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Transcript
```19
Magnetism
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 ﬁeld 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.
2.
3.
4.
5.
6.
7.
8.
in the positive z-direction (+z)
in the negative z-direction (−z)
in the positive x-direction (+x)
in the negative x-direction (−x)
in the positive y-direction (+y)
in the negative y-direction (−y)
It can move in any direction.
It is impossible for the force to be in the +x-direction when the magnetic ﬁeld 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 ﬁeld. So, for example, if velocity v and magnetic ﬁeld 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 ﬁeld is in the +z-direction. Therefore, 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 ﬁeld and
the charge’s velocity.
•
There are two directions perpendicular to any (non-parallel) magnetic ﬁeld 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
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 ﬁeld) 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 ﬁelds.
Question
In each of the following situations, point charge q moves in a uniform magnetic ﬁeld B. The strength of the
magnetic ﬁeld is indicated by the density of ﬁeld 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 ?
1
q
2
3
v
v
B
B
q
4
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
56157_19_ch19_p171-218.indd 172
v
B
5
v
q
q
B
q
v
B
1
2
3
4
5
1&3
2&4
1, 2, 3 & 4
1, 2, 3, 4 & 5
Cannot be determined
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Magnetism
173
Commentary
Purpose: To develop your understanding of how a magnetic ﬁeld 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 ﬁeld strength at its location in space. The cross product means
that the force is perpendicular to both the ﬁeld 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 ﬁeld.
Since the acceleration is always perpendicular to the velocity, the speed cannot change. Since the charge
has the same speed in all ﬁve 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 ﬁeld B is q(v × B).
•
The magnetic force on a moving point charge is always perpendicular to its direction of motion and to
the ﬁeld.
•
If the only force on a moving point charge is the magnetic force, its speed stays constant.
For Instructors Only
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 ﬁeld is much like the electric ﬁeld, and therefore, the answer
depends on the density of ﬁeld lines or the angle between the direction of motion and the ﬁeld.
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 discussion of displacement until then. One tactic is to present both questions in sequence without discussing or
revealing the answer to the ﬁrst, 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 ﬁelds.
Question
In each of the following situations, point charge q moves in a uniform magnetic ﬁeld B. The strength of the
magnetic ﬁeld is indicated by the density of ﬁeld 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 ?
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174
Chapter 19
1
q
2
3
v
v
B
B
q
4
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
v
B
5
v
q
q
B
q
v
B
1
2
3
4
5
1&3
2&4
1, 2, 3, 4 & 5
None of the above
Cannot be determined
Commentary
Purpose: To develop your understanding of how a magnetic ﬁeld 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 ﬁeld 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 = qvB sin (θ ) = qv⊥ B , where θ is the angle
between the velocity and the magnetic ﬁeld and v⊥ is the component of the velocity v perpendicular to the
magnetic ﬁeld B.
In situation (5), the velocity and magnetic ﬁeld are parallel to each other, so v⊥ = 0, and therefore the acceleration 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 ﬁeld B is q(v × B).
•
The magnitude of the magnetic force on a moving point charge can also be written
F = qvB sin (θ ) = qv⊥ B.
•
If a point charge’s velocity is parallel or anti-parallel to the magnetic ﬁeld, the magnetic force on it
is zero.
For Instructors Only
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 difﬁculty 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 ﬁeld lines are more dense. This may lead some students to
56157_19_ch19_p171-218.indd 174
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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.
1.
2.
3.
4.
5.
Which point charges move in circular paths? Describe the orientation of the path.
Of the charges moving in circular paths, order them from smallest to largest radius.
Which point charges move in helical paths? Describe the orientation of the path.
Of the charges moving in helical paths, order them from smallest to largest radius.
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 magnitude of the magnetic forces on the bar magnet (Fb) and on the charge q (Fq).
q
v=0
N
S
V
1.
2.
3.
4.
Fbar and Fq are both zero.
Fbar is zero and Fq is not zero.
Fbar is not zero and Fq is zero.
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 ﬁeld experiences a force. If the magnetic
ﬁeld 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.
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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
ﬁelds. So, the bar magnet experiences a force due to another magnetic ﬁeld. Since two magnets can attract
or repel, we know that a magnetic ﬁeld 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 ﬁeld.
•
A moving magnetic ﬁeld exerts a force on a stationary charge, just as a stationary magnetic ﬁeld exerts
a force on a moving charge.
For Instructors Only
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 Fbar and Fq 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 ﬁelds.
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 ﬁeld close to zero?
A
D
N
B
D
S
C
S
D
B
N
D
A
1.
2.
3.
4.
5.
6.
7.
8.
56157_19_ch19_p171-218.indd 176
A only
B only
C only
D only
A and B
A, B, and C
C and D
None of the above.
3/18/08 11:39:23 PM
Magnetism
177
Commentary
Purpose: To explore the superposition of magnetic ﬁelds.
Discussion: Consider the magnetic ﬁeld due to a single bar magnet. The magnetic ﬁeld at any point due to
the two magnets will be the sum of the magnetic ﬁelds due to each individual magnet (“superposition”).
Notice that the magnets have opposite orientations, so at point C, the ﬁeld lines due to one will be in the
opposite direction of the ﬁeld 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 ﬁelds will cancel exactly.
At other points, however, the ﬁelds won’t completely cancel. At points A, for example, the ﬁelds are antiparallel but not the same strength, so they’ll only partially cancel. At points B, ﬁelds have the same strength
but aren’t completely opposite in direction, so the net ﬁeld won’t be zero. And at D, the ﬁelds have neither
the same strength nor opposing directions. Thus, (3) is the best answer.
Key Points:
•
Magnetic ﬁelds obey superposition: the total magnetic ﬁeld from two sources is the vector sum of the
magnetic ﬁelds due to each individual source, for every point in space.
•
Two superposed ﬁelds 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 ﬁelds will cancel.
For Instructors Only
For this question, ﬁnding out students’ reasons for their answer is more important than their actual answers.
We recommend having students sketch the ﬁeld lines, and describe how the strength of the ﬁeld is related to
the ﬁeld line diagram.
A demonstration with iron ﬁlings 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 ﬁeld and a magnetic ﬁeld. Which of the
statements below is/are true?
A.
B.
C.
1.
2.
3.
4.
5.
6.
7.
8.
56157_19_ch19_p171-218.indd 177
The particle cannot accelerate in the direction of B.
The path of the particle must be a circle.
Any change in the particle’s kinetic energy is caused by the E ﬁeld.
Only A
Only B
Only C
Both A and B
Both A and C
Both B and C
All are true.
None are true.
3/18/08 11:39:23 PM
178
Chapter 19
Commentary
Purpose: To build your understanding of electric and magnetic forces.
Discussion: Proving a statement true can be difﬁcult; ﬁnding one counter-example that disproves a statement 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 ﬁeld 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 ﬁeld. 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 cannot contribute to any changes in kinetic energy. Statement C is true.
Key Points:
•
An electric ﬁeld acts on a moving charge by exerting a force in the direction of the ﬁeld.
•
A magnetic ﬁeld acts on a moving charge by exerting a force perpendicular to both the ﬁeld and the
charge’s velocity.
•
A magnetic ﬁeld can do no work on a moving charge.
•
The behavior of a moving charge in a combination of electric and magnetic ﬁelds depends on the
relative directions and strengths of the ﬁelds and on the initial velocity of the charge.
For Instructors Only
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 ﬁnd a counter-example to.
Discussing the proof of statement C provides a good opportunity to review the deﬁnition of work (and
reiterate that “force times distance” is an inadequate deﬁnition).
Question M1.07a
Description: Introducing charged particle motion in electric and magnetic ﬁeld combinations.
Question
A charge is released from rest in E and B ﬁelds. Both ﬁelds point along the x-axis. Which of the following
statements regarding the charge’s motion are correct?
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Magnetism
179
y
E
q
z
1.
2.
3.
4.
5.
6.
7.
8.
B
x
The charge will travel along a straight-line path.
The charge’s speed will change as it travels.
The charge will travel in a helical path.
The charge will travel in a helical path of increasing pitch.
The charge will travel in a circle in the x y plane.
1 and 2 only
2 and 4 only
None of the above
Commentary
Purpose: To check and reﬁne your understanding of electric and magnetic forces on moving particles.
Discussion: The full Lorentz force law is F = q ( E + v × B ). This implies that an electric ﬁeld exerts a force
on a charge in the direction of the ﬁeld (for a positive charge) or in the opposite direction (for a negative
charge). It also implies that a magnetic ﬁeld exerts a force that points perpendicularly to both the ﬁeld and
the charge’s velocity, and only if the charge’s velocity has a component perpendicular to the magnetic ﬁeld.
In this situation, the electric ﬁeld will cause the stationary charge to accelerate parallel to the magnetic
ﬁeld. Since the charge’s velocity starts with no component perpendicular to the magnetic ﬁeld, 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 ﬁeld exerts a force on a charge parallel to the ﬁeld, according to F = qE.
•
A magnetic ﬁeld exerts a force on a moving charge according to F = qv × B. If the charge has no
velocity component perpendicular to the ﬁeld, the force is zero.
For Instructors Only
This is the ﬁrst of two related questions.
Students who answer (3), (4), or (7) may be remembering the “fact” that charges in magnetic ﬁelds 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 ﬁrst moves in a straight line until it gets some speed, and then. . . .”)
56157_19_ch19_p171-218.indd 179
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180
Chapter 19
Question M1.07b
Description: Introducing charged particle motion in electric and magnetic ﬁeld combinations.
Question
A charge has an initial velocity parallel to the y-axis in E and B ﬁelds. Both ﬁelds point along the x-axis.
Which of the following statements regarding the charge’s motion are correct?
y
v
E
q
z
1.
2.
3.
4.
5.
6.
7.
8.
B
x
The charge will travel along a straight-line path.
The charge’s speed will change as it travels.
The charge will travel in a helical path.
The charge will travel in a helical path of increasing pitch.
The charge will travel in a circle in the x y plane.
1 and 2 only
2 and 4 only
None of the above
Commentary
Purpose: To check and reﬁne your understanding of electric and magnetic forces on moving particles.
Discussion: The full Lorentz force law is F = q ( E + v × B ). This implies that an electric ﬁeld exerts a force
on a charge in the direction of the ﬁeld (for a positive charge) or in the opposite direction (for a negative
charge). It also implies that a magnetic ﬁeld exerts a force that points perpendicularly to both the ﬁeld and
the charge’s velocity, and only if the charge’s velocity has a component perpendicular to the magnetic ﬁeld.
