29.1 Magnetic Fields and Forces 29.2 Magnetic Force Acting on a Current-Carrying Conductor 29.3 Torque on a Current Loop in a Uniform Magnetic Field 29.4 Motion of a Charged Particle in a Uniform Magnetic Field 29.5 Applications Involving Charged Particle

Magnetic Fields

CHAPTER OUTLINE

29.1 Magnetic Fields and Forces

29.2 Magnetic Force Acting on a Current-Carrying Conductor

29.3 Torque on a Current Loop in a Uniform Magnetic Field

29.4 Motion of a Charged Particle in a Uniform Magnetic Field

29.5 Applications Involving Charged Particles Moving in a Magnetic Field

29.6 The Hall Effect

▲ Magnetic fingerprinting allows fingerprints to be seen on surfaces that otherwise would not allow prints to be lifted. The powder spread on the surface is coated with an organic

material that adheres to the greasy residue in a fingerprint. A magnetic “brush” removes the excess powder and makes the fingerprint visible. (James King-Holmes/Photo Researchers, Inc.)

M any historians of science believe that the compass, which uses a magnetic needle,

was used in China as early as the 13th century B . C ., its invention being of Arabic or Indian origin. The early Greeks knew about magnetism as early as 800 B . C . They discov- ered that the stone magnetite (Fe 3 O 4 ) attracts pieces of iron. Legend ascribes the

name magnetite to the shepherd Magnes, the nails of whose shoes and the tip of whose staff stuck fast to chunks of magnetite while he pastured his flocks.

In 1269 a Frenchman named Pierre de Maricourt found that the directions of a needle near a spherical natural magnet formed lines that encircled the sphere and passed through two points diametrically opposite each other, which he called the poles of the magnet. Subsequent experiments showed that every magnet, regardless of its shape, has two poles, called north (N) and south (S) poles, that exert forces on other magnetic poles similar to the way that electric charges exert forces on one another. That is, like poles (N–N or S–S) repel each other, and opposite poles (N–S) attract each other.

The poles received their names because of the way a magnet, such as that in a compass, behaves in the presence of the Earth’s magnetic field. If a bar magnet is sus- pended from its midpoint and can swing freely in a horizontal plane, it will rotate until its north pole points to the Earth’s geographic North Pole and its south pole points to the Earth’s geographic South Pole. 1

In 1600 William Gilbert (1540–1603) extended de Maricourt’s experiments to a variety of materials. Using the fact that a compass needle orients in preferred directions,

he suggested that the Earth itself is a large permanent magnet. In 1750 experimenters used a torsion balance to show that magnetic poles exert attractive or repulsive forces on each other and that these forces vary as the inverse square of the distance between in- teracting poles. Although the force between two magnetic poles is otherwise similar to the force between two electric charges, electric charges can be isolated (witness the electron and proton) whereas a single magnetic pole has never been isolated. That is, magnetic poles are always found in pairs. All attempts thus far to detect an isolated magnetic pole have been unsuccessful. No matter how many times a permanent magnet is cut in two, each piece always has a north and a south pole. 2

Hans Christian

Oersted

The relationship between magnetism and electricity was discovered in 1819 when,

during a lecture demonstration, the Danish scientist Hans Christian Oersted found Danish Physicist and Chemist that an electric current in a wire deflected a nearby compass needle. (1777–1851) 3 In the 1820s,

Oersted is best known for observing that a compass needle deflects when placed near a wire

1 Note that the Earth’s geographic North Pole is magnetically a south pole, whereas its geo- carrying a current. This important graphic South Pole is magnetically a north pole. Because opposite magnetic poles attract each

discovery was the first evidence other, the pole on a magnet that is attracted to the Earth’s geographic North Pole is the magnet’s

of the connection between north pole and the pole attracted to the Earth’s geographic South Pole is the magnet’s south pole.

electric and magnetic 2 There is some theoretical basis for speculating that magnetic monopoles—isolated north or

phenomena. Oersted was also south poles—may exist in nature, and attempts to detect them are an active experimental field of

the first to prepare pure investigation.

aluminum. (North Wind Picture

Archives) The same discovery was reported in 1802 by an Italian jurist, Gian Dominico Romognosi, but

was overlooked, probably because it was published in an obscure journal.

CHAPTER 29• Magnetic Fields

further connections between electricity and magnetism were demonstrated indepen- dently by Faraday and Joseph Henry (1797–1878). They showed that an electric current can be produced in a circuit either by moving a magnet near the circuit or by changing the current in a nearby circuit. These observations demonstrate that a changing magnetic field creates an electric field. Years later, theoretical work by Maxwell showed that the reverse is also true: a changing electric field creates a mag- netic field.

This chapter examines the forces that act on moving charges and on current- carrying wires in the presence of a magnetic field. The source of the magnetic field is described in Chapter 30.

29.1 Magnetic Fields and Forces

In our study of electricity, we described the interactions between charged objects in terms of electric fields. Recall that an electric field surrounds any electric charge. In addition to containing an electric field, the region of space surrounding any moving electric charge also contains a magnetic field. A magnetic field also surrounds a mag- netic substance making up a permanent magnet.

Historically, the symbol B has been used to represent a magnetic field, and this is the notation we use in this text. The direction of the magnetic field B at any loca- tion is the direction in which a compass needle points at that location. As with the electric field, we can represent the magnetic field by means of drawings with mag- netic field lines.

