Download Magnetic Fields - Madison Public Schools

2/13/2017
AP PHYSICS 2
Over 2,500 years ago, ancient Chinese civilization
discovered that certain rocks - now called lodestones will attract each other, as well as pick up small bits of
iron.
UNIT 5
Magnetism and
Electromagnetic
Induction
CHAPTER 17
MAGNETISM
They soon learned that these rocks could
hung from a string and used for navigation.
PERMANENT MAGNETS
BAR MAGNET
N
S
HORSESHOE
MAGNET
N
be
Magnetic interaction
 If you bring the like poles of two magnets near
each other, they repel each other.
 If you bring opposite poles near each other, they
attract each other.
S
Magnetic interaction
 Magnets always have two poles.
 If you break a magnet into two pieces, each
piece still has two poles—a north pole and a
south pole.
Another way of stating this is that
there can be no magnetic monopoles
The physical laws of magnetism prove
that a single magnetic pole cannot exist.
All magnets will always have both a North
and a South pole.
1
2/13/2017
Magnetic and electrical interactions
are different
 Electrically charged objects do not interact with
magnets in the same way that magnets interact
with magnets.
 Magnetic poles are not electric charges.
Magnetic Fields
Just as electrically charged particles create
an electric field in the space around them,
magnets create a magnetic field that
distorts space around them.
E
How can we detect a magnetic field?
We need a test magnet.
COMPASS:
small magnet that has the
ability to rotate and always
align with a magnetic field.
B
Magnetic interaction
 A compass contains a
tiny magnet on a lowfriction pivot.
 The north pole of a
compass points toward
geographic north; the
south pole points toward
geographic south.
A compass measures the
direction of a magnetic field!
2
2/13/2017
What creates a
Magnetic Field?
Both orbital motion and the spinning
motion of every electron in an atom
produce magnetic fields.
2
What materials can be
magnetized?
Nickel
Cobalt
Iron
How can you make a
permanent magnet
weaker?
Heat it
Hit it
What creates a
Magnetic Field?
Magnetic fields are created by
ELECTRICALLY CHARGED
PARTICLES IN MOTION!
The fundamental nature of all magnetism is
the motion of charged particles.
How can you make a
permanent magnet?
Rub Fe-Ni-Co against a
strong magnet
Place Fe-Ni-Co in a
strong magnetic field
SOURCES OF
MAGNETIC FIELD
Permanent Magnets
Earth
Current-carrying wires
3
2/13/2017
SOURCE 1: PERMANENT MAGNETS
(MAGNETIC DOMAINS)
SOURCE 2: EARTH (currents in the
molten core of Earth)
Clusters of aligned atoms !!
Domains before
magnetization
Domains after
magnetization
SOURCE 3: Electric Current
carrying wires
Hans Christian Øersted
(1777 - 1851)
It’s all about the Øersteds.
DIRECTION OF MAGNETIC FIELD
SOURCE 1: Permanent Magnets
DIRECTION OF
MAGNETIC FIELD
Magnetic
Field Lines
The magnetic field lines of a permanent bar magnet points from
North to South. Easy way to remember: “Magnetic fields fly
South”
© Mrs. Rawding
4
2/13/2017
DIRECTION OF MAGNETIC FIELD
DIRECTION OF MAGNETIC FIELD
SOURCE 1: Permanent Magnets
SOURCE 1: Permanent Magnets
The north side of
a compass
needle will point
on the direction
of the magnetic
field
DIRECTION OF MAGNETIC FIELD
Direction of the magnetic field
SOURCE 1: Permanent Magnets
 We can use a compass to detect the direction of
the magnetic field at a particular location.
DIRECTION OF MAGNETIC FIELD
DIRECTION OF MAGNETIC FIELD
SOURCE 1: Permanent Magnets
SOURCE 2: Permanent Magnets
BAR MAGNETS
BAR MAGNET
S
N
N
S
S
N
5
2/13/2017
DIRECTION OF MAGNETIC FIELD
DIRECTION OF MAGNETIC FIELD
SOURCE 2: Permanent Magnets
SOURCE 1: Permanent Magnets
BAR MAGNETS
S
N
S
N
 We use a horseshoe magnet to generate a
magnetic field with almost parallel field lines
between the poles.
