Transcript Magnetism

Review on Magnetism
Chapter 28
Magnetism
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Magnetism
• Refrigerators are attracted to magnets!
Magnetism
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Where is Magnetism Used??
• Motors
• Navigation – Compass
• Magnetic Tapes
– Music, Data
• Television
– Beam deflection Coil
• Magnetic Resonance Imaging (MRI)
• High Energy Physics Research
Magnetism
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FE
(28 – 8)
FE  qE
Cathode
FB  qv  B
Anode
FB
Magnetism
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Consider a Permanent Magnet

B
N
S
The magnetic Field B goes from North to South.
Magnetism
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Units
F  Bqv Sin(θ )
Units :

F
N
N
B


qv Cm / s Amp  m
Magnetism
1 tesla  1 T  1 N/(A - m)
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Typical Representation
Magnetism
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A Look at the Physics

B
q

v

q B
There is NO force on
a charge placed into a
magnetic field if the
charge is NOT moving.
There is no force if the charge
moves parallel to the field.
• If the charge is moving, there
is a force on the charge,
perpendicular to both v and B.
F=qvxB
Magnetism
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The Lorentz Force
This can be summarized as:

 
F  qv  B
F
or:
F  qvBsin 
v
B
mq
 is the angle between B and V
Magnetism
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Nicer Picture
Magnetism
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The Wire in More Detail
Assume all electrons are moving
with the same velocity vd.
L
L
q  it  i
vd
F  qvd B  i
L
vd B  iLB
vd
vector :
F  iL  B
B out of plane of the paper
Magnetism
Vector L in the direction of the
motion of POSITIVE charge (i).
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(28 – 12)
Magnetic force on a straight wire in a uniform
magnetic field.
If we assume the more general case for which the
magnetic field B froms and angle  with the wire
the magnetic force equation can be written in vector
form as: FB  iL  B
FB  iL  B
B
i
dF
.
dFB = idL  B
FB  i  dL  B
Magnetism
Here L is a vector whose
magnitude is equal to the wire length L and
has a direction that coincides with that of the current.
The magnetic force magnitude FB  iLB sin 

dL
Magnetic force on a wire of arbitrary shape
placed in a non - uniform magnetic field.
In this case we divide the wire into elements of
length dL which can be considered as straight.
The magnetic force on each element is:
dFB = idL  B The net magnetic force on the
wire is given by the integral: FB  i  dL  B
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Current Loop
What is force
on the ends??
Loop will tend to rotate due to the torque the field applies to the loop.
Magnetism
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(28 – 13)
Side view
Top view
C
C
 net  iAB sin 
Fnet  0
Magnetism
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Dipole Moment Definition
Define the magnetic
dipole moment of
the coil m as:
m=NiA
=m x B
Magnetism
We can convert this
to a vector with A
as defined as being
normal to the area as
in the previous slide.
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  mB
U  mB
U  m B
U  m  B
Magnetic dipole moment :
The torque of a coil that has N loops exerted
by a uniform magnetic field B and carrries a
current i is given by the equation:   NiAB
We define a new vector m associated with the coil
which is known as the magnetic dipole moment of
the coil.
Magnetism
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(28 – 14)
R
L
The Hall effect
In 1879 Edwin Hall carried out an experiment in which
R
he was able to determine that conduction in metals is due
to the motion of negative charges (electrons). He was also
able to determine the concentration n of the electrons.
He used a strip of copper of width d and thickness . He passed
a current i along the length of the strip and applied a magnetic
field B perpendicular to the strip as shown in the figure. In the
L
presence of B the electrons experience a magnetic force FB that
L
R
pushes them to the right (labeled "R") side of the strip. This
accumulates negative charge on the R-side and leaves the left
side (labeled "L") of the strip positively charged. As a result
of the accumulated charge, an electric field E is generated as
shown in the figure so that the electric force balances the magnetic
force on the moving charges. FE  FB  eE  evd B 
E  vd B (eqs.1). From chapter 26 we have: J  nevd 
Magnetism
vd 
J
i
i


ne Ane
dne
(eqs.2)
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(28 – 15)
Motion of a charged
particle in a magnetic
Field
Magnetism
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Trajectory of Charged Particles
in a Magnetic Field
(B field points into plane of paper.)
+
+B
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v+
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+ F
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+ F +
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B
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Magnetism
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Trajectory of Charged Particles
in a Magnetic Field
(B field points into plane of paper.)
+
+B
+
v+
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+
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+
+
+ F
+
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+ F +
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B
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Magnetism
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Magnetic Force is a centripetal force
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Review of Rotational Motion

 = s / r  s =  r  ds/dt = d/dt r  v =  r
s
r
 = angle,  = angular speed,  = angular acceleration

at
ar
at = r 
tangential acceleration
ar = v2 / r radial acceleration
The radial acceleration changes the direction of motion,
while the tangential acceleration changes the speed.
Uniform Circular Motion
ar
 = constant  v and ar constant but direction changes

v
Magnetism
ar = v2/r = 2 r
KE = ½ mv2 = ½ mw2r2
F = mar = mv2/r = m2r
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Magnetism
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Radius of a Charged Particle
Orbit in a Magnetic Field
+B
+
+
v+
+
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
F
+
Magnetism
Centripetal
Force
=
Magnetic
Force
mv 2

 qvB
r

mv
r
qB
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Cyclotron Frequency
+B
+
v+
+
+
+
+
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+
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+
+
r
+
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+
F
+
Magnetism
The time taken to complete one
orbit is:
2r
T 
v
2 mv

v qB
1
qB
f  
T 2 m
qB
 c  2f 
m
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Mass Spectrometer
Smaller Mass
Magnetism
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Magnetism
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An Example
A 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
2mK
B
2 2
ed
Magnetism
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Problem Continued
From Before
r
mv
r
qB
1 2
2K
K  mv so v 
2
m
m 2K
2mK
r

d
2 2
eB m
e B
Solve for B :
2mK
B
e2d 2
Magnetism
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#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 2837
Magnetism
Problem
14.
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
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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
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?
Magnetism
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