Figure 29.1 shows how the magnetic field lines of a bar magnet can be traced with the aid of a compass. Note that the magnetic field lines outside the magnet point away from north poles and toward south poles. One can display magnetic field patterns of a bar magnet using small iron filings, as shown in Figure 29.2.

We can define a magnetic field B at some point in space in terms of the magnetic force F B that the field exerts on a charged particle moving with a velocity v, which we call the test object. For the time being, let us assume that no electric or gravitational fields are present at the location of the test object. Experiments on various charged particles moving in a magnetic field give the following results:

Properties of the magnetic force

• The magnitude F B of the magnetic force exerted on the particle is proportional to

on a charge moving in a mag-

the charge q and to the speed v of the particle.

netic field B

• The magnitude and direction of F B depend on the velocity of the particle and on

the magnitude and direction of the magnetic field B. • When a charged particle moves parallel to the magnetic field vector, the magnetic

force acting on the particle is zero.

At the Active Figures link at http://www.pse6.com, you can move the compass around

Active Figure 29.1 Compass needles can be used to

and trace the magnetic field

trace the magnetic field lines in the region outside a

lines for yourself.

bar magnet.

SECTION 29.1• Magnetic Fields and Forces

Henry Leap and Jim Lehman

Figure 29.2 (a) Magnetic field pattern surrounding a bar magnet as displayed with iron filings. (b) Magnetic field pattern between opposite poles (N–S) of two bar magnets. (c) Magnetic field pattern between like poles (N–N) of two bar magnets.

• When the particle’s velocity vector makes any angle ! ! 0 with the magnetic field,

the magnetic force acts in a direction perpendicular to both v and B; that is, F B is

perpendicular to the plane formed by v and B (Fig. 29.3a). • The magnetic force exerted on a positive charge is in the direction opposite the

direction of the magnetic force exerted on a negative charge moving in the same direction (Fig. 29.3b).

• The magnitude of the magnetic force exerted on the moving particle is propor- tional to sin !, where ! is the angle the particle’s velocity vector makes with the direction of B.

We can summarize these observations by writing the magnetic force in the form

F B "q v!B (29.1)

Vector expression for the magnetic force on a charged particle moving in a magnetic field

+q

(a)

(b) Figure 29.3 The direction of the magnetic force F B acting on a charged particle

moving with a velocity v in the presence of a magnetic field B. (a) The magnetic force is

perpendicular to both v and B. (b) Oppositely directed magnetic forces F B are exerted

on two oppositely charged particles moving at the same velocity in a magnetic field. The dashed lines show the paths of the particles, which we will investigate in Section 29.4.

CHAPTER 29• Magnetic Fields

(a)

(b)

Figure 29.4 Two right-hand rules for determining the direction of the magnetic force F B " q v ! B acting on a particle with charge q moving with a velocity v in a magnetic field B. (a) In this rule, the fingers point in the direction of v, with B coming out of your palm, so that you can curl your fingers in the direction of B. The direction of v!B , and the force on a positive charge, is the direction in which the thumb points. (b) In this rule, the vector v is in the direction of your thumb and B in the direction of

your fingers. The force F B on a positive charge is in the direction of your palm, as if you

are pushing the particle with your hand.

which by definition of the cross product (see Section 11.1) is perpendicular to both v and B. We can regard this equation as an operational definition of the magnetic field at some point in space. That is, the magnetic field is defined in terms of the force acting on a moving charged particle.

Figure 29.4 reviews two right-hand rules for determining the direction of the cross product v ! B and determining the direction of F B . The rule in Figure 29.4a depends on our right-hand rule for the cross product in Figure 11.2. Point the four fingers of your right hand along the direction of v with the palm facing B and curl them toward B. The extended thumb, which is at a right angle to the fingers, points

in the direction of v ! B. Because F B " q v ! B, F B is in the direction of your thumb if q is positive and opposite the direction of your thumb if q is negative. (If you need more help understanding the cross product, you should review pages 337 to 339, in- cluding Fig. 11.2.)

An alternative rule is shown in Figure 29.4b. Here the thumb points in the direc- tion of v and the extended fingers in the direction of B. Now, the force F B on a posi- tive charge extends outward from your palm. The advantage of this rule is that the force on the charge is in the direction that you would push on something with your hand—outward from your palm. The force on a negative charge is in the opposite direction. Feel free to use either of these two right-hand rules.

The magnitude of the magnetic force on a charged particle is

Magnitude of the magnetic

F B " ! q !vB sin ! (29.2)

force on a charged particle moving in a magnetic field

where ! is the smaller angle between v and B. From this expression, we see that F B is zero when v is parallel or antiparallel to B (! " 0 or 180°) and maximum when v is perpendicular to B (! " 90°).

There are several important differences between electric and magnetic forces: • The electric force acts along the direction of the electric field, whereas the mag-

netic force acts perpendicular to the magnetic field. • The electric force acts on a charged particle regardless of whether the particle is

moving, whereas the magnetic force acts on a charged particle only when the parti- cle is in motion.

SECTION 29.1• Magnetic Fields and Forces

• The electric force does work in displacing a charged particle, whereas the magnetic force associated with a steady magnetic field does no work when a particle is dis- placed because the force is perpendicular to the displacement.

From the last statement and on the basis of the work–kinetic energy theorem, we conclude that the kinetic energy of a charged particle moving through a magnetic field cannot be altered by the magnetic field alone. In other words, when a charged particle moves with a velocity v through a magnetic field, the field can alter the direction of the velocity vector but cannot change the speed or kinetic energy of the particle.

