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Chapter 27
Magnetism
Copyright © 2009 Pearson Education, Inc.
Velocity Selector
Lorentz force:
F  q  E  v  B 
If E   E yˆ
and v  v xˆ
and B   B zˆ
then v  B   vB  xˆ  zˆ    vB yˆ
so
F  q   E yˆ  vB yˆ   q   E  vB  yˆ
 0 if v 
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E
B
27-5 Torque on a Current Loop;
Magnetic Dipole Moment
The forces on opposite
sides of a current loop
will be equal and
opposite (if the field is
uniform and the loop is
symmetric), but there
may be a torque.
The torque is given by
  NIA  B
   NIAB sin
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27-5 Torque on a Current Loop;
Magnetic Dipole Moment
The quantity NIA is called the magnetic
dipole moment, μ:
  NIA
and     B
The potential energy of the loop
depends on its orientation in the field:
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27-5 Torque on a Current Loop;
Magnetic Dipole Moment
Example 27-12: Magnetic moment of a
hydrogen atom.
Determine the magnetic dipole moment of
the electron orbiting the proton of a
hydrogen atom at a given instant,
assuming (in the Bohr model) it is in its
ground state with a circular orbit of
radius r = 0.529 x 10-10 m. [This is a very
rough picture of atomic structure, but
nonetheless gives an accurate result.]
Copyright © 2009 Pearson Education, Inc.
Magnetic moment of a hydrogen atom.
ke 2 2 r
1
1

 
mr
T
T 2 r
ke 2 mv 2

v
2
r
r
ke 2
mr
e
e
I 
T 2 r
   IA 
e
2 r


ke 2
 r2
mr
1.6  10
e kr


2 m
2
 9.25  1024 A  m 2
2
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ke 2
mr
19

  9.0  10  0.529  10 
2
9
9.11  1031
10
27-5 Torque on a Current Loop;
Magnetic Dipole Moment
Example: A rectangular coil 5.40 cm X 8.50 cm
consists of 25 turns of wire and carries a current
of 15.0 mA. A 0.350-T magnetic field is applied
parallel to the plane of the coil.
a) Calculate the magnitude of the magnetic
dipole moment of the coil.
b) What is the magnitude of the torque acting on
the loop?
Copyright © 2009 Pearson Education, Inc.
27-5 Torque on a Current Loop;
Magnetic Dipole Moment
3


NIA

25
15.0

10
A   0.054 m  0.085 m 



a) coil
 1.72 103 A  m 2
b) Note: B  coil    coil  B  coil B sin 90o  coil B
 1.72 103 A  m 2   0.350 T 
 6.02 104 N  m
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27-6 Applications: Motors
An electric motor uses the torque on a
current loop in a magnetic field to turn
magnetic energy into kinetic energy.
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27-6 Applications:
Galvanometers
A galvanometer
takes advantage of
the torque on a
current loop to
measure current; the
spring constant is
calibrated so the
scale reads in
amperes.
Copyright © 2009 Pearson Education, Inc.
ConcepTest 27.7b Magnetic Force on a Loop II
1) move up
If there is a current in
2) move down
the loop in the direction
3) rotate clockwise
shown, the loop will:
4) rotate counterclockwise
5) both rotate and move
B field out of North
B field into South
N
S
N
S
ConcepTest 27.7b Magnetic Force on a Loop II
1) move up
If there is a current in
2) move down
the loop in the direction
3) rotate clockwise
shown, the loop will:
4) rotate counterclockwise
5) both rotate and move
Look at the north pole: here the
F
magnetic field points to the right and
the current points out of the page.
N
S
The right-hand rule says that the force
must point up. At the south pole, the
same logic leads to a downward force.
Thus the loop rotates clockwise.
F
27-7 Discovery and Properties of the
Electron
Electrons were first observed in cathode ray tubes.
These tubes had a very small amount of gas inside,
and when a high voltage was applied to the cathode,
some “cathode rays” appeared to travel from the
cathode to the anode.
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27-7 Discovery and Properties of the
Electron
The value of e/m for the cathode rays was
measured in 1897 using the apparatus below; it
was then that the rays began to be called
electrons.
Figure 27-30 goes here.
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27-7 Discovery and Properties of the
Electron
1
1) K 
mv 2  e V
2
e
v
2 V
m
Velocity Selector:
v
E

