56157 19 ch19 p171-218userhome.brooklyn.cuny.edu/kshum/documents/ism_chapter19.pdf · clicker...

19Magnetism

Clicker Questions

Question M1.01

Description: Introducing the directionality of the magnetic force on a charge.

Question

In a certain region of space there is a uniform magnetic fi eld pointing in the positive z-direction (+z). In what direction should a negative point charge move to experience a force in the positive x-direction (+x)?

1. in the positive z-direction (+z) 2. in the negative z-direction (−z) 3. in the positive x-direction (+x) 4. in the negative x-direction (−x) 5. in the positive y-direction (+y) 6. in the negative y-direction (−y) 7. It can move in any direction. 8. It is impossible for the force to be in the +x-direction when the magnetic fi eld is in the +z-direction.

Commentary

Purpose: To develop your understanding of the direction of the magnetic force on a moving point charge.

Discussion: The direction of the magnetic force is always perpendicular to both the velocity of the point charge and the direction of the magnetic fi eld. So, for example, if velocity v and magnetic fi eld B are in the xy plane, then the force is either in the +z or −z-direction, because the z-axis is perpendicular to the xy plane.

In this case, we want to have a force in the +x-direction and the magnetic fi eld is in the +z-direction. There-fore, we need to have the velocity be somewhere in the yz plane without being along the z-axis. Many possible directions of motion lie in the xy plane, but only two are among the listed answers: (5) and (6). Which of those is correct? Should the velocity be in the +y-direction or the −y-direction?

According to the right-hand rule, the vector v B× is in the +x-direction when v is in the +y-direction and B is in the +z-direction. But the magnetic force is q(v B× ). Since q is negative, the force is in the +x-direction when v is in the −y-direction and B is in the +z-direction.

Key Points:

• A magnetic force on a moving charge acts in a direction perpendicular to both the magnetic fi eld and the charge’s velocity.

• There are two directions perpendicular to any (non-parallel) magnetic fi eld and velocity directions; the right-hand rule tells you which one is meant by the cross product.

• The magnetic force on a negative charge is in the opposite direction of a magnetic force on a positive charge.

171

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172 Chapter 19

For Instructors Only

Students may be less conversant with the right-hand rule than you think! (Also, some left-handed people inadvertently use their left hands.)

Students choosing answer (5) are probably overlooking the sign of the charge.

You should elicit the reasoning of students choosing the correct as well as incorrect answers, since some students commit two errors that “cancel out.”

Students may improperly generalize from this result, thinking that only when the charge is moving along the y-axis does the force point along the x-axis. The minimal requirements (within the given constraints on force and magnetic fi eld) are that the x component of velocity is zero and the y component of velocity is negative; there are no restrictions on the z component of velocity. This point can be raised during discussion, or used as the basis for a follow-up question.

Question M1.02a

Description: Exploring charged particle dynamics in magnetic fi elds.

Question

In each of the following situations, point charge q moves in a uniform magnetic fi eld B. The strength of the magnetic fi eld is indicated by the density of fi eld lines. In each situation, the initial speed v of the charge is the same. For which situation(s) will the charge q travel the longest distance in a certain time T ?

q

vB q v B

qv B q

v B

2 31

4 5

q

vB

1. 1 2. 2 3. 3 4. 4 5. 5 6. 1 & 3 7. 2 & 4 8. 1, 2, 3 & 4 9. 1, 2, 3, 4 & 5 10. Cannot be determined


Magnetism 173

Commentary

Purpose: To develop your understanding of how a magnetic fi eld affects a moving charge.

Discussion: The magnetic force on a point charge is F = q(v B× ), where q is the amount of charge, v is the charge’s velocity, and B is the magnetic fi eld strength at its location in space. The cross product means that the force is perpendicular to both the fi eld and the velocity at all times. Since no other forces act on the point charge, the magnetic force is the net force, so the acceleration is also perpendicular to the velocity and magnetic fi eld.

Since the acceleration is always perpendicular to the velocity, the speed cannot change. Since the charge has the same speed in all fi ve situations, it must travel the same distance in all of them. (Note, however, that its displacement will not be the same for all the situations.)

Key Points:

• The magnetic force on a point charge q moving with velocity v in a magnetic fi eld B is q(v B× ).

• The magnetic force on a moving point charge is always perpendicular to its direction of motion and to the fi eld.

• If the only force on a moving point charge is the magnetic force, its speed stays constant.


It is important to have students who pick one of the other choices verbalize their reasons. This will reveal the nature of their misunderstanding or confusion about the magnetic force (or perhaps about Newtonian motion).

Some students might think that the magnetic fi eld is much like the electric fi eld, and therefore, the answer depends on the density of fi eld lines or the angle between the direction of motion and the fi eld.

Students who choose answer (5) may be interpreting the question as asking about displacement rather than distance, and correctly reasoning that the charge traveling in a straight line will have the largest magnitude of displacement. Students choosing (10) might also be thinking about displacement, but not realizing they can reason without having a time and other values to calculate with.

The next question in this set asks about the displacement. If you are using both questions, defer any discus-sion of displacement until then. One tactic is to present both questions in sequence without discussing or revealing the answer to the fi rst, and then discussing them together. Students misinterpreting this question as about displacement will likely realize their mistake when they see the next. This is good: it sensitizes them to the importance of paying attention to detail and to the precise meaning words have in physics.

Question M1.02b

Description: Exploring charged particle dynamics in magnetic fi elds.

Question

In each of the following situations, point charge q moves in a uniform magnetic fi eld B. The strength of the magnetic fi eld is indicated by the density of fi eld lines. In each situation, the initial speed v of the charge is the same. For which situation(s) will the charge q have the largest displacement in a certain time T ?


174 Chapter 19

q

vB q v B

qv B q

v B

2 31

4 5

q

vB

1. 1 2. 2 3. 3 4. 4 5. 5 6. 1 & 3 7. 2 & 4 8. 1, 2, 3, 4 & 5 9. None of the above 10. Cannot be determined

Commentary

Purpose: To develop your understanding of how a magnetic fi eld affects a moving charge.

Discussion: The magnetic force on a point charge is F = q(v B× ), where q is the amount of charge, v is the charge’s velocity, and B is the magnetic fi eld strength at its location in space. The cross product means that the force depends on the angle between v and B in a way that might seem unintuitive to you. Another way to write the magnetic force is to focus on its magnitude F q B q B= ( ) = ⊥v vsin θ , where θ is the angle between the velocity and the magnetic fi eld and v⊥ is the component of the velocity v perpendicular to the magnetic fi eld B.

In situation (5), the velocity and magnetic fi eld are parallel to each other, so v⊥ = 0, and therefore the accel-eration is zero and the charge moves in a straight line. This is the only situation in which the charge moves in a straight line. Since all of the charges move the same distance (as discussed in the previous question) and the rest follow curved paths, the charge in (5) must have the largest displacement.

Key Points:

• The magnetic force on a point charge q moving with velocity v in a magnetic fi eld B is q(v B× ).

• The magnitude of the magnetic force on a moving point charge can also be written F q B q B= ( ) = ⊥v vsin θ .

• If a point charge’s velocity is parallel or anti-parallel to the magnetic fi eld, the magnetic force on it is zero.


The magnetic force is one of the few quantities students encounter that depends on the perpendicular, rather than parallel, component of a vector, so they may have diffi culty learning to think about it.

It is also common for students to focus on magnitudes rather than directions and think the magnetic force is simply qvB, concluding that it is stronger when fi eld lines are more dense. This may lead some students to

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Magnetism 175

choose (3), (4), and (5). However, a larger force means that the circular path of q is smaller. Other students might pick (1) and (2).

In fact, it is impossible to determine which of the situations other than (5) has the largest displacement, because we’d need to know the time interval T and other values to determine where along their circular or helical trajectories the particles are at the end.

Additional Discussion Questions:

1. Which point charges move in circular paths? Describe the orientation of the path. 2. Of the charges moving in circular paths, order them from smallest to largest radius. 3. Which point charges move in helical paths? Describe the orientation of the path. 4. Of the charges moving in helical paths, order them from smallest to largest radius. 5. Compare the radii of the circular paths to the radii of the helical paths.

Question M1.03

Description: Extending understanding of the Lorentz force law and link to magnetism to Newton’s third law.

Question

A bar magnet moving with speed V passes below a stationary charge q. What can be said about the magni-tude of the magnetic forces on the bar magnet (F

b) and on the charge q (F

q).

V

v = 0q

N

S

1. Fbar

and Fq are both zero.

2. Fbar

is zero and Fq is not zero.

3. Fbar

is not zero and Fq is zero.

4. Fbar

and Fq are both non-zero.

Commentary

Purpose: To develop your understanding of the Lorentz force law, connecting it to Newton’s third law.

Discussion: We know that a charge moving through a magnetic fi eld experiences a force. If the magnetic fi eld moves past the charge, does the charge also experience a force?

Yes. In the magnet’s frame of reference, the charge is moving past it, and thus experiences a force. And since it experiences a force in one frame of reference, and both frames are inertial, it must experience a force in the other — in the original frame, with a stationary charge.


176 Chapter 19

According to Newton’s third law, if the charge experiences a force due to the bar, the bar must experience an equal-magnitude, opposite-direction force due to the charge. But what is the physical mechanism by which the charge exerts a force on the bar? A moving charge is a current, and currents create magnetic fi elds. So, the bar magnet experiences a force due to another magnetic fi eld. Since two magnets can attract or repel, we know that a magnetic fi eld can exert a force on a magnet.

Key Points:

• Newton’s third law holds for magnetic forces.

• A moving charge is a current, which creates a magnetic fi eld.

• A moving magnetic fi eld exerts a force on a stationary charge, just as a stationary magnetic fi eld exerts a force on a moving charge.


This is a good question for extending students’ understanding of magnetic forces into new territory, or for integrating their understanding of various magnetic-related forces.

Students who realize the charge will experience a force might, through blind faith in Newton’s second law, assert that F

bar and F

q are both non-zero without comprehending the mechanism by which the charge exerts

a force on the bar (or vice-versa). The question serves as a context and motivation for discussing this.

Critical students might wonder about our blithe statement that “since it experiences a force in one frame of reference, it must experience a force in the other.” Relativistically, the force may not be the same in both frames, but it must be nonzero.

Question M1.04

Description: Introducing or developing understanding of superposition of magnetic fi elds.

Question

Two identical bar magnets are placed rigidly and anti-parallel to each other as shown. At what locations, if any, is the net magnetic fi eld close to zero?

A

S N

A

CB

D

D

B

D

D

SN

1. A only 2. B only 3. C only 4. D only 5. A and B 6. A, B, and C 7. C and D 8. None of the above.


Magnetism 177

Commentary

Purpose: To explore the superposition of magnetic fi elds.

Discussion: Consider the magnetic fi eld due to a single bar magnet. The magnetic fi eld at any point due to the two magnets will be the sum of the magnetic fi elds due to each individual magnet (“superposition”). Notice that the magnets have opposite orientations, so at point C, the fi eld lines due to one will be in the opposite direction of the fi eld due to the other. Since the magnets are identical, C is the same distance from each magnet, and the lines are exactly anti-parallel at that point, the two fi elds will cancel exactly.

At other points, however, the fi elds won’t completely cancel. At points A, for example, the fi elds are anti-parallel but not the same strength, so they’ll only partially cancel. At points B, fi elds have the same strength but aren’t completely opposite in direction, so the net fi eld won’t be zero. And at D, the fi elds have neither the same strength nor opposing directions. Thus, (3) is the best answer.

Key Points:

• Magnetic fi elds obey superposition: the total magnetic fi eld from two sources is the vector sum of the magnetic fi elds due to each individual source, for every point in space.

• Two superposed fi elds will only cancel completely if they have the same magnitudes and exactly opposite directions.

• You can use the symmetries of a situation to deduce much about where fi elds will cancel.


For this question, fi nding out students’ reasons for their answer is more important than their actual answers.

We recommend having students sketch the fi eld lines, and describe how the strength of the fi eld is related to the fi eld line diagram.

A demonstration with iron fi lings on an overhead projector would be illuminating.

Question M1.06

Description: Developing understanding of electric and magnetic forces.

Question

A charged particle moves into a region containing both an electric fi eld and a magnetic fi eld. Which of the statements below is/are true?

A. The particle cannot accelerate in the direction of B. B. The path of the particle must be a circle. C. Any change in the particle’s kinetic energy is caused by the E fi eld. 1. Only A 2. Only B 3. Only C 4. Both A and B 5. Both A and C 6. Both B and C 7. All are true. 8. None are true.


178 Chapter 19

Commentary

Purpose: To build your understanding of electric and magnetic forces.

Discussion: Proving a statement true can be diffi cult; fi nding one counter-example that disproves a state-ment is often easier. Let’s consider each statement, one at a time.

Can the point charge accelerate in the direction of B? That is, can the net force on the charge point in the direction of B? Yes! If the velocity of the point charge is in the direction of B, the magnetic force is zero. So, if the electric fi eld is parallel to B, then the electric force is parallel to B also. Since the magnetic force is zero, the net force is in the direction of B, as is the acceleration. Statement A must be false.