From Equation 29.2, we see that the SI unit of magnetic field is the newton per coulomb-meter per second, which is called the tesla (T):

1T"1

The tesla

C#m/s

Because a coulomb per second is defined to be an ampere, we see that

N 1T"1 A#m

A non-SI magnetic-field unit in common use, called the gauss (G), is related to the

tesla through the conversion 1 T " 10 4 G. Table 29.1 shows some typical values of mag-

netic fields.

Quick Quiz 29.1 The north-pole end of a bar magnet is held near a posi- tively charged piece of plastic. Is the plastic (a) attracted, (b) repelled, or (c) unaf- fected by the magnet?

Quick Quiz 29.2

A charged particle moves with velocity v in a magnetic

field B. The magnetic force on the particle is a maximum when v is (a) parallel to B, (b) perpendicular to B, (c) zero.

Quick Quiz 29.3 An electron moves in the plane of this paper toward the top of the page. A magnetic field is also in the plane of the page and directed toward the right. The direction of the magnetic force on the electron is (a) toward the top of the page, (b) toward the bottom of the page, (c) toward the left edge of the page, (d) toward the right edge of the page, (e) upward out of the page, (f) downward into the page.

Table 29.1

Some Approximate Magnetic Field Magnitudes Source of Field

Field Magnitude (T)

Strong superconducting laboratory magnet

Strong conventional laboratory magnet

Medical MRI unit

Bar magnet

Surface of the Sun

Surface of the Earth

Inside human brain (due to nerve impulses)

CHAPTER 29• Magnetic Fields

Example 29.1 An Electron Moving in a Magnetic Field

An electron in a television picture tube moves toward the Because v ! B is in the positive z direction (from the right-

hand rule) and the charge is negative, F B is in the negative z axis (Fig. 29.5). Surrounding the neck of the tube are coils

front of the tube with a speed of 8.0 % 10 6 m/s along the x

direction.

of wire that create a magnetic field of magnitude 0.025 T, directed at an angle of 60° to the x axis and lying in the xy (B) Find a vector expression for the magnetic force on the

plane.

electron using Equation 29.1.

(A) Calculate the magnetic force on the electron using Solution We begin by writing a vector expression for the Equation 29.2.

velocity of the electron:

Solution Using Equation 29.2, we find the magnitude of v " (8.0 % 10 6 ˆi) m/s the magnetic force:

and one for the magnetic field:

F B " ! q !vB sin !

B " (0.025 cos 60&ˆi ' 0.025 sin 60&ˆj)T

19 C)(8.0 % 10 6 m/s)(0.025 T)(sin 60&)

" (0.013 ˆi ' 0.022 ˆj)T

2.8 % 10 $ 14 N The force on the electron, using Equation 29.1, is

F B "q v%B

" ($e)[(8.0 % 10 6 ˆi) m/s] % [(0.013ˆi ' 0.022 jˆ)T]

" ($e)[(8.0 % 10 6 ˆi) m/s] % [(0.013ˆi)T] ' ($e)[(8.0 % 10 6 ˆi) m/s] % [(0.022ˆj)T] " ($e)(8.0 % 10 6 m/s)(0.013 T)(ˆi ! ˆi ) ' ($e)(8.0 % 10 6 m/s)(0.022 T)(ˆi ! ˆj) " ($1.6 % 10 $ 19 C)(8.0 % 10 6

m/s)(0.022 T) ˆk

–e

B where we have used Equations 11.7a and 11.7b to eva- luate iˆ ! iˆ and iˆ ! jˆ . Carrying out the multiplication, we

find,

($2.8 % 10 $ 14 N)ˆk

This expression agrees with the result in part (A). The mag- the electron is in the negative z direction when v and B lie in

Figure 29.5 (Example 29.1) The magnetic force F B acting on

nitude is the same as we found there, and the force vector is the xy plane.

in the negative z direction.

29.2 Magnetic Force Acting on a Current-Carrying Conductor

If a magnetic force is exerted on a single charged particle when the particle moves through a magnetic field, it should not surprise you that a current-carrying wire also experiences a force when placed in a magnetic field. This follows from the fact that the current is a collection of many charged particles in motion; hence, the resultant force exerted by the field on the wire is the vector sum of the individual forces exerted on all the charged particles making up the current. The force exerted on the particles is transmitted to the wire when the particles collide with the atoms making up the wire.

Before we continue our discussion, some explanation of the notation used in this book is in order. To indicate the direction of B in illustrations, we sometimes present perspective views, such as those in Figure 29.5. If B lies in the plane of the page or is present in a perspective drawing, we use blue vectors or blue field lines with arrowheads. In non-perspective illustrations, we depict a magnetic field

SECTION 29.2• Magnetic Force Acting on a Current-Carrying Conductor

perpendicular to and directed out of the page with a series of blue dots, which

B out of page:

represent the tips of arrows coming toward you (see Fig. 29.6a). In this case, we label the field B out . If B is directed perpendicularly into the page, we use blue crosses, which represent the feathered tails of arrows fired away from you, as in Figure 29.6b. In this case, we label the field B in , where the subscript “in” indicates “into the page.” The same notation with crosses and dots is also used for other quantities that might be per- pendicular to the page, such as forces and current directions.