DIRECTION OF MAGNETIC FIELD
DIRECTION OF MAGNETIC FIELD
SOURCE 2: EARTH
SOURCE 3: Electric Current carrying wires
Earth acts as a
giant magnet, with
its magnetic south
pole close to its
geographic north
pole and its
magnetic north pole
close to its
geographic south
pole.
DIRECTION OF MAGNETIC FIELD
SOURCE 3: Electric Current carrying wires
A currentcarrying wire
creates a
magnetic field
that is in the
shape of
concentric
circular loops
around the wire!
A current-carrying wire contains electrons in
motion, and thus creates a magnetic field!
Magnetic Field
(B)
I
Electric current
(I)
6
2/13/2017
DIRECTION OF MAGNETIC FIELD
DIRECTION OF MAGNETIC FIELD
SOURCE 3: Electric Current carrying wires
SOURCE 3: Electric Current carrying wires
 Charged objects in
motion produce a
magnetic field; stationary
charged objects do not.
 The method for
determining the shape of
the B field produced by
the electric current in a
wire is called the righthand rule.
Right Hand Rule # 1
I
B
Point the thumb of
your right hand in
the direction of the
electric current
Your fingers will
wrap around the way
that the magnetic
field wraps around
the wire!
PHYSICAL QUANTITY:
MAGNETIC FIELD
• Definition:
• Symbol:
Region where a magnetic force
can be detected.
B
• Units:
B=
Tesla (T)
• Type of PQ:
0I
(2r)
Vector
1. Alternating Current
Working in Three Dimensions
2. Light
3. X-rays
4. Radio
5. Remote Control
6. Electric Motor
7. Robotics
This is an arrow pointing
out of the page.
This is an arrow pointing
into the page.
8. Laser
9 Wireless Communications
10. Limitless Free Energy
(Think of the tip of an
arrow pointing at you)
(Think of the tail of an arrow
pointing away from you)
11. Radar
7
2/13/2017
WHITEBOARD
I
What does the magnetic field
look like on each side of the wire?
I
B (into the paper)
WHITEBOARD
Draw the magnetic field lines that
show the B field surrounding the
current-carrying wire.
WHITEBOARD
Draw the magnetic field lines that
show the B field surrounding the
current-carrying wire.
I
I
The magnetic field will be tangent
to the magnetic field lines
B (out of the paper)
Whiteboard Warmup
Sketch the magnetic field of a loop of wire from a
cross-sectional view. (Imagine a donut cut in half
and looked at from the side)
I
I
8
2/13/2017
Use RHR #1 for each section of the loop, and then
use the Principle of Superposition!
Superposition Whiteboard
Two wires carrying equal currents are crossed, as shown above.
Determine the magnetic field in each of the labeled regions.
Current Events
Two parallel wires are each carrying a current of 0.8 Ampères
upward, as shown below. Calculate the magnitude and direction
of the magnetic field at points A, B and C shown below.
B=0T
10 cm
10 cm
6 cm 4 cm
B=0T
μ0 = 1.3 x 10-6 T*m/A
B=
10 cm
10 cm
m0 I
2p r
6 cm 4 cm
Vector superposition in
the third dimension!
BA = 2.2 x 10-6 T out of the page
BB = 0 T
BC = 3.1 x 10-6 T out of the page
The magnetic field produced by an electric
current in a long straight wire
 The magnitude of the magnetic field at a
perpendicular distance r from a long straight
current-carrying wire is expressed as:
𝐵𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑤𝑖𝑟𝑒 =
𝜇0 𝐼𝑤𝑖𝑟𝑒
2𝜋 𝑟
 The farther you move from the currentcarrying wire, the smaller the magnitude of
the magnetic field.
 The greater the current, the larger the
magnitude of the magnetic field.
9
2/13/2017
Magnetic fields produced by a straight wire
Vacuum/Magnetic permeability
 The constant μo is known as the magnetic
permeability. It is used when calculating the
magnetic field in a vacuum, although the value is
approximately the same for air.
 μ is the magnetic permeability of a substance
and replaces μo if the magnetic field is being
calculated inside a material.