Must the path of the point charge be in a circle? No! In the situation described above, the point charge would move in a straight line. Statement B must be false.

Can the magnetic force cause any change in kinetic energy? No! The direction of the magnetic force is always perpendicular to both the direction of motion and the direction of the magnetic fi eld. Therefore, the magnetic force is always perpendicular to the displacement of the point charge, even if the charge moves along a curved path. Thus, the work done by the magnetic force is always zero, so the magnetic force can-not contribute to any changes in kinetic energy. Statement C is true.

Key Points:

• An electric fi eld acts on a moving charge by exerting a force in the direction of the fi eld.

• A magnetic fi eld acts on a moving charge by exerting a force perpendicular to both the fi eld and the charge’s velocity.

• A magnetic fi eld can do no work on a moving charge.

• The behavior of a moving charge in a combination of electric and magnetic fi elds depends on the relative directions and strengths of the fi elds and on the initial velocity of the charge.


Students are most familiar with situations in which E and B are perpendicular, and may assume that here without explicitly realizing it. This may cause some to believe that statement A is true.

Statement B should be easy for students to fi nd a counter-example to.

Discussing the proof of statement C provides a good opportunity to review the defi nition of work (and reiterate that “force times distance” is an inadequate defi nition).

Question M1.07a

Description: Introducing charged particle motion in electric and magnetic fi eld combinations.

Question

A charge is released from rest in E and B fi elds. Both fi elds point along the x-axis. Which of the following statements regarding the charge’s motion are correct?


Magnetism 179

y

x

E

Bq

z

1. The charge will travel along a straight-line path. 2. The charge’s speed will change as it travels. 3. The charge will travel in a helical path. 4. The charge will travel in a helical path of increasing pitch. 5. The charge will travel in a circle in the x y plane. 6. 1 and 2 only 7. 2 and 4 only 8. None of the above

Commentary

Purpose: To check and refi ne your understanding of electric and magnetic forces on moving particles.

Discussion: The full Lorentz force law is F E v B= + ×( )q . This implies that an electric fi eld exerts a force on a charge in the direction of the fi eld (for a positive charge) or in the opposite direction (for a negative charge). It also implies that a magnetic fi eld exerts a force that points perpendicularly to both the fi eld and the charge’s velocity, and only if the charge’s velocity has a component perpendicular to the magnetic fi eld.

In this situation, the electric fi eld will cause the stationary charge to accelerate parallel to the magnetic fi eld. Since the charge’s velocity starts with no component perpendicular to the magnetic fi eld, and never gains one, the magnetic force will remain zero. Thus, the particle will simply accelerate along the x-axis. Statements (1) and (2) are both valid, so answer (6) is best.

Key Points:

• An electric fi eld exerts a force on a charge parallel to the fi eld, according to F = qE.

• A magnetic fi eld exerts a force on a moving charge according to F v B= ×q . If the charge has no velocity component perpendicular to the fi eld, the force is zero.


This is the fi rst of two related questions.

Students who answer (3), (4), or (7) may be remembering the “fact” that charges in magnetic fi elds move along helical paths. In a sense, that is true here, for the limiting case of a zero-radius helix.

A misunderstanding of the vector cross product or an inability to apply it reliably are common sources of error here.

Students choosing answer (8) should be encouraged to describe the motion they expect. (One answer that arises occasionally is “The charge fi rst moves in a straight line until it gets some speed, and then. . . .”)


180 Chapter 19

Question M1.07b

Description: Introducing charged particle motion in electric and magnetic fi eld combinations.

Question

A charge has an initial velocity parallel to the y-axis in E and B fi elds. Both fi elds point along the x-axis. Which of the following statements regarding the charge’s motion are correct?

y

x

E

Bq

v

z

1. The charge will travel along a straight-line path. 2. The charge’s speed will change as it travels. 3. The charge will travel in a helical path. 4. The charge will travel in a helical path of increasing pitch. 5. The charge will travel in a circle in the x y plane. 6. 1 and 2 only 7. 2 and 4 only 8. None of the above

Commentary

Purpose: To check and refi ne your understanding of electric and magnetic forces on moving particles.

Discussion: The full Lorentz force law is F E v B= + ×( )q . This implies that an electric fi eld exerts a force on a charge in the direction of the fi eld (for a positive charge) or in the opposite direction (for a negative charge). It also implies that a magnetic fi eld exerts a force that points perpendicularly to both the fi eld and the charge’s velocity, and only if the charge’s velocity has a component perpendicular to the magnetic fi eld.

The charge begins with a velocity in the y-direction, perpendicular to the magnetic fi eld. Therefore, the cross-product rule says that the magnetic force on it will be in the −z-direction. (If the charge is negative, the force will be in the +z-direction.) Since the magnetic force is perpendicular to the charge’s velocity, it does no work, and causes the velocity vector to change direction but not magnitude.

If there were no electric fi eld, the magnetic force would continue to bend the particle’s path without chang-ing its speed, causing it to move in a circle in the yz plane.

At the same time, the electric fi eld exerts a constant force on the charge in the +x-direction, parallel to the electric fi eld. (If the charge is negative, the force will be in the −x-direction.) This will cause it to experi-ence a constant acceleration in the x-direction.

Because of the cross product, the magnitude and direction of the magnetic force do not depend on any x velocity the particle might have. So, the particle moves in a circle in the yz plane while simultaneously accelerating in the x-direction. This results in the particle following a helical (spiral) trajectory of increas-ing pitch (distance between turns): answer (4). If the charge is positive, the helix will proceed to the right; if negative, it will proceed to the left.


Magnetism 181

Key Points:

• An electric fi eld exerts a force on a charge parallel to the fi eld, according to F = qE.

• A magnetic fi eld exerts a force on a moving charge according to F v B= ×q . If the charge has no velocity component perpendicular to the fi eld, the force is zero.

• When magnetic and electric fi elds are parallel, their effects on a charge’s motions are independent: the electric fi eld infl uences motion parallel to the fi elds, and the magnetic fi eld infl uences motion in the plane perpendicular to them.


This is the second of two related questions.

Students who know that the particle will travel in a helical path often forget or don’t realize that the pitch will change due to the electric fi eld’s infl uence, and consequently will answer (3).

Students are likely to struggle with the vector cross product and the direction the magnetic force will point as the particle’s trajectory curves.

Question M1.08

Description: Verbalizing and picturing magnetic fi eld lines.

Question

Consider a long thin straight wire with a current I. Which of the following statements about the magnetic fi eld lines is true?

A. Field lines are parallel to the wire. B. Field lines are perpendicular to the wire. C. Field lines are directed radially away from the wire. D. Field lines are circles centered on any point on the wire. 1. A only 2. B only 3. C only 4. D only 5. A and C only 6. B and D only 7. B and C only 8. None of them is true.


182 Chapter 19

Commentary

Purpose: To develop your ability to describe and picture magnetic fi eld lines.

Discussion: The magnetic fi eld lines due to current fl ow through a wire forms circles centered on the wire, perpendicular to the wire. One form of the right-hand rule for magnetic fi elds says that if you point your right thumb along a wire in the direction current fl ows, and curl your fi ngers, they indicate the direction that these circles of magnetic fi eld lines point. This result can be derived mathematically from the Biot-Savart law, but it’s worth memorizing so you can use it to reason about current and magnetic fi eld problems.

If the fi eld lines were parallel to the wire, they would not obey the direction indicated by the cross product in the Biot-Savart law.

If they pointed radially outward, they would have to end on the wire, and magnetic fi eld lines never begin or end; they always form loops.

Key Points:

• The magnetic fi eld lines generated by a current in a straight wire form circular loops around the wire, perpendicular to it and with a direction given by the right-hand rule.

• Magnetic fi eld lines always form closed loops, and never begin or end.


Students may have seen diagrams of the line geometry, and even be able to reproduce those diagrams, but be unable to describe them in words. This question provides an opportunity to practice relating graphical to verbal representations.

Question M1.09

Description: Introducing and understanding the force a magnetic fi eld exerts on a current-carrying wire.

Question

A very long wire lies in a plane with a short wire segment. The long wire carries current I, while the short wire of length L carries current i. The two wires are parallel to each other. Which of the following state-ments are true?

L

I

id

A. The direction of the magnetic force exerted by the long wire on the short wire is directed away from the long wire.

B. The magnitude of the force on the short wire is μ π0 2IiL d. C. The long wire experiences a force of exactly the same magnitude as the force experienced by the

short wire.


Magnetism 183

1. A only 2. B only 3. C only 4. A and B 5. A and C 6. B and C 7. A, B, and C 8. None of them are true

Commentary

Purpose: To explore and develop familiarity with the force a magnetic fi eld exerts on a current-carrying wire.

Discussion: To answer this question, we need to know the magnetic fi eld created by the long wire, and the force the short wire experiences because of that fi eld. We will assume that we can treat the very long wire as infi nitely long. (Why else would we be told it is “very long” and not given a variable for its length?)

The magnetic fi eld created by an infi nitely long, straight wire carrying current I is μ π0 2I d a distance d away from the wire. This result is worth remembering. It can be derived via the Biot-Savart law (compli-cated) or via Ampere’s law and a symmetry argument (simpler). The magnetic fi eld lines form circles surrounding the wire, with a direction given by the right-hand rule, so in the fi gure the fi eld will point into the page at every point along the short wire.

The force exerted on a segment of wire by a magnetic fi eld B (due to some source other than the segment itself) is F L B= ×i , where i is the current in the segment, L is the length of the segment, and the direction of the vector L is the direction the current is fl owing along the segment. (If B is not the same at every point along the segment — for example, if the segment were perpendicular to the very long wire in this question — you must use that equation separately for every infi nitesimal piece of the wire segment, and integrate.)

Putting our expression for the magnetic fi eld due to the long wire into this force equation, we fi nd that statement B is true.

How about the direction of the force? If the magnetic fi eld due to the long wire points into the page everywhere along the short segment, the right-hand rule indicates that the force exerted will be towards the long wire. So, statement A is false.

Statement C must be true, because Newton’s third law is valid for magnetic forces as well as all other forces. Furthermore, according to that law, the direction of the force on the long wire must be towards the short wire: equal magnitudes, opposite directions. So, the two wires attract each other.

The best answer to this question is therefore (6).

Key Points:

• The magnetic fi eld a distance d away from an infi nitely long wire carrying a current I is μ π0 2I d , circulating around the wire according to the right-hand rule.

• A segment of wire of length L carrying current i, located in an externally generated magnetic fi eld B, experiences a magnetic force F L B= ×i where the direction of L points along the wire in the direction of current fl ow.

• Two parallel current-carrying wires will attract each other if their currents fl ow in the same directions. This is called the pinch effect. (If the currents fl ow in opposite directions, the wires will repel.)


184 Chapter 19


This is a good question for students just encountering the force law for magnetic fi elds acting upon current-carrying wires.

Students claiming statement A is true (answers 1, 4, or 7) might be unable to apply the right-hand rule correctly or reliably. They might also have an intuitive reason for believing the short wire is repelled (for example, by analogy with like charges repelling), so getting them to explain their reasoning is important.

For students claiming statement B is false (answers 1, 3, 5, and 8), your next diagnostic step should be to ask what they think the magnitude of the force is. Those that think it is correct except for the factor of p might be accidentally remembering the magnetic fi eld at the center of a circle of current, rather than for an infi nite line.

One can reason about many of the factors in the answer: for example, the force should get stronger if either current is increased, so i and I must be in the numerator. A longer “short” segment provides more current to be acted upon by the fi eld, so L should be in the numerator as well. The effect should get weaker if the wires are farther apart, so some power of d must appear in the denominator. And μ0 appears in pretty much any magnetic fi eld expression. In fact, using dimensional analysis (i.e., considering units), one can fi gure out what the force magnitude must be to within a multiplicative constant. This exercise can be valuable for students to engage in or at least see you talk through: it helps them learn how to check their answers for reasonability, as well as believe that the mathematics in physics ought to be interpreted.

Confusion about variables is also likely, since this question uses d where r or R is often seen, and has both i and I. If such confusion occurs, we recommend taking the opportunity to make a strong point about the importance of understanding what the variables in defi nitions, laws, and other equations mean, of being able to use them comfortably with any choice of variables, and of being defensive about the fact that many letters are used with different meanings. To help students develop this facility, we recommend that you make a general effort to be inconsistent in your notation (between problems only, not within them, of course) and often at odds with the textbook’s conventions.

Students claiming statement C is false (1, 2, 4, or 8) should be engaged in a discussion to determine whether they really think Newton’s third law is violated (or just forgot about it), and if so, why. This could indicate a serious misunderstanding.

Question M2.01

Description: Reasoning with the Biot-Savart law and superposition of magnetic fi elds.

Question

In all cases the wire shown carries a current I. For which situation is the magnitude of the magnetic fi eld maximum at the point P?


Magnetism 185

P

PR

R

R

I

I

I

3

1 2

R/2P

R

I4

R/2

2πR

Commentary

Purpose: To reason with the Biot-Savart law and develop your intuition for the magnetic fi elds currents generate.