One can demonstrate the magnetic force acting on a current-carrying conductor by hanging a wire between the poles of a magnet, as shown in Figure 29.7a. For ease in

(a)

visualization, part of the horseshoe magnet in part (a) is removed to show the end face of the south pole in parts (b), (c), and (d) of Figure 29.7. The magnetic field is

B into page:

directed into the page and covers the region within the shaded squares. When the

current in the wire is zero, the wire remains vertical, as shown in Figure 29.7b.

However, when the wire carries a current directed upward, as shown in Figure 29.7c,

the wire deflects to the left. If we reverse the current, as shown in Figure 29.7d, the

wire deflects to the right.

Let us quantify this discussion by considering a straight segment of wire of length L

and cross-sectional area A, carrying a current I in a uniform magnetic field B, as shown in Figure 29.8. The magnetic force exerted on a charge q moving with a drift velocity (b)

B. To find the total force acting on the wire, we multiply the force q v d ! B Figure 29.6 (a) Magnetic field exerted on one charge by the number of charges in the segment. Because the volume lines coming out of the paper are

v d is q v d !

indicated by dots, representing the of the segment is AL, the number of charges in the segment is nAL, where n is the

tips of arrows coming outward. number of charges per unit volume. Hence, the total magnetic force on the wire of

(b) Magnetic field lines going into length L is

the paper are indicated by crosses, representing the feathers of arrows

going inward. We can write this expression in a more convenient form by noting that, from Equation

F B " (q v d ! B)nAL

27.4, the current in the wire is I " nqv d A. Therefore,

F B "I L!B (29.3)

Force on a segment of current-carrying wire in a uniform magnetic field

where L is a vector that points in the direction of the current I and has a magnitude equal to the length L of the segment. Note that this expression applies only to a straight segment of wire in a uniform magnetic field.

Figure 29.8 A segment of a current-carrying wire in a magnetic

Figure 29.7 (a) A wire suspended vertically between the poles of a magnet. (b) The field B. The magnetic force setup shown in part (a) as seen looking at the south pole of the magnet, so that the

exerted on each charge making up magnetic field (blue crosses) is directed into the page. When there is no current in

the current is q v d ! B and the net the wire, it remains vertical. (c) When the current is upward, the wire deflects to the

force on the segment of length left. (d) When the current is downward, the wire deflects to the right.

L is I L ! B.

CHAPTER 29• Magnetic Fields

Now consider an arbitrarily shaped wire segment of uniform cross section in

a magnetic field, as shown in Figure 29.9. It follows from Equation 29.3 that the

I magnetic force exerted on a small segment of vector length d s in the presence of a field B is

ds

(29.4) where d F B is directed out of the page for the directions of B and ds in Figure 29.9. We

d F B "Ids!B

can consider Equation 29.4 as an alternative definition of B. That is, we can define the magnetic field B in terms of a measurable force exerted on a current element, where the force is a maximum when B is perpendicular to the element and zero when B is

Figure 29.9 A wire segment of

parallel to the element.

arbitrary shape carrying a current I To calculate the total force F B acting on the wire shown in Figure 29.9, we integrate in a magnetic field B experiences a

magnetic force. The magnetic

Equation 29.4 over the length of the wire:

force on any segment ds is I ds ! B

s!B (29.5) should use the right-hand rule to

and is directed out of the page. You

F B "I

confirm this force direction. where a and b represent the end points of the wire. When this integration is carried out, the magnitude of the magnetic field and the direction the field makes with the vector ds may differ at different points.

We now treat two interesting special cases involving Equation 29.5. In both cases, the magnetic field is assumed to be uniform in magnitude and direction.

Case 1.

A curved wire carries a current I and is located in a uniform magnetic field B, as shown in Figure 29.10a. Because the field is uniform, we can take B outside the integral in Equation 29.5, and we obtain

F B "I # " d s $ ! B (29.6)

But the quantity " a b d s represents the vector sum of all the length elements from a to b. From the law of vector addition, the sum equals the vector L(, directed from a to b. Therefore, Equation 29.6 reduces to

F B "I L( ! B (29.7) From this we conclude that the magnetic force on a curved current-carrying wire

in a uniform magnetic field is equal to that on a straight wire connecting the end points and carrying the same current.

Figure 29.10 (a) A curved wire carrying a current I in a uniform magnetic field. The total magnetic force acting on the wire is equivalent to the force on a straight wire of length L( running between the ends of the curved wire. (b) A current-carrying loop of arbitrary shape in a uniform magnetic field. The net magnetic force on the loop is zero.

SECTION 29.2• Magnetic Force Acting on a Current-Carrying Conductor

Case 2. An arbitrarily shaped closed loop carrying a current I is placed in a uniform magnetic field, as shown in Figure 29.10b. We can again express the magnetic force

acting on the loop in the form of Equation 29.6, but this time we must take the vector sum of the length elements ds over the entire loop:

F B "I d s ! B

Because the set of length elements forms a closed polygon, the vector sum must be zero. This follows from the procedure for adding vectors by the graphical method.

Because % ds " 0, we conclude that F B " 0; that is, the net magnetic force acting on

any closed current loop in a uniform magnetic field is zero.

Quick Quiz 29.4 The four wires shown in Figure 29.11 all carry the same current from point A to point B through the same magnetic field. In all four parts of the figure, the points A and B are 10 cm apart. Rank the wires according to the magni- tude of the magnetic force exerted on them, from greatest to least.

