Download Physics 21

Physics
Chapter 21:
Magnetism
Magnetism
☺Magnets
☺Caused by the Polarization of Iron
Molecules
☺Material Containing Iron (Fe)
Magnetism
☺Polarization
☺The Alignment of Charged Molecules Within
the Material
N
S
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S
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N
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S
N
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N
S
Magnetism
☺Magnetic Field (B)
☺Surrounds Magnet
☺Possess Magnitude and Direction
☺“The Direction of the Magnetic Field at
Any Point in Space is the Direction
Indicated by the North Pole of a Small
Compass Needle Placed at that Point”
Magnetism
☺Magnetic Field (B)
Magnetism
☺Magnetic Field
☺Magnetic Field Lines
Currently 11.50 Between Geographic
North Pole and the Magnetic North Pole
Magnetism
☺Electromagnetism
☺Current Passing through a Wire
Creates a Magnetic Field
Magnetism
☺Electromagnetism
Magnetism
☺Electromagnetism
☺Right Hand Rule
Magnetism
☺Electromagnetism
☺Single Loop
Magnetism
☺Electromagnetism
☺Multiple Loops
Magnetism
☺Forces in Magnetic Fields
☺Magnetic Forces on a Charge
☺The Charge Must Be Moving
☺Magnetic Force Does Not Act On a Stationary
Charge
☺The Velocity of the Moving Charge Must
Have a Component that is Perpendicular to
the Direction of the Magnetic Field (Right
Hand Rule)
Magnetism
☺Forces in Magnetic Fields
☺Right Hand Rule (1)
☺Extend Your Right Hand so Your Fingers
Represent the Direction of the Magnetic Field and
Your Thumb Represents the Velocity of the Charge
☺Your Palm Then Indicates the Direction of the
Maximum Magnetic Force Acting on a Positive
Charge
☺If the Charge is Negative, the Back of Your Hand
Indicates the Direction of the Maximum Magnetic
Force Acting on the Charge
Magnetism
☺Forces in Magnetic Fields
Magnetism
☺Forces in Magnetic Fields
☺Magnitude
☺F = Magnitude of Magnetic Force on the Test Charge
☺q = Test Charge
☺v = Velocity of the Test Charge
☺q = Angle of the Test Charge Velocity in Relation to the
Magnetic Force
F
B
q(v sin q )
Magnetism
☺Forces in Magnetic
Fields
☺Magnitude (Units)
F
(newton )(second )

