forceexerted on each side is , where l is thelength of as each side of the loop. Because none ofthese quantities changes, the magnitude of the forceexerted on the horizontal sides stays the same. (b) SeeFigure S28.5ab. The lever arm of the force exerted onthe horizontal sides become smaller as the loop rotatesand so the torque caused by these forces decreases. (c)In the initial position, the vertical sides are parallel tothe magnetic field, which means the magnetic force ex-erted on them is zero (Eq. 27.7, withq = 0). After the loop has rotated 90°, the vertical sidesare perpendicular to and thus subject to an outwardmagnetic force. For 0 < q < 90°, the magnitude of theforce is again given by Eq. 27.7. As you can verify usingthe right-hand force rule, the forces exerted on the ver-tical sides are directed outward along the rotation axisand so cause no torque. (d) See Figure S28.5c. Becausethe top and bottom of the loop are now reversed, thedirection of the torque reverses.

28.6 See Figures S28.6, where the circular loop is ap-proximated by a series of vertical and horizontal seg-ments. The vertical segments experience no force, but

FB � � I �lB sin�

BÆ

FBmax � � I �lB

P2

P1

P5

P3

P4

(b)

(a)

–eF→w

BeF

→eBw

F→1B2

F→1B2

F→2B1

F→2B1

v→

B→

B→

B→

–

I

I

I

Figure S28.1

B1

B2 B→

12

(a)

B1

B2→ B

12

(b)

Figure S28.2

Solutions 33

the horizontal ones do. All the horizontal segments addup to the same length as two sides of a square strad-dling the circle, and thus Eq. 27.4, , tellsyou that the magnitude of the magnetic force exertedon the horizontal segments is the same as the force ex-erted on the horizontal side of the square loop. Some

FBmax � � I �lB

horizontal segments are closer to the rotation axis,however, and so the moment arms of the forces actingon these closer segments are smaller than the momentarms of the forces acting on the square loop. Thereforethe torque on the circular loop is smaller than that onthe square loop.

28.7 With the magnetic field on the left stronger thanthat on the right, . Therefore the vectorsum of the forces exerted on these two sides is nonze-ro and points upward. The effect is a clockwise rota-tion due to the torque and an upward acceleration dueto the upward vector sum of the forces.

28.8 Increasing the current increases the magnitude ofthe magnetic field and so the line integral increases.

28.9 (a) If the direction of the current is reversed, thedirection of the magnetic field is reversed, and so thealgebraic sign of the line integral is reversed. (b) Be-cause, the value of the line integral of the magnetic fieldaround a closed path does not depend on the positionof the wire inside the path, the second wire by itselfgives rise to the same line integral as the first wire.Adding the second wire thus doubles the value of theline integral. (c) Reversing the current though the sec-ond wire flips the sign of the line integral. If its valuewas C, it is –C after the current is reversed. The sum ofthe two line integrals is thus C + (–C) = 0.

28.10 If the path tilts, you can break it down into arcs,radial, segments, and axial segments, which are seg-ments parallel to the wire. Because the magnetic fieldis perpendicular to the axial segments, they don’t con-tribute to the line integral. For the same reason, ofcourse, the radial segments contribute nothing. So the

FBleft � FB

right

axis

Figure S28.6

wire

current out

Rr

r

Ampèrian loop 2

Ampèrian loop 1

Figure S28.12 Cross section through a current-carrying wire.

θ

B B

(a) (b)

mm

pivot pivot

leverarm

leverarm

current outof page

current outof page

current outof page

current into page

current into page

current into page

F→B

F→B

F→B

F→B

→

θ

B

(c)

m

pivot

F→B

F→B

→

→

Figure S28.5

34 CHAPTER 28 Magnetic fields of charged particles in motion Solutions

line integral again is that along all the arcs, which isequivalent to a single complete circular path.

28.11 (a) Reversing the direction of the currentthrough wire 1 changes the sign of its contributionfrom negative to positive, so that adding the contribu-tions of wires 1 and 3 yields a positive value. (b) Theloop does not encircle wire 2, and so reversing the cur-rent through wire 2 does not affect the answer to Ex-ercise 28.2. (c) Reversing the direction of the loopinverts the sign of all the contributions to the line inte-gral, but because the integral is zero, nothing changes.

28.12 See Figure S28.12. Inside the wire, use Am-pèrian path 1 of radius r < R. For this loop,

,

but the loop encircles only part of the cross section soencircles only part of the current I through the wire.The area of the cross-section of the wire is pR2, and thecross-sectional area of the Ampèrian path 1 is pr2, mak-ing the fraction of the wire cross section enclosed bythe loop (pr2)/(pR2) = r2/R2. Thus, the right side of Am-père’s law is . Ampère’s law thusyields , or

. Outside the wire, use Ampèrianpath 2 in Figure S28.12. For this loop,

.