Figure 29.11 (Quick Quiz 29.4) Which wire experiences the greatest magnetic force?

Quick Quiz 29.5

A wire carries current in the plane of this paper toward

the top of the page. The wire experiences a magnetic force toward the right edge of the page. The direction of the magnetic field causing this force is (a) in the plane of the page and toward the left edge, (b) in the plane of the page and toward the bottom edge, (c) upward out of the page, (d) downward into the page.

Example 29.2 Force on a Semicircular Conductor

A wire bent into a semicircle of radius R forms a closed on the straight portion of the wire and on the curved circuit and carries a current I. The wire lies in the xy

portion.

plane, and a uniform magnetic field is directed along the positive y axis, as shown in Figure 29.12. Find the

Solution The magnetic force F 1 acting on the straight por- magnitude and direction of the magnetic force acting

tion has a magnitude F 1 " ILB " 2IRB because L " 2R and

CHAPTER 29• Magnetic Fields

B To find the magnetic force F 2 acting on the curved part, we use the results of Case 1. The magnetic force on the

I curved portion is the same as that on a straight wire of length 2R carrying current I to the left. Thus, F

2 " ILB " 2IRB. The direction of F 2 is into the page based on the right-hand rule for the cross product L ! B. Because the wire lies in the xy plane, the two forces on the loop can be expressed as

I F 1 " 2IR Bˆ k

Figure 29.12 (Example 29.2) The net magnetic force acting on a closed current loop in a uniform magnetic field is zero. In

F 2 " $ 2IR Bˆ k the setup shown here, the magnetic force on the straight

portion of the loop is 2IRB and directed out of the page, and the magnetic force on the curved portion is 2IRB directed into

The net magnetic force on the loop is the page.

' F"F 1 'F 2 " 2IR Bˆ k $ 2IR Bˆ k"0

the wire is oriented perpendicular to B. The direction of F 1 is out of the page based on the right-hand rule for the cross

Note that this is consistent with Case 2, because the wire product L ! B.

forms a closed loop in a uniform magnetic field.

29.3 Torque on a Current Loop in a Uniform

Magnetic Field

I In the preceding section, we showed how a magnetic force is exerted on a current-

a # carrying conductor placed in a magnetic field. With this as a starting point, we now $ show that a torque is exerted on a current loop placed in a magnetic field. The results

I of this analysis will be of great value when we discuss motors in Chapter 31.

I " Consider a rectangular loop carrying a current I in the presence of a uniform mag- netic field directed parallel to the plane of the loop, as shown in Figure 29.13a. No

b magnetic forces act on sides ! and " because these wires are parallel to the field;

(a)

hence, L ! B " 0 for these sides. However, magnetic forces do act on sides # and $ because these sides are oriented perpendicular to the field. The magnitude of these forces is, from Equation 29.3,

F 2 "F 4 " IaB

The direction of F 2 , the magnetic force exerted on wire #, is out of the page in the view shown in Figure 29.13a, and that of F 4 , the magnetic force exerted on wire $, is into the

page in the same view. If we view the loop from side " and sight along sides # and $,

F 4 we see the view shown in Figure 29.13b, and the two magnetic forces F 2 and F 4 are di-

rected as shown. Note that the two forces point in opposite directions but are not di- rected along the same line of action. If the loop is pivoted so that it can rotate about

(b)

Figure 29.13 (a) Overhead view of a rectangular current loop in a

point O, these two forces produce about O a torque that rotates the loop clockwise. The uniform magnetic field. No mag-

magnitude of this torque ) max is

netic forces are acting on sides !

" IabB lel to B. Forces are acting on sides

and " because these sides are paral-

) max "F 2 'F 4 " (IaB)

' (IaB)

# and $, however. (b) Edge view of the loop sighting down sides # and

where the moment arm about O is b/2 for each force. Because the area enclosed by the $ shows that the magnetic forces F 2 loop is A " ab, we can express the maximum torque as

(29.8) clockwise. The purple dot in the left

and F 4 exerted on these sides create a torque that tends to twist the loop

) max " IAB

This maximum-torque result is valid only when the magnetic field is parallel to the plane circle represents current in wire #

coming toward you; the purple cross of the loop. The sense of the rotation is clockwise when viewed from side ", as indicated in the right circle represents current

in Figure 29.13b. If the current direction were reversed, the force directions would also in wire $ moving away from you.

reverse, and the rotational tendency would be counterclockwise.

SECTION 29.3• Torque on a Current Loop in a Uniform Magnetic Field

– sin b θ θ O 2

At the Active Figures link

Active Figure 29.14 An end view of the loop in

at http://www.pse6.com, you

Figure 29.13b rotated through an angle with

can choose the current in the

respect to the magnetic field. If B is at an angle !

loop, the magnetic field, and

with respect to vector A, which is perpendicular to

the initial orientation of the loop

the plane of the loop, the torque is IAB sin ! where

and observe the subsequent

F 4 the magnitude of A is A, the area of the loop.

motion.

Now suppose that the uniform magnetic field makes an angle ! * 90° with a line perpendicular to the plane of the loop, as in Figure 29.14. For convenience, we assume

that B is perpendicular to sides # and $. In this case, the magnetic forces F 1 and F 3

exerted on sides ! and " cancel each other and produce no torque because they pass µ

through a common origin. However, the magnetic forces F 2 and F 4 acting on sides #

and $ produce a torque about any point. Referring to the end view shown in Figure

29.14, we note that the moment arm of F 2 about the point O is equal to (b/2) sin !. Likewise, the moment arm of F 4 about O is also (b/2) sin !. Because F 2 "F 4 " IaB, the

magnitude of the net torque about O is

b b )"F 2 sin ! ' F 4 sin !