The charge begins with a velocity in the y-direction, perpendicular to the magnetic ﬁeld. 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 ﬁeld, the magnetic force would continue to bend the particle’s path without changing its speed, causing it to move in a circle in the yz plane.
At the same time, the electric ﬁeld exerts a constant force on the charge in the +x-direction, parallel to the
electric ﬁeld. (If the charge is negative, the force will be in the −x-direction.) This will cause it to experience 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 increasing 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.
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Magnetism
181
Key Points:
•
An electric ﬁeld exerts a force on a charge parallel to the ﬁeld, according to F = qE.
•
A magnetic ﬁeld exerts a force on a moving charge according to F = qv × B. If the charge has no
velocity component perpendicular to the ﬁeld, the force is zero.
•
When magnetic and electric ﬁelds are parallel, their effects on a charge’s motions are independent: the
electric ﬁeld inﬂuences motion parallel to the ﬁelds, and the magnetic ﬁeld inﬂuences motion in the
plane perpendicular to them.
For Instructors Only
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 ﬁeld’s inﬂuence, 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 ﬁeld lines.
Question
Consider a long thin straight wire with a current I. Which of the following statements about the magnetic
ﬁeld lines is true?
A.
B.
C.
D.
1.
2.
3.
4.
5.
6.
7.
8.
56157_19_ch19_p171-218.indd 181
Field lines are parallel to the wire.
Field lines are perpendicular to the wire.
Field lines are directed radially away from the wire.
Field lines are circles centered on any point on the wire.
A only
B only
C only
D only
A and C only
B and D only
B and C only
None of them is true.
3/18/08 11:39:24 PM
182
Chapter 19
Commentary
Purpose: To develop your ability to describe and picture magnetic ﬁeld lines.
Discussion: The magnetic ﬁeld lines due to current ﬂow through a wire forms circles centered on the wire,
perpendicular to the wire. One form of the right-hand rule for magnetic ﬁelds says that if you point your
right thumb along a wire in the direction current ﬂows, and curl your ﬁngers, they indicate the direction that
these circles of magnetic ﬁeld 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 ﬁeld problems.
If the ﬁeld 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 ﬁeld lines never begin
or end; they always form loops.
Key Points:
•
The magnetic ﬁeld 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 ﬁeld lines always form closed loops, and never begin or end.
For Instructors Only
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 ﬁeld 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 statements are true?
I
d
i
L
A.
B.
C.
56157_19_ch19_p171-218.indd 182
The direction of the magnetic force exerted by the long wire on the short wire is directed away from
the long wire.
The magnitude of the force on the short wire is μ0 IiL 2π d.
The long wire experiences a force of exactly the same magnitude as the force experienced by the
short wire.
3/18/08 11:39:25 PM
Magnetism
1.
2.
3.
4.
5.
6.
7.
8.
183
A only
B only
C only
A and B
A and C
B and C
A, B, and C
None of them are true
Commentary
Purpose: To explore and develop familiarity with the force a magnetic ﬁeld exerts on a current-carrying wire.
Discussion: To answer this question, we need to know the magnetic ﬁeld created by the long wire, and the
force the short wire experiences because of that ﬁeld. We will assume that we can treat the very long wire
as inﬁnitely long. (Why else would we be told it is “very long” and not given a variable for its length?)
The magnetic ﬁeld created by an inﬁnitely long, straight wire carrying current I is μ0 I 2π d a distance d
away from the wire. This result is worth remembering. It can be derived via the Biot-Savart law (complicated) or via Ampere’s law and a symmetry argument (simpler). The magnetic ﬁeld lines form circles
surrounding the wire, with a direction given by the right-hand rule, so in the ﬁgure the ﬁeld will point into
the page at every point along the short wire.
The force exerted on a segment of wire by a magnetic ﬁeld B (due to some source other than the
segment itself) is F = iL × B, 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 ﬂowing 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 inﬁnitesimal piece of the wire segment,
and integrate.)
Putting our expression for the magnetic ﬁeld due to the long wire into this force equation, we ﬁnd that
statement B is true.
How about the direction of the force? If the magnetic ﬁeld 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 ﬁeld a distance d away from an inﬁnitely long wire carrying a current I is μ0 I 2π 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 ﬁeld B,
experiences a magnetic force F = iL × B where the direction of L points along the wire in the direction
of current ﬂow.
•
Two parallel current-carrying wires will attract each other if their currents ﬂow in the same directions.
This is called the pinch effect. (If the currents ﬂow in opposite directions, the wires will repel.)
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Chapter 19
For Instructors Only
This is a good question for students just encountering the force law for magnetic ﬁelds acting upon currentcarrying 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 ﬁeld at the center of a circle of current, rather than for an
inﬁnite 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 ﬁeld, 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 ﬁeld expression. In fact, using dimensional analysis (i.e., considering units), one can ﬁgure
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 deﬁnitions, 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 ﬁelds.
Question
In all cases the wire shown carries a current I. For which situation is the magnitude of the magnetic ﬁeld
maximum at the point P?
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Magnetism
1
2
2πR
I
R
P
R
I
3
185
R
4
I
R
R/2
P
I
P
R/2
Commentary
Purpose: To reason with the Biot-Savart law and develop your intuition for the magnetic ﬁelds currents
generate.
Discussion: First, let’s review some basic results that are very useful for thinking about magnetic ﬁelds due
A distance R from an inﬁnitely long straight wire carrying a current I, the magnetic ﬁeld induced by the
current is μ0 I 2π R. Similarly, the magnetic ﬁeld at the center of a current-carrying circular wire of radius
R is μ0 I 2 R. Both of these can be derived from the Biot-Savart law. (The ﬁeld for an inﬁnite wire can also
be derived from Ampere’s law.)
The ﬁeld 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 ﬁeld contribution is in the same direction and has the same magnitude
for each inﬁnitesimal piece of the wire, so the contributions add up without any components that cancel. If
you have half as many inﬁnitesimal segments, you get half as much total ﬁeld strength. So, each half-circle
contributes μ0 I 4π R.
The magnetic ﬁeld due to a ﬁnite 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 ﬂow 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 ﬁeld due to the current
arrangements shown.
Case 4 must also produce a stronger ﬁeld 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 ﬁeld at P. The other
half is the same for both cases, and the small straight segments in case 4 ﬂow directly towards or away
from P and don’t create any ﬁeld there at all.
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Chapter 19
Case 4 must have a larger ﬁeld at P than case 3, because both involve similar geometry, but for case 3 the
two semicircular currents ﬂow in opposite directions around P, so their contributions will partially cancel.
For case 4, the ﬁeld 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 ﬁeld at the center of a
circular current of radius R is larger (by a factor of p ) than the ﬁeld a distance R away from an inﬁnite
wire having the same current. Since case 2 has only a ﬁnite segment of wire, the ﬁeld it creates will be
smaller than for an inﬁnite 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 inﬁnitesimal elements are the same distance away from point P. Since
the contribution to the ﬁeld from each element drops off as 1/r, the ﬁeld at P due to 1 must be larger than
in case 2.
Thus, we know case 4 must have the largest magnetic ﬁeld at P.
Key Points:
•
We can compare the magnetic ﬁelds due to many different current arrangements without actually
calculating them.
•
Knowing expressions for the magnetic ﬁelds created by various standard current arrangements
(inﬁnite lines, circular loops, etc.) is helpful.
•
The Biot-Savart law lets you reason qualitatively about magnetic ﬁelds and the variables they
depend on.
For Instructors Only
Students should be encouraged (or admonished) to reason to the answer, rather than determining expressions for the ﬁeld 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 magnetic ﬁeld strength falls off as 1/R; that a circular loop around a point creates a stronger ﬁeld than an inﬁnite
line at the same distance; and that a point inside a closed current path generally experiences a stronger ﬁeld
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 ﬁeld strength at P.
Question M2.02
Description: Reasoning with the Biot-Savart law and superposition of magnetic ﬁelds.
Question
Order the following situations according to the magnitude of the magnetic ﬁeld at the point P. Order from
highest to lowest.
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Magnetism
1.
2.
3.
4.
5.
6.
187
ABCD
BDAC
DABC
None of the above
Commentary
Purpose: To reason with the Biot-Savart law and develop your intuition for the magnetic ﬁelds currents
generate.
Discussion: From the Biot-Savart law, we can ﬁnd that the magnetic ﬁeld at the center of a circular current
loop of radius R is μ0 I 2 R, and the magnetic ﬁeld a distance R away from an inﬁnitely long, straight
current is μ0 I 2π R, where I is the current in both cases.
Similarly, the Biot-Savart law lets us reason that the magnetic ﬁeld due to half of a circular current loop will
be one-half the value for a complete circle, and the ﬁeld due to a “semi-inﬁnite” line (half an inﬁnite line,
starting at the point on the inﬁnite line closest to P) will be one-half the value for an inﬁnite line.
The other piece of information we’ll need to reason about the current arrangements in this problem is that
the magnetic ﬁeld 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 ﬂows, and curl your ﬁngers around the wire, your
ﬁngers will point the direction of the magnetic ﬁeld.
Once we know these facts, we can reason about the current arrangements shown in the question. Case
A consists of one inﬁnite line (equivalent to two semi-inﬁnite lines) and one circular loop (equivalent to
two semicircles), with current ﬂowing such that the magnetic ﬁelds due to the two oppose each other.
We’ll notate this as 2C − 2L, meaning the ﬁeld strength at P is due to two semicircles minus two opposing
semi-inﬁnite lines.
Case B consists of two semi-inﬁnite lines and one semicircle, such that the ﬁelds from all three are in the
same direction and add without cancellation. Thus, the ﬁeld for case B is 1C + 2L. Similarly, the ﬁeld 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 ﬁeld strength as: BC (2C + 4L) > BD (1.5C + 2L) > BB
(1C + 2L) > BA (2C − 2L).
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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 noting that L = C/p, substituting this into the inequality, and simplifying: 1C + 2L > 2C − 2 L → 4 > p QED.
Key Points:
•
We can compare the magnetic ﬁelds due to many different current arrangements without actually
calculating them.
•
Knowing expressions for the magnetic ﬁelds created by various standard current arrangements
(inﬁnite lines, circular loops, etc.) is helpful, as is knowing the ﬁeld values for “standard” fractions
of these arrangements.
