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General Physics (PHY 2140)
Lecture 12
¾ Electricity and Magnetism
9Magnetism
9Magnetic fields and force
9Application of magnetic forces
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 19
10/1/2003
1
Department of Physics and Astronomy
announces the Fall 2003 opening of
The Physics Resource Center
on Monday, September 22 in
Room 172 of Physics Research Building.
Hours of operation:
Mondays, Tuesdays, Wednesdays
Thursdays and Fridays
11 AM to 6 PM
11 AM to 3 PM
Undergraduate students taking PHY2130-2140 will be able to get assistance in this
Center with their homework, labwork and other issues related to their physics course.
The Center will be open: Monday, September 22 to Wednesday, December 10, 2003.
10/1/2003
2
Lightning Review
Last lecture:
1. DC circuits
9 Kirchoff’s rules
9 RC circuit
n
∑I
i =1
n
i
= 0, ∑Vi = 0
(
i =1
q = Q 1 − e−t / RC
)
q = Qe− t / RC
2. Magnetism
9 Magnets
Review Problem: The three light bulbs in the circuit all
have the same resistance. Given that brightness is
proportional to power dissipated, the brightness of bulbs
B and C together, compared with the brightness of bulb
A, is
1. twice as much.
2. the same.
3. half as much.
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3
Last lecture: Magnetic Field
Convenient to describe the interaction at a distance
between magnets with the notion of magnetic field.
Magnetic objects are surrounded a magnetic field.
Moving electrical charges are also surrounded by a
magnetic field (in addition to the electrical field).
A vector quantity: magnitude and direction…
The letter B is used to represent magnetic fields.
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4
Magnetic Field Direction
The magnetic field direction (of a magnet bar) can
studied with a small compass.
N
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1
S
5
Magnetic Field Lines
N
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1
S
6
Applications: A “bit” of history
IBM introduced the first
hard disk in 1957, when
data usually was stored on
tapes. It consisted of 50
platters, 24 inch diameter,
and was twice the size of
a refrigerator.
It cost $35,000 annually in leasing fees (IBM would not
sell it outright). It’s total storage capacity was 5 MB, a
huge number for its time!
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7
Magnetic Field of the Earth
A small magnetic bar should be said to have north and
south seeking poles. The north of the bar points towards
the North of the Earth.
The geographic north corresponds to a south
magnetic pole and the geographic south
corresponds to a magnetic north.
The configuration of the Earth magnetic resemble that of
a (big) magnetic bar one would put in its center.
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8
Magnetic Field of the Earth
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9
Magnetic Field of the Earth
Near the ground, the field is NOT parallel to the surface
of the Earth.
„
The angle between the direction of the magnetic field and the
horizontal is called dip angle.
The north and south magnetic pole do not exactly
correspond to the south and north geographic north.
„
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South magnetic pole found (in 1832) to be just north of Hudson
bay in Canada – 1300 miles from the north geographical pole.
10
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11
More on the Magnetic Field of the Earth
The difference between the geographical north and the direction
pointed at by a compass changes from point to point and is called
the magnetic declination.
Source of the field : charge-carrying convection currents in the core
of the earth.
„
In part related to the rotation of the earth
The orientation of the field “flips” and changes over time – every few
million years…
„
Basalt rocks
Other planets (e.G. Jupiter) are found to have a magnetic field.
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12
Mini-quiz
You travel to Australia for a business trip and bring
along your American-made compass. Does the
compass work correctly in Australia???
• No problem using the compass in Australia.
• North pole of the compass will be attracted to the
South geographic pole…
• The vertical component of the field is different
(opposite) but that cannot be detected with normal
operation of the compass.
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19.3 Magnetic Fields
Stationary charged particles do NOT interact with a
magnetic field.
Charge moving through a magnetic field experience a
magnetic force.
Value of the force is maximum when the charge moves
perpendicularly to the field lines.
Value of the force is zero when the charge moves
parallel to the field lines.
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Magnetic Fields in analogy with Electric Fields
Electric Field:
„ Distribution of charge creates an electric field E(r)
in the surrounding space.