 Tesla (T )
q(v sin q ) (coulomb )( meter )
F
B
q(v sin q )
Magnetism
☺Forces in Magnetic Fields
☺Magnitude (Units)
☺Tesla
☺A Test Charge Perpendicular to B
Experiences 1N @ 1m/s
☺For Much Smaller Quantities
☺1 Gauss = 1x10-4 Tesla
F
B
q(v sin q )
Magnetism
☺ The Motion of a Charged Particle in a Magnetic Field
☺ Remember: In an Electric Field (like a parallel plate capacitor) a
Positively Charged Particle Entering Perpendicular to the Field
will Move Toward the Negatively Charged Plate
☺ In a Magnetic Field, a Positively Charged Particle Entering
Perpendicular to the Field will Move Between the Sources of the
Field due to the Horizontal Position of the Magnetic Force
Particle Will Come Out of the Screen Toward You
++++++++++++++++
N
v
E
F
----------------
q0
F
v q
0
B
S
Magnetism
☺The Motion of a Charged Particle in a
Magnetic Field
☺While an Electric Field will Accelerate a
Charged Particle Placed Stationary Within the
Field, a Magnetic Field Cannot Change the
Kinetic Energy of the Particle
☺In A Magnetic Field, Particle Speed is
Constant
Magnetism
☺ The Motion of a Charged Particle in a Magnetic Field
☺ Circular Trajectory
☺ The Magnetic Force is Perpendicular to the Magnetic Field
☺ RHR1 Suggests a Change in Trajectory of a Particle Due to the
Perpendicular Force
☺ This Causes the Force to Always be Directed to the Center of a
Circular Path of the Particle
☺ Centripetal Force is then Exerted on the Particle (Fc)
Magnetism
☺ The Motion of a Charged Particle in a Magnetic Field
☺ Circular Trajectory
x
x
x
x
x
x
x
x
x
x
F x
x
x +q0 x
x
+q0 x
Magnetism
☺ The Motion of a Charged
Particle in a Magnetic
Field
☺ Circular Trajectory
mv
Fc 
r
2
x
x
x
x
x
x
x
x
x
x
F x
x
x +q0 x
x
+q0 x
Magnetism
☺ The Motion of a Charged
Particle in a Magnetic
Field
☺ Circular Trajectory
mv
Fc 
r
2
mv
r
qB
x
x
x
x
x
x
x
x
x
x
F x
x
x +q0 x
x
+q0 x
Magnetism
☺The Force on a Current in a Magnetic Field
☺Current Traveling through a Wire Within a
Magnetic Field is Affected by the Field
Magnetism
☺The Force on a Current in a Magnetic Field
☺Force Magnitude
☺I = Current
☺L = Length of Wire in Meters
☺B = Magnetic Field Magnitude
F  BIL sin q
Magnetism
☺The Force on a Charged Particle in a
Magnetic Field
F  BIL sin q
qv
I
L
 qv 
F  B  L sin q
 L
F  Bqv sin q
Magnetism
☺Problem
☺Conventional current is running through a
wire. Sketch the magnetic field that the
current generates.
Magnetism
☺Solution
Magnetism
☺Problem
☺This figure shows the end view of an
electromagnet with the current as shown.
What is the direction of the magnetic field
inside the loop and outside the loop?
Magnetism
☺Solution
Magnetism
☺Problem
☺A current-carrying wire is placed between the
poles of a magnet. What is the direction of the
force on the wire?
Magnetism
☺Solution
Magnetism
☺Problem
☺The current through a wire 0.80m long is 5.0A.
The wire is perpendicular to a 0.60T magnetic
field. What is the magnitude of the force on the
wire?
Magnetism
☺Solution
F  BIL sin q  (0.60T )(5.0 A)(0.80m)  2.4 N
Magnetism
☺Problem
☺A wire 35cm long is parallel to a 0.53T uniform
magnetic field. The current through the wire is
4.5A. What force acts on the wire?
Magnetism
☺Solution
☺The wire is parallel to the magnetic field. No
force acts on the wire.
☺q  0
F  BIL sin q  (0.53T )( 4.5 A)(0.35m)(sin 0)  0
Magnetism
☺Problem
☺A wire 625m long is in a 0.40T magnetic field.
A 1.8N force acts on the wire. What current is
in the wire?
Magnetism
☺Solution
☺L = 625m
☺B = 0.40T
☺F = 1.8N
☺q  90 (assume perpendicular if angle is not
given)
F
1.8 N
I

 7.2mA
BL sin q (0.40T )(625m)(sin 90)
Magnetism
☺Problem
☺The force on a 0.80m wire that is perpendicular
to Earth’s magnetic field is 0.12N. What is the
current in the wire?
Magnetism
☺Solution
☺L = 0.80m
☺B = 5x10-5T
☺F = 0.12N
☺q  90
F
0.12 N
I