Discussion: First, let’s review some basic results that are very useful for thinking about magnetic fi elds due to steady currents.

A distance R from an infi nitely long straight wire carrying a current I, the magnetic fi eld induced by the current is μ π0 2I R. Similarly, the magnetic fi eld at the center of a current-carrying circular wire of radius R is μ0 2I R. Both of these can be derived from the Biot-Savart law. (The fi eld for an infi nite wire can also be derived from Ampere’s law.)

The fi eld at the center of a half-circle of current has a magnitude of one-half the value for a whole circle of the same radius. This is because the fi eld contribution is in the same direction and has the same magnitude for each infi nitesimal piece of the wire, so the contributions add up without any components that cancel. If you have half as many infi nitesimal segments, you get half as much total fi eld strength. So, each half-circle contributes μ π0 4I R.

The magnetic fi eld due to a fi nite line segment of current is zero for points directly in line with (ahead of or behind) the current, because the angular part of the cross product in the Biot-Savart law involves the sine of the angle between the current’s direction of fl ow and the displacement vector from the current location to the point in question.

Now, we can use these results to reason about the relative strengths of the magnetic fi eld due to the current arrangements shown.

Case 4 must also produce a stronger fi eld than case 1, because both involve two semicircular currents, but for case 3 one of the semicircles has a smaller radius and therefore creates a stronger fi eld at P. The other half is the same for both cases, and the small straight segments in case 4 fl ow directly towards or away from P and don’t create any fi eld there at all.


186 Chapter 19

Case 4 must have a larger fi eld at P than case 3, because both involve similar geometry, but for case 3 the two semicircular currents fl ow in opposite directions around P, so their contributions will partially cancel. For case 4, the fi eld due to the two semicircular segments are both in the same direction.

Case 4 must also be larger than case 2. This can be reasoned two ways. First, the fi eld at the center of a circular current of radius R is larger (by a factor of p ) than the fi eld a distance R away from an infi nite wire having the same current. Since case 2 has only a fi nite segment of wire, the fi eld it creates will be smaller than for an infi nite wire and, therefore, less than case 1, which has already been shown to be less than case 4. The second way is more straightforward. The length of the wire segment in case 1 is the same as for case 2, but all of the infi nitesimal elements are the same distance away from point P. Since the contribution to the fi eld from each element drops off as 1/r, the fi eld at P due to 1 must be larger than in case 2.

Thus, we know case 4 must have the largest magnetic fi eld at P.

Key Points:

• We can compare the magnetic fi elds due to many different current arrangements without actually calculating them.

• Knowing expressions for the magnetic fi elds created by various standard current arrangements (infi nite lines, circular loops, etc.) is helpful.

• The Biot-Savart law lets you reason qualitatively about magnetic fi elds and the variables they depend on.


Students should be encouraged (or admonished) to reason to the answer, rather than determining expres-sions for the fi eld magnitude for all four cases. This can be accomplished by limiting the time students have to decide upon their answers, and — when they complain that they haven’t had enough time — telling them they should be reasoning qualitatively, not calculating.

Some general points that you should try to help students appreciate through this question include: that mag-netic fi eld strength falls off as 1/R; that a circular loop around a point creates a stronger fi eld than an infi nite line at the same distance; and that a point inside a closed current path generally experiences a stronger fi eld than a point outside a similar path.

A good follow-up question, perhaps as an informal rhetorical question for students to ponder as they leave class, asks students to well-order the four cases according to increasing magnetic fi eld strength at P.

Question M2.02

Description: Reasoning with the Biot-Savart law and superposition of magnetic fi elds.

Question

Order the following situations according to the magnitude of the magnetic fi eld at the point P. Order from highest to lowest.


Magnetism 187

1. ABCD 2. ADBC 3. BDAC 4. CADB 5. DABC 6. None of the above

Commentary

Purpose: To reason with the Biot-Savart law and develop your intuition for the magnetic fi elds currents generate.

Discussion: From the Biot-Savart law, we can fi nd that the magnetic fi eld at the center of a circular current loop of radius R is μ0 2I R, and the magnetic fi eld a distance R away from an infi nitely long, straight current is μ π0 2I R, where I is the current in both cases.

Similarly, the Biot-Savart law lets us reason that the magnetic fi eld due to half of a circular current loop will be one-half the value for a complete circle, and the fi eld due to a “semi-infi nite” line (half an infi nite line, starting at the point on the infi nite line closest to P) will be one-half the value for an infi nite line.

The other piece of information we’ll need to reason about the current arrangements in this problem is that the magnetic fi eld lines created by a current make circular loops around the current. If you point the thumb of your right hand along the wire in the direction current fl ows, and curl your fi ngers around the wire, your fi ngers will point the direction of the magnetic fi eld.

Once we know these facts, we can reason about the current arrangements shown in the question. Case A consists of one infi nite line (equivalent to two semi-infi nite lines) and one circular loop (equivalent to two semicircles), with current fl owing such that the magnetic fi elds due to the two oppose each other. We’ll notate this as 2C − 2L, meaning the fi eld strength at P is due to two semicircles minus two opposing semi-infi nite lines.

Case B consists of two semi-infi nite lines and one semicircle, such that the fi elds from all three are in the same direction and add without cancellation. Thus, the fi eld for case B is 1C + 2L. Similarly, the fi eld at P is 2C + 4L for case C. For case D, it is 1.5C + 2L.

So, we can rank the cases in order of decreasing fi eld strength as: BC (2C + 4L) > B

D (1.5C + 2L) > B

B

(1C + 2L) > BA (2C − 2L).


188 Chapter 19

In order to place the last case (case A), we had to determine that 1C + 2L > 2C − 2L. We can show this by not-ing that L = C/p, substituting this into the inequality, and simplifying: 1C 2L C L+ > −2 2 → 4 > p QED.

Key Points:

• We can compare the magnetic fi elds due to many different current arrangements without actually calculating them.

• Knowing expressions for the magnetic fi elds created by various standard current arrangements (infi nite lines, circular loops, etc.) is helpful, as is knowing the fi eld values for “standard” fractions of these arrangements.

• The Biot-Savart law lets you reason qualitatively about magnetic fi elds and the variables they depend on.


Students should be strongly pushed to reason qualitatively about the ranking of the cases, rather than deriving expressions for each. (Some quantitative work is necessary to place case A, but that is best done in terms of an inequality.) This problem is about reasoning with magnetic fi eld generation — involving qualitative features of the Biot-Savart law, symmetry, and superposition — not about calculating fi elds for various current loop shapes.

You might want to open a discussion of what “portions” of these standard current arrangements (loops and infi nite lines) are easy to deduce the fi eld for, and what require detailed calculation. For example, since all points on a circular current loop contribute the same amount to the net magnetic fi eld at the center, we can fi nd the fi eld due to any fraction of a circle by taking that fraction of the fi eld due to an entire circle. For lines, however, only infi nite and semi-infi nite lines (beginning at the point of closest approach to P) are simple to determine; any other fragment requires integrating the Biot-Savart law.

Students often have diffi culty determining what current arrangements Ampere’s law can or cannot be productively applied to. Although Ampere’s law is not particularly useful for answering this question (except as a way to fi nd the fi eld due to an infi nite line), students may be wondering about it, so a discussion may be warranted.

This question works well with Question M2.01: the reasoning is similar, but the application is different enough that one may be used to introduce and develop the ideas and the other to reveal (to students as much as to the instructor) whether students grasp and can apply them.

Question M2.03a

Description: Developing ability to apply the Biot-Savart law, and recognize where it (rather than Ampere’s law) is required.

Question

The diagram shows a circular wire loop of radius R carrying current I. What is the direction of the magnetic fi eld, B, at the center of the loop?


Magnetism 189

closest to you

farthest from you

Top View

R

I

I

1. Left 2. Right 3. Up 4. Down 5. None of the above

Commentary

Purpose: To develop your understanding of the Biot-Savart law.

Discussion: According to the Biot-Savart law, an infi nitesimal segment of current I with length ds creates an infi nitesimal magnetic fi eld d I d rB s r= ( ) ×μ π0

34 at a point if r is the displacement vector from the current element to the point in question. For a current-carrying wire, one can fi nd the total magnetic fi eld by adding up (integrating) the contributions from all of the infi nitely many infi nitesimal current segments. Note the cross product: this means that the magnetic fi eld created will have a direction that is perpendicular to both the current fl ow and the displacement vector r. This means the magnetic fi eld from that segment must form circular loops around the segment. The direction can be determined from the right-hand rule: place your thumb pointing along the current in the direction of positive current fl ow, and curl your fi ngers partially, and your fi ngers will show you the direction in which the magnetic fi eld points as it circles around the segment.

You can use this to determine that for any infi nitesimal segment of the circular current path shown, the infi nitesimal magnetic fi eld at the origin will point “upward” (out of the page in the right-hand diagram). If you add up all the contributions from the entire loop, therefore, the total magnetic fi eld at the center must point upward as well. Answer (3).

Key Points:

• The Biot-Savart law describes the magnetic fi eld due to current fl ow.

• The magnetic fi eld due to a current loop is the sum of the magnetic fi eld contributions from each infi nitesimal segment of the loop.

• Current fl owing through a wire causes magnetic fi eld lines to circle around the wire, with a direction given by the “point the thumb, curl the fi ngers” version of the right-hand rule.


This is a good question for wading into the vector calculus of the Biot-Savart law. Students frequently remain confused about the direction of r, whether it points from the source point to the fi eld point or vice-versa and subsequently have diffi culty reconciling the various forms of the right-hand rule.

For students who already know that the magnetic fi eld lines due to a straight wire forms circles around the wire, and can fi nd the direction, the question is simpler and can be approached by a common-sense super-position argument. (The question is even easier for those familiar with magnetic dipoles.)


190 Chapter 19

This question is a good lead-in to the next one, which asks students to determine the magnitude of the fi eld. While that may seem straightforward and trivial, it remains diffi cult for students who are not yet comfortable with vectors and calculus.

Question M2.03b

Description: Developing ability to apply the Biot-Savart law, and recognize where it (rather than Ampere’s law) is required.

Question

The diagram shows a circular wire loop of radius R carrying current I. What is the magnitude of the mag-netic fi eld, B, at the center of the loop?

closest to you

farthest from you

Top View

R

I

I

1. 0 2. μ0 I/4p R 3. μ0 I/2p R 4. μ0 I/4R 5. μ0 I/2R 6. None of the above.

Commentary

Purpose: To develop your understanding of the Biot-Savart law and help you distinguish it from Ampere’s law.

Discussion: According to the Biot-Savart law, an infi nitesimal segment of current I with length ds creates an infi nitesimal magnetic fi eld d I d rB s r= ( ) ×μ π0

34 at a point if r is the displacement vector from the current element to the point in question. For a current-carrying wire, one can fi nd the total magnetic fi eld by adding up (integrating) the contributions from all of the infi nitely many infi nitesimal current segments. All infi nitesimal segments around the circular current loop in this problem create magnetic fi elds pointing in the “up” direction (as discussed in the previous question), so we can fi nd the magnitude of dB due to one segment and integrate to fi nd the total magnitude.

The magnitude dB we get when we put our known values and angles into the Biot-Savart law is μ π0

24( ) I ds R . All segments make exactly the same contribution to the fi eld at the origin (same distance, same relative angles). Since the value of dB we found is the contribution per length ds of the loop, and the total length of the loop is 2p R, the total fi eld at the origin must be 2 20π μR dB I R = : answer (5).

Key Points:

• The Biot-Savart law describes the magnetic fi eld due to infi nitesimal current element.

• The magnetic fi eld due to a current loop is the sum of the magnetic fi eld contributions from each infi nitesimal segment of the loop.


Magnetism 191

• If all the segments create fi eld contributions with the same direction and magnitude at some point, we don’t actually have to solve an integral to fi nd the total fi eld.

• It is important to recognize when you can fi nd the magnetic fi eld due to steady currents using Ampere’s law and when you must use the Biot-Savart law.


How best to explain and discuss this question depends on your students’ mathematical sophistication and experience with the Biot-Savart law. It provides a good situation for working quantitatively with the law while avoiding ugly integrals.

A general discussion of why Ampere’s law is not useful here may be in order. Students often have a diffi cult time recognizing when they can use Ampere’s law and when they must use the Biot-Savart law. The magnetic fi eld due to a long wire is often found using the Biot-Savart law to demonstrate that the result is the same as when Ampere’s law is used. The signifi cance of this demonstration is frequently lost on many students. Even when they can assert that symmetry is required to use Ampere’s law, they remain uncertain what the symmetry statement applies to.

Answer (3) might indicate that students are recalling the magnetic fi eld due to a straight infi nite wire and not incorrectly applying the Biot-Savart law.

QUICK QUIZZES

1. (b). The force that a magnetic fi eld exerts on a charged particle moving through it is given by F q B q B= = ⊥v vsinθ , where B⊥ is the component of the fi eld perpendicular to the particle’s velocity. Since the particle moves in a straight line, the magnetic force (and hence B⊥, since qv ≠ 0) must be zero.