•
The Biot-Savart law lets you reason qualitatively about magnetic ﬁelds and the variables they
depend on.
For Instructors Only
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 ﬁeld generation — involving
qualitative features of the Biot-Savart law, symmetry, and superposition — not about calculating ﬁelds for
various current loop shapes.
You might want to open a discussion of what “portions” of these standard current arrangements (loops and
inﬁnite lines) are easy to deduce the ﬁeld for, and what require detailed calculation. For example, since all
points on a circular current loop contribute the same amount to the net magnetic ﬁeld at the center, we can
ﬁnd the ﬁeld due to any fraction of a circle by taking that fraction of the ﬁeld due to an entire circle. For
lines, however, only inﬁnite and semi-inﬁnite 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 difﬁculty 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 ﬁnd the ﬁeld due to an inﬁnite 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
ﬁeld, B, at the center of the loop?
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Magnetism
189
Top View
farthest from you
R
I
closest to you
I
1.
2.
3.
4.
5.
Left
Right
Up
Down
None of the above
Commentary
Purpose: To develop your understanding of the Biot-Savart law.
Discussion: According to the Biot-Savart law, an inﬁnitesimal segment of current I with length ds creates
an inﬁnitesimal magnetic ﬁeld dB = ( μ0 4π ) I ds × r r 3 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 ﬁnd the total magnetic ﬁeld by
adding up (integrating) the contributions from all of the inﬁnitely many inﬁnitesimal current segments. Note
the cross product: this means that the magnetic ﬁeld created will have a direction that is perpendicular to
both the current ﬂow and the displacement vector r. This means the magnetic ﬁeld 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 ﬂow, and curl your ﬁngers partially, and
your ﬁngers will show you the direction in which the magnetic ﬁeld points as it circles around the segment.
You can use this to determine that for any inﬁnitesimal segment of the circular current path shown, the
inﬁnitesimal magnetic ﬁeld 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 ﬁeld at the center must
point upward as well. Answer (3).
Key Points:
•
The Biot-Savart law describes the magnetic ﬁeld due to current ﬂow.
•
The magnetic ﬁeld due to a current loop is the sum of the magnetic ﬁeld contributions from each
inﬁnitesimal segment of the loop.
•
Current ﬂowing through a wire causes magnetic ﬁeld lines to circle around the wire, with a direction
given by the “point the thumb, curl the ﬁngers” version of the right-hand rule.
For Instructors Only
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 ﬁeld point or
vice-versa and subsequently have difﬁculty reconciling the various forms of the right-hand rule.
For students who already know that the magnetic ﬁeld lines due to a straight wire forms circles around the
wire, and can ﬁnd the direction, the question is simpler and can be approached by a common-sense superposition argument. (The question is even easier for those familiar with magnetic dipoles.)
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190
Chapter 19
This question is a good lead-in to the next one, which asks students to determine the magnitude of the
ﬁeld. While that may seem straightforward and trivial, it remains difﬁcult 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 magnetic ﬁeld, B, at the center of the loop?
Top View
farthest from you
R
I
closest to you
I
1.
2.
3.
4.
5.
6.
0
μ0 I/4p R
μ0 I/2p R
μ0 I/4R
μ0 I/2R
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 inﬁnitesimal segment of current I with length ds creates
an inﬁnitesimal magnetic ﬁeld dB = ( μ0 4π ) I ds × r r 3 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 ﬁnd the total magnetic ﬁeld
by adding up (integrating) the contributions from all of the inﬁnitely many inﬁnitesimal current segments.
All inﬁnitesimal segments around the circular current loop in this problem create magnetic ﬁelds pointing
in the “up” direction (as discussed in the previous question), so we can ﬁnd the magnitude of dB due to one
segment and integrate to ﬁnd the total magnitude.
The magnitude dB we get when we put our known values and angles into the Biot-Savart law is
( μ0 4π ) I ds R2 . All segments make exactly the same contribution to the ﬁeld 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 ﬁeld at the origin must be 2π R dB = μ0 I 2 R : answer (5).
Key Points:
•
The Biot-Savart law describes the magnetic ﬁeld due to inﬁnitesimal current element.
•
The magnetic ﬁeld due to a current loop is the sum of the magnetic ﬁeld contributions from each
inﬁnitesimal segment of the loop.
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Magnetism
•
If all the segments create ﬁeld contributions with the same direction and magnitude at some point,
we don’t actually have to solve an integral to ﬁnd the total ﬁeld.
•
It is important to recognize when you can ﬁnd the magnetic ﬁeld due to steady currents using
Ampere’s law and when you must use the Biot-Savart law.
191
For Instructors Only
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 difﬁcult
time recognizing when they can use Ampere’s law and when they must use the Biot-Savart law. The magnetic
ﬁeld 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 signiﬁcance 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 ﬁeld due to a straight inﬁnite wire and
not incorrectly applying the Biot-Savart law.
QUICK QUIZZES
1.
(b). The force that a magnetic ﬁeld exerts on a charged particle moving through it is given by
F = qvB sin θ = qvB⊥ , where B⊥ is the component of the ﬁeld 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 ﬁeld on a charge is proportional to the charge’s velocity relative to the ﬁeld. 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 ﬁeld is given
by τ = BIA sin θ . Note that this torque depends on the strength of the ﬁeld, 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 ﬁeld. 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 l = ( μ0 I1 I 2 ) ( 2π d ) . 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 magnitudes but opposite directions.
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192
Chapter 19
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 ﬁeld 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 ﬁeld is 55°. The
magnitude of the magnetic force experienced by the electron is then
)
)
)
F = q vB sin θ = (1.6 × 10 −19 C ( 2.5 × 10 6 m s ( 0.10 × 10 −4 T sin 55° = 3.3 × 10 −18 N
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 l sin θ = w , or B = w I l sin θ . The magnitude of the magnetic ﬁeld is a minimum when
θ = 90° and sin θ = 1. Thus,
Bmin =
1.0 × 10 −2 N
w
=
= 0.20 T
I l ( 0.10 A ) ( 0.50 m )
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 ﬁeld is θ = 30.0°. Thus, the magnitude of the torque
experienced by this coil containing N = 10 turns is
τ = BIAN sin θ = ( 0.010 T) ( 2.0 A ) ⎡⎣( 0.20 m ) ( 0.30 m ) ⎤⎦ (10 ) sin 30.0° = 6.0 × 10 −3 N ⋅ m
meaning that (c) is the correct choice.
4.
A charged particle moving perpendicular to a magnetic ﬁeld experiences a centripetal force of
magnitude Fc = m v 2 r = qvB and follows a circular path of radius r = mv qB . The speed of this
proton must be
v=
qBr (1.6 × 10
=
m
−19
)
C ( 0.050 T) (1.0 × 10 −3 m
1.67 × 10
−27
kg
) = 4.8 × 10
3
ms
and choice (e) is the correct answer.
5.
The magnitude of the magnetic ﬁeld at distance r from a long straight wire carrying current I is
B = μ0 I 2π r . Thus, for the described situation,
B=
( 4π × 10
−7
)
T ⋅ m /A (1 A )
2π ( 2 m )
= 1 × 10 −7 T
making (d) the correct response.
6.
The force per unit length between this pair of wires is
)
−7
F μ0 I1 I 2 ( 4π × 10 T ⋅ m A ( 3 A )
=
=
= 9 × 10 −7 N ∼ 1 × 10 −6 N
l
2π d
2π ( 2 m )
2
and (d) is the best choice for this question.
7.
The magnitude of the magnetic ﬁeld inside the speciﬁed solenoid is
⎛ N⎞
⎛ 120 ⎞
B = μ0 nI = μ0 ⎜ ⎟ I = ( 4π × 10 −7 T ⋅ m A ⎜
( 2.0 A ) = 6.0 × 10 −4 T
⎝ l⎠
⎝ 0.50 m ⎟⎠
)
which is choice (e).
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Magnetism
193
8.
The magnitude of the magnetic force experienced by a charged particle in a magnetic ﬁeld is
given by F = qvB sin θ , where v is the speed of the particle and θ is the angle between the direction of the particle’s velocity and the direction of the magnetic ﬁeld. If either v = 0 [choice (e)]
or sin θ = 0 [choice (c)], this force has zero magnitude. All other choices are false, so the correct
9.
The force that a magnetic ﬁeld exerts on a moving charge is always perpendicular to both the
direction of the ﬁeld 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 momentum 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 ﬁrst enters the ﬁeld, it experiences a force
directed upward, toward the top of the page. This will deﬂect 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 circular path, with no change in speed, as long as it is in the ﬁeld. After completing a half circle, the
proton will exit the ﬁeld traveling toward the left. The correct answer is choice (d).
11.
The contribution made to the magnetic ﬁeld 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 equidistant from the two wires, and the wires carry the same magnitude currents, these two oppositely
directed contributions to the magnetic ﬁeld have equal magnitudes and cancel each other. Therefore, the total magnetic ﬁeld at point P is zero, making (a) the correct answer for this question.
12.
The magnetic ﬁeld 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 ﬁeld 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 ﬁeld 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 ﬁeld 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 contributions 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 ﬁeld due the upper wire will have a greater magnitude than
will the inward directed ﬁeld due to the lower wire. Thus, the resultant ﬁeld 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 ﬁeld due to each wire is directed into the page, so their magnitudes add and the
resultant ﬁeld cannot be zero at any point in this region. This means that choice (b) is false. In
region III, the ﬁeld 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 ﬁelds due to the
two wires will have equal magnitudes, canceling each other and producing a zero resultant ﬁeld.
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 ﬁeld B is τ = BIA sin θ .
The line perpendicular to the plane of each coil is also perpendicular to the direction of the
magnetic ﬁeld (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 = (1 m ) ( 2 m ) = 2 m 2 . Coil B is elliptical with semi-major axis a = 0.75 m and
56157_19_ch19_p171-218.indd 193
3/18/08 11:39:39 PM
194
Chapter 19
semi-minor axis b = 0.5 m, giving an area AB = π ab or AB = π ( 0.75 m ) ( 0.5 m ) = 1.2 m 2.
1
1
Coil C is triangular with area AC = ( base ) ( height ) , or AC = (1 m ) ( 3 m ) = 1.5 m 2 . Thus,
2
2
AA > AC > AB, meaning that τ A > τ C > τ B and choice (b) is the correct answer.
15.
According to right-hand rule number 2, the magnetic ﬁeld 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 ﬁeld
is given by B = μ0 I 2π r , so choices (b) and (d) are false. Choice (a) is correct about both the
magnitude and direction of the ﬁeld and is the correct answer for the question.
