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Magnetism
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Magnetism
Chapter 21
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Chapter 21
Magnets and Magnetic Fields
Magnets
• Magnets attract iron-containing objects.
• Magnets have two distinct poles called the north pole
and the south pole. These names are derived from a
magnet’s behavior on Earth.
• Like poles of magnets repel each other; unlike poles
attract each other.
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Properties of Magnets
 Magnets have two
opposite poles.
— north
— south
 Magnets exert forces on
each other.
 The forces depend on the
alignment of the poles.
Magnets and Magnetic Fields
Magnets have two ends – poles – called
north and south.
Like poles repel; unlike poles attract.
Magnets and Magnetic Fields
However, if you cut a magnet in half, you don’t
get a north pole and a south pole – you get two
smaller magnets.
Chapter 19
Magnetic Fields
• A magnetic field is a region in which a
magnetic force can be detected.
• Magnetic field lines can be drawn with
the aid of a compass.
Magnets and Magnetic Fields
Magnetic fields can be visualized using
magnetic field lines, which are always closed
loops.
Chapter 21
Magnets and Magnetic Fields
The Direction of a Magnetic Field
Field lines are conventionally drawn FROM the
north pole TO the south pole.
The direction of a magnetic field at any
location is the direction that the north pole
of a compass at that location would point.
(The arrow head represents the north pole of a
magnet and points to the nearest south pole).
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Properties of magnets
 Materials that make good
permanent magnets are
called hard magnets.
 Steel, which contains iron
and carbon, is a common
and inexpensive material
used to create hard
magnets.
 Materials that lose their
magnetism quickly are
called soft magnets.
The Magnetic Field of the Earth
How do we use Earth’s
magnetic field to tell
direction?
The Magnetic Field of the Earth
 As early as 500 B.C. people
discovered that some naturally
occurring materials— such as
lodestone and magnetite—have
magnetic properties.
The Magnetic Field of the Earth
 As early as 500 B.C. people
discovered that some naturally
occurring materials— such as
lodestone and magnetite—have
magnetic properties.
 By 1200, explorers from Italy were
using a compass to guide ocean
voyages beyond the sight of land.
The Magnetic Field of the Earth
 As early as 500 B.C. people
discovered that some naturally
occurring materials— such as
lodestone and magnetite—have
magnetic properties.
 By 1200, explorers from Italy were
using a compass to guide ocean
voyages beyond the sight of land.
 (show bar magnet)
The Magnetic Field of the Earth
 The Earth’s magnetic
poles are defined by the
planet’s magnetic field.
 That means the south
magnetic pole of the
planet is near the north
geographic pole.
 When you use a compass,
the north-pointing end of
the needle points toward
a spot near (but not
exactly at) the Earth’s
geographic north pole.
Chapter 21
Magnets and Magnetic Fields
Earth’s Magnetic Field
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Chapter 21
Magnets and Magnetic Fields
Magnetic Fields, continued
• Earth’s magnetic field is similar to that of a bar
magnet.
• The magnetic south pole is near the Geographic
North Pole. The magnetic north pole is near the
Geographic South Pole.
• Magnetic declination is a measure of the difference
between true north and north indicated by a
compass.
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The magnetic field produced by certain
arrangements of bar magnets are represented in
the diagrams shown below
MAGNETS AND MAGNETIC
FIELDS
 The magnetic field lines drawn to represent the magnetic field
produced by certain arrangements of bar magnets are represented in
the diagrams shown below
Notice the similarity
between the lines of this
slide and the previous slide
particle distribution
The Why & How of Magnets
 The sources of nearly all
magnetic effects in matter
are the electrons in atoms.
 There are two ways in which
electrons create magnetism:
1. Electrons’ motion around
the nucleus makes the
entire atom a small
magnet.
2. Electrons themselves act
as though they were
magnets.
The Why & How of Magnets
 All atoms have electrons, so you might think that all
materials should be magnetic, but there is great
variability in the magnetic properties of materials.
 The electrons in some atoms align to cancel out
one another’s magnetic influence.
 While all materials show some kind of magnetic
effect, the magnetism in most materials is too weak
to detect without highly sensitive instruments.
Iron, Cobalt, Nickel
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Magnesium
Note: Magnetism is a
property not just of the
chemical make-up of a
material, but of its crystalline
structure and microscopic
organization.