2. (c). The magnetic force exerted by a magnetic fi eld on a charge is proportional to the charge’s veloc-ity relative to the fi eld. If the charge is stationary, as in this situation, there is no magnetic force.

3. (c). The torque that a planar current loop will experience when it is in a magnetic fi eld is given by τ θ= BIA sin . Note that this torque depends on the strength of the fi eld, the current in the coil, the area enclosed by the coil, and the orientation of the plane of the coil relative to the direction of the fi eld. However, it does not depend on the shape of the loop.

4. (a). The magnetic force acting on the particle is always perpendicular to the velocity of the particle, and hence to the displacement the particle is undergoing. Under these conditions, the force does no work on the particle and the particle’s kinetic energy remains constant.

5. (a) and (c). The magnitude of the force per unit length between two parallel current carrying wires is F I I dl = ( ) ( )μ π0 1 2 2 . The magnitude of this force can be doubled by doubling the magnitude of the current in either wire. It can also be doubled by decreasing the distance between them, d, by half. Thus, both choices (a) and (c) are correct.

6. (b). The two forces are an action-reaction pair. They act on different wires and have equal magni-tudes but opposite directions.


192 Chapter 19

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. The electron moves in a horizontal plane in a direction of 35° N or E, which is the same as 55° E of N. Since the magnetic fi eld at this location is horizontal and directed due north, the angle between the direction of the electron’s velocity and the direction of the magnetic fi eld is 55°. The magnitude of the magnetic force experienced by the electron is then

F q B= = ×( ) ×( ) ×− −v sin . . .θ 1 6 10 2 5 10 0 10 1019 6 C m s 44 1855 3 3 10 T N( ) = × −sin .°

The right-hand rule number 1 predicts a force directed upward, away from the Earth’s surface for a positively charged particle moving in the direction of the electron. However, the negatively charged electron will experience a force in the opposite direction, downward toward the Earth’s surface. Thus, the correct choice is (d).

2. If the magnitude of the magnetic force on the wire equals the weight of the wire, then BI wl sinθ = , or B w I= l sinθ . The magnitude of the magnetic fi eld is a minimum when θ θ= =90 1° and sin . Thus,

B

w

Imin

.

..= = ×

( )( ) =−

l

1 0 10

0 500 20

2 N

0.10 A m T

and (a) is the correct answer for this question.

3. The z-axis is perpendicular to the plane of the loop, and the angle between the direction of this normal line and the direction of the magnetic fi eld is θ = 30 0. °. Thus, the magnitude of the torque experienced by this coil containing N = 10 turns is

τ θ= = ( )( ) ( )( )BIAN sin . . . .0 010 2 0 0 20 0 30 T A m m⎡⎡⎣ ⎤⎦( ) = × ⋅−10 30 0 6 0 10 3sin . .° N m

meaning that (c) is the correct choice.

4. A charged particle moving perpendicular to a magnetic fi eld experiences a centripetal force of magnitude F m r q Bc = =v v2 and follows a circular path of radius r m qB= v . The speed of this proton must be

v = =

×( )( ) ×( )− −qBr

m

1 6 10 0 050 1 0 10

1

19 3. . .

.

C T m

667 104 8 1027

3

×= ×− kg

m s.

and choice (e) is the correct answer.

5. The magnitude of the magnetic fi eld at distance r from a long straight wire carrying current I is B I r= μ π0 2 . Thus, for the described situation,

B =

× ⋅( )( )( ) = ×

−−4 10 1

2 21 10

77

ππ

T m A A

m T

/

making (d) the correct response.

6. The force per unit length between this pair of wires is

F I I

dl= =

× ⋅( )( )( ) = ×

−μπ

ππ

0 1 2

7

2

4 10 3

2 29

T m A A

m

2

110 1 107 6− −× N N∼

and (d) is the best choice for this question.

7. The magnitude of the magnetic fi eld inside the specifi ed solenoid is

B nI

NI= = ⎛

⎝⎜⎞⎠⎟ = × ⋅( )⎛−μ μ π0 0

74 10120

0 50l T m A

m.⎝⎝⎜⎞⎠⎟ ( ) = × −2 0 6 0 10 4. . A T

which is choice (e).


Magnetism 193

8. The magnitude of the magnetic force experienced by a charged particle in a magnetic fi eld is given by F q B= v sinθ , where v is the speed of the particle and θ is the angle between the direc-tion of the particle’s velocity and the direction of the magnetic fi eld. If either v = 0 [choice (e)] or sinθ = 0 [choice (c)], this force has zero magnitude. All other choices are false, so the correct answers are (c) and (e).

9. The force that a magnetic fi eld exerts on a moving charge is always perpendicular to both the direction of the fi eld and the direction of the particle’s motion. Since the force is perpendicular to the direction of motion, it does no work on the particle and hence does not alter its speed. Because the speed is unchanged, both the kinetic energy and the magnitude of the linear momen-tum will be constant. Correct answers among the list of choices are (d) and (e). All other choices are false.

10. By the right-hand rule number 1, when the proton fi rst enters the fi eld, it experiences a force directed upward, toward the top of the page. This will defl ect the proton upward, and as the proton’s velocity changes direction, the force changes direction always staying perpendicular to the velocity. The force, being perpendicular to the motion, causes the particle to follow a circu-lar path, with no change in speed, as long as it is in the fi eld. After completing a half circle, the proton will exit the fi eld traveling toward the left. The correct answer is choice (d).

11. The contribution made to the magnetic fi eld at point P by the lower wire is directed out of the page, while the contribution due to the upper wire is directed into the page. Since point P is equi-distant from the two wires, and the wires carry the same magnitude currents, these two oppositely directed contributions to the magnetic fi eld have equal magnitudes and cancel each other. There-fore, the total magnetic fi eld at point P is zero, making (a) the correct answer for this question.

12. The magnetic fi eld due to the current in the vertical wire is directed into the page on the right side of the wire and out of the page on the left side. The fi eld due to the current in the horizontal wire is out of the page above this wire and into the page below the wire. Thus, the two contributions to the total magnetic fi eld have the same directions at points B (both out of the page) and D (both contributions into the page), while the two contributions have opposite directions at points A and C. The magnitude of the total magnetic fi eld will be greatest at points B and D where the two contributions are in the same direction, and smallest at points A and C where the two contribu-tions are in opposite directions and tend to cancel. The correct choices for this question are (a) and (c).

13. Any point in region I is closer to the upper wire which carries the larger current. At all points in this region, the outward directed fi eld due the upper wire will have a greater magnitude than will the inward directed fi eld due to the lower wire. Thus, the resultant fi eld in region I will be nonzero and out of the page, meaning that choice (d) is a true statement and choice (a) is false. In region II, the fi eld due to each wire is directed into the page, so their magnitudes add and the resultant fi eld cannot be zero at any point in this region. This means that choice (b) is false. In region III, the fi eld due to the upper wire is directed into the page while that due to the lower wire is out of the page. Since points in this region are closer to the wire carrying the smaller current, there are points in this region where the magnitudes of the oppositely directed fi elds due to the two wires will have equal magnitudes, canceling each other and producing a zero resultant fi eld. Thus, choice (c) is true and choice (e) is false. The correct answers for this question are choices (c) and (d).

14. The torque exerted on a single turn coil carrying current I by a magnetic fi eld B is τ θ= BIA sin . The line perpendicular to the plane of each coil is also perpendicular to the direction of the magnetic fi eld (i.e., θ = 90°). Since B and I are the same for all three coils, the torques exerted on them are proportional to the area A enclosed by each of the coils. Coil A is rectangular with area AA

2 m m m= ( )( ) =1 2 2 . Coil B is elliptical with semi-major axis a = 0 75. m and


194 Chapter 19

semi-minor axis b = 0 5. m, giving an area A abB = π or AB2 m m m= ( )( ) =π 0 75 0 5 1 2. . . .

Coil C is triangular with area A base heightC = ( )( )1

2, or AC

2 m m m= ( )( ) =1

21 3 1 5. . Thus,

A A AA C B> > , meaning that τ τ τA C B> > and choice (b) is the correct answer.

15. According to right-hand rule number 2, the magnetic fi eld at point P due to the current in the wire is directed out of the page, meaning that choices (c) and (e) are false. The magnitude of this fi eld is given by B I r= μ π0 2 , so choices (b) and (d) are false. Choice (a) is correct about both the magnitude and direction of the fi eld and is the correct answer for the question.

16. The magnetic fi eld inside a solenoid, carrying current I, with N turns and length L, is

B nIN

LI= = ⎛

⎝⎜⎞⎠⎟μ μ0 0 . Thus, B

N I

LA = μ0 A

A

, BN I

LBB

A

AA= =μ0

2

1

2, and B

N I

LBC

A

AA=

( )=

μ0 2

24 .

Therefore, we see that B B BC A B> > , and choice (d) gives the correct rankings.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

2. No. The force that a constant magnetic fi eld exerts on a charged particle is dependent on the velocity of that particle. If the particle has zero velocity, it will experience no magnetic force and cannot be set in motion by a constant magnetic fi eld.

4. Straight down toward the surface of Earth.

6. The magnet causes domain alignment in the iron such that the iron becomes magnetic and is attracted to the original magnet. Now that the iron is magnetic, it can produce an identical effect in another piece of iron.

8. The magnet produces domain alignment in the nail such that the nail is attracted to the magnet. Regardless of which pole is used, the alignment in the nail is such that it is attracted to the magnet.

10. No. The magnetic fi eld created by a single current loop resembles that of a bar magnet — strongest inside the loop, and decreasing in strength as you move away from the loop. Neither is the fi eld uniform in direction — the magnetic fi eld lines loop through the loop.

12. Near the poles the magnetic fi eld of Earth points almost straight downward (or straight upward), in the direction (or opposite to the direction) the charges are moving. As a result, there is little or no magnetic force exerted on the charged particles at the pole to defl ect them away from Earth.

14. The loop can be mounted on an axle that can rotate. The current loop will rotate when placed in an external magnetic fi eld for some arbitrary orientation of the fi eld relative to the loop. As the current in the loop is increased, the torque on it will increase.

16. (a) The blue magnet experiences an upward magnetic force equal to its weight. The yellow magnet is repelled by the red magnets by a force whose magnitude equals the weight of the yellow magnet plus the magnitude of the reaction force exerted on this magnet by the blue magnet.

(b) The rods prevent motion to the side and prevent the magnets from rotating under their mutual torques. Its constraint changes unstable equilibrium into stable.

(c) Most likely, the disks are magnetized perpendicular to their fl at faces, making one face a north pole and the other a south pole. The yellow magnet has a pole on its lower face which


Magnetism 195

is the same as the pole on the upper faces of the red magnets. The pole on the lower face of the blue magnet is the same as that on the upper face of the yellow magnet.

(d) If the upper magnet were inverted the yellow and blue magnets would attract each other and stick fi rmly together. The yellow magnet would continue to be repelled by and fl oat above the red magnets.

PROBLEM SOLUTIONS

19.1 Consider a three-dimensional coordinate system with the xy plane in the plane of this page, the +x-direction toward the right edge of the page and the +y-direction toward the top of the page. Then, the z-axis is perpendicular to the page with the +z-direction being upward, out of the page. The magnetic fi eld is directed in the +x-direction, toward the right.

(a) When a proton (positively charged) moves in the +y-direction, the right-hand rule number 1

gives the direction of the magnetic force as into the page or in the −z-direction .

(b) With rv in the −y-direction, the right-hand rule number 1 gives the direction of the force

on the proton as out of the page, in the + -directionz .

(c) When the proton moves in the +x-direction, parallel to the magnetic fi eld, the magnitude of

the magnetic force it experiences is F q B= ( ) =v sin 0 0° , or there is a zero force in this case .

19.2 (a) For a positively charged particle, the direction of the force is that predicted by the right-hand rule number one. These are:

(a′) in plane of page and to left (b′) into the page

(c′) out of the page (d′) in plane of page and toward the top

(e′) into the page (f′) out of the page

(b) For a negatively charged particle, the direction of the force is exactly opposite what the right-hand rule number 1 predicts for positive charges. Thus, the answers for part (b) are

reversed from those given in part (a) .

19.3 Since the particle is positively charged, use the right-hand rule number 1. In this case, start with the fi ngers of the right hand in the direction of

rv and the thumb pointing in the direction of

rF.

As you start closing the hand, the fi ngers point in the direction of rB after they have moved 90°.

The results are

(a) into the page (b) toward the right (c) toward bottom of page

19.4 Hold the right hand with the fi ngers in the direction of rv so that as you close your hand, the

fi ngers move toward the direction of rB. The thumb will point in the direction of the force (and

hence the defl ection) if the particle has a positive charge. The results are

(a) toward top of page (b) out of the page , since the charge is negative.