16.
The magnetic ﬁeld inside a solenoid, carrying current I, with N turns and length L, is
μ (2NA ) I
μ N I
μ N I 1
⎛ N⎞
= 4 BA .
B = μ0 nI = μ0 ⎜ ⎟ I . Thus, BA = 0 A , BB = 0 A = BA , and BC = 0
⎝ L⎠
2 LA
2
LA
LA 2
Therefore, we see that BC > BA > BB , and choice (d) gives the correct rankings.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
No. The force that a constant magnetic ﬁeld 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 ﬁeld.
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 ﬁeld 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 ﬁeld
uniform in direction — the magnetic ﬁeld lines loop through the loop.
12.
Near the poles the magnetic ﬁeld 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 deﬂect 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 ﬁeld for some arbitrary orientation of the ﬁeld 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 ﬂat faces, making one face a
north pole and the other a south pole. The yellow magnet has a pole on its lower face which
56157_19_ch19_p171-218.indd 194
3/18/08 11:39:40 PM
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 ﬁrmly together. The yellow magnet would continue to be repelled by and ﬂoat
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 ﬁeld is directed in the +x-direction, toward the right.
(a)
(b)
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 .
r
With v 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 +z-direction .
19.2
(c)
When the proton moves in the +x-direction, parallel to the magnetic ﬁeld, the magnitude of
the magnetic force it experiences is F = qvB sin ( 0°) = 0, or there is a zero force in this case .
(a)
For a positively charged particle, the direction of the force is that predicted by the right-hand
rule number one. These are:
(b)
19.3
(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
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) .
into the page
(b)
toward the right
(c)
toward bottom of page
r
Hold the right hand with the ﬁngers inr the direction of v so that as you close your hand, the
ﬁngers move toward the direction of B. The thumb will point in the direction of the force (and
hence the deﬂection) if the particle has a positive charge. The results are
(a)
(c)
56157_19_ch19_p171-218.indd 195
in plane of page and to left
Since the particle is positively charged, use the right-hand rule number 1. In this case, start with
r
r
the ﬁngers of the right hand in the direction of v and the thumb pointing
in the direction of F.
r
As you start closing the hand, the ﬁngers point in the direction of B after they have moved 90°.
The results are
(a)
19.4
(a′)
toward top of page
θ = 180° ⇒ zero force
(b)
out of the page , since the charge is negative.
(d)
into the page
3/19/08 3:38:30 AM
196
19.5
Chapter 19
(a)
The proton experiences maximum force when it moves perpendicular to the magnetic ﬁeld,
and the magnitude of this maximum force is
)
)
Fmax = qvB sin 90° = (1.60 × 10 −19 C ( 6.00 × 10 6 m s (1.50 T) (1) = 1.44 × 10 −12 N
19.6
Fmax 1.44 × 10 −12 N
=
= 8.62 × 1014 m s 2
m p 1.67 × 10 −27 kg
(b)
amax =
(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 electron would be much greater than that of the proton because of the mass of the electron is
much smaller.
From F = qvB sin θ , the magnitude of the force is found to be
)
)
)
F = (1.60 × 10 −19 C ( 6.2 × 10 6 m s ( 50.0 × 10 −6 T sin ( 90.0°) = 4.96 × 10 −17 N
r
Using the right-hand ruleur (ﬁngers point westward in direction of v , so they move downward
toward the direction of B 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,
m
v2
qBr
= qvB sin 90°, or v =
where r = RE + 1000 km = 7.38 × 10 6 m
r
m
v=
(1.60 × 10
−19
)
)
C ( 4.00 × 10 −8 T ( 7.38 × 10 6 m
1.67 × 10 −27 kg
)=
2.83 × 10 7 m s
r
r
r
If v is toward the west and B is northward, F will be directed downward as required.
19.8
The speed attained by the electron is found from
v=
(a)
2 e ( ΔV )
=
m
1
mv 2 = q ( ΔV ) , or
2
)
2 (1.60 × 10 −19 C ( 2 400 V)
9.11 × 10
−31
kg
= 2 .90 × 10 7 m s
Maximum force occurs when the electron enters the region perpendicular to the ﬁeld.
Fmax = q vB sin 90°
)
)
= (1.60 × 10 −19 C ( 2 .90 × 10 7 m s (1.70 T) = 7.90 × 10 −12 N
(b)
Minimum force occurs when the electron enters the region parallel to the ﬁeld.
Fmin = q vB sin 0° = 0
56157_19_ch19_p171-218.indd 196
3/18/08 11:39:41 PM
Magnetism
19.9
197
The magnitude of the magnetic force acting on the electron is F = q vB sin 90° = me a, so the
magnitude of the magnetic ﬁeld is given by
B=
)
)
)
−31
16
2
me a ( 9.11 × 10 kg ( 4.0 × 10 m s
=
= 1.5 × 10 −2 T
ev
(1.60 × 10−19 C (1.5 × 10 7 m s
)
To determine the direction of the ﬁeld, employ a variation of right-hand rule number 1. Hold your
right hand ﬂat with the ﬁngers 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 ﬁngers will point out of the page after they have moved
90°. This would be the correct direction for the magnetic ﬁeld if the particle were positively
charged. Since the electron is a negative particle, the actual direction of the ﬁeld is opposite that
predicted by the right-hand rule, or it is directed into the page (the –z-direction) .
19.10
The force on a single ion is
F1 = qvB sin θ
)
= (1.60 × 10 −19 C ( 0.851 m s ) ( 0.254 T) sin ( 51.0°) = 2 .69 × 10 −20 N
The total number of ions present is
ions ⎞
⎛
22
3
N = ⎜ 3.00 × 10 20
⎟ (100 cm = 3.00 × 10
⎝
cm 3 ⎠
)
Thus, assuming all ions move in the same direction through the ﬁeld, the total force is
)
)
F = N ⋅ F1 = ( 3.00 × 10 22 ( 2 .69 × 10 −20 N = 806 N
19.11
Gravitational force:
)
)
Fg = mg = ( 9.11 × 10 −31 kg ( 9.80 m s 2 = 8.93 × 10 −30 N downward
Electric force:
)
Fe = qE = ( −1.60 × 10 −19 C ( −100 N C ) = 1.60 × 10 −17 N upward
Magnetic force:
)
)
)
Fm = qvB sin θ = ( −1.60 × 10 −19 C ( 6.00 × 10 6 m s ( 50.0 × 10 −6 T sin ( 90.0°)
= 4.80 × 10 −17 N in direction opposite right hand rule prediction
Fm = 4.80 × 10 −17 N downward
19.12
56157_19_ch19_p171-218.indd 197
Hold the right hand with the ﬁngers in the direction of the current so, as you close the hand, the
ﬁngers move toward the direction of the magnetic ﬁeld. 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
3/18/08 11:39:42 PM
198
19.13
Chapter 19
From F = BI L sin θ , the magnetic ﬁeld is
B=
F L
0.12 N m
=
= 8.0 × 10 −3 T
I sin θ (15 A ) sin 90°
r
r
r
The direction of B must be the + z-direction to have F in the –y-direction when I is in the
+x-direction.
19.14
(a)
F = BIL sin θ = ( 0.28 T ) ( 3.0 A ) ( 0.14 m ) sin 90° = 0.12 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 ﬁeld.
19.15
Use the right-hand rule number 1, holding your right hand with the ﬁngers in the direction of the
current and the thumb pointing in the direction of the force. As you close your hand, the ﬁngers
will move toward the direction of the magnetic ﬁeld. The results are
(a)
19.16
into the page
(b)
toward the right
(c)
toward the bottom of the page
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
F
⎛ m⎞
= BI sin θ = ⎜ ⎟ g
⎝ l⎠
l
⎛ m⎞ g
B=⎜ ⎟
⎝ l ⎠ I sin θ
giving
To ﬁnd the minimum possible ﬁeld, the magnetic ﬁeld should be perpendicular to the current
(θ = 90.0°). Then,
⎡
g
g ⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞ ⎤ 9.80 m s 2
⎛ m⎞
Bmin = ⎜ ⎟
= 0.245 T
= ⎢ 0.500
⎥
⎝ l ⎠ I sin 90.0° ⎣
cm ⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ ⎦ ( 2.00 A ) (1)
To ﬁnd the direction of the ﬁeld, hold the right hand with the thumb pointing upward (direction of
the force) and the ﬁngers pointing southward (direction of current). Then, as you close the hand,
the ﬁngers point eastward. The magnetic ﬁeld should be directed eastward .
19.17
F = BIL sin θ = ( 0.300 T ) (10.0 A ) ( 5.00 m ) sin ( 30.0° ) = 7.50 N
19.18
(a)
The magnitude is
)
F = BIL sin θ = ( 0.60 × 10 −4 T (15 A ) (10.0 m ) sin ( 90°) = 9.0 × 10 −3 N
r
r
r
F is perpendicular to B. Using the right-hand rule number 1, the orientation of F is found
to be 15° above the horizontal in the northward direction .
(b)
)
F = BIL sin θ = ( 0.60 × 10 −4 T (15 A ) (10.0 m ) sin (165°) = 2 .3 × 10 −3 N
and, from the right-hand rule number 1, the direction is horizontal and due west .
56157_19_ch19_p171-218.indd 198
3/19/08 3:38:31 AM
Magnetism
19.19
199
r
For minimum ﬁeld, B should be perpendicular to the wire. If the force is to be northward, the
ﬁeld 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 sin 90° = μ k ( mg ) , or
2
μk ( m L ) g ( 0.200 ) (1.00 g cm ) ( 9.80 m s ) ⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞
B=
=
⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ = 0.131 T
I sin 90°
(1.50 A ) (1.00 )
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,
r
Fm
L
I=
= BI =
mg
, or
L
( m L ) g = ( 0.040
B
kg m ) ( 9.80 m s 2
3.60 T
)=
0.109 A
From the right-hand rule number 1, the current must be to the right if the force is to be upward
when the magnetic ﬁeld is into the page.
19.21
(a)
The magnetic force must be directed upward and 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 sin θ = mg, and for minimum ﬁeld
θ = 90°. Thus,
Bmin
)
2
mg ( 0.015 kg ) ( 9.80 m s
= 0.20 T
=
=
IL
( 5.0 A ) ( 0.15 m )
ur
For the magnetic force to be directed upward when the current is toward the left, B must
be directed out of the page .
