Place the thumb of your right hand along the direc- 123docz.net

2. Curl your fi ngers; they will then give the direction of BS as the fi eld lines encircle the current.

Right-hand rule number 2: Finding the direction of the magnetic force on a moving charge q.

1. Point the fi ngers of your right hand along the direction of vS.

Figure 20.18 Example 20.3.

Finding the direction of F

S B. A

Sv vS

ⴙq

Figure 20.19 Concept Check 20.4.

What is the direction of ?BS

B?S

S Sv 1

2. Curl your fi ngers in the direction of BS. Always curl through the smallest angle that connects Sv and BS.

3. If q is positive, the magnetic force on q is parallel to your thumb. If q is negative, the magnetic force is in the opposite direction.

2 0. 4 | M A G N E T I C F O R C E O N A N E L E C T R I C C U R R E N T

An electric current consists of a collection of moving charges, so our result for the magnetic force on a single moving charge (Eq. 20.1) can be used to fi nd the magnetic force on a current-carrying wire. This force is important in many applications, including electric motors. Consider a current-carrying wire placed in an external magnetic fi eld as sketched in Figure 20.20A, with BS constant and perpendicular to the wire.

This magnetic fi eld is produced by some external source; it is not produced by the current in the wire. We can calculate the magnetic force on the wire due to this external fi eld by adding up the magnetic forces on all the moving charges in the wire.

Let’s focus on the segment of the wire of length L in Figure 20.20B. The current in this segment is

I5 DQ

Dt (20.9) where Q is the electric charge that passes by one end of the wire segment in a time t; this is just our usual relation between charge and current from Chapter 19.

Applying Equation 20.1, the magnetic force on this moving charge is

FB5qvB5 1DQ2vB (20.10) The velocity of the charge is just

v5 L Dt Inserting this into Equation 20.10 gives

FB5 1DQ2vB5 1DQ2L Dt B

Using the relation between current and charge in Equation 20.9, we get FB5 DQ

Dt LB5ILB

This force on the moving charge is really a force on the wire, so we arrive at Fon wire5ILB

In Figure 20.20, we assumed the magnetic fi eld is perpendicular to the wire. If BS makes an angle u with the wire, one fi nds

Fon wire5ILB sin u (20.11)

The direction of Fon wire is given by (you guessed it) a right-hand rule as illustrated in Figure 20.20C. Begin with the fi ngers of your right hand in the direction of the current and curl them in the direction of the fi eld. Your thumb then points in the direction of FSon wire, downward in Figure 20.20C. The external magnetic fi eld pulls this wire downward as long as the current is directed to the right. The magnetic force on a current is due to the force on a moving charge, so the right-hand rule used here is another example of right-hand rule number 2, with the direction of Sv for a moving charge replaced by the direction of the current.

Magnetic force on a current- carrying wire. The direction is given by right-hand rule number 2.

Figure 20.20 A A current- carrying wire in an external magnetic fi eld. B The current in the wire is due to the motion of charge Q along the wire. C The magnetic force on this moving charge produces a force on the wire. The direction of the force is given by right-hand rule number 2.

Place the fi ngers of your right hand in the direction of the current and wrap your fi ngers in the direction of BS. Your thumb then points in the direction of the magnetic force.

I I

Fon wire

CO N C E P T C H E C K 20. 5 | Finding the Magnetic Force on a Wire

The wire in Figure 20.21 carries a current along the z direction. If there is an external magnetic fi eld in the y direction, is the direction of the magnetic force on the wire along (a) x, (b) x, (c) y, (d) y, (e) z, or (f) z?

20.4 | MAGNETIC FORCE ON AN ELECTRIC CURRENT 657 Figure 20.21 Concept Check 20.5.

y z

E X A M P L E 2 0 . 4 Magnetic Force between Two Wires

Consider two parallel wires, each carrying a current I as sketched in Figure 20.22A.

Find the direction of the magnetic force that the top wire exerts on the bottom one.

Then fi nd the direction of the force that the bottom wire exerts on the top one.

RECOGNIZE T HE PRINCIPLE

To fi nd the force exerted on the bottom wire, we must fi rst determine the direction of the magnetic fi eld produced by the top wire, which we can do by using right-hand rule number 1. We can then apply right-hand rule number 2 to fi nd the force produced by this fi eld on the bottom wire.

SK E TCH T HE PROBLEM

Figure 20.22B shows the direction of the magnetic fi eld produced by the top wire at the location of the bottom wire as well as the application of right-hand rule number 2 to fi nd the force on the bottom wire. Figure 20.22C shows the application of right- hand rule number 2 to fi nd the force on the top wire.

