Section 24.5 Magnetic Fields Exert Forces on Moving Charges Download

Transcript
Section 24.5 Magnetic Fields Exert
Forces on Moving Charges
© 2015 Pearson Education, Inc.
Magnetic Fields
• Sources of Magnetic Fields
• You already know that a moving charge is the “creator” of a
magnetic field.
• Effects of Magnetic Fields
• If a moving charge “experiences” a magnetic field, it there will be a
force on the charge.
• Sound familiar?!?!
Magnetic Fields Exert Forces on Moving
Charges
• Magnetic fields also exert forces on moving charged
particles and on electric currents in wires.
There is no magnetic
force on a charged
particle at rest.
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There is no magnetic force on
a charged particle moving
parallel to a magnetic field.
Slide 24-3
Magnetic Fields Exert Forces on Moving
Charges
As the angle α between the velocity and the magnetic field
increases, the magnetic force also increases. The force is
greatest when the angle is 90°. The magnetic force is always
perpendicular to the plane containing and .
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Slide 24-4
Magnetic Force on Particles
• A magnetic force is exerted on a particle within a magnetic
field only if
• the particle has a charge.
• the charged particle is moving….
• …with at least a portion of its velocity perpendicular to the
magnetic field.
Slide 24-5
Magnetic Force on a Charged Particle
moving
charge
• magnitude:
•
•
•
•
in a
magnetic
field
vv
FB = qv B sin θ
q: charge in Coulombs
v: speed in meters/second
B: magnetic field in Tesla
Θ: angle between v and B
• direction: Right Hand Rule
with an angle not equal to 0
or 180 degrees
Magnetic Fields Exert Forces on Moving
Charges
• We determine the correct direction of the force using the
right-hand rule for forces.
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Slide 24-7
The right hand rule to determine
magnetic force
This right hand rule is a little different
than what your book uses, but I think
it’s easier to remember. You must
keep your hand in this configuration
and turn your whole wrist!!
F
v
•
•
•
Point in the direction of the velocity.
Turn your hand so that your middle
finger (or three remaining fingers)
point in the direction of the field.
Your thumb gives you the direction of
the force.
B
Slide 24-8
Magnetic Fields Exert Forces on Moving
Charges
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Slide 24-9
QuickCheck 24.15
The direction of the magnetic force on the proton is
A.
B.
C.
D.
E.
To the right.
To the left.
Into the screen.
Out of the screen.
The magnetic force is zero.
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Slide 24-10
QuickCheck 24.15
The direction of the magnetic force on the proton is
A.
B.
C.
D.
E.
To the right.
To the left.
Into the screen.
Out of the screen.
The magnetic force is zero.
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Slide 24-11
QuickCheck 24.16
The diagram shows a top view of an electron beam passing
between the poles of a magnet. The beam will be deflected
A.
B.
C.
D.
Toward the north pole of the magnet.
Toward the south pole of the magnet.
Out of the plane of the figure
Into the plane of the figure.
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Slide 24-12
QuickCheck 24.16
The diagram shows a top view of an electron beam passing
between the poles of a magnet. The beam will be deflected
A.
B.
C.
D.
Toward the north pole of the magnet.
Toward the south pole of the magnet.
Out of the plane of the figure
Into the plane of the figure.
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Slide 24-13
QuickCheck 24.17
A beam of positively charged particles passes between the
poles of a magnet as shown in the figure; the force on the
particles is noted in the figure. The magnet’s north pole is
on the _____, the south pole on the _____.
A. Left, right
B. Right, left
C. There’s not enough
information to tell.
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Slide 24-14
QuickCheck 24.17
A beam of positively charged particles passes between the
poles of a magnet as shown in the figure; the force on the
particles is noted in the figure. The magnet’s north pole is
on the _____, the south pole on the _____.
A. Left, right
B. Right, left
C. There’s not enough
information to tell.
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Slide 24-15
QuickCheck 24.18
The direction of the magnetic force on the electron is
A.
B.
C.
D.
E.
Upward.
Downward.
Into the screen.
Out of the screen.
The magnetic force is zero.
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Slide 24-16
QuickCheck 24.18
The direction of the magnetic force on the electron is
A.
B.
C.
D.
E.
Upward.
Downward.
Into the screen.