„ Field exerts a force F=q E(r) on a charge q at r
Magnetic Field:
„ Moving charge or current creates a magnetic field
B(r) in the surrounding space.
„ Field exerts a force F on a charge moving q at r
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15
Strength of the Magnetic Field
Define the magnetic field, B, at a given point in space in
terms of the magnetic force imparted on a moving
charge at that point.
Observations show that the force is proportional to
„
„
„
„
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The field
The charge
The velocity of the particle
The sine of the angle between the field and the direction of the
particle’s motion.
16
Strength and direction of the Magnetic Force on a charge
in motion
F
F = qvB sin θ
B
+q
v
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Magnetic Field Magnitude
F
B=
qv sin θ
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Magnetic Field Units
[F] = newton
[v] = m/s
[q] = C
[B] = tesla (T).
„
„
„
„
Also called weber (Wb) per square meter.
1 T = 1 Wb/m2.
1 T = 1 N s m-1 C-1.
1 T = 1 N A-1 m-1.
CGS unit is the Gauss (G)
„
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1 T = 104 G.
19
Right Hand Rule
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Provides a convenient trick to remember the
spatial relationship between F, v, and B.
Consider the motion of positive charge
Direction of force reversed if negative charge.
20
Example: Proton traveling in Earth’s
magnetic field.
A proton moves with a speed of 1.0 x 105 m/s through the Earth’s magnetic
field which has a value of 55 µT a particular location. When the proton moves
eastward, the magnetic force acting on it is a maximum, and when it moves
northward, no magnetic force acts on it. What is the strength of the magnetic
force? And what is the direction of the magnetic field?
V = 1.0 x 105 m/s
B = 55 µT
F = qvBsinθ
(
)(
)
(
F = 1.6×10−19C 8.0×106m/ s ( 2.5T) sin60o
)
−12
= 2.8×10 N
Northward or southward.
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19.4 Magnetic Force on Current-carrying
conductor.
A magnetic force is exerted on a single charge in motion
through a magnetic field.
That implies a force should also be exerted on a
collection of charges in motion through a conductor I.e. a
current.
And it does!!!
The force on a current is the sum of all elementary
forces exerted on all charge carriers in motion.
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19.4 Magnetic Force on Current
If B is directed into the page
we use blue crosses
representing the tail of arrows
indicating the direction of the
field,
If B is directed out of the page,
we use dots.
If B is in the page, we use lines
with arrow heads.
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x x x x
x x x x x
x x x x x x
x x x x x
x x x x
. . . .
. . . . .
. . . . . .
. . . . .
. . . .
23
Force on a wire carrying current
in a magnetic field.
Bin
x x x x
x x x x x
x x x x x x
x x x x x
x x x x
I=0
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Bin x x x x
x x x x x
x x x x x x
x x x x x
x x x x
I
Bin x x x x
x x x x x
x x x x x x
x x x x x
x x x x
I
24
Force on a wire carrying
current in a magnetic field.
x
Ax
x
x
x x x x
x x x x x
x x vx x x
d
q
x x x x x
x x x x x
x x x x x
x x x x x
x
x
x
x
Fmax = ( qvd B )( nAl )
I = nqvd A
Fmax = BIl
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Magnetic Field and Current
at right angle from each
other.
25
Force on a wire carrying current in a magnetic field.
General Case: field at angle θ relative to current.
Fmax = BIl sin θ
B
B sin θ
θ
I
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26
Voice Coil
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27
Mini-Quiz
In a lightning strike, there is a rapid flow of negative
charges from a cloud to the ground. In what direction is
a lightning strike deflected by the Earth’s magnetic
field?
Reasoning:
Negative charge flow down.
Positive Current upward.
B field direction Geo South to Geo North
Answer:
Force towards the west.
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I
28
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29
Example: Wire in Earth’s B Field
A wire carries a current of 22 A from east to west. Assume that at this location
the magnetic field of the earth is horizontal and directed from south to north,
and has a magnitude of 0.50 x 10-4 T. Find the magnetic force on a 36-m length
of wire. What happens if the direction of the current is reversed?
