Aurora or polar
lightis
observable in a
few places on
Earth close to
the Arctic and
Antarctic
regions.
2
3.
3
Auroras sway alongthe vibrating field lines of the earth, as
a result of electrons being flung to Earth, then off the
planet’s magnetosphere, and back again.
It is a demonstration of how charged particles behave in a
magnetic field.
4.
4
What happens toa charged
particle as it moves in a
magnetic field?
5.
5
● interaction betweenelectrically-charged particles
resulting from their motion
● causes the attraction and repulsion of magnetic
materials
● force that results from the interaction between the
different magnetic fields formed by moving charges
Magnetic Force
6.
6
Magnetic force hasfour distinct characteristics:
1. The magnitude of the magnetic force is proportional to
the magnitude of the charge.
2. The magnitude of the magnetic force is proportional to
the magnitude of the magnetic field.
Magnetic Force on Moving Charges
7.
7
Magnetic force hasfour distinct characteristics:
3. The magnetic force is affected by the charged particle’s
velocity.
4. Magnetic force is perpendicular to the field and the
velocity.
Magnetic Force on Moving Charges
8.
8
Magnetic Force onMoving Charges
Perpendicularity of magnetic force to velocity and the magnitude of
the magnetic field
13
When two chargedparticle that have
equal but opposite magnitudes are
moving with equal velocity in the same
field, the forces on the charged
particles are also equal in magnitude,
but opposite in direction.
11.
14
Magnetic Force
Calculations
The previousequation can
be used to derive the
equation for the magnetic
force on a wire carrying a
current I.
Magnetic Force on Moving Charges
A current-carrying wire
16
● The SIunit for magnetic field is tesla
(T).
● Another common unit for field is
gauss (G), equivalent to 10-4 T.
● Magnetic force is measured in
newtons (N).
14.
17
An 8.12-μC chargedparticle with a speed of 40.0 m/s
is found in a uniform magnetic field with magnitude
0.75 T. Determine the magnitude of the magnetic
force exerted on the charged particle if the particle is
moving perpendicular to the field.
15.
18
An 8.12-μC chargedparticle with a speed of 40.0 m/s
is found in a uniform magnetic field with magnitude
0.75 T. Determine the magnitude of the magnetic
force exerted on the charged particle if the particle is
moving perpendicular to the field.
The magnitude of the magnetic force is equal to
2.44✕10-4 N.
16.
19
19
A 7.55-μC chargedparticle with a speed
of 30.50 m/s is found in a uniform
magnetic field with magnitude 1.2 T.
Solve for the magnitude of the magnetic
force exerted on the charged particle if
the particle is moving perpendicular to
the field.
17.
20
A 100-cm wirethat carries an electric current of 0.25
A perpendicular to the field has experienced a
magnetic force of 5.50✕10-3 N. Solve for the
magnitude of the magnetic field that caused this
force.
18.
21
A 100-cm wirethat carries an electric current of 0.25
A perpendicular to the field has experienced a
magnetic force of 5.50✕10-3 N. Solve for the
magnitude of the magnetic field that caused this
force.
The magnitude of the magnetic field is 0.022 T.
19.
22
22
An 88.30-cm wirethat carries an electric
current of 1.35 A perpendicular to the
field has experienced a magnetic force
of 9.0✕10-3 N. Solve for the magnitude of
the magnetic field that caused this
force.
20.
23
A 1.13✕10-19-C negativecharge is travelling with a
speed of 3.5✕107 m/s through a uniform field with
magnitude 0.076 T. The charge has a mass of 1.95✕10-
31 kg. Determine the angle the velocity vector makes
with the field if the particle’s acceleration is
1.25✕1017 m/s2.
21.
24
A 1.13✕10-19-C negativecharge is travelling with a
speed of 3.5✕107 m/s through a uniform field with
magnitude 0.076 T. The charge has a mass of 1.95✕10-
31 kg. Determine the angle the velocity vector makes
with the field if the particle’s acceleration is
1.25✕1017 m/s2.
The angle between the velocity vector and the
magnetic field is 4.65°.
22.
25
25
A 2.5✕10-19-C negativecharge is
travelling with a speed of 4.0✕107 m/s
through a uniform field with magnitude
0.03 T. The charge has a mass of 5.45✕10-
31 kg. Determine the angle the velocity
vector makes with the field if the
particle’s acceleration is 2.3✕1017 m/s2.
