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## AP Physics Chapter 19 Magnetism

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Title: AP Physics Chapter 19 Magnetism

1
AP Physics Chapter 19 Magnetism
2
Chapter 19 Magnetism
• 19.1 Magnets, Magnetic Poles, and Magnetic Field
Direction
• 19.2 Magnetic Field Strength and Magnetic Force
• 19.3 Electromagnetism The Source of Magnetic
Fields
• 19.4 Omitted
• 19.5 Magnetic Forces on Current-Carrying Wires
• 19.6 Applications of Electromagnetism
• 19.7 Omitted

3
Homework for Chapter 19
• HW 19.A p.626-627 6,9-13,15,24,25,28,29.
• HW 19.B p.629-63152-56,58,61,62,75,76,78,79.

4
19.1 Magnets, Magnetic Poles, and Magnetic Field
Direction
5
static cling
bar magnet
electromagnet
electric generator
6
• poleforce law or law of poles
- like magnetic poles repel each other, and
unlike magnetic poles attract each other.
• The north pole of a compass is the north-seeking
end .
• dipole - poles always occur in
pairs, never singly (monopole). If you break a
magnet in half, you end up with two smaller
dipole magnets.
• magnetic field (B) - The
direction of a magnetic field (or B field) at any
location is the direction that the north pole of
a compass at that location would point.
• Magnetic field lines always point
• from north to south .
• Magnetism and electricity are two
• aspects of a fundamental force, the
• electromagnetic force .

7
19.2 Magnetic Field Strength and Magnetic Force
8
• A magnetic field can exert forces only on moving
charges.
• When a charged particle enters a magnetic field,
the particle experiences a force that is evident
because the charge is deflected from its original
path.
• The electron beam in a cathode ray tube (made
visible by fluorescent paper) is normally
horizontal between the end electrodes but is
deflected here because of the magnet.
• We will use the convention
• X X X X
• X X X X
• is into the page and
• ? ? ? ?
• ? ? ? ?
• is out of the page.

9
On Gold Sheet
Right Hand Force Rule For a positively charged
particle the force is in the direction your palm
is facing for a negatively charged particle, the
force is in the direction of the back of your
hand.
10
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11
Example 19.1 An electron moves with a speed of
4.0 x 106 m/s along the x-axis. It enters a
region where there is a uniform magnetic field
of 2.5 T, directed at an angle of 60 to the x
axis and lying in the xy plane. Calculate the
initial force and acceleration of the electron.
12
A larger mass will have a larger radius.
Fc mac qvB mv2 r
13
Example 19.2 A proton has a speed of 4.5 x 106
m/s in a direction perpendicular to a uniform
magnetic field, and the proton moves in a circle
of radius 0.20 m. What is the magnitude of the
magnetic field?
14
• Check for Understanding
• When the ends of two bar magnets are near each
other, they attract one another. The ends must be
• one north, the other south
• one south, the other north
• both north
• both south
• either a or b
• 2. If you look directly down on the S pole of a
bar magnet, the magnetic field points
• a) to the right
• b) to the left
• c) away from you
• d) toward you

15
• Check for Understanding
• A proton moves vertically upward in a uniform
magnetic field and deflects to the right as you
watch it. What is the magnetic field direction?
• directly away from you
• directly toward you
• to the right
• to the left
• Answer b, according to the right hand rule

16
Check for Understanding
17
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18
Check for Understanding
19
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20
19.3 Electromagnetism The Source of Magnetic
Fields
21
Although a bit unlikely, the idea is that by
jarring the domains in the presence of the
earths magnetic field, they will align with it.
22
On Gold Sheet
Right-Hand Source Rule If a current-carrying
wire is grasped with the right hand with the
extended thumb pointing in the direction of the
current (I), the curled fingers indicate the
circular sense of the magnetic field direction.
23
? Since magnetic field is a vector, you must use
vector addition to find the net
field if there are contributions from
two or more sources.
24
Example 19.3 What current is required for a long
straight wire to produce a magnetic field of
magnitude equal to the strength of the Earths
magnetic field of about 5.0 x 10-5 T at a
location 2.5 cm from the wire?
25
• Example 19.5 Two long parallel wires carry
currents of 20 A and 5.0 A in opposite
directions. The wires are separated by 0.20 m.
• What is the magnetic field midway between the two
wires?
• At what point between the wires are the magnetic
fields from the two wires the same?