(c)
19.22
If the ﬁeld exceeds 0.20 T, the upward magnetic force exceeds the downward force of
gravity, so the wire accelerates upward .
The magnitude of the magnetic force exerted on a current-carrying conductor in a magnetic ﬁeld
is given by F = BIL sin θ , where B is the magnitude of the ﬁeld, 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
ﬁeld. In this case,
F = ( 0.390 T) ( 5.00 A ) ( 2.80 m ) sin θ = ( 5.46 N ) sin θ
56157_19_ch19_p171-218.indd 199
(a)
If θ = 60.0°, then sin θ = 0.866 and F = 4.73 N
(b)
If θ = 90.0°, then sin θ = 1.00 and F = 5.46 N
(c)
If θ = 120°, then sin θ = 0.866 and F = 4.73 N
3/18/08 11:39:43 PM
200
19.23
Chapter 19
For each segment, the magnitude of the force is given by F = BI L sin θ , 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
0.056 6
parallel to xz
plane at 45° to
both +x- and
+z-directions
da
19.24
0.400 2
90.0°
The magnitude of the force is
F = BIL sin θ = ( 5.0 × 10 −5 N ) ( 2.2 × 10 3 A ) ( 58 m ) sin 65° = 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 ﬁeld is τ = BIAN sin θ , and maximum torque occurs
when the ﬁeld is directed parallel to the plane of the loop ( θ = 90°). Thus,
2
τ max = ( 0.50 T ) ( 25 × 10 −3 A ) ⎡π ( 5.0 × 10 −2 m ) ⎤ ( 50 ) sin 90° = 4.9 × 10 −3 N ⋅ m
⎣
⎦
19.26
The magnitude of the torque is τ = NBIA sin θ , where θ is the angle between the ﬁeld and the
perpendicular to the plane of the loop. The circumference of the loop is 2π r = 2 .00 m, so the
1.00 m
1
and the area is A = π r 2 = m 2 .
π
π
Thus,
19.27
⎛1
⎞
τ = (1) ( 0.800 T) (17.0 × 10 −3 A ) ⎜ m 2 ⎟ sin 90.0° = 4.33 × 10 −3 N ⋅ m
⎝π
⎠
The area is A = π ab = π ( 0.200 m ) ( 0.150 m ) = 0.094 2 m 2 . Since the ﬁeld is parallel to the plane
of the loop, θ = 90.0° and the magnitude of the torque is
τ = NBIA sin θ
= 8 ( 2 .00 × 10 −4 T ) ( 6.00 A ) ( 0.094 2 m 2 ) 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 ﬁeld and the perpendicular to the plane of the loop is
θ = 90.0° − 30.0° = 60.0°. Then, the magnitude of the torque is
τ = NBIA sin θ = 100 ( 0.80 T ) (1.2 A ) [( 0.40 m ) ( 0.30 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 .
56157_19_ch19_p171-218.indd 200
3/19/08 3:38:32 AM
Magnetism
19.29
(a)
The torque exerted on a coil by a uniform magnetic ﬁeld is τ = BIAN sin θ , with maximum
torque occurring when θ = 90°. Thus, the current in the coil must be
I=
(b)
201
τ max
0.15 N ⋅ m
=
= 0.56 A
BAN ( 0.90 T) ⎡( 3.0 × 10 −2 m ) ( 5.0 × 10 −2 m ) ⎤ ( 200 )
⎣
⎦
If I has the value found above and θ is now 25°, the torque on the coil is
τ = BIAN sin θ = ( 0.90 T ) ( 0.56 A ) [( 0.030 m ) ( 0.050 m )] ( 200 ) sin 25° = 0.064 N ⋅ m
19.30
The resistance of the loop is
R=
−8
ρ L (1.70 × 10 Ω ⋅ m ) ( 8.00 m )
=
= 1.36 × 10 −3 Ω
A
1.00 × 10 −4 m 2
and the current in the loop is I =
ΔV
0.100 V
=
= 73.5 A
R 1.36 × 10 −3 Ω
The magnetic ﬁeld exerts torque τ = NBIA sin θ on the loop, and this is a maximum when
sin θ = 1. Thus,
τ max = NBIA = (1) ( 0.400 T ) ( 73.5 A ) ( 2 .00 m ) = 118 N ⋅ m
2
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 = 90.0° − θ . The number of turns on the loop is
N=
L
4.00 m
=
= 10.0
circumference 4 ( 0.100 m )
The torque about the z-axis due to gravity is
⎛s
⎞
τ g = mg ⎜ cos θ ⎟ , where s = 0.100 m is the length
⎝2
⎠
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 m = NBIA sin θ . At
)
equilibrium, τ m = τ g or NBI ( s 2 sin θ = mg ( s cos θ ) 2 .
This reduces to
tan θ =
)
( 0.100 kg ) ( 9.80 m s2
mg
=
= 14.4
2 NBIs 2 (10.0 ) ( 0.010 0 T) ( 3.40 A ) ( 0.100 m )
Since tan θ = tan ( 90.0° − φ ) = cot φ , the angle the loop makes with the vertical at
equilibrium is φ = cot −1 (14.4 ) = 3.97° .
continued on next page
56157_19_ch19_p171-218.indd 201
3/19/08 3:38:33 AM
202
Chapter 19
(b)
At equilibrium,
τ m = NBI ( s 2 ) sin θ
= (10.0 ) ( 0.010 0 T ) ( 3.40 A ) ( 0.100 m ) sin ( 90.0° − 3.97° )
2
= 3.39 × 10 −3 N ⋅ m
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
+ x -direction . 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 − z-direction .
(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 + z-direction .
(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 ﬁeld. As given by right-hand rule 1, this places it
in the yz plane at 130° counterclockwise from the +y-axis . The force acting on segment
b-c tends to rotate it counterclockwise about the x-axis, so the torque is in the +x -direction .
(e)
The loop tends to rotate counterclockwise about the x-axis .
(f)
μ = IAN = ( 0.900 A ) ⎡⎣( 0.500 m ) ( 0.300 m ) ⎤⎦ (1) = 0.135 A ⋅ m 2
(g)
(h)
19.33
(a)
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 ﬂows 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 ﬁeld is θ = 90.0° + 40.0° = 130° .
τ = μ B sin θ = ( 0.135 A ⋅ m 2 ) (1.50 T) sin (130°) = 0.155 N ⋅ m
The magnetic force acting on the electron provides the centripetal acceleration, holding the
electron in the circular path. Therefore, F = q vB sin 90° = me v 2 r , or
r=
(b)
)
)
The time to complete one revolution around the orbit (i.e., the period) is
T=
56157_19_ch19_p171-218.indd 202
)
)
−31
7
me v ( 9.11 × 10 kg (1.5 × 10 m s
=
= 0.043 m = 4.3 cm
eB
(1.60 × 10−19 C ( 2.0 × 10−3 T
distance traveled 2π r 2π ( 0.0043 m )
=
=
= 1.8 × 10 −8 s
constant speed
1.5 × 10 7 m s
v
3/19/08 3:38:34 AM
Magnetism
19.34
19.35
)
203
)
(a)
F = qvB sin θ = (1.60 × 10 −19 C ( 5.02 × 10 6 m s ( 0.180 T) sin ( 60.0°) = 1.25 × 10 −13 N
(b)
a=
F 1.25 × 10 −13 N
=
= 7.50 × 1013 m s 2
m 1.67 × 10 −27 kg
For the particle to pass through with no deﬂection, 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
Fm = Fe , or qvB = qE , 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).
mv 2
In the deﬂection chamber, the magnetic force supplies the centripetal acceleration, so qvB =
,
r
mv m ( E B ) mE
or r =
=
.
=
qB
qB
qB 2
Using the given data, the radius of the path is found to be
( 2 .18 × 10
(1.60 × 10
−26
r=
19.37
−19
)
)
kg ( 950 V m )
C ( 0.930 T)
2
= 1.50 × 10 −4 m = 0.150 mm
1
From conservation of energy, ( KE + PE ) f = ( KE + PE )i , we ﬁnd that mv 2 + qV f = 0 + qVi ,
2
or the speed of the particle is
(
2 q Vi − V f
v=
m
)=
2 q ( ΔV )
=
m
)
2 (1.60 × 10 −19 C ( 250 V)
2 .50 × 10
−26
kg
= 5.66 × 10 4 m s
The magnetic force supplies the centripetal acceleration giving qvB =
or
19.38
r=
)
)
−26
4
mv ( 2 .50 × 10 kg ( 5.66 × 10 m s
=
= 1.77 × 10 −2 m = 1.77 cm
qB
(1.60 × 10−19 C ( 0.500 T)
)
Since the centripetal acceleration is furnished by the magnetic force acting on the ions,
mv 2
mv
or the radius of the path is r =
. Thus, the distance between the impact points
qvB =
r
qB
(that is, the difference in the diameters of the paths followed by the U 238 and the U 235 isotopes) is
Δd = 2 ( r238 − r235 ) =
=
or
56157_19_ch19_p171-218.indd 203
mv 2
r
2v
( m238 − m235 )
qB
2 ( 3.00 × 10 5 m s
)
kg
⎡
( 238 u − 235 u ) ⎛⎜⎝ 1.66 × 10 −27 ⎞⎟⎠ ⎤⎥
⎢
−19
u ⎦
C
T
1
.
60
×
10
0
.