Out of the screen.
The magnetic force is zero.
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Slide 24-17
QuickCheck 24.19
Which magnetic field causes the observed force?
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Slide 24-18
QuickCheck 24.19
Which magnetic field causes the observed force?
C.
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Slide 24-19
Sample Problem
•
What is the magnetic force exerted on a 3.0 µC charge
moving north at 300,000 m/s in a magnetic field of 200
mT if the field is directed
a) North.
b) South.
c) East.
d) West.
Slide 24-20
Sample Problem
•
What is the magnetic force exerted on a 3.0 µC charge
moving north at 300,000 m/s in a magnetic field of 200
mT if the field is directed
a) North.
b) South.
c) East.
d) West.
FB = qvB sin θ
(
)
(
)
FB = 3 × 10 −6 C (300,000 m/s ) 200 × 10 −3 T sin (0°)
FB = 0
FB = qvB sin θ
(
)
(
)
FB = 3 ×10−6 C (300,000 m/s) 200 ×10−3 T sin (180°)
FB = 0
FB = qvB sin θ
(
)
(
)
FB = 3 ×10 −6 C (300,000 m/s ) 200 ×10 −3 T sin (90°)
FB = 0.18 N, down (toward Earth)
FB = qvB sin θ
(
)
(
)
FB = 3 ×10−6 C (300,000 m/s) 200 ×10 −3 T sin (90°)
FB = 0.18 N, up (toward sky)
Sample Problem
• Calculate the magnitude and direction of the magnetic
force in the situation below.
FB = qvB sin θ
(
)
(
)
FB = 3 ×10 −6 C (300,000 m/s ) 200 ×10−3 T sin (34°)
FB = 0.101 N, out of the page
Conceptual Example 24.6 Determining the
force on a moving electron
An electron is moving to the right in a magnetic field that
points upward, as in FIGURE 24.26. What is the direction
of the magnetic force?
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Slide 24-23
Magnetic forces…
• are always orthogonal (at right angles) to the plane
established by the velocity and magnetic field vectors.
• can accelerate charged particles by changing their
direction.
• can cause charged particles to move in circular or helical
paths.
Slide 24-24
Magnetic forces cannot…
• do work on charged particles. Why?
• The force is always perpendicular to the motion.
• What are the implications of this?
• They cannot change the speed or kinetic energy of charged
particles.
Slide 24-25
This means magnetic forces…
• …are centripetal!
• Remember that centripetal acceleration is
2
v
ac =
r
• Therefore, centripetal force is
mv
ΣFc = mac =
r
2
Slide 24-26
Paths of Charged Particles in Magnetic Fields
• When we studied the motion of objects subject to a force
that was always perpendicular to the velocity, the result
was circular motion at a constant speed. For example, a
ball moved at the end of a string moved in a circle due to
the perpendicular force of tension in the string.
• For a charged particle moving in a magnetic field, the
magnetic force is always perpendicular to and so it
causes the particle to move in a circle.
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Slide 24-27
Paths of Charged Particles in Magnetic Fields
• A particle moving
perpendicular to a
uniform magnetic field
undergoes uniform
circular motion at
constant speed.
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Slide 24-28
Paths of Charged Particles in Magnetic Fields
• Derive an equation for the radius of orbit for a charged
particle in a magnetic field.
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Slide 24-29
Sample Problem
• What is the orbital radius of a proton moving at 20,000 m/s
perpendicular to a 40 T magnetic field?
Slide 24-30
Sample Problem
• What is the orbital radius of a proton moving at 20,000 m/s
perpendicular to a 40 T magnetic field?
mac = ΣFc
v2
m = FB
r
v2
m = qvB sin θ
r
v
m = qB sin θ
r
mv
1.67 ×10 − 27 kg (20,000 m/s )
−6
r=
=
=
5
.
22
×
10
m
−19
qB sin θ
1.6 ×10 C (40 T )sin (90°)
(
(
)
)
Paths of Charged Particles in Magnetic Fields
• The motion of a charged particle when its velocity is
neither parallel nor perpendicular to the magnetic field:
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Slide 24-32
Paths of Charged Particles in Magnetic Fields
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Slide 24-33
Paths of Charged Particles in Magnetic Fields
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Slide 24-34
Paths of Charged Particles in Magnetic Fields
• High-energy particles stream out from the sun in the solar
wind, some of which becomes trapped in the earth’s
magnetic field.
