magnetic field chapter 28. magnetism2 refrigerators are attracted to magnets!
TRANSCRIPT
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Magnetic fieldMagnetic field
Chapter 28Chapter 28
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Magnetism 2
Magnetism
• Refrigerators are attracted to magnets!
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Magnetism 3
Where is Magnetism Used??Where is Magnetism Used??
• Motors• Navigation – Compass• Magnetic Tapes
– Music, Data
• Television– Beam deflection Coil
• Magnetic Resonance Imaging (MRI)
• High Energy Physics Research
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Magnetism 4
EF
BF
Cathode
Anode
EF qE
BF qv B
(28 – 8)
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Magnetism 5
Consider a Permanent Magnet
N S
B
The magnetic Field B goes from North to South.
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Magnetism 6
Units
N/(A.m) 1 T 1 tesla1
m.Amp
N
/
:
)
sCm
N
qv
FB
Units
Bqv Sin(θF
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Magnetism 7
Typical Representation
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Magnetism 8
A Look at the Physics
B
q
There is NO force ona charge placed into amagnetic field if thecharge is NOT moving.
Bq
v
• If the charge is moving, thereis a force on the charge,perpendicular to both v and B.
F = q v x B
There is no force if the chargemoves parallel to the field.
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Magnetism 9
The Lorentz Force
This can be summarized as: F q B v
vF
Bqm
or:
F qvBsin
is the angle between B and V
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Magnetism 10
Nicer Picture
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Magnetism 11
The Wire in More Detail
B out of plane of the paper
Assume all electrons are moving with the same velocity vd.
(i). charge POSITIVE ofmotion
theofdirection in the
:
L
BLF
Vector
i
vector
iLBBvv
LiBqvF
v
Litiq
dd
d
d
L
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Magnetism 12
. dL
B
dF
i
If we assume the more general case for which the
magnetic field froms and angle with the wire
the magnetic force equation can be writ
B
Magnetic force on a straight wire in a uniform
magnetic field.
ten in vector
form as: Here is a vector whose
magnitude is equal to the wire length and
has a direction that coincides with that of the current.
The magnetic force magnitude
B
B
F iL B L
L
F
sin
In this case we divide the wire into elements of
length which can be considered as straight.
The magnetic f
iLB
dL
Magnetic force on a wire of arbitrary shape
placed in a non - uniform magnetic field.
orce on each element is:
The net magnetic force on the
wire is given by the integral:
B
B
dF idL B
F i dL B
=
BF iL B
BdF idL B=
BF i dL B
(28 – 12)
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Magnetism 13
Current Loop
Loop will tend to rotate due to the torque the field applies to the loop.
What is forceon the ends??
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Magnetism 14
Magnetic Force on a Current Loop
I
N S
F=BIL
F=BIL
L
F
F
B
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Magnetism 15
Magnetic Force on a Current Loop
IF=BIL
Simplified view:
dL
B
F=BIL
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Magnetism 16
Magnetic Force on a Current Loop Torque & Electric Motor
IF=BIL
Simplified view:
dL
Torque BILd
IAB
2
2sin
sin
B
F=BIL
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Magnetism 17
Magnetic Force on a Current LoopTorque & Electric Motor
I F=BIL
dL
IA B
B
for a current loop
F=BIL
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Magnetism 18
C
C
Top viewSide view
sinnet iAB
0netF
(28 – 13)
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Magnetism 19
By analogy with electric dipoles, for which:
The expression,
implies that a current loop acts as a magnetic dipole!Here is the magnetic dipole moment, and
Magnetic Force on a Current Loop Torque & Magnetic Dipole
IA B
p E
IA
B
(Torque on acurrent loop)
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Magnetism 20
Dipole Moment DefinitionDefine the magneticdipole moment ofthe coil as:
=NiA= x B
We can convert thisto a vector with Aas defined as being normal to the area asin the previous slide.