IDENT IF Y T HE REL AT IONSHIPS A ND SOLV E

From right-hand rule number 1, the magnetic fi eld produced by the top wire is directed into the plane of the drawing near the bottom wire as shown in Fig- ure 20.22B. (See also Fig. 20.6.) Applying right-hand rule number 2 then gives a magnetic force on the bottom wire directed upward, and the bottom wire is attracted to the top wire.

Now consider the force on the top wire. The magnetic fi eld produced by the bottom wire in the vicinity of the top wire is directed out of the plane of the drawing (Fig. 20.22C), and applying right-hand rule number 2 shows that the force exerted on the top wire is directed downward, toward the bottom wire. Hence, the force on the top wire is downward and the force on the bottom wire is upward. Two wires carrying parallel currents are attracted to each other.

What does it mean?

The magnetic forces exerted by each wire on the other are an action–reaction pair of forces. Hence, in accord with Newton’s third law (the action–reaction principle), these forces must be in opposite directions, which is precisely what we have found. You should expect that these forces must have equal magnitudes; we’ll show this in Section 20.7.

CO N C E P T C H E C K 20.6 | How Does the Force Depend on the Direction of I?

Consider again the two wires in Figure 20.22 where the currents are parallel and both directed to the right. Suppose the currents are instead in opposite directions, with Itop to the right and Ibottom to the left. Which of the following state- ments is correct?

(a) The force on the bottom wire is still upward in Figure 20.22, but the force on the top wire is now also upward.

(b) The force on the bottom wire is now downward and the force on the top wire is upward, so the overall effect is a repulsive force.

Figure 20.22 Example 20.4.

A Two parallel current-carrying wires. B The magnetic force on the bottom wire. C The magnetic force on the top wire.

I A

C B

Fon bot

I BS Fon top

2 0. 5 | T O R Q U E O N A C U R R E N T L O O P A N D M A G N E T I C M O M E N T S

In Section 20.2, we saw that when a bar magnet is placed in a uniform magnetic fi eld, there is a torque on the magnet (Fig. 20.11). A magnetic fi eld can also produce a torque on a current loop. Figure 20.23 shows a square current loop with sides of length L carrying a current I in a constant magnetic fi eld. In Figure 20.23A, the fi eld is parallel to the plane of the loop and the forces on all four sides of the loop are shown. These forces can all be found using right-hand rule 2. In sides 2 and 4, the current is either parallel or antiparallel to BS, so the forces are zero. The forces on sides 1 and 3, denoted FS1 and SF3, respectively, are in opposite directions in Figure 20.23A and produce a torque around the loop’s axis. The lever arms for these two forces are both L/2, and magnitude of this torque is

t 5F1aL

2b 1F3aL 2b

The force on each wire segment is F1 F3 ILB (from Eq. 20.11), so t 5F1aL

2b 1F3aL

2b 52ILBaL 2b

t 5IL2B (20.12)

Figure 20.23B shows the forces on the loop when its plane is perpendicular to the magnetic fi eld. Now there is a nonzero force on all four sides of the loop; the four forces are of equal magnitude F1 F2 F3 F4 ILB but are in different directions, and the total force on the loop is zero. The total torque is now also zero.

Comparing parts A and B of Figure 20.23, we fi nd that the torque on a current tends to align the plane of the loop perpendicular to BS.

So far, we have considered the torque when the plane of the loop is perpendicular or parallel to the fi eld. When the angle between these two directions is u (Fig. 20.23C), the torque is

t 5IL2B sin u

The factor of L2 is just the area of the square loop. For loops with different shapes, including circular, the torque on a loop of area A is

t 5IAB sin u (20.13)

The torque on a current loop is very similar to the torque on a bar magnet; both the loop and the magnet tend to “line up” with the magnetic fi eld. Both act as a magnetic moment, with the direction of the magnetic moment either along the axis of the bar magnet or perpendicular to the current loop as shown in Figure 20.24.

S 1

S 3

S 4

F 2 S

2 0

4 0 FS1 FS3u ILB B

S SB

L I

Rotation axis of loop

I u u u

A B C

Rotation axis

Side view of current loop

L 2

4 3

3 2

Figure 20.23 Torque on a current loop. A The magnetic forces on the sides of this loop tend to rotate the loop clockwise as viewed from the front, along the axis. B When the plane of the loop is perpendicular to BS, the torque is zero. C The torque on a current loop depends on the angle u the fi eld makes with the direction perpendicular to the loop.

Figure 20.24 A The magnetic moment of a bar magnet is along the axis of the magnet, whereas

B the magnetic moment of a current loop is perpendicular to the loop. In both cases, a magnetic fi eld produces a torque that tends to align the magnetic moment with the fi eld.