26
• Check for Understanding
• A long, straight wire is parallel to the ground
and carries a steady current to the east. At a
point directly below the wire, what is the
direction of the magnetic field the wire
produces?
• north
• east
• south
• west
• Answer a, according to the right hand source
rule.
• 2. A long, straight current-carrying wire is
oriented vertically. On its east side, the field
it creates points south. What is the current
direction?
• a) up
• b) down
• Answer b, according to the right hand source
rule.

27
Homework for Chapter 19
• HW 19.A p.626-627 6,9-13,15,24,25,28,29.

28
19.5 Magnetic Forces on Current-Carrying Wires
29
no current in the wire
X
X
X
30
On Gold Sheet
31
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32
a) Use the right hand force rule to determine the
direction of force. Point the thumb in the
direction of the conventional current I and the
fingers in the direction of the B-field. The
force F is the direction of the palm.
b) Here the current is flowing in the opposite
direction. Point the thumb in the direction of
the current I and the fingers in the direction of
B. F is the direction of the palm.
33
• Forces exist between two parallel
current-carrying wires. This is because the
magnetic field produced by the current in one
wire exerts a force on the other wire.
• If the currents are in the same direction, the
forces attract. If the currents are in opposite
directions, the forces repel. Use the right hand
rule for

34
Example 19.6 A wire carries a current of 6.0 A
in a direction 60 with respect to the direction
of a magnetic field of 0.75 T. Find the magnitude
of the magnetic force on a 0.50 length of the
wire.
35
• Torque on a Current-Carrying Loop
• A magnetic field can exert torque on a
current-carrying loop.
• Suppose that the loop in figure a is free to
rotate about an axis passing through opposite
sides. There are no net forces or torques on the
pivot sides of the loop. When these sides are
parallel to the B field, the force on them is
zero. At any other angle to the field, the forces
on them are equal and opposite in the plane of
the loop and so produce no net force or net
torque.
• The other two sides of the loop do produce a net
torque, which tends to rotate the loop.

36
Torque on a Current Carrying Coil ? NIAB sin
? (not on gold sheet) Where ? is the
torque N is the number of loops in the
coil I is the current A is the area of
the loop. It can be any shape. B is the
magnetic field ? is the angle between the
normal to the plane of the loop and the
B-field.
37
• Example 19.7 A circular loop of wire radius 0.50
m is in a uniform magnetic field of 0.30 T. The
current in the loop is 2.0 A. Find the magnitude
of the torque when
• the plane of the loop is parallel to the magnetic
field,
• the plane of the loop is perpendicular to the
magnetic field,
• the plane of the loop is at 30 to the magnetic
field.

38
Example 19.8 Two long, straight wires separated
by a distance of 0.30 m carry currents in the
same direction. If the current in one wire is 10
A and the current in the other 8.0 A, find the
magnitude and direction of the forces per unit
length (per meter) between the wires. What if the
currents are in opposite directions?
39
• Check for Understanding
• A long, straight, horizontal wire is located on
the equator and carries a current directed toward
the east. What is the direction of the force on
it due to the Earths magnetic field?
• east
• west
• south
• upward
• Answer d, according to the right hand force
rule

40
Check for Understanding
41
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42
Check for Understanding
43
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44
19.6 Applications of Electromagnetism
45
electron is not moving
no current in the wire
X
X
46
• The dc Motor
• An electrical motor is a device that converts
electrical energy into mechanical energy.
• A pivoted, current carrying coil with N loops in
a magnetic field will rotate freely, but for only
a half-cycle, or through a maximum angle of 180.
• Recall that ? NAIB sin ?, and when the
magnetic field is perpendicular to the plane of
the coil (sin ? 0), the torque is zero and the
coil is in equilibrium.
• To provide for continuous rotation, the current
is reversed every half turn so that the
torque-producing forces are reversed. This is
done by means of a split-ring commutator.
• Contact brushes provide a path for current.