Because the loop encircles all of the current I throughthe wire, , and so the magnetic field out-side the wire is the same as that for a long thin wire:

.

28.13 (a) Figure S28.13 shows a head-on view of themagnetic fields of the two sheets separately and theirsum. The magnetic fields add up to zero between thesheets. Below and above the sheets, the magnetic field

�O BÆ.dlÆ

� Boutside2�r

�O BÆ.dlÆ

� Binside �O dl � Binside 2�r

Boutside � oI2�r

oIenc � oI

Binside(2�r) � o(r2R2)I

Binside � oIr2�R2

oIenc � o(r2R2)I

points to the left and its magnitude is twice that of a sin-gle sheet: . (b) With the currentthrough the lower sheet reversed, its magnetic field isalso reversed. Now the magnetic fields add up to zerobelow and above the sheets. Between them the magnet-ic field is to the right and its magnitude is twice that ofa single sheet: .

28.14 Outside the toroid, you use Ampèrian path 1 inFigure S28.14. The left side of Ampère’s law is stillgiven by Eq. 28.7, but the loop now encircles both thecurrent upward through the inside of each winding andthe current downward through the outside of eachwinding. The enclosed current is thus NI – NI = 0, mak-ing B outside the toroid zero.Inside the toroid’s inner radius, use Ampèrian path

2. This loop encloses no current, so here, too, B = 0.

28.15 For this wire the integration in Example 28.6extends from x = 0 to :x � �

oK

2(12 oK) � oK

B→

B→

B→ = 0

B→

+ =

Figure S28.13

Ampèrian loop 1

Ampèrian loop 2

Figure S28.14

Solutions 35

.

28.16 (a) Using our result from Example 28.7 with f= 2p :

.

(b) The outer arc spans an angle f = p/2 and has a ra-dius 2R. Using our result from Example 28.7 again, itcontributes a magnetic field

.

The inner arc also spans an angle of f = p/2, but it hasradius R, and the current through it runs in the oppo-site direction so the magnetic field due to this arc is di-rected out of the page. For this arc,the magnitude ofthe magnetic field is

.

The two straight segments do not contribute becausethey point straight toward P. Because Binner, whichpoints out of the page, is larger than Bouter, whichpoints into the page, the magnetic field

is out of the page rather than intothe page. The magnitude of the magnetic field is

28.17 (a) The direction of the magnetic force is givenby the vector products in Eq. 28.23. Figure S28.17ashows the vectors appearing in this expression. Youmust first evaluate The direction of the re-sulting vector is found by curling the fingers of yourright hand from to which tells you points into the plane of the drawing (Figure S28.17b).

vÆ

1 r̂ 12, vÆ

1 � r̂ 12

B � oId4� ��

0

dx[x2�d2]32 �

BÆ

� BÆ

inner � BÆ

outer

B � oId4� � 1d2

x[x2�d2]12 �

x��

x�0�

oI4�d

vÆ

1 � r̂ 12.

B � oI(2�)

4�R�

oI2R

Bouter � oI(�2)

4�(2R)�

oI16R

Binner � oI(�2)

4�(R)�

oI8R

B � oI8R � oI16R � oI16R.

Next evaluate the vector product between andthe vector product you just found. The right-hand vec-tor product rule tells you this yields a vector that pointsupward from proton 2 to proton 1. Thus, points up-ward. Applying the same procedure to find youfind that it points downward, telling you that the mag-netic interaction is attractive, like that of two parallelcurrent-carrying wires.

(b) Because the angles involved in the vector productsin Eq. 28.23 are all 90°

.

The magnitude of the electric force is given byCoulomb’s law:

.

The ratio is

.

Substituting numerical values, you find that this ratio is10–6.

vÆ

2

FÆB12

FÆB12

F12B �

o

4�

q1q2r122 v1v2

F12B

F12E �

� o

4� �q1q2r122 v1v2

� 14��o

�q1q2r122

� o�ov1v2

F12E �

14��o

q1q2r122

1

2

v1

v2

r̂12

v1 x r̂12

v1 x r̂12

1

2

v1

v2

r̂12

v2 x (v1 x r̂12)

1

2

v1

v2

r̂12

(a) (b) (c)

→

→ → →

→ →

→ →

→

→

Figure S28.17

magnetic fields of charged particles in motion

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