The Why & How of Magnets
 In diamagnetic materials,
the electrons are oriented
so their individual
magnetic fields cancel
each other out within an
atom.
 Individual atoms in
paramagnetic materials
are magnetic but the atoms
themselves are randomly
arranged so the overall
magnetism of a sample is
zero.
But when paramagnetic
materials are placed in a
magnetic field, the atoms align
so that the material is weakly
magnetic.
The Why & How of Magnets
 A small group of metals
have very strong magnetic
properties, including iron,
nickel, and cobalt.
 These metals are the best
known examples of
ferromagnetic materials.
 Atoms with similar
magnetic orientations line
up with neighboring atoms
in groups called magnetic
domains.
Induced Magnetism
" Metals that are ferromagnetic have atoms
that behave like mini magnets, but on a
grand scale they cancel out. When placed in
external magnetic field, these can be lined up
and the metal then becomes magnetized.
Magnetizing Iron
Magnetic domains in iron nails are
induced to align by proximity of
the strong magnet
Each nail becomes itself a magnet,
which in turn magnetizes the nail
below it, forming a chain.
When the strong magnet is
removed, most of the domains unalign and the nails lose most of
their magnetization.
Demagnetizing Iron
Magnetic domains can be scrambled by
heating the iron, striking it with great force,
or other disruptions of alignment.
Magnetized
Test tube of
un-magnetized
iron filings
Demagnetized
SHAKE
S
S
N
Magnetizer
N
Demagnetizing Iron
Iron nail is attracted to
the large magnet due to
alignment of domains in
the nail.
Heat the nail to a high
temperature and the
domains become
randomized so the nail is
no longer attracted to
the magnet.
The Why & How of Magnets
 Magnetic domains in a ferromagnetic material will always
orient themselves to attract a permanent magnet.
— If a north pole approaches, domains grow that have south
poles facing out.
— If a south pole approaches, domains grow that have north
poles facing out.
Chapter 21
Magnets and Magnetic Fields
Magnetic Domains
• Magnetic Domain
A region composed of a group of atoms whose
magnetic fields are aligned in the same direction is
called a magnetic domain.
• Some materials can be made into permanent
magnets.
– Soft magnetic materials (for example iron) are
easily magnetized but tend to lose their
magnetism easily.
– Hard magnetic materials (for example nickel) tend
to retain their magnetism.
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Names of some magnetic materials
Examples of ferromagnetic materials: iron,
cobalt, nickel
Examples of paramagnetic materials:
aluminum, manganese, chromium, oxygen,
sodium, lithium
Examples of diamagnetic materials: water,
pyrolitic graphite, copper, gold, bismuth
Diamagnetism
• Diamagnetism is the property of an object or material
which causes it to create a magnetic field in
opposition to an externally applied magnetic field.
Diamagnetism
• Diamagnetism is the property of an object or material
which causes it to create a magnetic field in
opposition to an externally applied magnetic field.
• Diamagnetism is believed to be due to quantum
mechanics (and is understood in terms of Landau
levels) and occurs because the external field alters the
orbital velocity of electrons around their nuclei, thus
changing the magnetic dipole moment. According to
Lenz’s law, the field of these electrons will oppose
the magnetic field changes provided by the applied
field.
On July 21, 1820, Hans Oersted, professor of
physics at Copenhagen, delivered a lecture to
students. By chance, a wire (connected to a
power source) was nearly parallel to & above a
compass that happened to be on the table. When
the circuit was closed, the needle swung around
almost perpendicular to the current carrying wire,
as if gripped by a powerful magnet.
Andre Marie Ampere heard the news &
immediately performed some experiments of his
own. Within two weeks, Ampere showed that a
current-carrying wire always induces a magnetic
force that acts at right angles to the current.
Chapter 21
Magnetism from Electricity
Magnetic Field of a Current-Carrying Wire
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Chapter 21
Magnetism from Electricity
Magnetic Field of a Current-Carrying Wire
Negative Current
Zero Current
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Positive Current
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Direction of B
45
Magnetic Field Produced by a
Current-Carrying Wire
Chapter 21
Magnetism from Electricity
The Right-Hand Rule
• Right Hand Rule
– Grasp the wire in your
right hand
– Point your thumb in
the direction of the
current
– Your fingers will curl
in the direction of the
field
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Chapter 21
Magnetism from Electricity
Magnetic Field of a Current-Carrying Wire
Negative Current
Zero Current
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Positive Current
Resources
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Chapter 21
Magnetism from Electricity
Magnetic Field of a Current Loop
• Solenoids produce a strong magnetic field by
combining several loops.
