Transcript Section 1

2.2 MOTION OF CHARGED PARTICLES IN
MAGNETIC FIELDS
Magnetic Force on a Particle
The force on a charged particle moving through a
magnetic field is given by:
𝐹 = π‘žπ‘£π΅π‘ π‘–π‘›πœƒ
𝐹 = force (newtons, 𝑁)
π‘ž = charge of particle (coulombs, 𝐢)
𝑣 = velocity of particle (metre per second, π‘š. 𝑠 βˆ’1 )
𝐡 = magnetic field (tesla, 𝑇)
πœƒ = angle between the field and velocity (degrees, °)
Magnetic Force on a Particle
Hence, in a uniform magnetic field, there is no magnetic
force on a particle if:
The particle has no
charge
The particle is
stationary
(𝑣 = 0)
The particle moves
parallel to the field
(πœƒ = 0° or 180°)
Right Hand Rule (for Charged Particles)
B magnetic field
F force
(negative charge )
Out of back of hand
F force
(positive charge)
Out of palm
v velocity
Example 2
State the direction of the force on the particle in
each of the following situations.
up
out of
the page
out of
the page
no force
Example 1
Calculate the force on an electron fired at
4.0 × 106 π‘š. 𝑠 βˆ’1 at 60° to a magnetic field
with strength 1.5T.
v
-
B
60°
Motion of a Charged Particle at πŸ—πŸŽ° to the Field
A charged particle will experience a force of constant
magnitude at right angles to its velocity.
This force provides a centripetal acceleration,
making the particle move in uniform circular motion.
Motion of a Charged Particle at πŸ—πŸŽ° to the Field
Motion in Magnetic Field Simulation
Motion of a Charged Particle at πŸ—πŸŽ° to the Field
F
F
F
Loop generating
the magnetic field
Derivation: Radius of the Circular Path
πΉπ‘π‘–π‘Ÿπ‘π‘’π‘™π‘Žπ‘Ÿ = πΉπ‘šπ‘Žπ‘”π‘›π‘’π‘‘π‘–π‘
π‘šπ‘Ž = π‘žπ‘£π΅π‘ π‘–π‘›πœƒ
π‘šπ‘£ 2
= π‘žπ‘£π΅π‘ π‘–π‘›πœƒ
π‘Ÿ
π‘šπ‘£
= π‘žπ΅ π‘€β„Žπ‘’π‘› πœƒ = 90°, π‘ π‘–π‘›πœƒ = 1
π‘Ÿ
π‘šπ‘£ = π‘žπ΅π‘Ÿ
π‘šπ‘£
π‘Ÿ=
π‘žπ΅
Note: Increasing π‘š or 𝑣 will
increase the radius. Increasing π‘ž
or 𝐡 will decrease the radius.
Example 3
An electron enters a 0.4 T
uniform magnetic field moving
at 2.0 × 108 m.s-1 as shown.
a) Find the force it experiences
as it enters the field.
F
b) Draw the path taken by the
electron on the diagram.
c) Find the radius of the path
it takes.
Example 4
A
positive
B
negative
D
C
negative
no charge
Particles moving at other angles to the field
Particles moving at other angles to the field
The particle only
experiences a force
perpendicular to the
field.
Therefore, its motion
parallel to the field
remains constant.
Application: The Cyclotron
A cyclotron is a device used to accelerate ions
(charged particles) to high speeds.
Application: The Cyclotron
These ions can then be
used to create
radioactive isotopes.
The isotopes are used
for medical diagnosis
and treatment
(commonly PET scans).
SAHMRI in Adelaide
A vial of radioactive glucose produced
by the $4 million cyclotron.
evacuated chamber
ion source
magnet
magnetic
field
D-shaped
electrodes
(dees)
magnet
target
electric field
(between
the dees)
deflecting
magnet
Ion Production
A heated filament
produces electrons by
thermionic emission
1 electron less
positive ion
Ion Production
Modern cyclotrons use an
electric arc (a plasma) to
produce the ions.
How a Cyclotron Works
Ion source
The magnetic field
bends the path of
the ion into a
semi-circle.
The electric field
accelerates the
ion each time it
crosses the gap.
The dees. One is
positively charged
and the other is
negatively charged.
How a Cyclotron Works
Every time the ion crosses
the electric field (the gap
between the dees), it is
accelerated and gains
kinetic energy.
βˆ†πΎ = π‘Š
= π‘žβˆ†π‘‰
Example 5
dees
Find the direction of
the magnetic field if
the ion source is
(a) positive
out of the page
(b) negative
into the page
F
v
Practice
Find the direction of the magnetic field if the ion source is
positive.
magnetic field
is into the page
F
v
Derivation: Period of the Ion’s Circular Motion
π‘Ÿ=
π‘šπ‘£
π‘žπ΅
but
2πœ‹π‘Ÿ
π‘š
𝑇
βˆ΄π‘Ÿ=
π‘žπ΅
π‘š2πœ‹π‘Ÿ
βˆ΄π‘Ÿ=
π‘žπ΅π‘‡
2πœ‹π‘šπ‘Ÿ
∴ π‘Ÿπ‘‡ =
π‘žπ΅
2πœ‹π‘š
βˆ΄π‘‡=
π‘žπ΅
𝑣=
2πœ‹π‘Ÿ
𝑇
This shows that the time it
takes the ions to circle the
cyclotron once is independent
of their speed.
Hence, the potential difference
alternates at a constant frequency
𝑓=
1
𝑇
=
π‘žπ΅
2πœ‹π‘š
Derivation: Kinetic Energy of Emerging Ions
π‘šπ‘£
π‘žπ΅π‘Ÿ
π‘Ÿ=
βŸΉπ‘£=
π‘žπ΅
π‘š
1
𝐾 = π‘šπ‘£ 2
2
1
π‘žπ΅π‘Ÿ
∴𝐾= π‘š
2
π‘š
2
2 2 2
π‘š π‘ž 𝐡 π‘Ÿ
∴𝐾= ×
2
π‘š2
π‘ž2 𝐡2 π‘Ÿ 2
∴𝐾=
2π‘š
This shows that the final
kinetic energy of the ions
does not depend on the
electric field but rather the
magnetic field and the radius
of the cyclotron.