Download Magnetic Field Interactions

Chapter 28 – Magnetic Field
Interactions
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
1
Right Hand Rule WS 10 #1-7
 The right hand rule is used in order to determine the direction in which the
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vector resulting from the cross product would act.
Consider the following cross product.
To find the direction of vector C, you would start by pointing the fingers of
your right hand in the direction of the first vector (A) with your palm open in
the direction in which vector B points.
Next, curl your fingers towards the second vector (B).
Finally, extend your thumb. The direction of C is in the same direction in
which your thumb points.
A
C  A B
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B
C
2
Right Hand Rule WS 10 #1-7
 In what direction would vector Z point?
 Start by pointing the fingers of your right
hand in the direction of the first vector (X)
with your palm open in the direction in which
vector Y points.
 Curl your fingers towards the second vector
(Y).
 Finally, extend your thumb. The direction of
Z is in the same direction in which your
thumb points.
 In which direction would vector W point?
z  x y
w  y x
z
x
y
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
3
Forces on a Charged Particle in a Magnetic Field
 You use the cross product in order to determine which direction a
charged particle would move in a magnetic field.
 The equation used to determine the magnitude and direction of this
force is
F  qv  B
 In this equation, q is the magnitude of the charge, v is the speed of
the charge, and B is the magnitude of the magnetic field.
 We will use the cross product to determine the direction of the force
(the direction in which the charge would accelerate) on this moving
charge in the given magnetic field.
 This direction is found using the Right Hand Rule as discussed
earlier.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
4
Forces on a Charged Particle in a Magnetic Field WS 10 #8
 When a charged particle moves through a magnetic field, its magnetic field
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interacts with the uniform magnetic field.
As a result, its path is deflected.
We use the right hand rule in order to determine the direction of the force
(acceleration) acting on the particle.
Lets exam the resulting path of the positive charge shown below once it
encounters the magnetic field.
As we saw, the direction of the force acting on the charge was down;
therefore, the particle moved down and out of the magnetic field.
v
F  qv  B
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
B
F
5
Forces on a Charged Particle in a Magnetic Field WS 10 #9
 Lets now exam the path for a negatively charged particle moving through the
same uniform magnetic field.
 The right hand rule shows us the direction of the cross product is also down.
 However, the negative sign included with the negative charge would reverse
the direction of the force applied to the particle.
 As a result, the particle would move up and out of the magnetic field.
F
F  qv  B
v
B
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
6
Magnetic field WS 10 #10 & 11
 We used the right hand rule in order to determine the direction that a charged
particle in motion would be deflected as it passes through a magnetic field.
 The magnitude of the force acting on such a particle can be found using the
following equation.
 If this force is already known, then we can find the magnitude of the magnetic
field by solving this equation for B.
 The unit for the magnetic field is the Tesla defined as follows.
F  qv  B
F  q vB sin 
F
B
q v sin 

 

N   N   N 


T    m    C


Am


C
m
s
s

 

 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
 
7
Cross Product Review
F  qv  B
F  q vB sin 

 
A  B  Axiˆ  Ay ˆj  Az kˆ  Bxiˆ  By ˆj  Bz kˆ

A  B  Axiˆ  Bxiˆ  Axiˆ  By ˆj  Axiˆ  Bz kˆ 
Ay ˆj  Bx iˆ  Ay ˆj  By ˆj  Ay ˆj  Bz kˆ 
Az kˆ  Bxiˆ  Az kˆ  By ˆj  Az kˆ  Bz kˆ
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8
Cross Product Review
 
 
 
A B  ˆj  iˆ   A B  ˆj  ˆj   A B  ˆj  kˆ  
A  B  Ax Bx iˆ  iˆ  Ax By iˆ  ˆj  Ax Bz iˆ  kˆ 
y
x
y


y

y
z


Az Bx kˆ  iˆ  Az By kˆ  ˆj  Az Bz kˆ  kˆ

iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  0
iˆ  ˆj   ˆj  iˆ  kˆ
ˆj  kˆ   kˆ  ˆj  iˆ
kˆ  iˆ  iˆ  kˆ  ˆj


 




 

A  B  Ax By iˆ  ˆj  Ax Bz iˆ  kˆ  Ay Bx ˆj  iˆ  Ay Bz ˆj  kˆ  Az Bx kˆ  iˆ  Az By kˆ  ˆj
A  B  Ax By kˆ  Ax Bz ˆj  Ay Bx kˆ  Ay Bziˆ  Az Bx ˆj  Az Byiˆ
A  B   Ay Bz  Az By  iˆ   Az Bx  Ax Bz  ˆj   Ax By  Ay Bx  kˆ
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
9

Magnetic field WS 10 #12
 A charged particle from problem number
eight (q = 3.25 x 10-6 C) moves at a velocity
given below.
 Determine the force acting on this charged
particle if the magnetic field is given by the
equation below.