B=0.50 x 10-4 T.
I = 22 A
l = 36 m
Fmax = BIl
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Fmax = BIl
(
= 0.50 ×10−4 T
) ( 22 A)( 36m )
= 4.0 ×10−2 N
30
19.5 Torque on a Current Loop
Imagine a current loop in a magnetic field as follows:
B
I
B
b
a
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F
F
a/2
F
F
31
B
I
B
F
F
b
a
a/2
F
F
F1 = F2 = BIb
τ max = F1 a2 + F2 a2 = ( BIb ) a2 + ( BIb ) a2
τ max = BIba = BIA
τ = BIA sin θ
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32
In a motor, one has “N” loops of current
τ = NBIA sin θ
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33
Example: Torque on a circular loop in a
magnetic field
A circular loop of radius 50.0 cm is oriented at an
angle of 30.0o to a magnetic field of 0.50 T. The
current in the loop is 2.0 A. Find the magnitude of the
torque.
r = 0.500 m
θ = 30o
B = 0.50 T
I = 2.0 A
N=1
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30.0o
B
τ = NBIA sin θ
2

= ( 0.50T )( 2.0 A ) π ( 0.50 m )  sin 30.0 o


τ = 0.39 Nm
34
19.6 Galvanometer/Applications
Device used in the construction
of ammeters and voltmeters.
Scale
Current loop
or coil
Magnet
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Spring
35
Galvanometer used as Ammeter
Typical galvanometer have an internal resistance of the order of
60 W - that could significantly disturb (reduce) a current
measurement.
Built to have full scale for small current ~ 1 mA or less.
Must therefore be mounted in parallel with a small resistor or
shunt resistor.
Galvanometer
60 Ω
Rp
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36
Galvanometer
60 Ω
Rp
• Let’s convert a 60 W, 1 mA full scale galvanometer to an
ammeter that can measure up to 2 A current.
• Rp must be selected such that when 2 A passes through the
ammeter, only 0.001 A goes through the galvanometer.
( 0.001A)( 60Ω) = (1.999 A) Rp
Rp = 0.03002Ω
• Rp is rather small!
• The equivalent resistance of the circuit is also small!
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37
Galvanometer used as Voltmeter
• Finite internal resistance of a galvanometer must also
addressed if one wishes to use it as voltmeter.
• Must mounted a large resistor in series to limit the current
going though the voltmeter to 1 mA.
• Must also have a large resistance to avoid disturbing
circuit when measured in parallel.
Rs
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Galvanometer
60 Ω
38
Rs
Galvanometer
60 Ω
Maximum voltage across galvanometer:
∆Vmax = ( 0.001A)( 60Ω) = 0.06V
Suppose one wish to have a voltmeter that can measure
voltage difference up to 100 V:
100V = ( 0.001A) ( Rp + 60Ω)
Rp = 99940Ω
10/1/2003
Large resistance
39
19.7 Motion of Charged Particle in magnetic field
Consider positively charge
particle moving in a uniform
magnetic field.
Suppose the initial velocity of
the particle is perpendicular to
the direction of the field.
Then a magnetic force will be
exerted on the particle and
make follow a circular path.
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×
×
×
q
v
×
×
×
×F ×
×
×
×
×
×
×
×
r
Bin
× ×
×
×
×
×
×
×
×
×
×
×
×
40
The magnetic force produces a centripetal acceleration.
mv2
F = qvB =
r
The particle travels on a circular trajectory with a radius:
mv
r=
qB
10/1/2003
41
Example: Proton moving in uniform magnetic field
A proton is moving in a circular orbit of radius 14 cm in a uniform
magnetic field of magnitude 0.35 T, directed perpendicular to the
velocity of the proton. Find the orbital speed of the proton.
r = 0.14 m
B = 0.35 T
m = 1.67x10-27 kg
q = 1.6 x 10-19 C
mv
r=
qB
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qBr
v=
m
(
=
)
(
1.6 ×10−19 C ( 0.35T ) 14 ×10−2 m
(
1.67 ×10−27 kg
)
)
= 4.7 ×106 m s
42