• A solenoid is a long, helically wound coil of insulated
wire.
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Electric Currents Produce
Magnetic Fields
The direction of the
field is given by a
right-hand rule.
" So we have learned that moving
charges (currents) exert a force on a
compass needle (a magnet).
" So we have learned that moving
charges (currents) exert a force on a
compass needle (a magnet).
~moving charges exert force on magnets~
" So we have learned that moving
charges (currents) exert a force on a
compass needle (a magnet).
~moving charges exert force on magnets~
" It follows, from Newton’s Third Law,
that magnets (or a magnetic field)
ought to exert forces on currents
(moving charged particles).
" So we have learned that moving
charges (currents) exert a force on a
compass needle (a magnet).
~moving charges exert force on magnets~
" It follows, from Newton’s Third Law,
that magnets (or a magnetic field)
ought to exert forces on currents
(moving charged particles).
~magnetic fields exert force on moving charges~
Chapter 21
Magnetic Force
Charged Particles in a Magnetic Field
• A charge moving through a magnetic field experiences
a force proportional to the charge, velocity, and the
magnetic field.
F = qvB
or
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Defining Magnetic Field, B
Fe
E=
q
" Recall that the electric field, E, is
defined as electric force per unit charge
exerted on a moving charged particle.
€
" Similarly, B used to be defined as the
magnetic force per unit pole
" NOW we define B in terms of the force
FB
exerted on a moving charged particle. B =
qv
Chapter 21
Magnetic Force
Force on a Charge Moving in a Magnetic Field
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Chapter 21
Magnetic Force
Charged Particles in a Magnetic Field,
continued
• The direction of the magnetic force on a moving
charge is always perpendicular to both the magnetic
field and the velocity of the charge.
• An alternative right-hand rule can be used to find the
direction of the magnetic force.
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Chapter 21
Magnetic Force
Right-Hand Palm Rule: Force on a Moving Charge
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CONVENTIONS
Certain CONVENTIONS have been adopted in order to
represent the direction of the magnetic field and the
current in a wire. A magnetic field directed into the paper
is represented by a group of x's, while a magnetic field
out of the paper is represented by a group of dots.
Magnetic field
line in plane of
paper
Determine the direction of
the magnetic force in the
following diagrams:
1)
v
(a proton)
No force
Force up page
.
F
v
F
-e
v
-e
B
B
Note: Force is in opposite direction if charge
is negative (instead of conventional positive
charge).
Determine the direction of
the magnetic force in the
following diagram:
Force up page
-
B
Force into page
Force into page
The only component of v that contributes to the magnetic
force is the component that is perpendicular to B.
θ = angle between v and B
B
B
B
B
Center-seeking (centripetal) Force
B
B
A charge moving through a magnetic
field follows a circular path.
Centripetal Force
Remember,
centripetal force
causes a change in
direction, not a
change in speed (so
magnitude of v stays
the same).
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Units of Magnetic Field


The SI unit of magnetic field is the
Tesla (T)
€
Another (non-SI) unit is a Gauss (G)
4
 1 T = 10 G
FB
B=
qv
A Few Typical B Values

Conventional laboratory magnets


Superconducting magnets


2.5 T
30 T
Earth’s magnetic field

5 x 10-5 T
For each of the following 3 particles, state
whether the particle’s charge is positive,
negative, or zero.
For each of the following 3 particles,
state whether the particle’s charge is
positive, negative, or zero.
Solution
Particle 3 follows the path given by the righthand rule. So #3 is positive.
Particle 1 is opposite that, so #1 is negative.
Particle 2 is un-deflected so there must be no
force on it, which means the charge is 0.
Chapter 21
RE-CAP
The Right-Hand Rule
• Right Hand Rule
– Grasp the wire in your
right hand
– Point your thumb in
the direction of the
current
– Your fingers will curl
in the direction of the
field
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Chapter 21
Magnetic Force
Right-Hand Palm Rule: Force on a Moving Charge
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Chapter 21
Magnetic Force
Example Problem #1
Particle in a Magnetic Field
A proton moving east experiences a force of
8.8 × 10–19 N upward due to the Earth’s
magnetic field. At this location, the field has a
magnitude of 5.5 × 10–5 T to the north. Find
the speed of the particle.