" IaB # sin !

' IaB # sin ! $ " IabB sin !

" IAB sin ! I where A " ab is the area of the loop. This result shows that the torque has its

Figure 29.15 Right-hand rule maximum value IAB when the field is perpendicular to the normal to the plane of the

for determining the direction of loop (! " 90°), as we saw when discussing Figure 29.13, and is zero when the field is

the vector A. The direction of the magnetic moment " is the same as

parallel to the normal to the plane of the loop (! " 0). the direction of A.

A convenient expression for the torque exerted on a loop placed in a uniform magnetic field B is

#"I A!B (29.9)

Torque on a current loop in a magnetic field

where A, the vector shown in Figure 29.14, is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop. We determine the direction of A using the right-hand rule described in Figure 29.15. When you curl the fingers of your right hand in the direction of the current in the loop, your thumb points in the direction of A. As we see in Figure 29.14, the loop tends to rotate in the direction of decreasing values of ! (that is, such that the area vector A rotates toward the direction of the magnetic field).

The product I A is defined to be the magnetic dipole moment " (often simply called the “magnetic moment”) of the loop:

""I A (29.10)

Magnetic dipole moment of a

The SI unit of magnetic dipole moment is ampere-meter 2 (A # m 2 ). Using this definition, current loop we can express the torque exerted on a current-carrying loop in a magnetic field B as

#$"!B

Torque on a magnetic moment in a magnetic field

Note that this result is analogous to Equation 26.18, # " p ! E, for the torque exerted on an electric dipole in the presence of an electric field E, where p is the electric dipole moment.

CHAPTER 29• Magnetic Fields

Although we obtained the torque for a particular orientation of B with respect to the loop, the equation # " " ! B is valid for any orientation. Furthermore, although we derived the torque expression for a rectangular loop, the result is valid for a loop of any shape.

If a coil consists of N turns of wire, each carrying the same current and enclosing the same area, the total magnetic dipole moment of the coil is N times the magnetic dipole moment for one turn. The torque on an N-turn coil is N times that on a one- turn coil. Thus, we write # " N" loop ! B"" coil ! B.

In Section 26.6, we found that the potential energy of a system of an electric dipole in an electric field is given by U " $ p ! E. This energy depends on the orientation of the dipole in the electric field. Likewise, the potential energy of a system of a magnetic dipole in a magnetic field depends on the orientation of the dipole in the magnetic field and is given by

Potential energy of a system

U"$"% B (29.12)

of a magnetic moment in a magnetic field

From this expression, we see that the system has its lowest energy U min "$ +B when " points in the same direction as B. The system has its highest energy U max "' +B when " points in the direction opposite B.

Quick Quiz 29.6 Rank the magnitudes of the torques acting on the rectan- gular loops shown edge-on in Figure 29.16, from highest to lowest. All loops are identi- cal and carry the same current.

Figure 29.16 (Quick Quiz 29.6) Which current loop (seen edge-on) experiences the greatest torque? (Quick Quiz 29.7) Which current loop (seen edge-on) experiences the greatest net force?

Quick Quiz 29.7 Rank the magnitudes of the net forces acting on the rec- tangular loops shown in Figure 29.16, from highest to lowest. All loops are identical and carry the same current.

Example 29.3 The Magnetic Dipole Moment of a Coil

A rectangular coil of dimensions 5.40 cm % 8.50 cm consists (B) What is the magnitude of the torque acting on the of 25 turns of wire and carries a current of 15.0 mA. A 0.350-T

loop?

magnetic field is applied parallel to the plane of the loop. Solution Because B is perpendicular to " coil , Equation

(A) Calculate the magnitude of its magnetic dipole moment.

29.11 gives

Solution Because the coil has 25 turns, we modify Equation )"+ coil B " (1.72 % 10 $ 3 A#m 2 )(0.350 T )

29.10 to obtain

6.02 % 10 $ 4 N#m

+ " NIA " (25)(15.0 % 10 $ coil 3 A)(0.054 0 m)(0.085 0 m)

1.72 % 10 $ 3 A#m 2

SECTION 29.4• Motion of a Charged Particle in a Uniform Magnetic Field

Example 29.4 Satellite Attitude Control

Many satellites use coils called torquers to adjust their orien- Solution We once again apply Equation 29.11, recognizing tation. These devices interact with the Earth’s magnetic field

that the maximum torque is obtained when the magnetic to create a torque on the spacecraft in the x, y, or z direc-

dipole moment of the torquer is perpendicular to the tion. The major advantage of this type of attitude-control

Earth’s magnetic field:

system is that it uses solar-generated electricity and so does ) max " +B " (250 A#m 2 )(3.0 % 10 $ not consume any thruster fuel. 5 T)

If a typical device has a magnetic dipole moment of

7.5 % 10 $ 250 A # m 3 , what is the maximum torque applied to a satellite " N#m when its torquer is turned on at an altitude where the mag-

nitude of the Earth’s magnetic field is 3.0 % 10 $ 5 T?