B

v  2.23 103 iˆ  4.76 106 ˆj  1.14 103 kˆ m / s


B  1.25iˆ  0.05kˆ T
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
10
Field Representations
There are three ways we represent the presence of
a field (electric, magnetic, fluid flow, ….) in
drawings expressing physical quantities.
The first way is by drawing the field lines in the
plane of the paper.
Here is an example of a field moving to the right.
The other two ways are best visualized by thinking
of an arrow.
If the arrow was coming towards you, then you
would see a dot.
If the arrow was going away from, then you would
see an x.
This figure is of a field coming out of the plane of
the paper towards you.
This figure is of a field going into the plane of the
paper away from you.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
11
Magnetic Field Lines WS 10 #13
 Around every magnet there are many invisible lines known as magnetic field
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lines.
These lines influence other magnets and charged particles that may be in the
vicinity of the magnet.
The direction of these lines at any point is the direction of the magnetic field
acting at that point.
The density of the lines represents the strength of the magnetic field.
For example, the higher the density of the magnetic field lines, the stronger the
magnetic field.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
12
Other Magnetic Fields WS 10 #13
 The magnetic field generated by a
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horseshoe magnetic are like the one
shown to the right.
At other times we may wish to have a
uniform, parallel magnetic field.
Magnets like the one below where the
North and South poles are parallel
will generate a parallel magnetic field
as shown.
Label the North and South poles on
these two magnets.
Remember, the magnetic field lines
go from the North pole to the South
pole.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
13
WS 10 #14
 An electron (m = 9.109 x 10-31 kg,
q = -1.602 x 10-19 C) with 4.9 MeV
of kinetic energy moves through a
uniform magnetic field (B = 1.35
T) as shown.
 Label the appropriate poles
(represented by the parallelograms)
shown as North and South.
 What is the force (magnitude and
direction) acting on the electron
traveling through this uniform
magnetic field.
 What is the acceleration
experienced by this electron?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
14
Magnetic Force on a Current Carrying Wire WS 10 #15
 We previously learned that magnetic fields interact with
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moving electric charges (tiny magnets).
We will now examine the force exerted on a moving
electric charge by a magnetic field.
The direction of this force is found by applying the right
hand rule to the following equation.
In this equation, the vector L is in the same direction as the
current flowing through the wire.
Which way would the wire to the right bend if we turned on
the following magnetic field?
It would bend to the right.
The magnitude of the force acting on the electrons may be
found using the following equation.
F  iL  B
F  iLB sin 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
15
Magnetic Force on a Current Carrying Wire
 Which way would the wire to the right bend if we turned on
the following magnetic field?
 It would bend to the left.
F  iL  B
F  iLB sin 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
16
Torque on a Current Loop WS 10 #18
 When an electric motor is turned on, electricity flows
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through the brushes into the commutators finally ending
up in the wires of the armature.
This electric flow establishes an electric field that
interacts with the magnetic field causing the armature to
rotate.
On the figure below the yellow dots represent wires
where the electrons are flowing into the page, and the
orange dots represent wires where the electrons are
flows out of the page.
The direction of the resulting force on these points may
be found using the below cross product.
The force at the top is to the left and the force at the
bottom is to the right.
These two forces produce a counterclockwise torque on
the motor causing the motor to rotate.
F  iL  B
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
FTop
I Bot
I Top
FBot
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WS 10 #19
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A battery is connected to two horizontal wires as shown.
A third wire (m = 18.2 mg, l = 2.15 cm) rest across the horizontal wires.
A magnetic field (B = 71.15 mT) is turned on.
The battery delivers a constant current of 6.20 mA.
Which direction will the third wire move?
What will its speed be after 77.7 mS?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
18
WS 10 #20
 A conducting wire (l = 85.0 cm, m = 17.5 g) hangs from two conducting springs
connected to the positive (left) and negative (right) terminals of a DC power supply.
 A magnetic field of 1.18 T acts on this wire segment.
 What is the current needed through the wire that will remove the tension in the two
conducting springs?
0.0
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
0.0
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Discovery of the Electron – Thomson’s Apparatus WS 10 #21
 The device below, known as a cathode ray tube, is similar to the one used by JJ
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Thomson in order to discover the electron.
This device demonstrates the fact that both electric and magnetic fields interact with
charged particles (in this case electrons).
Here is how it works.
A voltage difference is applied across two discs generating an electric field between