Direction of magnetic moment

Bar magnet

Direction of magnetic moment

Current loop

B S

The magnetic torque tends to align the magnetic moment parallel to BS, and the strength of the torque depends on the magnitude of the magnetic moment. For a current loop, the magnitude of the magnetic moment equals IA, the product of the current and loop area.

CO N C E P T C H E C K 20.7 | Torque on a Current Loop

Figure 20.25 shows a current loop in a magnetic fi eld. Is there a torque on the loop and, if so, in what direction?

(a) There is a torque that tends to rotate the loop in direction (a), clockwise.

(b) There is a torque that tends to rotate the loop in direction (b), counter- clockwise.

2 0. 6 | M O T I O N O F C H A R G E D PA R T I C L E S I N T H E P R E S E N C E O F E L E C T R I C A N D M A G N E T I C F I E L D S

We considered the magnetic force on a charged particle in a magnetic fi eld in Sec- tion 20.3 and found that various types of motion, including circular and helical trajectories, are possible. Also of interest is the motion of an electric charge in the presence of a combination of magnetic and electric fi elds. We’ll analyze a few such cases and describe some applications in this section.

The Mass Spectrometer

Scientists often want to separate ions according to their mass or charge. For example, archaeologists use a technique called carbon dating to determine an object’s age.

This technique is based on the observations that there are two isotopes of carbon atoms—denoted 12C and 14C (called “carbon 12” and “carbon 14,” respectively)—

and that the relative amounts of 12C and 14C in certain types of materials such as an animal bone can be used to determine the bone’s age. (See Section 30.5.) One way to distinguish 12C from 14C is through the difference in their masses, which can be done using a mass spectrometer.

Figure 20.26 shows the essential parts of a mass spectrometer. The charges are typically ions such as 12C or 14C that enter from the left, and for simplicity we’ll assume they all have the same speed v. The ions then pass into a region in which the magnetic fi eld is perpendicular to the velocity. In Figure 20.26, the magnetic fi eld is out of the plane and perpendicular to the plane of the drawing, producing a magnetic force that causes the ions to move in a circle (as in Fig. 20.15B). The radius of this circle is (Eq. 20.6)

r5 mv

qB (20.14) where v is the velocity and q is the charge of the ion.

Suppose the incoming ions are of two types, 12C and 14C, with different masses m12 and m14. According to Equation 20.14, the different values of m lead to different values of r, so the two types of ions follow different circular arcs through the mass spectrometer. A mass spectrometer has an ion detector positioned at a certain radius r as shown in Figure 20.26. The value of either v or B is then adjusted until the detector gives a maximum ion current, indicating that the ion’s circular trajec- tory has the radius r of the detector. For our hypothetical example with 12C and

14C, the mass spectrometer could be used to measure the ion currents for 12C and

14Cseparately and thereby determine the relative amounts of the two isotopes in an object, which would then be used to fi nd the object’s age.

Another application of mass spectrometers is to “fi ngerprint” unknown molecules. Molecules of unknown composition are fi rst ionized—by bombarding them

20.6 | MOTION OF CHARGED PARTICLES IN THE PRESENCE OF ELECTRIC AND MAGNETIC FIELDS 659 Figure 20.25 Concept Check 20.7.

(a) (b) 30 B

Figure 20.26 Design of a mass spectrometer.

Incoming ions

Ion detector

S Sv

with electrons, for instance—producing several different types of ions that are then fed to a mass spectrometer. The spectrometer then measures the values of v, B, and r corresponding to each ion. Equation 20.14 can be rearranged to give

q m 5 v

Because v, B, and r are all measured, the mass spectrometer gives q/m, the ratio of charge to mass for the ions. Chemists use this information to deduce the composition of each ion and hence the composition of the original molecule.

E X A M P L E 2 0 . 5 Using a Mass Spectrometer

A mass spectrometer (Fig. 20.26) is used in a carbon dating experiment. The incoming ions are a mixture of12C and 14Chaving a velocity v 1.0 105 m/s, and the magnetic fi eld is B 0.10 T. The ion detector is fi rst positioned to fi nd the value of r for

12Cand then moved to fi nd the value of r for14C. How far must the detector move?

RECOGNIZE T HE PRINCIPLE

The location of the detector depends on r, the radius of the ion’s circular path. The velocity v and fi eld B are given, and the ions 12Cand14C each have a charge e.

Equation 20.14 gives the value of r in terms of v, B, and m, the mass of the ion. If we can also fi nd the mass m of each ion, we can therefore calculate r.

SK E TCH T HE PROBLEM

Figure 20.26 shows the problem. The ions follow a circular arc from where they enter the mass spectrometer to where they enter the detector.