B
e
2 V
m
E  B
2
e E 1
  
m  B  2 V
mv 2
2) E  0  e v B 
r
e
v
 
m Br
Velocity Selector:
v
E
e
E
  2
B
m Br
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V
27-7 Discovery and Properties of the
Electron
Millikan measured the electron charge directly
shortly thereafter, using the oil-drop apparatus
diagrammed below, and showed that the
electron was a constituent of the atom (and not
an atom itself, as its mass is far too small).
The currently accepted
values of the electron
mass and charge are
m = 9.1 x 10-31 kg
e = 1.6 x 10-19 C
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27-8 The Hall Effect
When a current-carrying wire
is placed in a magnetic field,
there is a sideways force on
the electrons in the wire. This
tends to push them to one
side and results in a potential
difference from one side of the
wire to the other; this is called
the Hall effect. The emf differs
in sign depending on the sign
of the charge carriers; this is
how it was first determined
that the charge carriers in
ordinary conductors are
negatively charged.
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27-9 Mass Spectrometer
All the atoms
reaching the second
magnetic field will
have the same speed;
their radius of
curvature will depend
on their mass.
E
v
B
mv 2 m E
q v B' 

r
r B
qBB ' r
m
E
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27-9 Mass Spectrometer
Example 27-14: Mass spectrometry.
Carbon atoms of atomic mass 12.0 u are found to be
mixed with another, unknown, element. In a mass
spectrometer with fixed B′, the carbon traverses a
path of radius 22.4 cm and the unknown’s path has a
26.2-cm radius. What is the unknown element?
Assume the ions of both elements have the same
charge.
Copyright © 2009 Pearson Education, Inc.
27-9 Mass Spectrometer
Example 27-14: Mass spectrometry.
Carbon atoms of atomic mass 12.0 u are found to be
mixed with another, unknown, element. In a mass
spectrometer with fixed B′, the carbon traverses a
path of radius 22.4 cm and the unknown’s path has a
26.2-cm radius. What is the unknown element?
Assume the ions of both elements have the same
charge.
 qBB ' 
 qBB ' 
m12  
r12 and m?  
r?


 E 
 E 
m?
r?
r?
26.2


 m?  m12 
12 u  14.0 u (= nitrogen)
m12 r12
r12
22.4
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 27
• Magnets have north and south poles.
• Like poles repel, unlike attract.
• Unit of magnetic field: tesla.
• Electric currents produce magnetic fields.
• A magnetic field exerts a force on an electric
current:
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Summary of Chapter 27
• A magnetic field exerts a force on a moving
charge:
• Torque on a current loop:
• Magnetic dipole moment:
Copyright © 2009 Pearson Education, Inc.
Chapter 28
Sources of Magnetic Field
Copyright © 2009 Pearson Education, Inc.
Units of Chapter 28
• Magnetic Field Due to a Straight Wire
• Force between Two Parallel Wires
• Definitions of the Ampere and the Coulomb
• Ampère’s Law
• Magnetic Field of a Solenoid and a Toroid
• Biot-Savart Law
• Magnetic Materials – Ferromagnetism
• Electromagnets and Solenoids – Applications
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Units of Chapter 28
• Magnetic Fields in Magnetic Materials;
Hysteresis
• Paramagnetism and Diamagnetism
Copyright © 2009 Pearson Education, Inc.
28-1 Magnetic Field Due to a Straight
Wire
The magnetic field due to a
straight wire is inversely
proportional to the distance
from the wire:
0 I
2 r
0 I ˆ
B
 rˆ
2 r
B
[near a long straight wire]
The constant μ0 is called the
permeability of free space,
and has the value (exactly)
μ0 = 4π x 10-7 T·m/A.
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ˆ
28-1 Magnetic Field Due to a Straight
Wire
Example 28-1: Calculation of B
B
near a wire.
An electric wire in the wall of a
building carries a dc current of
25 A vertically upward. What is
the magnetic field due to this
current at a point P 10 cm due
north of the wire?
Copyright © 2009 Pearson Education, Inc.
ConcepTest 28.1a Magnetic Field of a Wire I
If the currents in these wires have
1) direction 1
the same magnitude but opposite
2) direction 2
3) direction 3
directions, what is the direction of
4) direction 4
the magnetic field at point P?
5) the B field is zero
1
P
4
2
3
ConcepTest 28.1a Magnetic Field of a Wire I
If the currents in these wires have
1) direction 1
the same magnitude but opposite
2) direction 2
directions, what is the direction of
the magnetic field at point P?
3) direction 3
4) direction 4
5) the B field is zero
1
P
Using the right-hand rule, we
can sketch the B fields due
to the two currents. Adding
them up as vectors gives a
total magnetic field pointing
downward.
4
2
3
ConcepTest 28.3 Current Loop
1) left
What is the direction of the
2) right
magnetic field at the center
3) zero
(point P) of the square loop
4) into the page
of current?
5) out of the page
I
P
ConcepTest 28.3 Current Loop
1) left
What is the direction of the
magnetic field at the center
(point P) of the square loop
of current?
2) right
3) zero
4) into the page
5) out of the page
Use the right-hand rule for each
wire segment to find that each
I
segment has its B field pointing
out of the page at point P.
P
28-1 Magnetic Field Due to a Straight
Wire
Example 28-2: Magnetic field midway between two
currents.
Two parallel straight wires 10.0 cm apart carry
currents in opposite directions. Current I1 = 5.0 A is
out of the page, and I2 = 7.0 A is into the page.
Determine the magnitude and direction of the
magnetic field halfway between the two wires.
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28-1 Magnetic Field Due to a Straight
Wire
Conceptual Example 28-3: Magnetic field due to
four wires.
This figure shows four long parallel wires which
carry equal currents into or out of the page. In
which configuration, (a) or (b), is the magnetic
field greater at the center of the square?
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28-2 Force between Two Parallel Wires
The magnetic field produced
at the position of wire 2 due
to the current in wire 1 is
0 I1 ˆ ˆ
B1 
1  d 21
2 d
The force this field exerts
on a length l2 of wire 2 is
0 I1 I 2 ˆ
ˆ  dˆ
 F21  I 2 2  B1 