• The particles spiral in helical trajectories along the earth’s
magnetic field lines. When they enter the atmosphere at
the poles, they ionize gas, creating the aurora.
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Slide 24-35
Sample Problem
• An electric field of 2000 N/C is directed to the south. A
proton is traveling at 300,000 m/s to the west. What is the
magnitude and direction of the force on the proton?
Describe the path of the proton. Ignore gravitational
effects.
F = qE
(
)
F = 1.6 ×10 −19 C (2000 NC )
v
E
F = 3.2 ×10 −16 N
The force is south. (Since the charge is positive,
it will experience a force in the direction of the
field.) Since the horizontal velocity is
unchanged, the proton will follow a parabolic
path downward.
Slide 24-36
Sample Problem
• A magnetic field of 2000 mT is directed to the south. A proton
is traveling at 300,000 m/s to the west. What is the magnitude
and direction of the force on the proton? Describe the path of
the proton. Ignore gravitational effects.
F = qvB sin θ
(
)
(
)
F = 1.6 ×10 −19 C (300,000 ms ) 2000 ×10-3 T sin (90°)
v
F = 9.6 ×10 −14 N
The force is up (out of the page). (Since the
charge is positive, use the right hand rule.) It
will move in a circular path.
B
Slide 24-37
Sample Problem
• How would you arrange a magnetic field and an electric field
so that a charged particle of velocity v would pass straight
through without deflection?
Slide 24-40
QuickCheck 24.20
Which magnetic field (if it’s the correct strength) allows the
electron to pass through the charged electrodes without
being deflected?
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Slide 24-41
QuickCheck 24.20
Which magnetic field (if it’s the correct strength) allows the
electron to pass through the charged electrodes without
being deflected?
E.
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Slide 24-42
Electric and Magnetic Fields Together
B
e-
E
This electron will experience an upward force from the electric field (opposite the direction
of the field) and a downward force from the magnetic field (left hand rule).
Slide 24-43
Sample Problem
• It is found that protons traveling at 20,000 m/s pass
undeflected through the velocity filter below. What is
the magnitude and direction of the magnetic field
between the plates?
e
20,000 m/s
0.02 m
400 V
Solution on next page.
Slide 24-44
Solution
Undeflected means that ΣF = 0.
ΣF = 0 = FB − Fe
V = Ed
V 400 V
E= =
= 20,000 mV
d 0.02 m
Fe = FB
qE = qvB sin θ
qE
qvB sin θ
=
qv sin θ
qv sin θ
E
=B
v sin θ
(
20,000 NC )
B=
= 1 T, into the page
(20,000 ms )sin 90°
Slide 24-45
Electromagnetic Flowmeters
• An electromagnetic flowmeter is a device that can be used
to measure the blood flow in an artery.
• It applies a magnetic field across the artery, which
separates the positive and negative ions in the blood.
• The flowmeter measures the potential difference due to the
separation of the ions.
• The faster the blood’s ions are moving, the greater the
forces separating the ions become, therefore generating a
higher voltage.
• Therefore, the measured voltage is proportional to the
velocity of the blood.
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Slide 24-46
Electromagnetic Flowmeters
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Slide 24-47
Electromagnetic Flowmeters
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Slide 24-48
Don’t Try It Yourself: Magnets and Television
Screens
The image on a cathode-ray tube
television screen is drawn by an
electron beam that is steered by
magnetic fields from coils of wire.
Other magnetic fields can also
exert forces on the moving
electrons. If you place a strong
magnet near the TV screen, the
electrons will be forced along altered trajectories and will strike
different places on the screen than they are supposed to, producing an
array of bright colors. (The magnet can magnetize internal
components and permanently alter the image, so do not do this to your
television!)
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Slide 24-57
Section 24.6 Magnetic Fields Exert
Forces on Currents
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The Form of the Magnetic Force on a Current
• We learned that the magnetic
field exerts no force on a
charged particle moving
parallel to a magnetic field.
• If a current-carrying wire is
parallel to a magnetic field,
we also find that the force on
it is zero.
• There is a force on a currentcarrying wire that is
perpendicular to a magnetic
field.