(c) θ = ° ⇒180 zero force (d) into the page


196 Chapter 19

19.5 (a) The proton experiences maximum force when it moves perpendicular to the magnetic fi eld, and the magnitude of this maximum force is

F q Bmax sin . .= = ×( ) ×( )−v 90 1 60 10 6 00 10 119 6° C m s .. .50 1 1 44 10 12 T N( )( ) = × −

(b) aF

mpmax

max ..= = ×

×= ×

−

−

1 44 10

108 62

12 N

1.67 kg27 11014 m s2

(c) Since the magnitude of the charge of an electron is the same as that of a proton, the

force experienced by the electron would havee the same magnitude , but would be in the

opposite direction due to the negative charge of the electron. The acceleration of the elec-tron would be much greater than that of the proton because of the mass of the electron is much smaller.

19.6 From F q B= v sinθ, the magnitude of the force is found to be

F = ×( ) ×( ) ×( )− −1 60 10 6 2 10 50 0 1019 6 6. . . si C m s T nn . .90 0 4 96 10 17°( ) = × − N

Using the right-hand rule (fi ngers point westward in direction of rv , so they move downward

toward the direction of Bur

as you close the hand, the thumb points southward. Thus, the direction of the force exerted on a proton (a positive charge) is toward the south .

19.7 The gravitational force is small enough to be ignored, so the magnetic force must supply the needed centripetal acceleration. Thus,

mr

q Bv

v2

90= sin °, or v = q B r

m where r RE= + = ×1000 km 7.38 10 m6

v =

×( ) ×( ) ×( )− −1 60 10 4 00 10 7 38 10

1

19 8 6. . .

.

C T m

667 102 83 1027

7

×= ×− kg

m s.

If rv is toward the west and

rB is northward,

rF will be directed downward as required.

19.8 The speed attained by the electron is found from 1

22m q Vv = ( )Δ , or

v =

( ) =×( )( )

×

−

−

2 2 1 60 10 2 400

9 11 10

19

31

e V

m

Δ .

.

C V

kg m s= ×2 90 107.

(a) Maximum force occurs when the electron enters the region perpendicular to the fi eld.

F q Bmax

C m s

=

= ×( ) ×( )−

v sin

. .

90

1 60 10 2 90 10 119 7

°

.. .70 7 90 10 12 T N( ) = × −

(b) Minimum force occurs when the electron enters the region parallel to the fi eld.

F q Bmin = =v sin 0 0°


Magnetism 197

19.9 The magnitude of the magnetic force acting on the electron is F q B m ae= =v sin 90° , so the magnitude of the magnetic fi eld is given by

Bm a

ee= =

×( ) ×( )×

−

v

9 11 10 4 0 10

1 60 1

31 16. .

.

kg m s2

00 1 5 101 5 10

19 72

−−

( ) ×( ) = × C m s

T.

.

To determine the direction of the fi eld, employ a variation of right-hand rule number 1. Hold your right hand fl at with the fi ngers extended in the direction of the electron’s velocity (toward the top of the page) and the thumb in the direction of the magnetic force (toward the right edge of the page). Then, as you close your hand, the fi ngers will point out of the page after they have moved 90°. This would be the correct direction for the magnetic fi eld if the particle were positively charged. Since the electron is a negative particle, the actual direction of the fi eld is opposite that predicted by the right-hand rule, or it is directed into the page (the – -direction)z .

19.10 The force on a single ion is

F q B1

191 60 10 0 851 0 254

=

= ×( )( )(−

v sin

. . .

θ

C m s T)) ( ) = × −sin . .51 0 2 69 10 20° N

The total number of ions present is

N = ×⎛

⎝⎜⎞⎠⎟ ( ) = ×3 00 10 100 3 00 1020 2. .

ions

cm cm3

3 22

Thus, assuming all ions move in the same direction through the fi eld, the total force is

F N F= ⋅ = ×( ) ×( ) =−

122 203 00 10 2 69 10 806. . N N

19.11 Gravitational force:

F mgg = = ×( )( ) = ×− −9 11 10 9 80 8 93 1031 30. . . kg m s 2 NN downward

Electric force:

F qEe = = − ×( ) −( ) = ×− −1 60 10 1 60 1019 17. . C 100 N C N upward

Magnetic force:

F q Bm = = − ×( ) ×( )−v sin . . .θ 1 60 10 6 00 10 50 019 6 C m s ××( ) ( )= ×

−

−

10 90 0

4 80 10

6

17

T

N in direct

sin .

.

°

iion opposite right hand rule prediction

Fm = 44 80 10 17. × − N downward

19.12 Hold the right hand with the fi ngers in the direction of the current so, as you close the hand, the fi ngers move toward the direction of the magnetic fi eld. The thumb then points in the direction of the force. The results are

(a) to the left (b) into the page (c) out of the page

(d) toward top of page (e) into the page (f) out of the page


198 Chapter 19

19.13 From F BI L= sinθ, the magnetic fi eld is

B

F L

I= = ( ) °

= × −

sin

.

sin.

θ0 12

908 0 10 3 N m

15 A T

The direction of rB must be the -direction+ z to have

rF in the –y-direction when

rI is in the

+x-direction.

19.14 (a) F BIL= = ( )( )( ) =sin . . . sin .θ 0 28 3 0 0 14 90 0 1T A m ° 22 N

(b) Neither the direction of the magnetic field nor that of the current is given . Both must be

known before the direction of the force can be determined. In this problem, you can only say that the force is perpendicular to both the wire and the fi eld.

19.15 Use the right-hand rule number 1, holding your right hand with the fi ngers in the direction of the current and the thumb pointing in the direction of the force. As you close your hand, the fi ngers will move toward the direction of the magnetic fi eld. The results are

(a) into the page (b) toward the right (c) toward the bottom of the page

19.16 In order to just lift the wire, the magnetic force exerted on a unit length of the wire must be directed upward and have a magnitude equal to the weight per unit length. That is, the magnitude is

FBI

mg

l l= = ⎛

⎝⎜⎞⎠⎟sinθ giving B

m g

I= ⎛

⎝⎜⎞⎠⎟l sinθ

To fi nd the minimum possible fi eld, the magnetic fi eld should be perpendicular to the currentθ =( )90 0. ° . Then,

Bm g

Imin sin ..= ⎛

⎝⎜⎞⎠⎟ °

=⎛

l 90 00 500

g

cm

1 kg

10 g3⎝⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ ( )

10 9 802 cm

1 m

m s

2.00 A

2.

110 245( ) = . T

To fi nd the direction of the fi eld, hold the right hand with the thumb pointing upward (direction of the force) and the fi ngers pointing southward (direction of current). Then, as you close the hand, the fi ngers point eastward. The magnetic fi eld should be directed eastward .

19.17 F BIL= = ( )( )( )sin sin .θ 0 1 5 30 0.300 T 0.0 A .00 m °(( ) = 7 50. N

19.18 (a) The magnitude is

F BIL= = ×( )( )( )−sin sinθ 0 1 1 90.60 10 T 5 A 0.0 m4 °(( ) = × −9 0 10 3. N

rF is perpendicular to

rB. Using the right-hand rule number 1, the orientation of

rF is found

to be 1 above the horizontal in the northward d5° iirection .

(b) F BIL= = ×( )( )( )−sin sinθ 0 1 1 165.60 10 T 5 A 0.0 m4 °°( ) = × −2 3 10 3. N

and, from the right-hand rule number 1, the direction is horizontal and due west .


Magnetism 199

19.19 For minimum fi eld, rB should be perpendicular to the wire. If the force is to be northward, the

fi eld must be directed downward .

To keep the wire moving, the magnitude of the magnetic force must equal that of the kinetic friction force. Thus, BI L mgksin 90° = ( )μ , or

B

m L g

Ik=( )

°=

( )( )(μsin

. . .

90

0 200 1 00 9 80 g cm m s2 ))( )( )

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜1 50 1 00

1

. . A

kg

10 g

10 cm

1 m3

2 ⎞⎞⎠⎟

= 0 131. T

19.20 To have zero tension in the wires, the magnetic force per unit length must be directed upward and equal to the weight per unit length of the conductor. Thus,

rFm

LB I

mg

L= = , or

Im L g

B=

( )=

( )( )=

0 040 9 80

3 600 109

. .

..

kg m m s

T

2

AA

From the right-hand rule number 1, the current must be to the right if the force is to be upward when the magnetic fi eld is into the page.

19.21 (a) The magnetic force must be directed upward aand its magnitude must equal mg , the weight

of the wire. Then, the net force acting on the wire will be zero and it can move upward at

constant speed.

(b) The magnitude of the magnetic force must be BI L mgsinθ = , and for minimum fi eldθ = 90°. Thus,

Bmg

I Lmin

. .

.= =

( )( )( )

0 015 9 80

0 15

kg m s

5.0 A m

2

(( ) = 0 20. T

For the magnetic force to be directed upward when the current is toward the left, Bur

must be directed out of the page .

(c) If the fi eld exceeds 0.20 T, the upward magnetic force exceeds the downward force of

gravity, so the wire accelerates upward .

19.22 The magnitude of the magnetic force exerted on a current-carrying conductor in a magnetic fi eld is given by F BIL= sinθ, where B is the magnitude of the fi eld, L is the length of the conductor, I is the current in the conductor, and θ is the angle the conductor makes with the direction of the fi eld. In this case,

F = ( )( )( ) = ( )0 390 5 00 2 80 5 46. . . sin . sin T A m Nθ θθ

(a) If θ θ= = =60 0 0 866 4 73. , sin . .° then and NF

(b) If θ θ= = =90 0 1 00 5 46. , sin . .° then and NF

(c) If θ θ= = =120 0 866 4 73°, sin . . then and NF


200 Chapter 19

19.23 For each segment, the magnitude of the force is given by F BI L= sinθ, and the direction is given by the right-hand rule number 1. The results of applying these to each of the four segments are summarized below.

Segment L (m) q F (N) Direction

ab 0.400 180° 0 _

bc 0.400 90.0° 0.040 0 negative x

cd 0 400 2. 45.0° 0.040 0 negative z

da 0 400 2. 90.0° 0.056 6

parallel to xz plane at 45° to both +x- and +z-directions

19.24 The magnitude of the force is

F BIL= = ×( ) ×( )( )−sin . . sinθ 5 0 10 2 2 10 58 65 3N A m 55 5 8° = . N

and the right-hand rule number 1 shows its direction to be into the page .

19.25 The torque on a current loop in a magnetic fi eld is τ θ= BIAN sin , and maximum torque occurs when the fi eld is directed parallel to the plane of the loop (θ = 90°). Thus,

τ πmax . .= ( ) ×( ) ×( )⎡

⎣⎤⎦

− −0 50 25 10 5 0 10 53 2 2T A m 00 90 4 9 10 3( ) = × ⋅−sin .° N m

19.26 The magnitude of the torque is τ θ= NBIA sin , where θ is the angle between the fi eld and the perpendicular to the plane of the loop. The circumference of the loop is 2 2 00πr = . m, so the

radius is r = 1 00. m

π and the area is A r= =π

π2 1

m2.

Thus, τπ

= ( )( ) ×( )⎛⎝⎜

⎞⎠⎟

−1 0 800 17 0 101

903. . sin T A m2 .. .0 4 33 10 3° = × ⋅− N m

19.27 The area is A ab= = ( )( ) =π π 0 200 0 150 0 094 2. . . m m m2. Since the fi eld is parallel to the plane of the loop, θ = °90 0. and the magnitude of the torque is

τ θ=

= ×( )( )(−

NBIAsin

. . .8 2 00 10 6 00 0 094 24 T A m2 )) = × ⋅−sin . .90 0 9 05 10 4° N m

The torque is directed to make the left-hand side of the loop move toward you and the right-hand side move away.

19.28 Note that the angle between the fi eld and the perpendicular to the plane of the loop isθ = − =90 0 30 0 60 0. . .° ° °. Then, the magnitude of the torque is

τ θ= = ( )( ) ( )NBIAsin . .100 0 1 0 40 0 30.80 T .2 A m m(( )[ ] = ⋅sin .60 0 10° N m

With current in the –y-direction, the outside edge of the loop will experience a force directed out of the page (+z-direction) according to the right-hand rule number 1. Thus, the loop will rotate

clockwise as viewed from above .


Magnetism 201

19.29 (a) The torque exerted on a coil by a uniform magnetic fi eld is τ θ= BIAN sin , with maximum torque occurring when θ = 90°. Thus, the current in the coil must be

IBAN

= = ⋅( ) ×( ) ×−

τmax .

. . .

0 15

0 90 3 0 10 5 0 12

N m

T m 00 2000 56

2−( )⎡⎣ ⎤⎦( )=

m A.

(b) If I has the value found above and θ is now 25°, the torque on the coil is

τ θ= = ( )( ) ( )BIAN sin . . . .0 90 0 56 0 030 0 050T A m m(( )[ ]( ) = ⋅200 25 0 064sin .° N m

19.30 The resistance of the loop is

R

L

A= =

× ⋅( )( )×

=−

−

ρ 1 70 10 8 00

1 00 101

8

4

. .

.

m m

m2

Ω..36 10 3× − Ω

and the current in the loop is IV

R= =

×=−

ΔΩ

0 100

1 36 1073 53

.