23.
26
How do wedetermine the
magnetic force acting on a
moving charge?
24.
27
Motion of ChargedParticles in
a Magnetic Field
● Newton’s Laws of Motion is
still applicable
● “Particle’s motion as
influenced by the field alone
may be treated with constant
speed and kinetic energy.”
Magnetic Force on Moving Charges
Circular motion of a charged
particle in a uniform
magnetic field
30
Motion of ChargedParticles in a Magnetic Field
● if charge is positive, the particle’s revolution is
counterclockwise
● if charge is negative, the particle revolves clockwise
around the orbit
Magnetic Force on Moving Charges
28.
31
Motion of ChargedParticles in a Magnetic Field
● Cyclotron radius (gyroradius or radius of
gyration) is the radius of the circular path traveled
by the particle in a uniform magnetic field
● Cyclotron frequency is the number of revolutions
per unit time taken by the particle around the path
Magnetic Force on Moving Charges
29.
32
Motion of Charged
Particlesin a Magnetic
Field
A cyclotron can accelerate
subatomic charged particles
using an alternating electric
field (in a spiral path).
Magnetic Force on Moving Charges
Cyclotron
30.
33
Uniform vs. Non-UniformMagnetic Fields
Magnetic Force on Moving Charges
A helical uniform magnetic field and a non-uniform field (magnetic
bottle)
31.
34
Magnetic Force onMoving Charges
Uniform vs. Non-Uniform Magnetic Fields
● In a uniform magnetic field, the field lines are
parallel to each other
● In a non-uniform magnetic field, field lines may
converge or diverge.
● An example of a non-uniform field is the magnetic
field of Earth
32.
35
How does magneticforce affect
the speed, and kinetic energy of
a moving particle?
33.
36
Identify whether eachstatement is true or false.
1. Magnetic force results from the interaction between
electrically-charged particles that are stationary.
1. The magnitude of the magnetic force is proportional to
the magnitude of the magnetic field.
1. When the charged particle is negative, the direction of
the magnetic force is similar to the one the right-hand
rule suggests.
34.
37
Identify what isbeing referred to in each statement.
1. The charged particle’s motion as influenced by the field
alone may be treated as motion with __________ speed
and kinetic energy.
2. The motion of a charged particle as influenced by a
magnetic field is characterized as __________
acceleration.
3. The radius of the circular path traversed by the particle
in a uniform magnetic field is called the __________ .
35.
38
● Magnetic force(measured in newtons) is the
interaction between electrically-charged particles
resulting from their motion. It causes the
attraction and repulsion of magnetic materials.
● The magnitude of the magnetic force is
proportional to the magnitude of the charge.
36.
39
● The magnitudeof the magnetic force is
proportional to the magnitude of the magnetic
field.
● The magnetic force is affected by the charged
particle’s velocity.
● The magnetic force and the magnetic field do not
have similar directions. Instead, magnetic force is
perpendicular to the field and the velocity.
37.
40
● A particle’smotion as influenced by the field alone
may be treated as motion with constant speed
and kinetic energy.
● The radius of the circular path traversed by the
particle in a uniform magnetic field is called the
cyclotron radius. It is also called gyroradius or
radius of gyration.
38.
41
● The numberof revolutions per unit of time taken
by the particle around the circular path is called
the cyclotron frequency.
● Uniform magnetic fields produce parallel field
lines. Non-uniform fields have unequal
distributions of field magnitudes throughout, thus
their field lines are either diverging or converging.
39.
42
Concept Formula Description
Magnetic
Force
where
●is the magnetic force on a moving
charged particle
● q is the particle’s charge
● is the particle’s velocity
● is the magnetic field
● is the angle between the velocity
vector and the field vector
Use this formula
to solve for the
magnetic force
of a moving
charged particle.
40.
43
Concept Formula Description
Magnetic
Forcein a
Uniform Wire where
● F is the magnitude of the
magnetic force
● B is the magnitude of the
magnetic field
● I is the current flowing through
the wire
● L is the length of the wire
● is the angle between the velocity
vector and the field vector
Use this formula
to solve for the
magnetic force in
a uniform wire.