Example 29.5 The D’Arsonval Galvanometer

An end view of a D’Arsonval galvanometer (see Section We can substitute this expression for + in Equation (1) to

28.5) is shown in Figure 29.17. When the turns of wire mak-

obtain

ing up the coil carry a current, the magnetic field created by the magnet exerts on the coil a torque that turns it (along

(NIA)B $ ,- " 0

with its attached pointer) against the spring. Show that the angle of deflection of the pointer is directly proportional to

NAB

I the current in the coil.

, Solution We can use Equation 29.11 to find the torque ) m

Thus, the angle of deflection of the pointer is directly pro- that the magnetic field exerts on the coil. If we assume that

portional to the current in the loop. The factor NAB/, tells the magnetic field through the coil is perpendicular to the

us that deflection also depends on the design of the meter. normal to the plane of the coil, Equation 29.11 becomes

) m " +B

(This is a reasonable assumption because the circular cross section of the magnet ensures radial magnetic field lines.) This magnetic torque is opposed by the torque due to the spring, which is given by the rotational version of Hooke’s law, ) s "$ ,-, where , is the torsional spring constant and - is the angle through which the spring turns. Because the coil does not have an angular acceleration when the pointer

is at rest, the sum of these torques must be zero: (1)

) m ' ) s " +B $ ,- " 0

Equation 29.10 allows us to relate the magnetic moment of Coil the N turns of wire to the current through them: Spring

Figure 29.17 (Example 29.5) Structure of a moving-coil

+ " NIA

galvanometer.

29.4 Motion of a Charged Particle in a Uniform Magnetic Field

In Section 29.1 we found that the magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the velocity of the particle and that consequently the work done by the magnetic force on the particle is zero. Now consider the special case of a positively charged particle moving in a uniform magnetic field with the initial velocity vector of the particle perpendicular to the field. Let us assume that the direction of the magnetic field is into the page, as in Figure 29.18. As the particle changes the direction of its velocity in response to the magnetic force, the magnetic force remains perpendicular to the velocity. As we found in Section 6.1, if the force is always perpendicular to the velocity, the path of

CHAPTER 29• Magnetic Fields

× B in

the particle is a circle! Figure 29.18 shows the particle moving in a circle in a plane perpendicular to the magnetic field.

q + The particle moves in a circle because the magnetic force F B is perpendicular to v ×

r and B and has a constant magnitude qvB. As Figure 29.18 illustrates, the rotation is

counterclockwise for a positive charge. If q were negative, the rotation would be F B

clockwise. We can use Equation 6.1 to equate this magnetic force to the product of the

particle mass and the centripetal acceleration:

' F " ma c

mv

F B " q vB "

Active Figure 29.18 When the

mv

(29.13) velocity of a charged particle is

r"

qB

perpendicular to a uniform magnetic field, the particle moves

That is, the radius of the path is proportional to the linear momentum mv of the parti- in a circular path in a plane

cle and inversely proportional to the magnitude of the charge on the particle and to perpendicular to B. The magnetic

force F B acting on the charge is the magnitude of the magnetic field. The angular speed of the particle (from Eq.

always directed toward the center

10.10) is

of the circle.

At the Active Figures link

at http://www.pse6.com, you can adjust the mass, speed, and charge of the particle and

The period of the motion (the time interval the particle requires to complete one rev-

the magnitude of the magnetic

olution) is equal to the circumference of the circle divided by the linear speed of the

field to observe the resulting

particle:

circular motion.

These results show that the angular speed of the particle and the period of the circular motion do not depend on the linear speed of the particle or on the radius of the orbit. The angular speed . is often referred to as the cyclotron frequency because charged

y particles circulate at this angular frequency in the type of accelerator called a cyclotron, which is discussed in Section 29.5.

Helical If a charged particle moves in a uniform magnetic field with its velocity at some +q

path arbitrary angle with respect to B, its path is a helix. For example, if the field is directed +

in the x direction, as shown in Figure 29.19, there is no component of force in the x direction. As a result, a x "

0, and the x component of velocity remains constant. However, the magnetic force q v ! B causes the components v y and v z to change in

B time, and the resulting motion is a helix whose axis is parallel to the magnetic field. z

The projection of the path onto the yz plane (viewed along the x axis) is a circle. (The

projections of the path onto the xy and xz planes are sinusoids!) Equations 29.13 to Active Figure 29.19 A charged

29.15 still apply provided that v is replaced by v ! " √ v y 2 'v z 2 .

particle having a velocity vector that has a component parallel to

a uniform magnetic field moves in a helical path.

At the Active Figures link

Quick Quiz 29.8

A charged particle is moving perpendicular to a magnetic

at http://www.pse6.com, you can adjust the x component of

field in a circle with a radius r. An identical particle enters the field, with v perpendicu-

the velocity of the particle and

lar to B, but with a higher speed v than the first particle. Compared to the radius of the

observe the resulting helical

circle for the first particle, the radius of the circle for the second particle is (a) smaller

motion.

(b) larger (c) equal in size.

Quick Quiz 29.9

A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. The magnitude of the magnetic field is increased. Com- pared to the initial radius of the circular path, the radius of the new path is (a) smaller (b) larger (c) equal in size.