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Chapter 21
Magnetic Force
Sample Problem, continued
Particle in a Magnetic Field
Given:
q = 1.60 × 10–19 C
B = 5.5 × 10–5 T
Fmagnetic = 8.8 × 10–19 N
Unknown:
v=?
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Chapter 21
Magnetic Force
Sample Problem, continued
Particle in a Magnetic Field
Use the definition of magnetic field strength.
Rearrange to solve for v.
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Example Problem #2
A beam of electrons travels to the right at 3.0 x 106 m/s
through a uniform magnetic field of 4.0 x 10-2 T at
right angles to the field (coming out of the page).
What is the force acting on each electron
(magnitude & direction)?
Example Problem #2
A beam of electrons travels to the right at 3.0 x 106 m/s
through a uniform magnetic field of 4.0 x 10-2 T at
right angles to the field (coming out of the page).
What is the force acting on each electron
(magnitude & direction)?
Example Problem #2
A beam of electrons travels at 3.0 x 106 m/s through a
uniform magnetic field of 4.0 x 10-2 T at right angles
to the field. What is the force acting on each electron
(magnitude & direction)?
Example Problem #2
A beam of electrons travels at 3.0 x 106 m/s through a
uniform magnetic field of 4.0 x 10-2 T at right angles
to the field. What is the force acting on each electron
(magnitude & direction)?
Solve for the unknown:
F = qvB
F = (1.60 x 10-19 C) (3.0 x 106 m/s)(4.0 x 10-2 T )
F = 1.90 x 10-14 N up the page
Force on a Current-Carrying Wire
If conducting charges move through a wire with an
average speed v, the time required for them to move
from one end of the wire to the other (length, l) is
Δt = l /v (since velocity = distance/Δt = l/Δt).
The amount of charge that flows through the wire in
this time is q = IΔt = Il /v. Therefore, the force
exerted on the wire is
F = q vB = (Il /v) vB
so
F = Il B or BIl
Since current, I , is
just moving charge, q,
we can use sine in
our formula F = B I l
as well!
Force on an Electric Current in a Magnetic
Field; Definition of B
The force on the wire depends on the
current, the length of the wire, the magnetic
field, and its orientation.
This equation defines the magnetic field B.
Chapter 21
Magnetic Force
Force on a Current-Carrying Wire in a
Magnetic Field
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Magnetic Force
Moving charges in an electric current
experience a force due to magnetic field.
N
N
Magnetic Force on Charges
Moving negative electric charges deflected
by magnetic fields.
Chapter
19a CONVENTIONAL POSITIVE
Force on
Current-Carrying Wire in a Magnetic Field
Force on an Electric Current in a Magnetic
Field; Definition of B
The force on the wire depends on the
current, the length of the wire, the magnetic
field, and its orientation.
This equation defines the magnetic field B.
Chapter 21
Magnetic Force
Example Problem 3
Force on a Current-Carrying Conductor
A wire 36 m long carries a current of 22 A from east
to west. If the magnetic force on the wire due to
Earth’s magnetic field is downward (toward Earth)
and has a magnitude of 4.0 × 10–2 N, find the
magnitude and direction of the magnetic field at this
location.
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Chapter 21
Magnetic Force
Example Problem, continued
Force on a Current-Carrying Conductor
Given:
I = 22 A
Fmagnetic = 4.0 × 10–2 N
Unknown:
B=?
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Chapter 21
Magnetic Force
Example Problem, continued
Force on a Current-Carrying Conductor
Use the equation for the force on a current-carrying
conductor perpendicular to a magnetic field.
Rearrange to solve for B.
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Chapter 21
Magnetic Force
Example Problem, continued
Force on a Current-Carrying Conductor
Using the right-hand rule to find the direction of B,
face north with your thumb pointing to the west (in the
direction of the current) and the palm of your hand
down (in the direction of the force). Your fingers point
north. Thus, Earth’s magnetic field is from south to
north.
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Example #4: Magnetic Force on a current-carrying wire.
A wire carrying a 30-A
current has a length  of 12 cm
between the pole
faces of a magnet at an
angle θ = 60°, as shown.