600
(
)⎣
(
)
Δd = 3.11 × 10 −2 m = 3.11 cm
3/18/08 11:39:46 PM
204
19.39
Chapter 19
In the perfectly elastic, head-on collision between the α -particle and the initially stationary proton, conservation of momentum requires that m p v p + mα vα = mα v0 while conservation of kinetic
energy also requires that v0 − 0 = − vα − v p or v p = vα + v0. Using the fact that mα = 4 m p and
combining these equations gives
)
(
)
(
(
)
m p ( vα + v0 ) + 4 m p vα = 4 m p v0
and
vα = 3v0 5
or
v p = ( 3v0 5 ) + v0 = 8 v0 5
vα =
Thus,
3
3⎛5 ⎞ 3
v0 = ⎜ v p ⎟ = v p
5
5⎝8 ⎠ 8
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
v 2p
mpvp mpvp
qpvp B = mp
or
R=
=
R
qp B
eB
and
19.40
qα vα B = mα
vα2
r
or
r=
(
)(
)
4 m p 3v p 8
mα vα
3 ⎛ mpvp ⎞
3
=
= ⎜
=
R
4 ⎝ eB ⎟⎠
4
qα B
( 2e) B
A charged particle follows a circular path when it moves perpendicular to the magnetic ﬁeld. The
magnetic force acting on the particle provides the centripetal acceleration, holding the particle in
the circular path. Therefore, F = qvB sin 90° = m v 2 r . Since the kinetic energy is K = mv 2 2, we
rewrite the force as F = qvB sin 90° = 2 K r, and solving for the speed v gives v =
19.41
(a)
2K
.
qBr
Within the velocity selector, the electric and magnetic ﬁelds exert forces in opposite directions on charged particles passing through. For particles having a particular speed, these
forces have equal magnitudes, and the particles pass through without deﬂection. The selected
speed is found from Fe = qE = qvB = Fm , giving v = E B. In the deﬂection chamber, the
selected particles follow a circular path having a diameter of d = 2 r = 2 mv qB. Thus, the
mass to charge ratio for these particles is
m Bd
Bd
B 2 d ( 0.093 1 T) ( 0.396 m )
=
=
= 2.08 × 10 −7 kg C
=
=
q 2v 2 ( E B ) 2 E
2 ( 8 250 V m )
2
(b)
If the particle is doubly ionized (i.e., two electrons have been removed from the neutral
atom), then q = 2 e and the mass of the ion is
⎛ m⎞
m = ( 2 e ) ⎜ ⎟ = 2 (1.60 × 10 −19 C ( 2.08 × 10 −7 kg C = 6.66 × 10 −26 kg
⎝ q⎠
)
(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. ≈
56157_19_ch19_p171-218.indd 204
)
m 6.66 × 10 −26 kg
=
= 39.9, suggesting that the element is calcium .
m p 1.67 × 10 −27 kg
3/18/08 11:39:47 PM
Magnetism
19.42
205
Since the path is circular, the particle moves perpendicular to the magnetic ﬁeld, and the
v2
mv
magnetic force supplies the centripetal acceleration. Hence, m
= qvB, or B =
. But
r
qr
the momentum is given by p = mv = 2 m ( KE ) , and the kinetic energy of this proton is
⎛ 1.60 × 10 −19 J ⎞
−12
KE = (10.0 × 10 6 eV ⎜
⎟⎠ = 1.60 × 10 J. We then have
1 eV
⎝
)
B=
19.43
)
(1.60 × 10−19 C ( 5.80 × 1010 m
)
)=
7.88 × 10 −12 T
−7
4
μ0 I ( 4π × 10 T ⋅ m A ) (1.00 × 10 A )
=
= 2 .00 × 10 −5 T = 20.0 μ T
2π r
2π (100 m )
toward the left
(b)
out of page
(c)
lower left to upper right
The magnetic ﬁeld at distance r from a long conducting wire is B = μ0 I 2π r . Thus, if
B = 1.0 × 10 −15 T at r = 4.0 cm, the current must be
I=
19.46
)
2 (1.67 × 10 −27 kg (1.60 × 10 −12 J
Imagine grasping the conductor with the right hand so the ﬁngers curl around the conductor in the
direction of the magnetic ﬁeld. The thumb then points along the conductor in the direction of the
current. The results are
(a)
19.45
qr
=
Treat the lightning bolt as a long, straight conductor. Then, the magnetic ﬁeld is
B=
19.44
2 m ( KE )
)
−15
2π rB 2π ( 0.040 m ) (1.0 × 10 T
=
= 2.0 × 10 −10 A
4π × 10 −7 T ⋅ m A
μ0
Model the tornado as a long, straight, vertical conductor and imagine grasping it with
the right hand so the ﬁngers 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 = μ0 I 2π r as
I=
19.47
)
From B = μ0 I 2π r, the required distance is
r=
56157_19_ch19_p171-218.indd 205
)
3
−8
2π rB 2π ( 9.00 × 10 m (1.50 × 10 T
=
= 675 A
4π × 10 −7 T ⋅ m A
μ0
−7
μ0 I ( 4π × 10 T ⋅ m A ) ( 20 A )
=
= 2 .4 × 10 −3 m = 2 .4 mm
2π B
2π (1.7 × 10 −3 T)
3/19/08 3:38:35 AM
206
19.48
Chapter 19
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 ﬁeld to be out of the page and negative into the page.
(a)
At the point half way between the two wires,
⎡μ I
μ I ⎤
μ
Bnet = − B1 − B2 = − ⎢ 0 1 + 0 2 ⎥ = − 0 ( I1 + I 2 )
2π r
⎣ 2π r1 2π r2 ⎦
( 4π × 10 T ⋅ m A) (10.0 A) = − 4.00 × 10
2π ( 5.00 × 10 m )
−7
=−
At point P1 ,
Bnet =
(c)
−5
T
Bnet = 40.0 μ T into the page
or
(b)
−2
μ0
2π
⎡ I1 I 2 ⎤
⎢+ r − r ⎥
⎣ 1
2 ⎦
4π × 10 −7 T ⋅ m A ⎡ 5.00 A
5.00 A ⎤
⎢⎣ 0.100 m − 0.200 m ⎥⎦ = 5.00 μ T out of page
2π
μ 0 ⎡ I1 I 2 ⎤
− +
2π ⎢⎣ r1 r2 ⎥⎦
4π × 10 −7 T ⋅ m A ⎡ 5.00 A
5.00 A ⎤
=
−
+
⎢
2π
⎣ 0.300 m 0.200 m ⎥⎦
At point P2 ,
Bnet
Bnet = + B1 − B2 =
Bnet = − B1 + B2 =
= 1.67 μ T out of page
19.49
The distance from each wire to point P is given by
r=
1
2
( 0.200 m )2 + ( 0.200 m )2
= 0.141 m
At point P, the magnitude of the magnetic ﬁeld
produced by each of the wires is
B=
−7
μ0 I ( 4π × 10 T ⋅ m A ) ( 5.00 A )
=
= 7.07 μ T
2π r
2π ( 0.141 m )
Carrying currents into the page, the ﬁeld A produces
at P is directed to the left and down at – 1 35°, while B
creates a ﬁeld to the right and down at – 45°. Carrying currents toward you, C produces a ﬁeld
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 ﬁeld is then
Bnet = 4 ( 7.07 μ T ) sin 45.0° = 20.0 μ T toward the bottom of the page
56157_19_ch19_p171-218.indd 206
3/18/08 11:39:48 PM
Magnetism
19.50
207
Call the wire carrying a current of 3.00 A wire 1 and the other wire 2. Also, choose the line running from wire 1 to wire 2 as the positive x-direction.
(a)
At the point midway between the wires, the ﬁeld due to
each wire is parallel to the y-axis and the net ﬁeld is
Bnet = + B1y − B2 y = μ0 ( I1 − I 2 ) 2π r
Bnet =
Thus,
−7
T⋅m A
2π ( 0.100 m )
) ( 3.00 A − 5.00 A) = − 4.00 × 10
−6
T
Bnet = 4.00 μ T toward the bottom of the page
or
(b)
( 4π × 10
At point P, r1 = ( 0.200 m ) 2 and B1 is directed at θ1 = +135°.
The magnitude of B1 is
B1 =
−7
μ0 I1 ( 4π × 10 T ⋅ m A ) ( 3.00 A )
=
= 2 .12 μ T
2π r1
2π 0.200 2 m
(
)
The contribution from wire 2 is in the –x-direction and
has magnitude
B2 =
−7
μ0 I 2 ( 4π × 10 T ⋅ m A ) ( 5.00 A )
= 5.00 μ T
=
2π r2
2π ( 0.200 m )
Therefore, the components of the net ﬁeld at point P are:
Bx = B1 cos 135° + B2 cos 180°
= ( 2 .12 μ T) cos 135° + ( 5.00 μ T) cos 180° = −6.50 μ T
By = B1 sin 135° + B2 sin 180° = ( 2 .12 μ T) sin 135° + 0 = +1.50 μ T
and
Therefore, Bnet = Bx2 + By2 = 6.67 μ T at
⎛ B ⎞
⎛ 6.50 μ T ⎞
= 77.0°
θ = tan −1 ⎜ x ⎟ = tan −1 ⎜
⎝ 1.50 μ T ⎟⎠
⎝ By ⎠
ur
Bnet = 6.67 μ T at 77.0° to the left of vertical
or
19.51
Call the wire along the x-axis wire 1 and the other wire 2. Also, choose the positive direction for
the magnetic ﬁelds at point P to be out of the page.
At point P, Bnet = + B1 − B2 =
or
Bnet =
( 4π × 10
−7
μ 0 I1 μ 0 I 2 μ 0 ⎛ I1 I 2 ⎞
−
=
−
2π r1 2π r2 2π ⎜⎝ r1 r2 ⎟⎠
)
T ⋅ m A ⎛ 7.00 A 6.00 A ⎞
−7
−
⎜⎝
⎟ = +1.67 × 10 T
2π
3.00 m 4.00 m ⎠
Bnet = 0.167 μ T out of the page
56157_19_ch19_p171-218.indd 207
3/18/08 11:39:49 PM
208
19.52
Chapter 19
(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 ﬂowing in the positive x-direction, the right-hand rule
2 gives the direction of the magnetic ﬁeld above the wire as being toward the left, or
in the −y-direction.
(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 z-direction .
(c)
Since the proton moves with constant velocity, a zero net force acts on it. Thus, the
magnitude of the magnetic force must equal that of the gravitational force .
(d)
ΣFz = maz = 0 ⇒ Fm = Fg or qvB = mg where B = μ0 I 2π d . This gives
q vμ 0 I
qvμ0 I
.
= mg, or the distance the proton is above the wire must be d =
2π d
2π mg
(e)
d=
qvμ0 I (1.60 × 10
=
2π mg
d = 5.40 × 10
19.53
−2
−19
)
)
)
C ( 2.30 × 10 4 m s ( 4π × 10 −7 T ⋅ m A (1.20 × 10 −6 A
2π (1.67 × 10
−27
)
kg ( 9.80 m s
2
)
)
m = 5.40 cm
(a)
From B = μ0 I 2π r , observe that the ﬁeld is inversely proportional to the distance from the
conductor. Thus, the ﬁeld will have one-tenth its original value if the distance is increased
by a factor of 10. The required distance is then r ′ = 10 r = 10 ( 0.400 m ) = 4.00 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 ﬁeld. Therefore, Bnet = B1 − B2, or
Bnet =
−7
⎞
1
1
μ0 I ⎛ 1 1 ⎞ ( 4π × 10 T ⋅ m A ) ( 2 .00 A ) ⎛
=
−
−
⎟
⎜
⎜
2π
2π ⎝ r1 r2 ⎠
⎝ 0.398 5 m 0.401 5 m ⎟⎠
= 7.50 × 10 −9 T = 7.50 nT
(c)
Call r the distance from cord center to ﬁeld point P
and 2 d = 3.00 mm the distance between centers of
the conductors.