1
21
2 d 2 2
0 I1 I 2 ˆ

2 d12  force attractive
2 d

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
28-2 Force between Two Parallel Wires
Parallel
currents
attract;
antiparallel
currents repel.
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28-2 Force between Two Parallel Wires
Example 28-4. Force between
two current-carrying wires.
The two wires of a 2.0-m-long
appliance cord are 3.0 mm apart
and carry a current of 8.0 A dc.
Calculate the force one wire
exerts on the other.
Copyright © 2009 Pearson Education, Inc.
28-2 Force between Two Parallel Wires
Example 28-5: Suspending a wire with a current.
A horizontal wire carries a current I1 = 80 A 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.
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28-3 Definitions of the Ampere and
the Coulomb
The ampere is officially defined in terms of
the force between two current-carrying
wires:
One ampere is defined as that current flowing in
each of two long parallel wires 1 m apart, which
results in a force of exactly 2 x 10-7 N per meter
of length of each wire.
The coulomb is then defined as exactly
one ampere-second.
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28-4 Ampère’s Law
Ampère’s law relates the
magnetic field around a
closed loop to the total
current flowing through
the loop:
This integral is taken
around the edge of the
closed loop.
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28-4 Ampère’s Law
Using Ampère’s law to find
the field around a long
straight wire:
Use a circular path with the
wire at the center; then B is
tangent to dl at every point.
The integral then gives
so B = μ0I/2πr, as before.
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28-4 Ampère’s Law
Example 28-6: Field inside
and outside a wire.
A long straight cylindrical wire
conductor of radius R carries a
current I of uniform current density
in the conductor. Determine the
magnetic field due to this current
at (a) points outside the conductor
(r > R) and (b) points inside the
conductor (r < R). Assume that r,
the radial distance from the axis, is
much less than the length of the
wire.
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28-4 Ampère’s Law
Example 28-8: A nice use for Ampère’s law.
Use Ampère’s law to show that in any region of
space where there are no currents the
magnetic field cannot be both unidirectional
and nonuniform as shown in the figure.
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28-4 Ampère’s Law
Solving problems using Ampère’s law:
• Ampère’s law is only useful for solving
problems when there is a great deal of
symmetry. Identify the symmetry.
• Choose an integration path that reflects the
symmetry (typically, the path is along lines
where the field is constant and perpendicular
to the field where it is changing).
• Use the symmetry to determine the direction
of the field.
• Determine the enclosed current.
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28-5 Magnetic Field of a Solenoid
A solenoid is a coil of wire containing
many loops. To find the field inside, we use
Ampère’s law along the path indicated in
the figure.
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28-5 Magnetic Field of a Solenoid
The field is zero outside the solenoid,
and the path integral is zero along the
vertical lines, so the field is (n is the
number of loops per unit length)
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28-5 Magnetic Field of a Solenoid
Example 28-9: Field inside a solenoid.
A thin 10-cm-long solenoid used for fast
electromechanical switching has a total of
400 turns of wire and carries a current of
2.0 A. Calculate the field inside near the
center.
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Questions?
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