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General Physics (PHY 2140)
Lecture 13
 Electricity and Magnetism
Magnetism
Application of magnetic forces
 Ampere’s law
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 19
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Lightning Review
Last lecture:
1. Magnetism
 Magnetic field
 Magnetic force on a moving particle
 Magnetic force on a current
F  qvB sin 
F  BIl sin 
Review Problem:
How does the aurora borealis (the
Northern Lights) work?
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Magnetic Field of the Earth
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Review Problem 2
How does your credit card work?
The stripe on the back of a credit card is a magnetic stripe, often called a
magstripe. The magstripe is made up of tiny iron-based magnetic particles in a
plastic-like film. Each particle is really a tiny bar magnet about 20-millionths of
an inch long.
The magstripe can be "written" because the tiny bar magnets can be
magnetized in either a north or south pole direction. The magstripe on the back
of the card is very similar to a piece of cassette tape .
A magstripe reader (you may have seen one hooked to someone's PC at a
bazaar or fair) can understand the information on the three-track stripe.
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Review Example 1: Flying duck
A duck flying horizontally due north at 15 m/s passes over Atlanta, where the
magnetic field of the Earth is 5.0×10-5T in a direction 60° below a horizontal
line running north and south. The duck has a positive charge of 4.0×10-8C.
What is the magnetic force acting on the duck?
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Review Example 2: 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
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19.5 Torque on a Current Loop
Imagine a current loop in a magnetic field as follows:
I
B
B
F
F
a/2
b
F
F
a
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I
B
B
F
F
a/2
b
F
F
a
F1  F2  BIb
 max  F1 a2  F2 a2   BIb  a2   BIb  a2
 max  BIba  BIA
  BIAsin 
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In a motor, one has “N” loops of current
  NBIAsin 
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Example 1 : 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
 = 30o
B = 0.50 T
I = 2.0 A
N=1
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30.0o
B
  NBIA sin 
2

  0.50T  2.0 A   0.50m   sin 30.0o


  0.39 Nm
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Example 2: triangular loop
A 2.00m long wire carrying a current of 2.00A forms a 1 turn loop in the
shape of an equilateral triangle. If the loop is placed in a constant magnetic
field of magnitude 0.500T, determine the maximum torque that acts on it.
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19.6 Galvanometer/Applications
Device used in the construction
of ammeters and voltmeters.
Scale
Current loop
or coil
Magnet
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Spring
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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 W
Rp
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Galvanometer
60 W
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 60W   1.999 A Rp
R p  0.03002W
• Rp is rather small!
• The equivalent resistance of the circuit is also small!
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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 W
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Rs
Galvanometer
60 W
Maximum voltage across galvanometer:
Vmax   0.001A 60W  0.06V
Suppose one wish to have a voltmeter that can measure
voltage difference up to 100 V:
100V   0.001A  Rp  60W 
Rp  99940W
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Large resistance
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19.7 Motion of Charged Particle in magnetic field
Consider positively charge
 
particle moving in a uniform
magnetic field.
q
Suppose the initial velocity of
the particle is perpendicular to  v 
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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


F 

r
Bin
 













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The magnetic force produces a centripetal acceleration.
mv 2
F  qvB 
r
The particle travels on a circular trajectory with a radius:
mv
r
qB
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Example 1 : 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
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Example 2:
Consider the mass spectrometer. The electric field between the plates of the
velocity selector is 950 V/m, and the magnetic fields in both the velocity
selector and the deflection chamber have magnitudes of 0.930 T. Calculate the
radius of the path in the system for a singly charged ion with mass m=2.18×1026 kg.
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19.8 Magnetic Field of a long straight wire
Danish scientist Hans Oersted (1777-1851) discovered
somewhat by accident that an electric current in a wire
deflects a nearby compass needle.
In 1820, he performed a simple experiment with many
compasses that clearly showed the presence of a
magnetic field around a wire carrying a current.
I=0
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I
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Magnetic Field due to Currents
The passage of a steady current in a wire produces a
magnetic field around the wire.


Field form concentric lines around the wire
Direction of the field given by the right hand rule.
If the wire is grasped in the right hand with the thumb in the
direction of the current, the fingers will curl in the direction of
the field.

Magnitude of the field
I
o I
B
2 r
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Magnitude of the field
o I
B
2 r
I
r
B
o called the permeability of free space
o  4 10 Tm / A
7
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Ampere’s Law
Consider a circular path surrounding a current, divided
in segments l, Ampere showed that the sum of the
products of the field by the length of the segment is
equal to o times the current.
 B l   I
Andre-Marie Ampere
I
o
r
l
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B
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Consider a case where B is constant and uniform.
 B l  B  l  B 2 r   I
o
Then one finds:
o I
B 
2 r
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