The magnetic field is
approximately uniform at
0.90 T. We ignore the field
beyond the pole pieces.
What is the force on the wire?
Copyright © 2009 Pearson Education, Inc.
Copyright © 2009 Pearson Education, Inc.
= 30 A (0.12 m) (0.90 T) (0.866)
= 2.8 N
Copyright © 2009 Pearson Education, Inc.
= 30 A (0.12 m) (0.90 T) (0.866)
= 2.8 N
Direction:
Copyright © 2009 Pearson Education, Inc.
= 30 A (0.12 m) (0.90 T) (0.866)
= 2.8 N
Direction: into the page
Copyright © 2009 Pearson Education, Inc.
Chapter 21
Magnetic Force
Force Between Parallel Conducting Wires
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Force Between Two Wires I
I2
I1
d
106
Force Between Two Wires II
I1
I2
107
Current same
direction—force
attractive.
Current opposite direction
—force repulsive.
108
Concept Check
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Chapter 21
Standardized Test Prep
Multiple Choice
•
Wire 1 carries current I1 and
creates magnetic field B1.
• Wire 2 carries current I2 and
creates magnetic field B2.
Q. What is the direction of the
magnetic field B1 at the
location of wire 2?
F. to the left
G. to the right
H. into the page
J. out of the page
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Chapter 21
Standardized Test Prep
Multiple Choice, continued
•
Wire 1 carries current I1 and
creates magnetic field B1.
• Wire 2 carries current I2 and
creates magnetic field B2.
Q. What is the direction of the
magnetic field B1 at the
location of wire 2?
F. to the left
G. to the right
H. into the page
J. out of the page
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Chapter 21
Standardized Test Prep
Multiple Choice, continued
•
Wire 1 carries current I1 and
creates magnetic field B1.
• Wire 2 carries current I2 and
creates magnetic field B2.
Q2. What is the direction of the
force on wire 2 as a result of
B1?
A. to the left
B. to the right
C. into the page
D. out of the page
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Chapter 21
Standardized Test Prep
Multiple Choice, continued
•
Wire 1 carries current I1 and
creates magnetic field B1.
• Wire 2 carries current I2 and
creates magnetic field B2.
Q2. What is the direction of the
force on wire 2 as a result of
B1?
A. to the left
B. to the right
C. into the page
D. out of the page
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Chapter 21
Standardized Test Prep
Multiple Choice, continued
•
Wire 1 carries current I1 and
creates magnetic field B1.
• Wire 2 carries current I2 and
creates magnetic field B2.
Q3. What is the magnitude of the
magnetic force on wire 2?
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Chapter 21
Standardized Test Prep
Multiple Choice, continued
•
Wire 1 carries current I1 and
creates magnetic field B1.
• Wire 2 carries current I2 and
creates magnetic field B2.
Q3. What is the magnitude of the
magnetic force on wire 2?
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Example 5: Measuring a magnetic field.
A rectangular loop of wire hangs
vertically as shown. A magnetic field
B is directed horizontally,
perpendicular to the wire, and points
out of the page at all points. The
magnetic field is very nearly uniform
along the horizontal portion of wire
ab (length
 l = 10.0 cm) which is near
the center of the gap of a large
magnet producing the field. The top
portion of the wire loop is free of the
field. The loop hangs from a balance
which measures a downward
magnetic force (in addition to the
gravitational force) of F = 3.48 x 10-2
N when the wire carries a current I =
0.245 A. What is the magnitude of
the magnetic field B?
Copyright © 2009 Pearson Education, Inc.
Solution
The magnetic force on the left vertical section of wire
points to the left; the force on the vertical section on
the right points to the right. These two forces are
equal and opposite, so they cancel.
Copyright © 2009 Pearson Education, Inc.
Solution
The magnetic force on the left vertical section of wire
points to the left; the force on the vertical section on
the right points to the right. These two forces are
equal and opposite, so they cancel.
−2
F 3.48 × 10
B= =
= 1.42 T
Il .245(.100)
Copyright © 2009 Pearson Education, Inc.