Bnet =
μ0 I ⎛ 1
1 ⎞ μ0 I ⎛ 2 d ⎞
−
⎟
⎜⎝
⎟=
⎜
2π r − d r + d ⎠ 2π ⎝ r 2 − d 2 ⎠
7.50 × 10 −10 T =
so
( 4π × 10
−7
)
T ⋅ m A ( 2 .00 A ) ⎛ 3.00 × 10 −3 m ⎞
⎜⎝ r 2 − 2.25 × 10 −6 m 2 ⎟⎠
2π
r = 1.26 m
The ﬁeld of the two-conductor cord is weak to start with and falls off rapidly with distance.
(d)
56157_19_ch19_p171-218.indd 208
The cable creates zero ﬁeld at exterior points, since a loop in Ampère’s law encloses zero
total current.
3/18/08 11:39:50 PM
Magnetism
19.54
209
(a)
Point P is equidistant from the two wires, which carry
identical
ur currents.urThus, the contributions of the two
wires, Bupper and Blower, to the magnetic ﬁeld at P will
have equal magnitudes. The horizontal components of
these contributions will cancel, while the vertical
components add. The resultant ﬁeld will be vertical,
in the + y-direction .
(b)
The distance of each wire from point P is
ur
r = x 2 + d 2 , and the cosine of the angle that Bupper
ur
and Blower make with the vertical is cos θ = x r. The
ur
ur
magnitude of either Bupper or Blower is Bwire = μ0 I 2π r and the vertical components of either
of these contributions have values of
( Bwire )y = ( Bwire ) cos θ = ⎛⎜⎝
μ0 I ⎞ x μ0 Ix
⎟ =
2π r ⎠ r 2π r 2
The magnitude of the resultant ﬁeld at point P is then
BP = 2 ( Bwire )y =
(c)
19.55
μ0 Ix
μ0 I x
=
π r2
π ( x2 + d 2 )
The point midway between the two wires is the origin (0,0). From the above result for part
(b), the resultant ﬁeld at this midpoint is BP x= 0 = 0 . This is as expected, because righthand rule 2 shows that at the midpoint the ﬁeld due to the upper wire is toward the right,
while that due to the lower wire is toward the left. Thus, the two ﬁelds cancel, yielding a
zero resultant ﬁeld.
The force per unit length that one wire exerts on the other is F l = μ0 I1 I 2 2π d, where d is the
distance separating the two wires. In this case, the value of this force is
)
−7
F ( 4π × 10 T ⋅ m A ( 3.0 A )
= 3.00 × 10 −5 N m
=
l
2π ( 6.00 × 10 −2 m
2
)
Imagine these two wires lying side by side on a table with the two currents ﬂowing toward you,
wire 1 on the left and wire 2 on the right. Right-hand rule 2 shows the magnetic ﬁeld due to wire
1 at the location of wire 2 is directed vertically upward. Then, right-hand rule 1 gives the direction of the force experienced by wire 2, with its current ﬂowing through this ﬁeld, 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 l = μ0 I1 I 2 2π d .
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
I2 =
(b)
)
2π ( 4.0 × 10 −2 m
2π d ⎛ F ⎞
=
( 2.0 × 10−4 N m = 8.0 A
⎜ ⎟
μ0 I1 ⎝ l ⎠ ( 4π × 10 −7 T ⋅ m A ( 5.0 A )
)
)
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 .
continued on next page
56157_19_ch19_p171-218.indd 209
3/19/08 3:38:35 AM
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 l = μ0 I1 I 2 2π d , 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.
F μ 0 I1 I 2
=
= 0.080 N m, and the distance between the wires will be
L
2π d
Thus,
( 4π × 10−7 T ⋅ m A) ( 60.0 A) ( 30.0 A)
μ 0 I1 I 2
=
2π ( 0.080 N m )
2π ( 0.080 N m )
d=
= 4.5 × 10 −3 m = 4.5 mm
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
Fnet = +
or
Fnet =
1 ⎞
μ 0 I1 I 2 l
μ II l
μ I I l⎛1
− 0 1 2 = 0 1 2 ⎜ −
⎟
2π c
2π ( c + a )
2π ⎝ c c + a ⎠
( 4π × 10
−7
)
T ⋅ m A ( 5.00 A ) (10.0 A ) ( 0.450 m ) ⎛
1
1
⎞
−
⎜⎝
⎟
2π
0.100 m 0.250 m ⎠
= + 2 .70 × 10 −5 N = 2 .70 × 10 −5 N to the left
19.59
The magnetic ﬁeld inside of a solenoid is B = μ0 nI = μ0 ( N L ) I . Thus, the current in this
solenoid must be
I=
19.60
)
)
The magnetic ﬁeld inside of a solenoid is B = μ0 nI = μ0 ( N L ) I . Thus, the number of turns on
this solenoid must be
N=
56157_19_ch19_p171-218.indd 210
)
−3
−2
BL ( 2.0 × 10 T ( 6.0 × 10 m
=
= 3.2 A
μ0 N
( 4π × 10−7 T ⋅ m A ( 30 )
( 9.0 T) ( 0.50 m )
BL
=
= 4.8 × 10 4 turns
μ0 I ( 4π × 10 −7 T ⋅ m A ) ( 75 A )
3/19/08 3:38:36 AM
Magnetism
19.61
(a)
211
From R = ρ L A, the required length of wire to be used is
−3
⎡
R ⋅ A ( 5.00 Ω ) ⎣π ( 0.500 × 10 m
L=
=
1.7 × 10 −8 Ω ⋅ m
ρ
)
2
4⎤
⎦ = 58 m
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
N=
(b)
58 m
L
= 9.2 × 10 2 = 920
=
2π r 2π (1.00 × 10 −2 m
)
From B = μ0 nI , the number of turns per unit length on the solenoid is
n=
B
4.00 × 10 −2 T
=
= 7.96 × 10 3 turns m
μ0 I ( 4π × 10 −7 T ⋅ m A ( 4.00 A )
)
Thus, the required length of the solenoid is
L=
19.62
N
9.2 × 10 2 turns
=
= 0.12 m = 12 cm
n 7.96 × 10 3 turns m
The magnetic ﬁeld inside the solenoid is
m⎞ ⎤
turns ⎞ ⎛ 100 cm
⎡⎛
−2
B = μ0 nI1 = ( 4π × 10 −7 T ⋅ m A ⎢⎜ 30
⎟⎠ ⎜⎝
⎟⎠ ⎥ (15.0 A ) = 5.65 × 10 T
⎝
1
m
cm
⎦
⎣
)
Therefore, the magnitude of the magnetic force on any one of the sides of the square loop is
)
)
F = BI 2 L sin 90.0° = ( 5.65 × 10 −2 T ( 0.200 A ) ( 2 .00 × 10 −2 m = 2 .26 × 10 −4 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 qvB = mv 2 r . The magnetic
ﬁeld inside the solenoid is then found to be
B=
(b)
)
)
)
−31
4
mv ( 9.11 × 10 kg (1.0 × 10 m s
=
= 2 .8 × 10 −6 T = 2 .8 μ T
qr
(1.60 × 10−19 C ( 2 .0 × 10−2 m
)
From B = μ0 nI , the current is the solenoid is found to be
I=
B
2 .8 × 10 −6 T
=
μ0 n ( 4π × 10 −7 T ⋅ m A ⎡⎣( 25 turns cm ) (100 cm 1 m ) ⎤⎦
)
= 8.9 × 10 −4 A = 0.89 mA
56157_19_ch19_p171-218.indd 211
3/18/08 11:39:52 PM
212
19.64
Chapter 19
(a)
When switch S is closed, a total current NI (current I in a total
of N conductors) ﬂows toward the right through the lower side
of the coil. This results in a downward force of magnitude
Fm = B ( NI ) w being exerted on the coil by the magnetic ﬁeld,
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 ﬁeld must be directed out of the page toward the reader. For the system to
be restored to balance, it is necessary that
Fm = F ′
19.65
or
B ( NI ) w = mg , giving
B=
mg
NIw
(b)
The magnetic ﬁeld 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)
B=
(a)
The magnetic ﬁeld at the center of a circular current loop of radius R and carrying current I
is B = μ0 I 2 R. The direction of the ﬁeld at this center is given by right-hand rule number 2.
Taking out of the page (toward the reader) as positive the net magnetic ﬁeld at the common
center of these coplanar loops is
)
)
)
−3
2
mg ( 20.0 × 10 kg ( 9.80 m s
=
= 0.26 T
NIw ( 50 ) ( 0.30 A ) ( 5.0 × 10 −2 m
Bnet = B2 − B1 =
−7
μ0 I 2 μ0 I1 ( 4π × 10 T ⋅ m A ) ⎛
3.0 A
5.0 A ⎞
−
=
−
⎜⎝
⎟
−2
2 r2
2 r1
2
9.0 × 10 m 12 × 10 −2 m ⎠
= −5.2 × 10 −6 T = 5.2 μ T into the page
(b)
To have Bnet = 0, it is necessary that I 2 r2 = I1 r1 , or
⎛I ⎞
⎛ 3.0 A ⎞
r2 = ⎜ 2 ⎟ r1 = ⎜
(12 cm ) = 7.2 cm
⎝ 5.0 A ⎟⎠
⎝ I1 ⎠
19.66
Since the magnetic force must supply the centripetal acceleration, qvB = mv 2 r or the radius of
the path is r = mv qB .