47
• When current is supplied, one half-ring is
electrically positive and the other negative. The
coil and ring rotates.
• When the coil and ring have gone through half a
rotation, the half rings come in contact with the
opposite brushes. Their polarity is reversed, and
the current in the coil is in the opposite
direction. This changes the directions of the
torque. The process repeats and the spinning
continues.

For a real motor, the rotating shaft is called
the armature.
48
• Cathode Ray Tube (CRT)
• The cathode ray tube is a vacuum tube that is
used in an oscilloscope, such as those in some
laboratories.
• Electrons, negatively charged, are boiled off
a hot filament in an electron gun and accelerated
by a voltage applied between the cathode (-) and
anode ().
• The picture tube in older television sets and
computer monitors is also a cathode ray tube.
• Magnetic coils are usually used there to deflect
the electron beam.

Open this Cathode Ray Tube Simulation in
Explorer http//highered.mcgraw-hill.com/olcweb/
cgi/pluginpop.cgi?itswf100100/sites/dl/fr
ee/0072512644/117354/01_Cathode_Ray_Tube.swfCath
ode Ray Tube
49
• The Mass Spectrometer
• A mass spectrometer is a device used to measure
the mass of atoms or molecules. It is often used
to separate isotopes, or atoms of different
masses.
• Actually, the masses of ions are measured since
electric and magnetic fields have motional
effects only on charged particles. (An ion is an
atom or molecule with a net electric charge.)
• Ions with a known charge (q) are produced by
heating the substance to be analyzed.

50
• The resulting beam of ions introduced into the
mass spectrometer has a distribution of speeds.
Ions with a particular velocity are selected by
means of a velocity selector, made up of charged
plates and a magnetic field that allow particles
traveling at only that velocity to go
undeflected.
• The values of the E and B fields between the
plates of the velocity selector determine the
velocity.
• For a positively charged ion, the E-field
produces a downward force F qE. The B-field
produces an upward force F qvB1.
• If the beam is not deflected, the resultant
force must be zero, so
• qE qvB1
• or v E
• B1
• If the plates are parallel, E V/d. Since the
voltage
• and plate separation are controllable
quantities, a
• more practical version of the equation is
• v V ion speed in a velocity selector
• B1d

51
• Beyond the velocity selector, the beam passes
through a slit into another magnetic field (B2),
which is perpendicular to the direction of the
beam.
• The force due to this magnetic field (F qvB2)
is always perpendicular to the velocity of the
ions, which are therefore deflected along a
circular path.
• The magnetic force supplies the centripetal
force for this motion, and
• mv2 qvB2 so, m qdB1B2
r particle mass via
• r V mass spectrometer

52
• Example 19.9 In a mass spectrometer, a
single-charged particle has a speed of 1.0 x 106
m/s and enters a uniform magnetic field of 0.20
T. The radius of the circular orbit is 0.020 m.
• What is the mass of the particle?
• What is the kinetic energy of the particle?

53
• Check for Understanding
• A mass spectrometer
• can be used to determine the masses of atoms and
molecules
• requires charged particles
• can be use to determine relative abundances of
isotopes
• all of these
• 2. Why can a nearby magnet distort the display of
a computer monitor or television picture tube?
• Answer The magnetic force on the electron beam,
which prints pictures, causes the deflection of
the electrons.

54
Homework for Chapter 19
• HW 19.B p.629-63152-56,58,61,62,75,76,78,79.

55
Chapter 19 Formulas
56
v E V ion speed in a
velocity selector B1 B1d m
qdB1B2 r qB2r particle mass via V
v mass spectrometer