Chapter 21
Magnetism from Electricity
Magnetic Field Due to a Long Straight Wire
• Right Hand Rule
– Grasp the wire in your
right hand
– Point your thumb in
the direction of the
current
– Your fingers will curl
in the direction of the
field
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Magnetic Field Due to a Long Straight Wire
The field is inversely proportional to the
distance from the wire:
The constant µ0 is called the permeability of
free space, and has the value:
Magnetic Field Due to a Straight Wire
Experimentally
determined equation.
Current in wire
µ
I
0
B=
2π
r
Perpendicular
distance from wire to
point at which B is to
be determined.
The value of the constant µ0, which is called the permeability
of free space, is µ0/2π = 2 x 10-7 T m/A exactly.
121
Example 6
Calculation of B near a wire:
A vertical electric wire in the wall of a building
carries a dc current of 25 A upward. What is the
magnetic field at a point 10 cm due north of this
wire?
122
r
123
.
r
124
Using the formula for the magnetic field near a straight wire
125
Using the formula for the magnetic field near a straight wire
So we can obtain the magnetic field at 10cm away as
−7
(4 π × 10 T⋅ m / A)(25A)
=
(2π )(0.1m)
126
€
Using the formula for the magnetic field near a straight wire
So we can obtain the magnetic field at 10cm away as
−7
(4 π × 10 T⋅ m / A)(25A)
=
(2π )(0.1m)
= 5.0 x 10-5 T
€
127
Force Between Two Parallel Wires
128
Force between Two Parallel Wires
The magnetic field produced
at the position of wire 2 due to
the current in wire 1 is:
The force this field exerts on
a length l2 of wire 2 is:
Force Between Two Parallel Wires
Force per unit
length of wire
F µ0 I1I2
= 2π
d
l
(d or r can be used).
Distance between wires
130
d
131
Current same
direction—force
attractive.
Current opposite direction
—force repulsive.
132
Example 7
Force between two current-carrying wires:
The two wires of a 2.0-m-long appliance cord are 3.0
mm apart and carrying a current of 8.0 Amps (dc) in
opposite directions. Calculate the force between the
wires.
133
134
−7
2
(2.0 × 10 T⋅ m / A)(8.0A) (2.0m)
F=
−3
3.0 × 10 m
where
µ0
−7
= 2.0 × 10
Tm/A
2π
135
−7
2
(2.0 × 10 T⋅ m / A)(8.0A) (2.0m)
F=
−3
3.0 × 10 m
= 8.5 x 10-3 N
136
Example 8
Suspending a current with a current:
A horizontal wire carries a current I1 = 80 Amps, dc.
A second parallel wire 20 cm below it must carry
how much current, I2, so that it doesn’t fall due to
gravity? The lower wire has a mass of 0.12 g per
meter of length.
137
Magnetic field due
the current I1
138
This downward force must be balanced by the
magnetic force exerted on the wire by the first wire.
139
This downward force must be balanced by the
magnetic force exerted on the wire by the first wire.
µ0 I1I2 l
Fg = mg = FB =
2πd
€
140
This downward force must be balanced by the
magnetic force exerted on the wire by the first wire.
µ0 I1I2 l
Fg = mg = FB =
2πd
Fg mg FB µ0 I1I2
=
=
=
l
l
l
2πd
€
€
141
This downward force must be balanced by the
magnetic force exerted on the wire by the first wire.
Fg mg FB µ0 I1I2
=
=
=
l
l
l
2πd
Solving for I2…
142
This downward force must be balanced by the
magnetic force exerted on the wire by the first wire.
Fg mg FB µ0 I1I2
=
=
=
l
l
l
2πd
Solving for I2…
mg2πd
I2 =
lµ0 I1
143
€
This downward force must be balanced by the
magnetic force exerted on the wire by the first wire.
Fg mg FB µ0 I1I2
=
=
=
l
l
l
2πd
Solving for I2…
mg2πd
I2 =
lµ0 I1
−3
2
(0.12 × 10 kg /m)(9.8m /s )2π (0.20m)
=
−7
(4 π × 10 T⋅ m / A)(80A)
144
€
This downward force must be balanced by the
magnetic force exerted on the wire by the first wire.
µ0 I1I2 l2
Fg = mg = FB =
2πd
Solving for I2…
mg2πd
I2 =
l2 µ0 I1
−3
2
(0.12 × 10 kg /m)(9.8m /s )2π (0.20m)
=
−7
(4 π × 10 T⋅ m / A)(80A)
= 15 A
€
145
end
146