 3kA
5
BL sin q (5 x10 T )(0.80m)(sin 90)
Magnetism
☺Problem
☺A power line carries a 225A current from east
to west parallel to the surface of Earth. What is
the magnitude and direction of the force
resulting from Earth’s magnetic field acting on
each meter of the wire?
Magnetism
☺Solution
☺L = 1m
☺B = 5x10-5T
☺I = 225A
☺q  90
F
5
 IB  (225 A)(5 x10 T )  0.011N / m
L
Magnetism
☺Problem
☺A beta particle (high-speed electron) is
traveling at right angles to a 0.60T magnetic
field. It has a speed of 2.5x107 m/s. What size
force acts on the particle?
Magnetism
☺Solution
☺B = 0.60T
☺q = 1.6x10-19C
☺v = 2.5x107 m/s
F  Bqv  (0.60T )(1.6 x1019 C )(2.5x107 m / s)  2.4 x1012 N
Magnetism
☺Problem
☺A magnetic field of 16T acts in a direction due
west. An electron is traveling due south at
8.1x105 m/s. What are the magnitude and
direction of the force acting on the electron?
Magnetism
☺Solution
☺B = 16T
☺q = 1.6x10-19C
☺v = 8.1x105 m/s
F  Bqv  (16T )(1.6 x1019 C )(8.1x105 m / s)  2.1x1012 N
☺Upward, remember that electron flow is opposite of
current flow.
Magnetism
☺Problem
☺A wire carrying 15A of current has a length of
25cm in a magnetic field of 0.85T. What is the
force on the wire if it makes an angle with the
magnetic field lines of 45°?
Magnetism
☺Solution
☺B = 0.85T
☺L = 0.25m
☺I = 15A
F  BIL sin q  (0.85T )(15 A)(0.25m) sin 45  2.3N
Magnetism
☺Problem
☺A magnet attracts a nail, which, in turn,
attracts many small tacks. If the N-pole of the
permanent magnet is the top face, which end of
the nail is the N-pole?
N
S
Magnetism
☺Solution
☺Down
N
S
N
S
S
N
Magnetism
☺Problem
☺A student makes a magnet by winding wire
around a nail and connecting it to a battery, as
shown in the figure below. Which end of the
nail, the pointed end or the head, will be the
north pole?
Magnetism
☺Solution
☺The Pointed End
S
N
Magnetism
☺Problem
☺A wire 0.50m long carrying a current of 8.0A is
at right angles to a 0.40T magnetic field. How
strong a force acts on the wire?
Magnetism
☺Solution
☺B = 0.40T
☺L = 0.50m
☺I = 8.0A
F  BIL sin q  (0.40T )(8.0 A)(0.50m) sin 90  1.6 N
Magnetism
☺Problem
☺A wire 75cm long carrying a current of 6.0A is
at right angles to a uniform magnetic field. The
magnitude of the force acting on the wire is 0.60
N. What is the strength of the magnetic field?
Magnetism
☺Solution
☺F = 0.60N
☺L = 0.75m
☺I = 6.0A
F
(0.60 N )
B

 0.13T
IL sin q (6.0 A)(0.75m) sin 90
Magnetism
☺Problem
☺A copper wire 40cm long carries a current of
6.0A and weighs 0.35N. A certain magnetic
field is strong enough to balance the force of
gravity on the wire. What is the strength of the
magnetic field?
Magnetism
☺Solution
☺F = 0.35N
☺L = 0.4m
☺I = 6.0A
F
(0.35 N )
B

 0.145T
IL sin q (6.0 A)(0.4m) sin 90
Magnetism
☺Problem
☺An electron passes through a magnetic field at
right angles to the field at a velocity of 4.0x106
m/s. The strength of the magnetic field is 0.50T.
What is the magnitude of the force acting on
the electron?
Magnetism
☺Solution
☺B = 0.50T
☺q = 1.6x10-19C
☺v = 4x106 m/s
F  Bqv  (.50T )(1.6 x10
19
C )(4 x10 m / s)  3.2 x10
6
13
N
Magnetism
☺Problem
☺Doubly ionized helium atoms (alpha particles)
are traveling at right angles to a magnetic field
at a speed of 4.0x10–2 m/s. The force that acts
on each particle is 6.4x10-22N. What is the
magnitude of the magnetic field?
Magnetism
☺Solution
☺F = 6.4x10-22N
☺q = 1.6x10-19C (x2)
☺v = 4x10-2 m/s
F
6.4 x1022 N
2
B


5
x
10
T
19
2
qv (2)(1.6 x10 C )( 4 x10 m / s)
Magnetism
☺Problem
☺A force of 5.78x10–16 N acts on an unknown
particle traveling at a 90° angle through a
magnetic field. If the velocity of the particle is
5.65x104 m/s and the field is 3.20x10–2 T, how
many elementary charges does the particle
carry?
Magnetism
☺Solution
☺F = 5.78x10-16N
☺B = 3.20x10-2T
☺v = 5.65x104 m/s
F
5.78x1016 N
19
q

 3.20 x10 C
2
4
Bv (3.20 x10 T )(5.65 x10 m / s)
3.2 x10
19
e


C
2
19
 1.6 x10 C 
Magnetism
☺Homework:
☺Pages 782-783
☺31 (2.1x10-3m/s)
☺32 (0.3T)
☺38 (1.9x1014m/s2)