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Magnetism 21
U B U B
The torque of a coil that has loops exerted
by a uniform magnetic field and carrries a
current is given by the equation:
We define a new vector associated wi
N
B
i NiAB
Magnetic dipole moment :
th the coil
which is known as the magnetic dipole moment of
the coil.
B
(28 – 14)
U B
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Magnetism 22
L L
L
R R
R
In 1879 Edwin Hall carried out an experiment in which
he was able to determine that conduction in metals is due
to the motion of charges (electrons). He was also
able to determin
ne
The Hall
gat
e t
ive
ffec
e the concentration of the electrons.
He used a strip of copper of width and thickness . He passed
a current along the length of the strip and applied a magnetic
field perpendicular to the stri
n
d
i
B
p as shown in the figure. In the
presence of the electrons experience a magnetic force that
pushes them to the right (labeled "R") side of the strip. This
accumulates negative charge on the R-sid
BB F
e and leaves the left
side (labeled "L") of the strip positively charged. As a result
of the accumulated charge, an electric field is generated as
shown in the figure so that the electric force bala
E
nces the magnetic
force on the moving charges.
( ). From chapter 26 we have:
( )
E B d
d d
d
F F eE ev B
E v B J nev
J i iv
ne Ane dne
eqs.1
eqs.2 (28 – 15)
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MagnetismMagnetism 2323
Motion of a chargedMotion of a chargedparticle in a magneticparticle in a magnetic
FieldField
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Magnetism 24
Trajectory of Charged Particlesin a Magnetic Field
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
vB
F
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
vB
F
(B field points into plane of paper.)
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Magnetism 25
Trajectory of Charged Particlesin a Magnetic Field
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
vvB B
FF
(B field points into plane of paper.)
Magnetic Force is a centripetal force
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Magnetism 26
Review of Rotational Motion
atar
at = r tangential acceleration
ar = v2 / r radial acceleration
The radial acceleration changes the direction of motion,while the tangential acceleration changes the speed.
r s = s / r s = r ds/dt = d/dt r v = r
= angle, = angular speed, = angular acceleration
Uniform Circular Motion
= constant v and ar constant but direction changes
ar = v2/r = 2 rF = mar = mv2/r = m2r
KE = ½ mv2 = ½ mw2r2
v
ar
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Magnetism 27
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
Radius of a Charged ParticleOrbit in a Magnetic Field
vB
F
r
mr
q B
r mqB
v v
v
2
Centripetal Magnetic Force Force =
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Magnetism 28
Cyclotron Frequency
+ + + +
+ + + +
+ + + +
+ + + +
+ + + +
vB
F
r
The time taken to complete one orbit is:
qB
m
rT
v
v
2v
2
m
qBf
m
qB
Tf
c
2
2
1
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Magnetism 29
Mass Spectrometer
Smaller Mass
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Magnetism 30
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Magnetism 31
An ExampleA beam of electrons whose kinetic energy is K emerges from a thin-foil “window” at the end of an accelerator tube. There is a metal plate a distance d from this window and perpendicular to the direction of the emerging beam. Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field B such that
22
2
de
mKB
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Magnetism 32
Problem Continued
r
22
22
2
e
2mKB
:Bfor Solvee
22
eB
m
2
2
1
qB
mvr
Before From
d
dB
mK
m
Kr
m
KvmvK
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Magnetism 33
#14 Chapter 28 A metal strip 6.50 cm long, 0.850 cm wide, and 0.760 mm
thick moves with constant velocity through a uniform magnetic field B= 1.20mTdirected perpendicular to the strip, as shown in the Figure. A potential difference of 3.90 ηV is measured between points x and y across the strip. Calculate the speed v.
FIGURE 28-37
Problem 14.
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Magnetism 34
21. (a) Find the frequency of revolution of an electron with an energy of 100 eV in a uniform magnetic field of magnitude 35.0 µT . (b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field.
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Magnetism 35
39. A 13.0 g wire of length L = 62.0 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.440 T. What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?