© 2015 Pearson Education, Inc.
Slide 24-59
The right hand rule to determine
magnetic force
This right hand rule is a little different
than what your book uses, but I think
it’s easier to remember. You must
keep your hand in this configuration
and turn your whole wrist!!
F
v
•
•
•
Point in the direction of the
CURRENT (velocity of the
charges).
Turn your hand so that your middle
finger (or three remaining fingers)
point in the direction of the field.
Your thumb gives you the direction of
the force.
B
Slide 24-60
Calculating Magnetic Force on a Current
• Derive an equation that can be used to calculate magnetic
force on a current-carrying wire.
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Slide 24-61
Review
• What is the equation for a magnetic force on a moving
charge?
F = qvB sin θ
• What if there are many moving charges, like a current in a
wire?
 ∆x 
∆x represents the length of the wire (the
F = q  B sin θ
distance the charges move)
 ∆t 
q∆xB sin θ  q 
F=
=  ∆xB sin θ = I∆xB sin θ
∆t
 ∆t 
Slide 24-62
Magnetic Force on CurrentCarrying Wire
•
•
•
•
FB = IlB sin θ
I: current in Amps
ℓ: length in meters
B: magnetic field in Tesla
θ: angle between current and field
Slide 24-63
Sample Problem
• What is the force on a 100 m long wire bearing a 30
A current flowing north if the wire is in a downwarddirected magnetic field of 400 mT?
FB = IlB sin θ
(
)
FB = (30 Cs )(100 m ) 400 ×10−3 T sin 90°
FB = 1200 N
The force is west (use the right hand rule…remember we assume positive
charges are flowing in the wire).
Slide 24-65
Sample Problem
• A wire is in a magnetic field that is directed out of the
page. What is the magnetic field strength if the
current in the wire is 15 A and the force is downward
and has a magnitude of 40 N/m? What is the direction
of the current?
ANS: 2.67Slide
T, right
24-66
Which way will this loop of wire rotate
a) if the current is clockwise?
b) if the current is counterclockwise?
B
a) The right side of the loop will rotate out of the page.
b) The right side of the loop will rotate into the page.
Slide 24-67
QuickCheck 24.23
The horizontal wire can be levitated—held up against the
force of gravity—if the current in the wire is
A. Right to left.
B. Left to right.
C. It can’t be done with
this magnetic field.
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Slide 24-70
QuickCheck 24.23
The horizontal wire can be levitated—held up against the
force of gravity—if the current in the wire is
A. Right to left.
B. Left to right.
C. It can’t be done with
this magnetic field.
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Slide 24-71
Example 24.11 Magnetic force on a power line
A DC power line near the equator runs east-west. At this
location, the earth’s magnetic field is parallel to the ground,
points north, and has magnitude 50 µT. A 400 m length of
the heavy cable that spans the distance between two towers
has a mass of 1000 kg. What direction and magnitude of
current would be necessary to offset the force of gravity and
“levitate” the wire? (The power line will actually carry a
current that is much less than this; 850 A is a typical value.)
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Slide 24-72
Example 24.11 Magnetic force on a power line
(cont.)
First, we sketch a top
view of the situation, as in
FIGURE 24.38. The magnetic
force on the wire must be
opposite that of gravity. An
application of the right-hand
rule for forces shows that a
current to the east will result in
an upward force—out of the
page.
PREPARE
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Slide 24-73
Example 24.11 Magnetic force on a power line
(cont.)
SOLVE The
magnetic field is perpendicular to the current, so
the magnitude of the magnetic force is given by Equation
24.10. To levitate the wire, this force must be opposite to the
weight force but equal in magnitude, so we can write
mg = ILB
where m and L are the mass and length of the wire and B is
the magnitude of the earth’s field. Solving for the current,
we find
directed to the east.
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Slide 24-74
Example 24.11 Magnetic force on a power line
(cont.)
ASSESS The
current is much larger than a typical current, as
we expected.
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Slide 24-75
Forces Between Currents
• Because a current produces
a magnetic field, and a
magnetic field exerts a
force on a current, it
follows that two currentcarrying wires will exert
forces on each other.
• A wire carrying a current I1
will create a magnetic field
1.