SECTION 29.4• Motion of a Charged Particle in a Uniform Magnetic Field

Example 29.6

A Proton Moving Perpendicular to a Uniform Magnetic Field

A proton is moving in a circular orbit of radius 14 cm in a What If? What if an electron, rather than a proton, moves in uniform 0.35-T magnetic field perpendicular to the velocity

a direction perpendicular to the same magnetic field with this of the proton. Find the linear speed of the proton.

same linear speed? Will the radius of its orbit be different? Solution From Equation 29.13, we have

Answer An electron has a much smaller mass than a proton, so the magnetic force should be able to change its velocity

much easier than for the proton. Thus, we should expect the v"

qBr " (1.60 % 10 $ 19 C)(0.35 T)(0.14 m)

1.67 % 10 $ 27 kg

radius to be smaller. Looking at Equation 29.13, we see that r is proportional to m with q, B, and v the same for the electron

4.7 % 10 6 m/s as for the proton. Consequently, the radius will be smaller by the same factor as the ratio of masses m e /m p .

Example 29.7 Bending an Electron Beam Interactive

In an experiment designed to measure the magnitude of a uniform magnetic field, electrons are accelerated from rest through a potential difference of 350 V. The electrons travel along a curved path because of the magnetic force exerted on them, and the radius of the path is measured to be 7.5 cm. (Fig. 29.20 shows such a curved beam of electrons.) If the magnetic field is perpendicular to the beam,

(A) what is the magnitude of the field? Solution Conceptualize the circular motion of the

electrons with the help of Figures 29.18 and 29.20. We Henry Leap and Jim Lehman categorize this problem as one involving both uniform

Figure 29.20 (Example 29.7) The bending of an electron beam in a magnetic field.

circular motion and a magnetic force. Looking at Equation

29.13, we see that we need the speed v of the electron if we are to find the magnetic field magnitude, and v is not given. 7 v 1.11 % 10 m/s

1.5 % 10 8 rad/s Consequently, we must find the speed of the electron based

0.075 m

on the potential difference through which it is accelerated. To finalize this problem, note that the angular speed can Therefore, we also categorize this as a problem in conserva-

be represented as . " (1.5 % 10 8 rad/s)(1 rev/2/ rad) " tion of mechanical energy for an isolated system. To begin

2.4 % 10 7 rev/s. The electrons travel around the circle 24 analyzing the problem, we find the electron speed. For the

million times per second! This is consistent with the very isolated electron–electric field system, the loss of potential

high speed that we found in part (A). energy as the electron moves through the 350-V potential difference appears as an increase in the kinetic energy of

What If? What if a sudden voltage surge causes the accel- the electron. Because K " 0 and K f " i 1

2 m e v , we have

erating voltage to increase to 400 V? How does this affect the angular speed of the electrons, assuming that the

9: m e v 2 ' ($e) ∆V " 0

∆K ' ∆U " 0

magnetic field remains constant?

Answer The increase in accelerating voltage 0V will cause v"

9.11 % 10 e $ 31 √ the electrons to enter the magnetic field with a higher speed √ kg

2e ∆V

2(1.60 % 10 $ 19 C)(350 V)

v. This will cause them to travel in a circle with a larger

" 1.11 % 10 7 m/s

radius r. The angular speed is the ratio of v to r. Both v and r increase by the same factor, so that the effects cancel and

Now, using Equation 29.13, we find the angular speed remains the same. Equation 29.14 is an $ 31 7 m expression for the cyclotron frequency, which is the same as

e v " (9.11 % 10

kg)(1.11 % 10 m/s)

B" (1.60 % 10 $ 19

the angular speed of the electrons. The cyclotron frequency

depends only on the charge q, the magnetic field B, and the "

e r C)(0.075 m)

mass m e , none of which have changed. Thus, the voltage T

surge has no effect on the angular speed. (However, in real- ity, the voltage surge may also increase the magnetic field if

(B) What is the angular speed of the electrons? the magnetic field is powered by the same source as the accelerating voltage. In this case, the angular speed will

Solution Using Equation 29.14, we find that increase according to Equation 29.14.) At the Interactive Worked Example link at http://www.pse6.com, you can investigate the relationship between the radius

of the circular path of the electrons and the magnetic field.

CHAPTER 29• Magnetic Fields

Figure 29.22 The Van Allen

belts are made up of charged particles trapped by the Earth’s nonuniform magnetic field. The magnetic field lines are in blue and the particle paths in red.

When charged particles move in a nonuniform magnetic field, the motion is complex. For example, in a magnetic field that is strong at the ends and weak in the middle, such as Path of

that shown in Figure 29.21, the particles can oscillate back and forth between two particle

positions. A charged particle starting at one end spirals along the field lines until it reaches the other end, where it reverses its path and spirals back. This configuration is known as a magnetic bottle because charged particles can be trapped within it. The

+ magnetic bottle has been used to confine a plasma, a gas consisting of ions and electrons. Such a plasma-confinement scheme could fulfill a crucial role in the control of nuclear fusion, a process that could supply us with an almost endless source of energy. Unfortunately, the magnetic bottle has its problems. If a large number of particles are trapped, collisions between them cause the particles to eventually leak from the system.

Figure 29.21 A charged particle moving in a nonuniform

The Van Allen radiation belts consist of charged particles (mostly electrons and magnetic field (a magnetic bottle)

protons) surrounding the Earth in doughnut-shaped regions (Fig. 29.22). The parti- spirals about the field and

cles, trapped by the Earth’s nonuniform magnetic field, spiral around the field lines oscillates between the end points.

from pole to pole, covering the distance in just a few seconds. These particles originate The magnetic force exerted on