IDENT IF Y T HE REL AT IONSHIPS

The value of r in the mass spectrometer depends on the mass of the ion, so we need to fi nd the masses of 12Cand14C. The ion 12Ccontains six protons, six neutrons, and fi ve electrons. Consulting the table of fundamental constants on the inside front cover, we fi nd a total mass of

m1256mp16mn15me

m125611.67310227 kg2 1611.67310227 kg2 1519.11310231 kg2 m1252.0310226 kg

A similar calculation gives the mass of14C:

m1452.3310226 kg SOLV E

From Equation 20.14, the radius in the mass spectrometer for12Cis r125 m12v

and the charge is q 1.60 1019 C. Inserting the given values, we fi nd r125m12v

qB 5 12.0310226 kg2 11.03105 m/s2

11.60310219 C2 10.10 T2 50.13 m The same approach gives

r1450.14 m

The ion detector must move a distance equal to the difference in the diameters of the circular trajectories, so it must move a distance

Dr521r142r122 < 0.02 m or about 2 cm.

What does it mean?

The difference between r12 and r14 is not large, so to get a more accurate value we would need to employ more signifi cant fi gures in our calculation. Even so, we can see that the trajectories of the two ions would differ by about 2 cm, which can easily be measured in a modern mass spectrometer.

Hall Effect: How Do We Know If the Charge Carriers Are Positive or Negative?

An electric current is produced by moving electric charges. In Chapter 19, we saw that the value of the current is proportional to the product of the charge q and velocity v of the charge carriers (Eq. 19.6), or

I~qv (20.15)

Suppose we reverse the direction of the velocity, changing from v to v, and at the same time change the charge from q to q. The new current is

Inew~ 12q2 12v2 5qv

which equals the old value. A particular value of the current can thus be produced by a positive charge moving to the right (Fig. 20.27A) or by a negative charge moving to the left (Fig. 20.27B). Because the current in a metal is carried by electrons (negative charges), the picture in Figure 20.27B should apply, but how do we really know? Is there some measurement we can perform that distinguishes between these two possibilities?

The answer to this question was given more than a century ago by Edwin Hall.1 He suggested placing a current-carrying wire in a magnetic fi eld directed perpendicular to the current. Suppose the current is produced by positive charges moving to the right (Fig. 20.28A). If BS is directed into the plane of the drawing, the magnetic force on the moving charges is directed upward, causing the positively charged carriers to be defl ected toward the wire’s upper edge. The result is an excess of positive charge on the top edge of the wire and a defi cit of positive charge on the bottom, giving an electric potential difference between the top and bottom edges. In Figure 20.28A, the upper edge has a positive electric potential relative to the bottom, which can be measured by attaching a voltmeter to the wire’s top and bottom edges.

Now consider what happens if the current is carried instead by negative charges (electrons) moving to the left (Fig. 20.28B). Applying right-hand rule number 2 for negative charges, we fi nd that the magnetic force is again directed toward the top edge of the wire. The magnetic force is thus in the same direction as with the positive charge carriers in Figure 20.28A, and the electrons in Figure 20.28B are

20.6 | MOTION OF CHARGED PARTICLES IN THE PRESENCE OF ELECTRIC AND MAGNETIC FIELDS 661 Insight 20.2

APPLICATIONS OF MASS SPECTROMETERS TO GENOMICS Mass spectrometers are now being used in work on genomics and pro- teomics. For example, a protein of unknown sequence is fi rst “cut”

into fragments (e.g., peptides) using biochemical techniques. The fragments are then analyzed with a mass spectrometer that gives the ratios of q/m for the different ions (using Eq.

20.14). This information, together with the pattern of mass spectrometer intensities, is compared with mass spectrometer data on known peptides. With knowledge of the peptide

“identities,” the sequence of the original protein can then be determined.

1Hall had the idea for this experiment when he was a student! This question is still a good homework problem.

Figure 20.27 The Hall effect gives a way to measure if a current is caused by the motion of A positive charges traveling to the right or B negative charges traveling to the left. In both cases, the current is directed to the right.

ⴙq ⴙq

v I

ⴚq ⴚq

Positive charges moving to the right

Negative charges moving to the left

The current direction is the same.

Figure 20.28 Hall’s experiment. Motion of electric charges in a wire if the current is carried by

A positive charges traveling to the right, or B negative charges traveling to the left.

I B

Sv ⴙq V

ⴙ ⴙ ⴙ

ⴚ ⴚ ⴚ Positive charges

accumulate on the top edge.

Force on positive charge carriers is upward.

FB S

ⴙ ⴙ ⴙ ⴚ ⴚ ⴚ V I

Sv ⴚq

Negative charges accumulate on the top edge.

Force on negative charge carriers is again upward.

FB S

Place the thumb of your right hand along the direction of the current