(a)
The time for the electron to travel the semicircular path (of length π r ) is
t=
π ( 9.11 × 10 −31 kg )
π r π ⎛ mv ⎞ π m
=
=
= ⎜
v
v ⎝ qB ⎟⎠ qB (1.600 × 10 −19 C ) ( 0.010 0 T)
= 1.79 × 10 −9 s = 1.79 ns
continued on next page
56157_19_ch19_p171-218.indd 212
3/19/08 3:38:37 AM
Magnetism
(b)
213
The radius of the semicircular path is 2.00 cm. From r = mv qB , the momentum of the
electron is p = mv = qBr, and the kinetic energy is
KE = mv 2 =
1
2
)
−19
−2
q 2 B 2 r 2 (1.60 × 10 C ( 0.010 0 T) ( 2 .00 × 10 m
=
=
2m
2m
2 ( 9.11 × 10 −31 kg
( m v )2
2
2
)
)
2
1 keV
⎛
⎞
= 3.51 keV
KE = ( 5.62 × 10 −16 J ⎜
⎝ 1.60 × 10 −16 J ⎟⎠
)
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 ﬁeld direction, the ﬁeld at point P is
−7
μ0 ⎛ I1 I 2 ⎞ ( 4π × 10 T ⋅ m A ) ⎛ 5.00 A
3.00 A ⎞
=
−
B = B1 − B2 =
−
⎜⎝
⎟
2π ⎜⎝ r1 r2 ⎟⎠
2π
0.400 m 0.300 m ⎠
= 5.00 × 10 −7 T = 0.500 μ T out of the page
(b)
At 30.0 cm above the intersection of the wires, the ﬁeld
components are as shown at the right, where
By = − B1 = −
=−
and
μ 0 I1
2π r
( 4π × 10
Bx = B2 =
−7
)
T ⋅ m A ( 5.00 A )
2π ( 0.3300 m )
= −3.33 × 10 −6 T
−7
μ0 I 2 ( 4π × 10 T ⋅ m A ) ( 3.00 A )
=
= 2 .00 × 10 −6 T
2π r
2π ( 0.300 m )
The resultant ﬁeld is
⎛B ⎞
B = Bx2 + By2 = 3.89 × 10 −6 T at θ = tan −1 ⎜ y ⎟ = −59.0°
⎝ Bx ⎠
or
56157_19_ch19_p171-218.indd 213
r
B = 3.89 μ T at 59.0 ° clockwise from +x direction
3/18/08 11:39:54 PM
214
19.68
Chapter 19
For the rail to move at constant velocity, the net force acting on it must be zero. Thus, the magnitude of the magnetic force must equal that of the friction force giving BIL = μ k ( mg ) , or
B=
19.69
(a)
2
μk ( mg ) ( 0.100 ) ( 0.200 kg ) ( 9.80 m s )
=
= 3.92 × 10 −2 T
IL
(10.0 A ) ( 0.500 m )
Since the magnetic ﬁeld 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 ﬂow experience a magnetic deﬂection toward electrode A. Negative ions will experience a force deﬂecting them toward
electrode B. This separation of charges creates an electric ﬁeld directed from A toward B.
At equilibrium, the electric force caused by this ﬁeld must balance the magnetic force, so
qvB = qE = q ( ΔV d )
or
19.70
v=
ΔV
160 × 10 −6 V
=
= 1.333 m s
Bd ( 0.040 0 T) ( 3.00 × 10 −3 m
)
(b)
The magnetic ﬁeld is directed from N to S. If the charge carriers are negative moving in the
r
direction of v, 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
r
moving in the direction of v, 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 .
(a)
The magnetic force acting on the wire is directed upward and of magnitude
Fm = BIL sin 90° = BIL
Thus,
ay =
ay =
(b)
m
=
( 4.0 × 10
Fm − mg
BI
=
− g , or
m
(m L)
−3
5.0 × 10
Using Δy = v0 y t +
t=
56157_19_ch19_p171-218.indd 214
ΣFy
)
T ( 2 .0 A )
−4
kg m
− 9.80 m s 2 = 6.2 m s 2
1 2
ay t with v0 y = 0 gives
2
2 ( Δy )
=
ay
2 ( 0.50 m )
= 0.40 s
6.2 m s 2
3/18/08 11:39:54 PM
Magnetism
19.71
215
Label the wires 1, 2, and 3 as shown in Figure 1,
and let B1 , B2 , and B3 respectively represent the
magnitudes of the ﬁelds produced by the
currents in those wires. Also, observe that
θ = 45° in Figure 1.
(
)
At point A, B1 = B2 = μ0 I 2π a 2 or
B1 = B2 =
and B3 =
( 4π × 10
−7
)
T ⋅ m A ( 2 .0 A )
2π ( 0.010 m ) 2
= 28 μ T
Figure 1
( 4π × 10−7 T ⋅ m A) ( 2 .0 A) = 13 μ T
μ0 I
=
2π ( 3a )
2π ( 0.030 m )
These ﬁeld contributions are oriented as shown in Figure 2.
r
r
Observe that the horizontal components of B1 and B2 cancel
r
while their vertical components add to B 3. The resultant
ﬁeld at point A is then
BA = ( B1 + B2 ) cos 45° + B3 = 53 μ T, or
Figure 2
r
B A = 5 3 μ T directed toward the bottom of the page
At point B, B1 = B2 =
−7
μ0 I ( 4π × 10 T ⋅ m A ) ( 2 .0 A )
=
= 40 μ T
2π a
2π ( 0.010 m )
μ0 I
= 20 μ T . These contributions are oriented as
2π ( 2 a )
shown in Figure 3. Thus, the resultant ﬁeld at B is
and B3 =
r
r
B B = B 3 = 20 μ T directed toward the bottom of the page
(
Figure 3
)
At point C, B1 = B2 = μ0 I 2π a 2 = 28 μ T while
B3 = μ0 I 2π a = 40 μ T . These contributions are oriented
as shown in Figure 4. Observe that the horizontal
r
r
components of B1 and B2 cancel while their vertical
r
components add to oppose B 3 . The magnitude of the
resultant ﬁeld at C is
BC = ( B1 + B2 ) sin 45° − B3
Figure 4
= ( 56 μ T) sin 45° − 40 μ T = 0
56157_19_ch19_p171-218.indd 215
3/18/08 11:39:55 PM
216
19.72
Chapter 19
(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.
ΣFy = 0 ⇒ T cos 8.0° = mg
T=
or
mg
cos 8.0°
⎛ mg ⎞
ΣFx = 0 ⇒ Fm = T sin 8.0° = ⎜
sin 8.0°
⎝ cos 8.0° ⎟⎠
or
Fm = ( mg ) tan 8.0°. Thus,
I=
μ0 I 2 L
= ( mg ) tan 8.0° which gives
2π d
d ⎡⎣( m L ) g ⎤⎦ tan 8.0°
μ0 2π
Observe that the distance between the two wires is
d = 2 ⎡⎣( 6.0 cm ) sin 8.0° ⎤⎦ = 1.7 cm , so
I=
19.73
(1.7 × 10
−2
)
)
m ( 0.040 kg m ) ( 9.80 m s 2 tan 8.0°
2 .0 × 10
−7
T⋅m A
= 68 A
Note: We solve part (b) before part (a) for this problem.
(b)
Since the magnetic force supplies the centripetal acceleration for this particle, q vB = mv 2 r
or the radius of the path is r = mv qB. The speed of the particle may be written as
v = 2 ( KE ) m , so the radius becomes
r=
2 m ( KE )
qB
=
)
)
2 (1.67 × 10 −27 kg ( 5.00 × 10 6 eV (1.60 × 10 −19 J eV
(1.60 × 10
−19
)
C ( 0.050 0 T)
)
= 6.46 m
Consider the circular path shown at the right and
observe that the desired angle is
α = sin −1 ⎛
⎝
1.00 m ⎞
1.00 m ⎞
= sin −1 ⎛
= 8.90°
⎠
⎝
r
6.46 m ⎠
continued on next page
56157_19_ch19_p171-218.indd 216
3/19/08 3:38:37 AM
Magnetism
(a)
217
The constant speed of the particle is v = 2 ( KE ) m, so
the vertical component of the momentum as the particle
leaves the ﬁeld is
py = mv y = − mv sin α = − m
(
)
2 ( KE ) m sin α = − sin α
)
2 m ( KE )
)
py = − sin ( 8.90°) 2 (1.67 × 10 −27 kg ( 5.00 × 10 6 eV (1.60 × 10 −19 J eV
or
)
= − 8.00 × 10 −21 kg ⋅ m s
19.74
The force constant of the spring system is found from the elongation produced by the weight
acting alone.
)
)
−3
2
F mg (10.0 × 10 kg ( 9.80 m s
= 19.6 N m
=
=
x
x
0.50 × 10 −2 m
k=
The total force stretching the springs when the ﬁeld is turned on is
ΣFy = Fm + mg = kxtotal
Thus, the downward magnetic force acting on the wire is
Fm = kxtotal − mg
)
)
= (19.6 N m ) ( 0.80 × 10 −2 m − (10.0 × 10 −3 kg ( 9.80 m s 2
)
= 5.9 × 10 −2 N
Since the magnetic force is given by Fm = BIL sin 90°, the magnetic ﬁeld is
B=
19.75
)
)
(12 Ω ) ( 5.9 × 10 −2 N
Fm
Fm
=
=
= 0.59 T
IL ( ΔV R ) L ( 24 V) ( 5.0 × 10 −2 m
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 components with magnitudes of
v x = v0 x = 20.0 m s, and
)
v y = v02 y + 2 ay ( Δy ) = 0 + 2 ( −9.80 m s 2 ( −20.0 m ) = 19.8 m s
The velocity of the ball is perpendicular to the magnetic ﬁeld and, just before it reaches the
ground, has magnitude v = v x2 + v y2 = 28.1 m s. Thus, the magnitude of the magnetic force is
Fm = qvB sin θ
)
= ( 5.00 × 10 −6 C ( 28.1 m s ) ( 0.010 0 T) sin 90.0° = 1.41 × 10 −6 N
56157_19_ch19_p171-218.indd 217
3/18/08 11:39:56 PM
218
19.76
56157_19_ch19_p171-218.indd 218
Chapter 19
−7
μ0 I1 ( 4π × 10 T ⋅ m A ) ( 5.00 A )
= 1.00 × 10 −5 T
=
2π d
2π ( 0.100 m )
(a)
B1 =
(b)
F21
= B1 I 2 = (1.00 × 10 −5 T ( 8.00 A ) = 8.00 × 10 −5 N directed toward wire 1
l
(c)
B2 =
(d)
F12
= B2 I1 = (1.60 × 10 −5 T ( 5.00 A ) = 8.00 × 10 −5 N directed toward wire 2
l
)
−7
μ0 I 2 ( 4π × 10 T ⋅ m A ) ( 8.00 A )
= 1.60 × 10 −5 T
=
2π d
2π ( 0.100 m )
)
3/18/08 11:39:57 PM
```
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