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Slide 24-76
Forces Between Currents
• A second wire with current
I2 will experience the
magnetic force due to the
wire with current I1.
• Using the right-hand rule for
forces, we can see that when
I2 is in the same direction as
I1, the second wire is
attracted to the first wire.
• If they were in opposite
directions, the second wire
would be repelled.
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Slide 24-77
Forces Between Currents
• The magnetic field created
by the wire with current I2
will also exert an attractive
force on the wire with
current I1.
• The forces on the two wires
form a Newton’s third law
action/reaction pair.
• The forces due to the
magnetic fields of the wires
are directed in opposite
directions and must have the
same magnitude.
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Slide 24-78
Forces Between Currents
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Slide 24-79
Example 24.12 Finding the force between wires
in jumper cables
You may have used a set of jumper cables connected to a
running vehicle to start a car with a dead battery. Jumper
cables are a matched pair of wires, red and black, joined
together along their length. Suppose we have a set of jumper
cables in which the two wires are separated by 1.2 cm along
their 3.7 m (12 ft) length. While starting a car, the wires
each carry a current of 150 A, in opposite directions. What
is the force between the two wires?
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Slide 24-80
Example 24.12 Finding the force between wires
in jumper cables (cont.)
Our first step is to sketch the situation, noting
distances and currents, as shown in FIGURE 24.41. Let’s
find the force on the red wire; from the discussion above,
the force on the black wire has the same magnitude but is in
the opposite direction.
PREPARE
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Slide 24-81
Example 24.12 Finding the force between wires
in jumper cables (cont.)
The force on the red wire is found using a two-step process.
First, we find the magnetic field due to the current in the
black wire at the position of the red wire. Then, we find the
force on the current in the red wire due to this magnetic
field.
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Slide 24-82
Example 24.12 Finding the force between wires
in jumper cables (cont.)
SOLVE The
magnetic field at the position of the red wire,
due to the current in the black wire, is
According to the right-hand rule for fields, this magnetic
field is directed into the page. The magnitude of the force on
the red wire is then
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Slide 24-83
Example 24.12 Finding the force between wires
in jumper cables (cont.)
The direction of the force can be found using the right-hand
rule for forces. The magnetic field at the position of the red
wire is into the page, while the current is to the right. This
means that the force on the red wire is in the plane of the
page, directed away from the black wire. Thus the force
between the two wires is repulsive, as we expect when their
currents are directed oppositely.
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Slide 24-84
Example 24.12 Finding the force between wires
in jumper cables (cont.)
ASSESS These
wires are long, close together, and carry very
large currents. But the force between them is quite small—
much less than the weight of the wires. In practice, the
forces between currents are not an important consideration
unless there are many coils of wire, leading to a large total
force. This is the case in an MRI solenoid.
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Slide 24-85
Forces Between Current Loops
• Just as there is an attractive force
between parallel wires that have
currents in the same direction, there is
an attractive force between parallel
loops with currents in the same
direction.
• There is a repulsive force between
parallel loops with currents in opposite
directions.
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Slide 24-86
Forces Between Current Loops
• The field of a current loop is very similar to that of a bar
magnet.
• A current loop, like a bar magnet, is a magnetic dipole
with a north and a south pole.
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Slide 24-87
Forces Between Current Loops
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Slide 24-88
QuickCheck 24.24
The diagram below shows slices through two adjacent
current loops. Think about the force exerted on the loop on
the right due to the loop on the left. The force on the right
loop is directed
A.
B.
C.
D.
To the left.
Up.
To the right.
Down.
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Slide 24-89
QuickCheck 24.24
The diagram below shows slices through two adjacent
current loops. Think about the force exerted on the loop on
the right due to the loop on the left. The force on the right
loop is directed
A.
B.
C.
D.
To the left.
Up.
To the right.
Down.
© 2015 Pearson Education, Inc.
Slide 24-90
Example Problem
A 10 cm length of wire carries a current of 3.0 A. The wire
is in a uniform field with a strength of 5E-3 Tesla as in the
following diagram. What are the magnitude and direction of
the force on this segment of wire?
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Slide 24-91
Summary: General Principles
Text: p. 794
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Slide 24-92
Summary: Important Concepts
Text: p. 794
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Slide 24-93
Summary: Applications
Text: p. 794
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Slide 24-94