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Applied Physics
Lecture 14
 Electricity and Magnetism
Magnetism
 Ampere’s law
Applications of magnetic forces
Chapter 19
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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 the  v 
direction of the field.
Then a magnetic force will be
exerted on the particle…


F 

r
Bin
 

















Where is it directed?
… and make it follow a circular
path.
Remember that
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vF
2
The magnetic force produces a centripetal acceleration.
F  mac
mv 2
F  qvB 
2
2
v
ac 
r
sin   sin 90  1
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.
Given:
r = 0.14 m
B = 0.35 T
m = 1.67x10-27 kg
q = 1.6 x 10-19 C
Find:
v=?
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mv
r
qB
Recall that the proton’s radius would be
Thus
v
qBr
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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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
5
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 (second right-hand rule).

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
0  4 10 T  m A
7
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Ampere’s Law
Consider a circular path surrounding a current, divided
in segments Dl, 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 Dl   I
Andre-Marie Ampere
I
o
r
Dl
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B
8
Consider a case where B is constant and uniform:
 B Dl  B  Dl  B 2 r   I
o
Then one finds:
o I
B 
2 r
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l
1
B2
2
F1
I1
d
I2
o I 2
B2 
2 d
o I1 I 2l
 o I 2 
F1  B2 I1l  
I1l 

2 d
 2 d 
F1 o I1 I 2

l
2 d
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Force per
unit length
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Definition of the SI unit Ampere
F1 o I1 I 2

l
2 d
Used to define the SI unit of current called
Ampere.
If two long, parallel wires 1 m apart carry the same current, and the
magnetic force per unit length on each wire is 2x10-7 N/m, then the
current is defined to be 1 A.
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Example 1: Levitating a wire
Two wires, each having a weight per units length of 1.0x10-4 N/m, are
strung parallel to one another above the surface of the Earth, one
directly above the other. The wires are aligned north-south. When their
distance of separation is 0.10 mm what must be the current in each in
order for the lower wire to levitate the upper wire. (Assume the two
wires carry the same current).
1
2
l
d
I1
I2
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1
I1
B2
mg/l
2
l
Weight of wire per unit
length:
mg/l = 1.0x10-4 N/m
Wire separation:
d=0.1 m
I1 = I 2
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Two wires, each having a weight per
units length of 1.0x10-4 N/m, are strung
parallel to one another above the surface
of the Earth, one directly above the
other. The wires are aligned north-south.
When their distance of separation is 0.10
mm what must be the current in each in
order for the lower wire to levitate the
upper wire. (Assume the two wires carry
the same current).
F1
d
I2
F1 mg o I 2


l
l
2 d
1.0 10
4

N /m 
 
4 107 Tm A I 2
 2  0.10m 
I  7.1A
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19.10 Magnetic Field of a current loop
Magnetic field produced by a wire can be enhanced
by having the wire in a loop.
Dx1
B
I
Dx2
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19.11 Magnetic Field of a solenoid
Solenoid magnet consists of a wire coil with multiple
loops.
It is often called an electromagnet.
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Solenoid Magnet
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Field lines inside a solenoid magnet are parallel, uniformly spaced
and close together.
The field inside is uniform and strong.
The field outside is non uniform and much weaker.
One end of the solenoid acts as a north pole, the other as a south
pole.
For a long and tightly looped solenoid, the field inside has a value:
B  o nI
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Solenoid Magnet
B  o nI
n = N/l : number of (loop) turns per unit length.
I : current in the solenoid.
o  4 10 Tm / A
7
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Example: Magnetic Field inside a Solenoid.
Consider a solenoid consisting of 100 turns of wire and
length of 10.0 cm. Find the magnetic field inside when it
carries a current of 0.500 A.
N = 100
l = 0.100 m
I = 0.500 A
o  4 10 Tm / A
7
N 100turns
n 
 1000turns / m
l 0.10m


B  o nI  4 107 Tm / A 1000turns / m  0.500 A
B  6.28 104 T
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Comparison:
Electric Field vs. Magnetic Field
Source
Acts on
Force
Direction
Electric
Magnetic
Charges
Charges
F = Eq
Parallel E
Moving Charges
Moving Charges
F = q v B sin()
Perpendicular to v,B
Field Lines
Opposites
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Charges Attract
Currents Repel
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