..

V

A

The magnetic fi eld exerts torque τ θ= NBIA sin on the loop, and this is a maximum when sinθ = 1. Thus,

τmax T A m= = ( )( )( )( ) =NBIA 1 0 400 73 5 2 00 118

2. . . N m⋅

19.31 (a) Let θ be the angle the plane of the loop makes with the horizontal as shown in the sketch at the right. Then, the angle it makes with the vertical is φ θ= −90 0. ° . The number of turns on the loop is

NL

circumference= = ( ) =4 00

10 0.

. m

4 0.100 m

The torque about the z-axis due to gravity is

τ θg mgs= ⎛

⎝⎜⎞⎠⎟2

cos , where s = 0 100. m is the length

of one side of the loop. This torque tends to rotate the loop clockwise. The torque due to the magnetic force tends to rotate the loop counterclockwise about the z-axis and has magnitude τ θm NBIA= sin . At

equilibrium, τ τm g= or NBI s mg s2 2( ) = ( )sin cosθ θ . This reduces to

tan

. .

.θ = =

( )( )( )

mg

NBIs2

0 100 9 80

0 0

kg m s

2 10.0

2

110 0 3 40 0 10014 4

T A m( )( )( ) =. .

.

Since tan tan . cotθ φ φ= ° −( ) =90 0 , the angle the loop makes with the vertical at

equilibrium is φ = ( ) =−cot . .1 14 4 3 97° .

continued on next page


202 Chapter 19

(b) At equilibrium,

τ θm NBI s= ( )= ( )( )( )

2

10 0 0 010 0 3 40 0 10

sin

. . . .T A 00 90 0 3 97

3 39 10

2

3

m

N m

( ) −( )

= × ⋅−

sin . .

.

° °

19.32 (a) The current in segment a-b is in the +y-direction. Thus, by right-hand rule 1, the magnetic force on it is in the

+ -directionx . Imagine this force being concentrated at the center of segment a-b. Then, with a pivot at point a (a point on the x-axis), this force would tend to rotate the conductor a-b in a clockwise direction about the z-axis,

so the direction of this torque is in the -direction− z .

(b) The current in segment c-d is in the −y-direction, and the right-hand rule 1 gives the direction of the magnetic force as the −x-direction . With a pivot at point d (a point on the x-axis), this force would tend to rotate the conductor c-d counterclockwise about the z-axis, and the direction of this torque is in the -direction+ z .

(c) No. The torques due to these forces are along the z-axis and cannot cause rotation about the x-axis. Further, both the forces and the torques are equal in magnitude and opposite in direction, so they sum to zero and cannot affect the motion of the loop.

(d) The magnetic force is perpendicular to both the direction of the current in b-c (the +x-direction) and the magnetic fi eld. As given by right-hand rule 1, this places it

in the plane at 130° counterclockwise fryz oom the + -axisy . The force acting on segment

b-c tends to rotate it counterclockwise about the x-axis, so the torque is in the + -directionx .

(e) The loop tends to rotate counterclockwise about the -axisx .

(f) μ = = ( ) ( )( )⎡⎣ ⎤⎦( ) =IAN 0 900 0 500 0 300 1 0. . . . A m m 1135 A m2⋅

(g) The magnetic moment vector is perpendicular to the plane of the loop (the x y plane), and is therefore parallel to the z-axis. Because the current fl ows clockwise around the loop, the magnetic moment vector is directed downward, in the negative z-direction. This means that the angle between it and the direction of the magnetic fi eld is θ = + =90 0 40 0 130. .° ° ° .

(h) τ μ θ= = ⋅( )( ) ( ) =B sin . . sin .0 135 1 50 130 0 15 A m T2 ° 55 N m⋅

19.33 (a) The magnetic force acting on the electron provides the centripetal acceleration, holding the electron in the circular path. Therefore, F q B m re= =v vsin 90 2° , or

rm

eBe= =

×( ) ×( )×

−

−

v 9 11 10 1 5 10

1 60 10

31 7. .

.

kg m s119 32 0 10

0 043 4 3 C T

m cm( ) ×( ) = =−.. .

(b) The time to complete one revolution around the orbit (i.e., the period) is

T

r= = =distance traveled

constant speed

2 2 0π πv

.0043

1 5 101 8 107

8 m

m s s

( )×

= × −

..


Magnetism 203

19.34 (a) F q B= = ×( ) ×( )−v sin . . .θ 1 60 10 5 02 10 0 18019 6 C m s TT N( ) ( ) = × −sin . .60 0 1 25 10 13°

(b) aF

m= = ×

×= ×

−

−

1 25 10

107 50 10

1313.

. N

1.67 kg m s27

22

19.35 For the particle to pass through with no defl ection, the net force acting on it must be zero. Thus, the magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives

F Fm e= , or q B qEv = , which reduces to v = E B

19.36 The speed of the particles emerging from the velocity selector is v = E B (see Problem 35).

In the defl ection chamber, the magnetic force supplies the centripetal acceleration, so q Bm

rv

v=2

,

or rm

qB

m E B

qB

mE

qB= =

( )=v

2 .

Using the given data, the radius of the path is found to be

r =×( )( )×( )

−

−

2 18 10 950

0 93

26.

.

kg V m

1.60 10 C19 001 50 10 0 1502

4

T m mm

( )= × =−. .

19.37 From conservation of energy, KE PE KE PEf i

+( ) = +( ) , we fi nd that 1

202m qV qVf iv + = + ,

or the speed of the particle is

v =

−( )=

( ) =×( )( )−2 2 2 1 60 10 25019q V V

m

q V

mi f Δ . C V

22 .50 10 kg m s26×

= ×− 5 66 104.

The magnetic force supplies the centripetal acceleration giving q Bm

rv

v=2

or rm

qB= =

×( ) ×( )×

−

−

v 2 50 10 5 66 10

1 60 10

26 4. .

.

kg m s119

2

0 5001 77 10 1 77

C T m cm( )( ) = × =−

.. .

19.38 Since the centripetal acceleration is furnished by the magnetic force acting on the ions,

q Bm

rv

v=2

or the radius of the path is rm

qB= v

. Thus, the distance between the impact points

(that is, the difference in the diameters of the paths followed by the U238 and the U235 isotopes) is

Δd r rqB

m m= −( ) = −( )

=×

22

2 3 00 10

238 235 238 235

5

v

. m ss

C T u u

( )×( )( ) −( )−1 60 10 0 600

238 235 1 619. .

. 66 10 27×⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

− kg

u

or Δd = × =−3 11 10 3 112. . m cm


204 Chapter 19

19.39 In the perfectly elastic, head-on collision between the α-particle and the initially stationary pro-ton, conservation of momentum requires that m m mp pv v v+ =α α α 0 while conservation of kinetic

energy also requires that v v v v v v0 00− = − −( ) = +α αp p or . Using the fact that m mpα = 4 and combining these equations gives

m m mp p pv v v vα α+( ) + ( ) = ( )0 04 4 or v vα = 3 50

and v v v vp = ( ) + =3 5 8 50 0 0 Thus, v v v vα = = ⎛⎝⎜

⎞⎠⎟ =3

5

3

5

5

8

3

80 p p

After the collision, each particle follows a circular path in the horizontal plane with the magnetic force supplying the centripetal acceleration. If the radius of the proton’s trajectory is R, and that of the alpha particle is r, we have

q B mR

Rm

q Bp p pp p p

p

vv v

= = =2

or mm

eBp pv

and q B mr

rm

q Bα α αα α α

α

vv v

= = =2

or 44 3 8

2

3

4

3

4

m

e B

m

eBR

p p p p( )( )( ) =

⎛⎝⎜

⎞⎠⎟

=v v

19.40 A charged particle follows a circular path when it moves perpendicular to the magnetic fi eld. The magnetic force acting on the particle provides the centripetal acceleration, holding the particle in the circular path. Therefore, F q B m r= =v vsin 90 2° . Since the kinetic energy is K m= v2 2, we

rewrite the force as F q B K r= =v sin 90 2° , and solving for the speed v gives v = 2K

qBr.

19.41 (a) Within the velocity selector, the electric and magnetic fi elds exert forces in opposite direc-tions on charged particles passing through. For particles having a particular speed, these forces have equal magnitudes, and the particles pass through without defl ection. The selected speed is found from F qE q B Fe m= = =v , giving v = E B. In the defl ection chamber, the selected particles follow a circular path having a diameter of d r m qB= =2 2 v . Thus, the mass to charge ratio for these particles is

m

q

Bd Bd

E B

B d

E= = ( ) = =

( ) ( )2 2 2

0 0931 0 396

2

2 2

v. . T m

88 2502 08 10 7

V m kg C( ) = × −.

(b) If the particle is doubly ionized (i.e., two electrons have been removed from the neutral atom), then q e= 2 and the mass of the ion is

m em

q= ( )⎛

⎝⎜⎞⎠⎟

= ×( ) ×− −2 2 1 60 10 2 08 1019 7. . C kg C(( ) = × −6 66 10 26. kg

(c) Assuming this is an element, the mass of the ion should be roughly equal to the atomic weight multiplied by the mass of a proton (or neutron). This would give the atomic weight as

At. wt. kg

1.67 kg≈ = ×

×=

−

−

m

mp

6 66 10

1039 9

26

27

.. , suggesting that the element is calcium .


Magnetism 205

19.42 Since the path is circular, the particle moves perpendicular to the magnetic fi eld, and the

magnetic force supplies the centripetal acceleration. Hence, mr

q Bv

v2

= , or Bm

qr= v

. But

the momentum is given by p m m KE= = ( )v 2 , and the kinetic energy of this proton is

KE = ×( ) ×⎛⎝⎜

⎞⎠⎟

=−

10 0 101 60 10

1 60619

..

. eV J

1 eV×× −10 12 J. We then have

Bm KE

qr=

( )=

×( ) ×( )− −2 2 1 67 10 1 60 10

1

27 12. .

.

kg J

660 10 5 80 107 88 10

19 1012

×( ) ×( ) = ×−−

C m T

..

19.43 Treat the lightning bolt as a long, straight conductor. Then, the magnetic fi eld is

B

I

r= =

× ⋅( ) ×( )(

−μπ

ππ

0

7 4

2

4 10 1 00 10

2 100

T m A A

m

.

)) = × =−2 T T. .00 10 20 05 μ

19.44 Imagine grasping the conductor with the right hand so the fi ngers curl around the conductor in the direction of the magnetic fi eld. The thumb then points along the conductor in the direction of the current. The results are

(a) toward the left (b) out of page (c) lower left to upper right

19.45 The magnetic fi eld at distance r from a long conducting wire is B I r= μ π0 2 . Thus, if B = × −1 0 10 15. T at r = 4 0. cm, the current must be

I

rB= =( ) ×( )

× ⋅

−

−

2 2 0 040 1 0 10

100

15πμ

ππ

. . m T

4 T7 mm A A= × −2 0 10 10.

19.46 Model the tornado as a long, straight, vertical conductor and imagine grasping it with the right hand so the fi ngers point northward on the western side of the tornado (that is, at the observatory’s location). The thumb is directed downward, meaning that the

conventional current is downward or negativee charge flows upward .

The magnitude of the current is found from B I r= μ π0 2 as

I

rB= =×( ) ×( )

×

−

−

2 2 9 00 10 1 50 10

0

3 8πμ

ππ

. . m T

4 10 7 T m A A

⋅= 675

19.47 From B I r= μ π0 2 , the required distance is

rI

B= =

× ⋅( )( )×( )

−

−

μπ

ππ

032

20

2 1 7 10

4 10 T m A A

T

7

.== × =−2 4 10 2 43. . m mm


206 Chapter 19

19.48 Assume that the wire on the right is wire 1 and that on the left is wire 2. Also, choose the positive direction for the magnetic fi eld to be out of the page and negative into the page.

(a) At the point half way between the two wires,

B B BI

r

I

r rnet = − − = − +⎡

⎣⎢

⎤

⎦⎥ = −1 2

0 1

1

0 2

2

0

2 2 2

μπ

μπ

μπ

II I1 2

74 1010 0

+( )

= −× ⋅( )

×( )−

−

ππ

T m A

2 5.00 10 m2. A T( ) = − × −4 00 10 5.

or Bnet T into the page= 40 0. μ

(b) At point P1 , B B BI

r

I

rnet = + − = + −⎡

⎣⎢

⎤

⎦⎥1 2

0 1

1

2

22

μπ

Bnet

T m A

2

A

0.100 m

A

0.20= × ⋅ −

−4 10 5 00 5 007ππ

. .

00 m T out of page⎡

⎣⎢⎤⎦⎥

= 5 00. μ

(c) At point P2 , B B BI

r

I

rnet = − + = − +⎡

⎣⎢

⎤

⎦⎥1 2

0 1

1

2

22

μπ

Bnet

T m A

2

A

0.300 m

A

0.2= × ⋅ − +

−4 10 5 00 5 007ππ

. .