the discs.
A heated filament emits electrons which are accelerated by the electric field.
When the capacitor and the electromagnetic are off, these charges fly straight and
strike the screen.
0.0
0.0
0.0
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
20
Discovery of the Electron – Thomson’s Apparatus WS 10 #21
 The capacitor and the electromagnet are used to deviate the path of the electrons.
 Watch what happens when the electromagnet is turned on and the capacitor is
charged.
 As can be seen the magnetic field and the electric field deviate the path of the
electrons.
0.0
0.0
0.0
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
21
Cathode Ray Tube WS 10 #22-25
 In the setup below both the capacitor and the
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electromagnet are aligned in such a way as to deviate an
electron’s vertical position.
When both are off, the electron flies straight through and
hits in the screens center.
When the electric field generated by the capacitor is
directed up, the electron’s path is deflected downwards.
When the electric field generated by the capacitor
is directed down, the electron’s path is
deflected upwards.
Which way will the electron be deflected due
to the following magnetic field?
When the magnetic field direction is reversed,
so to is the deflection direction.
F  qv  B
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
22
Cathode Ray Tube WS 10 #26
 An electron is fired through the magnetic field shown and
strikes the screen as indicated in the top-center position.
 Afterwards, an electric field is generated across the plates
of a capacitor.
 Another electron is fired through both the magnetic field
and the electric field striking the screen in the dead center
position.
 Derive the equation needed in order to determine the
magnitude of the electric field.
F  qv  B
F  q0 E
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
23
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24
A
 A
v
B
F  qv  B
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25
Forces on a Charged Particle in a Magnetic Field
 A
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26
A
A
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27
Forces on a Charged Particle in a Magnetic Field
 A
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28
Forces on a Charged Particle in a Magnetic Field
 A
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29
Forces on a Charged Particle in a Magnetic Field
 A
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31
Magnetic Induction
 Realigning magnetic dipoles can also cause electrons to move.
 Flowing electrons from a power supply cause the domains in the vicinity of the
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first coil to align.
These domains in turn induce the other domains in the nail to realign.
Watch what happens to the electrons in the second coil and watch the light bulb as
the magnetic domains in the vicinity of the second coil realign.
As long as these domains were moving, electrons in the second coil moved
causing the light to go on.
Once the domains stop moving, the electrons stop moving, and the light goes out
even though the electrons are still moving in the first coil.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
32
Interactions Between Electric and Magnetic Fields.
 A wire that carries a current (flowing electrons) will bend when it passes through
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a magnetic field.
This image is a picture of a magnetic field traveling into the board.
Right now the current in the wire is off.
Lets turn off the magnetic field and turn on the current.
Now turn the magnetic field back on and watch what happens.
The wire bent. Why?
The magnetic field produced by moving electric charges interfered with the large
magnetic field causing the wire to bend.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
33
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
34
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
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© 2001-2005 Shannon W. Helzer. All Rights Reserved.
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12.5
0.0
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
0.285
0.0
37
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
38
Torque on a Current Loop
 When an electric motor is turned on, electricity flows
through the wire in the armature.
 This electricity interacts with the electric field
generating a force that causes the armature to rotate.
 On the figure below the blue dots represent wires
where the current is flowing into the page, and the
purple dots represent wires where the current flows
out of the page.
 The direction of the resulting for
FOut V
FIn H
FOut H
F  iL  B
FIn V
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
39
Torque on a Current Loop
 a
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40
Torque on a Current Loop
 a
FTop
FBot
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41
Torque on a Current Loop
 a
FTop
FBot
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
42
Torque on a Current Loop
 a
FTop
FBot
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
43
A
 A
F  qv  B
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
44
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
45
F
B
E
C
D
v
B
v
B
v
E
v
v
B
v
+
B
v
E
B
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
46
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
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© 2001-2005 Shannon W. Helzer. All Rights Reserved.
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