000 m

T out of page

⎡⎣⎢

⎤⎦⎥

= 1 67. μ

19.49 The distance from each wire to point P is given by

r = ( ) + ( ) =1

20 200 0 200 0 141

2 2. . . m m m

At point P, the magnitude of the magnetic fi eld produced by each of the wires is

B

I

r= =

× ⋅( )( )( ) =

−μπ

ππ

0

7

2

4 10 5 00

2 0 141

T m A A

m

.

.77 07. Tμ

Carrying currents into the page, the fi eld A produces at P is directed to the left and down at – 1 35°, while B creates a fi eld to the right and down at – 45°. Carrying currents toward you, C produces a fi eld downward and to the right at – 45°, while D’s contribution is down and to the left at – 135°. The horizontal components of these equal magnitude contributions cancel in pairs, while the vertical components all add. The total fi eld is then

Bnet T T toward the= ( ) =4 7 07 45 0 20 0. sin . .μ μ° bottom of the page


Magnetism 207

19.50 Call the wire carrying a current of 3.00 A wire 1 and the other wire 2. Also, choose the line run-ning from wire 1 to wire 2 as the positive x-direction.

(a) At the point midway between the wires, the fi eld due to each wire is parallel to the y-axis and the net fi eld is

B B B I I ry ynet = + − = −( )1 2 0 1 2 2μ π

Thus, Bnet

T m A

m A A=

× ⋅( )( ) −

−4 10

2 0 1003 00 5 00

7ππ .

. .(( ) = − × −4 00 10 6. T

or Bnet T toward the bottom of the page= 4 00. μ

(b) At point P, r1 0 200 2= ( ). m and B1 is directed at θ1 135= + °.

The magnitude of B1 is

BI

r10 1

1

7

2

4 10 3 00

2 0 200 2= =

× ⋅( )( )−μπ

π

π

T m A A

.

. mm T( ) = 2 12. μ

The contribution from wire 2 is in the –x-direction and has magnitude

B

I

r20 2

2

7

2

4 10 5 00

2 0 200= =

× ⋅( )( )−μπ

ππ

T m A A

m

.

.(( ) = 5 00. Tμ

Therefore, the components of the net fi eld at point P are:

B B Bx = +

= ( ) +

1 2135 180

2 12 135 5

cos cos

. cos

° °

° T .0μ 00 T Tμ μ( ) = −cos .180 6 50°

and B B By = + = ( ) + = +1 2135 180 2 12 135 0sin sin . sin° ° ° Tμ 11 50. Tμ

Therefore, B B Bx ynet T= + =2 2 6 67. μ at

θ μμ

=⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

⎞⎠⎟

− −tan tan.

.1 1 6 50

1 50

B

Bx

y

T

T== 77 0. °

or Bur

net T at 77.0° to the left of vert= 6 67. μ iical

19.51 Call the wire along the x-axis wire 1 and the other wire 2. Also, choose the positive direction for the magnetic fi elds at point P to be out of the page.

At point P, B B BI

r

I

r

I

r

I

rnet = + − = − = −1 20 1

1

0 2

2

0 1

1

2

22 2 2

μπ

μπ

μπ

⎛⎛⎝⎜

⎞⎠⎟

or Bnet

T m A A

m

A=× ⋅( )

−−4 10

2

7 00

3 00

6 00

4 0

7ππ

.

.

.

. 001 67 10 7

m T⎛

⎝⎜⎞⎠⎟ = + × −.

Bnet T out of the page= 0 167. μ


208 Chapter 19

19.52 (a) Imagine the horizontal x y plane being perpendicular to the page, with the positive x-axis coming out of the page toward you and the positive y-axis toward the right edge of the page. Then, the vertically upward positive z-axis is directed toward the top of the page. With the current in the wire fl owing in the positive x-direction, the right-hand rule 2 gives the direction of the magnetic fi eld above the wire as being toward the left, or

in the -direction.−y

(b) With the positively charged proton moving in the −x-direction (into the page), right-hand rule 1 gives the direction of the magnetic force on the proton as being directed toward the

top of the page, or upward, in the positive -directionz .

(c) Since the proton moves with constant velocity, a zero net force acts on it. Thus, the

magnitude of the magnetic force must equal tthat of the gravitational force .

(d) ΣF ma F Fz z m g= = ⇒ =0 or q B mgv = where B I d= μ π0 2 . This gives

q I

dmg

vμπ

0

2= , or the distance the proton is above the wire must be d

q I

mg= vμ

π0

2.

(e) dq I

mg= =

×( ) ×( ) ×−vμπ

π0

19 4

2

1 60 10 2 30 10 4 1. . C m s 00 1 20 10

2 1 67 10 9

7 6

27

− −

−

⋅( ) ×( )×( )

T m A A

kg

.

. .π 880 m s2( )

d = × =−5 40 10 5 402. . m cm

19.53 (a) From B I r= μ π0 2 , observe that the fi eld is inversely proportional to the distance from the conductor. Thus, the fi eld will have one-tenth its original value if the distance is increased

by a factor of 10. The required distance is then ′ = = ( ) =r r10 10 0 400 4 00. . .m m

(b) A point in the plane of the conductors and 40.0 cm from the center of the cord is located 39.85 cm from the nearer wire and 40.15 cm from the far wire. Since the currents are in opposite directions, so are their contributions to the net fi eld. Therefore, B B Bnet = −1 2, or

BI

r rnet

T m A = −

⎛⎝⎜

⎞⎠⎟

=× ⋅( )−μ

ππ

0

1 2

7

2

1 1 4 10 2 00. AA

2

1

0.398 5 m

1

0.401 5 m

T

( )−

⎛⎝⎜

⎞⎠⎟

= × −

π

7 50 10 9. == 7 50. nT

(c) Call r the distance from cord center to fi eld point P and 2 3 00d = . mm the distance between centers of the conductors.

B

I

r d r d

I d

r dnet =−

−+

⎛⎝⎜

⎞⎠⎟ =

−⎛⎝⎜

⎞⎠

μπ

μπ

0 02 22

1 1

2

2⎟⎟

7 50 10

4 10 2 00

2

3 00 1107

.. .× =

× ⋅( )( ) ×−−

T T m A Aπ

π00

2 25 10

3

2 6

−

−− ×⎛⎝⎜

⎞⎠⎟

m

m2r .

so r = 1 26. m

The fi eld of the two-conductor cord is weak to start with and falls off rapidly with distance.

(d) The cable creates zero fi eld at exterior points, since a loop in Ampère’s law encloses zero total current.


Magnetism 209

19.54 (a) Point P is equidistant from the two wires, which carry identical currents. Thus, the contributions of the two wires, B

urupper and B

urlower, to the magnetic fi eld at P will

have equal magnitudes. The horizontal components of these contributions will cancel, while the vertical components add. The resultant fi eld will be vertical, in the -direction+ y .

(b) The distance of each wire from point P is r x d= +2 2 , and the cosine of the angle that B

urupper

and Bur

lower make with the vertical is cosθ = x r . The

magnitude of either Bur

upper or Bur

lower is B I rwire = μ π0 2 and the vertical components of either of these contributions have values of

B BI

r

x

r

Ix

rywire wire( ) = ( ) = ⎛⎝⎜

⎞⎠⎟ =cosθ μ

πμπ

0 0

2 2 22

The magnitude of the resultant fi eld at point P is then

B BIx

r

I x

x dP y

= ( ) = =+( )2 0

202 2wire

μπ

μπ

(c) The point midway between the two wires is the origin (0,0). From the above result for part

(b), the resultant fi eld at this midpoint is BP x==

00 . This is as expected, because right-

hand rule 2 shows that at the midpoint the fi eld due to the upper wire is toward the right, while that due to the lower wire is toward the left. Thus, the two fi elds cancel, yielding a zero resultant fi eld.

19.55 The force per unit length that one wire exerts on the other is F I I dl = μ π0 1 2 2 , where d is the distance separating the two wires. In this case, the value of this force is

F

l=

× ⋅( )( )×( ) =

−

−

4 10 3 0

2 6 00 103

7 2

2

ππ

T m A A

m

.

..00 10 5× − N m

Imagine these two wires lying side by side on a table with the two currents fl owing toward you, wire 1 on the left and wire 2 on the right. Right-hand rule 2 shows the magnetic fi eld due to wire 1 at the location of wire 2 is directed vertically upward. Then, right-hand rule 1 gives the direc-tion of the force experienced by wire 2, with its current fl owing through this fi eld, as being to the

left, back toward wire 1. Thus, the force one wire exerts on the other is an attractive force.

19.56 (a) The force per unit length that parallel conductors exert on each other is F I I dl = μ π0 1 2 2 . Thus, if F l = × −2 0 10 4. N m, I1 5 0= . A, and d = 4 0. cm, the current in the second wire must be

Id

I

F2

0 1

2

7

2 2 4 0 10

4 10= ⎛

⎝⎜⎞⎠⎟ =

×( )× ⋅

−

−

πμ

ππl

. m

T mm A A N m A( )( ) ×( ) =−

5 02 0 10 8 04

.. .

(b) Since parallel conductors carrying currents in the same direction attract each other (see Section 19.8 in the textbook), the currents in these conductors which repel each other

must be in opposite directions .



210 Chapter 19

(c) The result of reversing the direction of either of the currents would be that the

force of interaction would change from a forrce of repulsion to an attractive force . The

expression for the force per unit length, F I I dl = μ π0 1 2 2 , shows that doubling either

of the currents would double the magnitude of the force of interacction .

19.57 In order for the system to be in equilibrium, the repulsive magnetic force per unit length on the top wire must equal the weight per unit length of this wire.

Thus, F

L

I I

d= =μ

π0 1 2

20 080. N m, and the distance between the wires will be

dI I= ( ) =

× ⋅( )( )−μπ

π0 1 2

7

2 0 080

4 10 60 0

.

.

N m

T m A A 330 0

2 0 080

4 5 10 4 53

.

.

. .

A

N m

m mm

( )( )

= × =−

π

19.58 The magnetic forces exerted on the top and bottom segments of the rectangular loop are equal in magnitude and opposite in direction. Thus, these forces cancel, and we only need consider the sum of the forces exerted on the right and left sides of the loop. Choosing to the left (toward the long, straight wire) as the positive direction, the sum of these two forces is

F

I I

c

I I

c a

I I

cnet = + −+( ) =μ

πμπ

μπ

0 1 2 0 1 2 0 1 2

2 2 2

1l l l −−+

⎛⎝⎜

⎞⎠⎟

1

c a

or Fnet

T m A A A m=

× ⋅( )( )( )(−4 10 5 00 10 0 0 4507π . . . ))−⎛

⎝⎜⎞⎠⎟

= + × =−

2

1

0 100

1

0 250

2 70 10 25

π . .

. .

m m

N 770 10 5× − N to the left

19.59 The magnetic fi eld inside of a solenoid is B nI N L I= = ( )μ μ0 0 . Thus, the current in this solenoid must be

IBL

N= =

×( ) ×( )× ⋅

− −

−μ π0

3 2

7

2 0 10 6 0 10

4 10

. . T m

T m AA A( )( ) =

303 2.

19.60 The magnetic fi eld inside of a solenoid is B nI N L I= = ( )μ μ0 0 . Thus, the number of turns on this solenoid must be

NBL

I= =

( )( )× ⋅( )( )−μ π0

7

9 0 0 50

4 10 75

. . T m

T m A A== ×4 8 104. turns


Magnetism 211

19.61 (a) From R L A= ρ , the required length of wire to be used is

LR A= ⋅ =

( ) ×( )⎡⎣

⎤⎦

×

−

−ρ

π5 00 0 500 10 4

1 7 10

3 2. .

.

mΩ88 58 m

mΩ ⋅

=

The total number of turns on the solenoid (that is, the number of times this length of wire will go around a 1.00 cm radius cylinder) is

NL

r= =

×( ) = × =−2

58

2 1 00 109 2 10 920

22

π π m

m..

(b) From B nI= μ0 , the number of turns per unit length on the solenoid is

nB

I= = ×

× ⋅( )( ) =−

−μ π0

24 00 10

4 007

.

..

T

4 10 T m A A7996 103× turns m

Thus, the required length of the solenoid is

LN

n= = ×

×= =9 2 10

7 96 100 12 12

2

3

.

..

turns

turns m m cm

19.62 The magnetic fi eld inside the solenoid is

B nI= = × ⋅( ) ⎛

⎝⎜⎞⎠⎟

−μ π0 174 10 30

100 T m A

turns

cm

cmm

1 m A T⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥( ) = × −15 0 5 65 10 2. .

Therefore, the magnitude of the magnetic force on any one of the sides of the square loop is

F BI L= ° = ×( )( ) ×−

2290 0 5 65 10 0 200 2 00 1sin . . . . T A 00 2 26 102 4− −( ) = × m N.

The forces acting on the sides of the loop lie in the plane of the loop, are perpendicular to the

sides, and are directed away from the interior of the loop. Thus, they tend to stretch the loop

but do not tend to rotate it. The torque acting on the loop is τ = 0 .

19.63 (a) The magnetic force supplies the centripetal acceleration, so q B m rv v= 2 . The magnetic fi eld inside the solenoid is then found to be

Bm

q r= =

×( ) ×( )×

−

−

v 9 11 10 1 0 10

1 60 10

31 4

1

. .

.

kg m s99 2

6

2 0 102 8 10 2 8

C m T T( ) ×( ) = × =−

−

.. . μ

(b) From B nI= μ0 , the current is the solenoid is found to be

IB

n= = ×

× ⋅( ) ( )−

−μ π0

62 8 10

25

. T

4 10 T m A turns cm7 1100 1

8 9 10 4

cm m

A 0.89 mA

( )⎡⎣ ⎤⎦

= × =−.


212 Chapter 19

19.64 (a) When switch S is closed, a total current NI (current I in a total of N conductors) fl ows toward the right through the lower side of the coil. This results in a downward force of magnitude F B NI wm = ( ) being exerted on the coil by the magnetic fi eld, with the requirement that the balance exert a upward force

′ =F mg on the coil to bring the system back into balance.

In order for the magnetic force to be downward, the right-hand rule number 1 shows that

the magnetic fi eld must be directed out of the page toward the reader. For the system to be restored to balance, it is necessary that

F F B NI w mgm = ′ ( ) = or , giving Bmg

NIw=

(b) The magnetic fi eld exerts forces of equal magnitudes and opposite directions on the two

sides of the coil. These forces cancel each other and do not afffect the balance of the coil. Hence the dimension of the sizes is not needed.

(c) Bmg

NIw= =

×( )( )( )(

−20 0 10 9 80

50 0 30

3. .

.

kg m s

A

2

)) ×( ) =−5 0 100 26

2..

m T

19.65 (a) The magnetic fi eld at the center of a circular current loop of radius R and carrying current I is B I R= μ0 2 . The direction of the fi eld at this center is given by right-hand rule number 2. Taking out of the page (toward the reader) as positive the net magnetic fi eld at the common center of these coplanar loops is

B B BI

r

I

rnet

T m A= − = − =

× ⋅( )−

2 10 2

2

0 1

1

7

2 2

4 10

2

3μ μ π .. .

.

0

10

5 0

10

5 2 1

A

9.0 m

A

12 m2 2×−

×⎛⎝⎜

⎞⎠⎟

= − ×

− −

00 5 26− = T T into the page. μ

(b) To have Bnet = 0, it is necessary that I r I r2 2 1 1= , or

rI

Ir2

2

11

3 012 7 2=

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ ( ) =.

. A

5.0 A cm cm

19.66 Since the magnetic force must supply the centripetal acceleration, q B m rv v= 2 or the radius of the path is r m qB= v .

(a) The time for the electron to travel the semicircular path (of length πr ) is

tr m

qB

m

qB= =

⎛⎝⎜

⎞⎠⎟

= =×( )−π π π π

v vv 9 11 10 31. kg

1.600 10 C T

s 1 ns

19×( )( )

= × =

−

−

0 010 0

1 79 10 799

.

. .



Magnetism 213

(b) The radius of the semicircular path is 2.00 cm. From r m qB= v , the momentum of the electron is p = mv = qBr, and the kinetic energy is

KE mm

m

q B r

m= =

( ) = =×( )−

12

22 2 2 2 19 2

2 2

1 60 10 0v

v . . C 0010 0 2 00 10

2 9 11

2 2 2 T m

10 kg31

( ) ×( )×( )

−

−

.

.

KE = ×( ) ×⎛⎝⎜

⎞⎠⎟ =−

−5 62 101

316. . J keV

1.60 10 J16 551 keV

19.67 Assume wire 1 is along the x-axis and wire 2 along the y-axis.

(a) Choosing out of the page as the positive fi eld direction, the fi eld at point P is

B B BI

r

I

r= − = −

⎛⎝⎜

⎞⎠⎟

=× ⋅( )−

1 20 1

1

2

2

7

2

4 10

2

μπ

π T m A

ππ5 00 3 00

5 00 10

. .

.

A

0.400 m

A

0.300 m−⎛

⎝⎜⎞⎠⎟

= × −− =7 0 500 T T out of the page. μ

(b) At 30.0 cm above the intersection of the wires, the fi eld components are as shown at the right, where

B BI

ry = − = −

= −× ⋅( )( )−

10 1

7

2

4 10 5 00

2 0

μπ

ππ

T m A A.

.33003 33 10 6

m T( ) = − × −.

and B BI

rx = = =× ⋅( )( )−

20 2

7

2

4 10 3 00

2 0 300

μπ

ππ

T m A A.

. m T( ) = × −2 00 10 6.

The resultant fi eld is

B B Bx y= + = × −2 2 63 89 10. T

at

θ =

⎛⎝⎜

⎞⎠⎟

= − °−tan 1 B

By

x

59 0.

or rB = °3 89. T at 59.0 clockwise from + direμ x cction


214 Chapter 19

19.68 For the rail to move at constant velocity, the net force acting on it must be zero. Thus, the magni-tude of the magnetic force must equal that of the friction force giving BIL mgk= ( )μ , or

B

mg

ILk=( ) =

( )( )( )μ 0 100 0 200 9 80

10 0

. . .

.

kg m s

2

AA m T( )( ) = × −

0 5003 92 10 2

..

19.69 (a) Since the magnetic fi eld is directed from N to S (that is, from left to right within the artery), positive ions with velocity in the direction of the blood fl ow experience a magnetic defl ec-tion toward electrode A. Negative ions will experience a force defl ecting them toward electrode B. This separation of charges creates an electric fi eld directed from A toward B. At equilibrium, the electric force caused by this fi eld must balance the magnetic force, so

q B qE q V dv = = ( )Δ

or v = = ×( ) ×( ) =

−

−

ΔV

Bd

160 10

3 001

6 V

0.040 0 T 10 m3..333 m s

(b) The magnetic fi eld is directed from N to S. If the charge carriers are negative moving in the direction of

rv, the magnetic force is directed toward point B. Negative charges build up at

point B, making the potential at A higher than that at B. If the charge carriers are positive moving in the direction of

rv, the magnetic force is directed toward A, so positive charges

build up at A. This also makes the potential at A higher than that at B. Therefore, the sign of

the potential difference does not depend on the charge of the ions .

19.70 (a) The magnetic force acting on the wire is directed upward and of magnitude

F BIL BILm = =sin 90°

Thus, aF

m

F mg

m

BI

m Lgy

y m= = − = ( ) −Σ

, or

ay =

×( )( )×

−−

−

4 0 10 2 0

5 0 109 80

3

4

. .

..

T A

kg m m s2 == 6 2. m s2

(b) Using Δy t a ty y= +v021

2 with v0 0y = gives

ty

ay

=( ) =

( ) =2 2 0 50

6 20 40

Δ .

..

m

m s s2


Magnetism 215

19.71 Label the wires 1, 2, and 3 as shown in Figure 1, and let B B B1 2 3, , and respectively represent the magnitudes of the fi elds produced by the currents in those wires. Also, observe that θ = 45° in Figure 1.

At point A, B B I a1 2 0 2 2= = ( )μ π or

B B1 2

74 10 2 0

2 0 010 228= =

× ⋅( )( )( )

=−π

π T m A A

m

.

.μμT

and BI

a30

7

2 3

4 10 2 0

2 0 030= ( ) =

× ⋅( )( )−μπ

ππ

T m A A

m

.

.(( ) = 13 Tμ

These fi eld contributions are oriented as shown in Figure 2.

Observe that the horizontal components of r rB B1 2 and cancel

while their vertical components add to rB3. The resultant

fi eld at point A is then

B B B BA = +( ) + =1 2 345 53cos ° Tμ , or

rBA = 5 T directed toward the bottom of the3 μ page

At point B, B BI

a1 20

7

2

4 10 2 0

2 0 010= = =

× ⋅( )( )−μπ

ππ

T m A A

m

.

.(( ) = 40 Tμ

and BI

a30

2 220= ( ) =μ

πμ T . These contributions are oriented as

shown in Figure 3. Thus, the resultant fi eld at B is

r rB BB = =3 20 T directed toward the bottom ofμ the page

At point C, B B I a1 2 0 2 2 28= = ( ) =μ π μ T while

B I a3 0 2 40= =μ π μ T . These contributions are oriented as shown in Figure 4. Observe that the horizontal

components of r rB B1 2 and cancel while their vertical

components add to oppose rB3. The magnitude of the

resultant fi eld at C is

B B B BC = +( ) −

= ( ) − =

1 2 345

56 45 40 0

sin

sin

°

° T Tμ μ

Figure 1

Figure 2

Figure 3

Figure 4


216 Chapter 19

19.72 (a) Since one wire repels the other, the currents must be in opposite directions .

(b) Consider a free body diagram of one of the wires as shown at the right.

ΣF T mgy = ⇒ =0 8 0 cos . °

or Tmg=

cos .8 0°

ΣF F Tmg

x m= ⇒ = = ⎛⎝⎜

⎞⎠⎟ °0 8 0

8 08 0 sin .

cos .sin .°

°

or F mgm = ( ) tan .8 0°. Thus, μ

π0

2

28 0

I L

dmg= ( ) °tan . which gives

I

d m L g=

( )⎡⎣ ⎤⎦ tan .8 0

20

°

μ π

Observe that the distance between the two wires is

d = ( )⎡⎣ ⎤⎦ =2 6 0 8 0 1 7. sin . . cm cm° , so

I =

×( )( )( ) °−1 7 10 0 040 9 80 8 0

2

2. . . tan . m kg m m s2

..0 1087× ⋅

=− T m A6 A

19.73 Note: We solve part (b) before part (a) for this problem.

(b) Since the magnetic force supplies the centripetal acceleration for this particle, q B m rv v= 2

or the radius of the path is r m qB= v . The speed of the particle may be written as

v = ( )2 KE m , so the radius becomes

rm KE

qB=

( )=

×( ) ×( )−2 2 1 67 10 5 00 10 1 627 6. . . kg eV 00 10

10 0 050 0

6 46

19×( )×( )( )

=

−

−

J eV

1.60 C T19 .

. m

Consider the circular path shown at the right and observe that the desired angle is

α = ⎛⎝

⎞⎠ = ⎛

⎝⎞⎠ =− −sin

.sin

..1 11 00 1 00

8m m

6.46 mr990°



Magnetism 217

(a) The constant speed of the particle is v = ( )2 KE m, so the vertical component of the momentum as the particle leaves the fi eld is

p m m m KE m m KEy y= = − = − ( )( ) = − ( )v v sin sin sinα α α2 2

or

py = − ( ) ×( ) ×( )−sin . . .8 90 2 1 67 10 5 00 1027 6° kg eV 11 60 10

8 00 10

19

21

.

.

×( )

= − × ⋅

−

−

J eV

kg m s

19.74 The force constant of the spring system is found from the elongation produced by the weight acting alone.

k

F

x

mg

x= = =

×( )( )×

−

−

10 0 10 9 80

0 50 10

3

2

. .

.

kg m s

2

mm N m= 19 6.

The total force stretching the springs when the fi eld is turned on is

ΣF F mg kxy m= + = total

Thus, the downward magnetic force acting on the wire is

F kx mgm = −

= ( ) ×( ) − ×−

total

N m m19 6 0 80 10 10 0 12. . . 00 9 80

5 9 10

3

2

−

−

( )( )

= ×

kg m s

N

2.

.

Since the magnetic force is given by F BILm = sin 90°, the magnetic fi eld is

BF

IL

F

V R Lm m= = ( ) =

( ) ×( )( )

−

ΔΩ12 5 9 10

24 5 0

2 N

V

.

. ××( ) =−100 59

2 m T.

19.75 The magnetic force is very small in comparison to the weight of the ball, so we treat the motion as that of a freely falling body. Then, as the ball approaches the ground, it has velocity compo-nents with magnitudes of

v vx x= =0 20 0. m s, and

v vy y ya y= + ( ) = + −( ) −( ) =0

2 2 0 2 9 80 20 0 19Δ . . . m s m2 88 m s

The velocity of the ball is perpendicular to the magnetic fi eld and, just before it reaches the

ground, has magnitude v v v= + =x y2 2 28 1. m s. Thus, the magnitude of the magnetic force is

F q Bm =

= ×( )( )( )−

v sin

. . .

θ

5 00 28 1 0 010 010 C m s T6 ssin . .90 0 1 41 10 6° = × − N


218 Chapter 19

19.76 (a) BI

d10 1

7

2

4 10 5 00

2 0 100= =

× ⋅( )( )(

−μπ

ππ

T m A A

m

.

. )) = × −1 00 10 5. T

(b) F

B I211 2

5 51 00 10 8 00 8 00 10l

= = ×( )( ) = ×− −. . . T A N directed toward wire 1

(c) BI

d20 2

7

2

4 10 8 00

2 0 100= =

× ⋅( )( )(

−μπ

ππ

T m A A

m

.

. )) = × −1 60 10 5. T

(d) F

B I122 1

5 51 60 10 5 00 8 00 10l

= = ×( )( ) = ×− −. . . T A N directed toward wire 2


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