𝐵𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑤𝑖𝑟𝑒
𝜇0 𝐼𝑤𝑖𝑟𝑒
=
2𝜋 𝑟
WHITEBOARD
Vacuum permeability
𝜇0 =
4x10
−7
Tm
𝐴
Magnetic fields produced by a loop or coil
Find the magnitude of the magnetic
field at a point located 0.8 m away
from a long, straight, wire carrying an
electric current of 2 A.
I
𝐵=
𝜇0 𝐼𝑤𝑖𝑟𝑒
2𝜋 𝑟
𝐵 = 5𝑥10−7 𝑇
Magnetic field due to electron motion
in an atom
 In an early model of the hydrogen atom, electron
motion was depicted as a circular electric
current.
 The magnetic field at the center of a currentcarrying loop of radius r is:
𝐵=
𝜇0 𝐼
2𝑟
 This motion also describes a magnetic dipole
moment for atoms, making this model
potentially useful for explaining magnetic
properties of materials.
𝜇0 𝐼
𝑎𝑡 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑙𝑜𝑜𝑝
2𝑟
𝑁𝜇0 𝐼
𝐵=
𝑎𝑡 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑐𝑜𝑖𝑙 𝑤𝑖𝑡ℎ 𝑁 𝑙𝑜𝑜𝑝𝑠
2𝑟
𝐵=
WHITEBOARD
 In the early 20th-century model of the hydrogen atom,
the electron was thought to move in a circle of radius
0.53x10–10 m, orbiting once around the nucleus every
1.5x 10–16 s. Determine the magnitude of the magnetic
field produced by the electron at the center of its circular
orbit and its dipole moment.
𝐵=
𝜇0 𝐼
2𝑟
𝐼=
−7
𝑞
∆𝑡
4x10 0.001
𝐵=
2 ∙ 0.53𝑥10−10
𝐼=
𝐵 = 12.65 𝑇
𝐼 = 1.067 𝑚𝐴
1.6𝑥10−19 𝐶
1.5𝑥10−16 𝑠
10
2/13/2017
WHITEBOARD
Magnetic fields produced by a solenoid
 Draw the magnetic field lines of a solenoid
connected to a battery.
𝐵=
𝑁 𝜇0 𝐼
𝑖𝑛𝑠𝑖𝑑𝑒 𝑠𝑜𝑙𝑒𝑛𝑜𝑖𝑑
𝐿
Current loops and bar magnets
 The B field produced by the current in a loop or a coil
and that produced by a bar magnet are very similar.
MAGNETIC FORCE ON
AN ELECTRIC
CURRENT CARRYING
WIRE
VIDEO
 Wire coils with current are known as electromagnets.
Magnetic force exerted by the magnetic field
on a current-carrying wire
 If a current-carrying wire is similar in some ways
to a magnet, then a magnetic field should exert a
magnetic force on a current-carrying wire similar
to the force it exerts on another magnet.
 A magnet sometimes pulls on a wire and
sometimes does not—the effect depends on the
relative directions of the B field and the current
in the wire.
RIGHT HAND RULE #2
B
I
FB
11
2/13/2017
MAGNETIC FORCE ON AN
ELECTRIC CURRENT CARRYING
WIRE
B
B
I
MAGNETIC FORCE ON AN
ELECTRIC CURRENT CARRYING
WIRE
B out of the paper
I
I
I
MAGNETIC FORCE ON AN
ELECTRIC CURRENT CARRYING
WIRE
B
I
B
I
I
I
I
I
I
I
12
2/13/2017
MAGNETIC FORCE ON AN
ELECTRIC CURRENT CARRYING
WIRE
MAGNETIC FORCE ON AN ELECTRIC
CURRENT CARRYING WIRE
B
I
B into the paper
I
I
I
I
I
I
I
B
MAGNETIC FORCE ON AN ELECTRIC
CURRENT CARRYING WIRE
I
I
Magnetic force exerted on an electric
current carrying wire
B
 The magnitude of the magnet force FBonW that a
uniform magnetic field B exerts on a straight
current-carrying wire of length L with current I is
FBonW = LBI sin
I
 Where  is the angle between the direction of the
B field and the current I. The direction of this
magnetic force is given by the right-hand rule for
the magnetic force.
I
B
WHITEBOARD
WHITEBOARD
You wonder if instead of supporting your clothesline with two
poles you could replace the poles and the clothes line with a
current carrying wire in Earth’s Magnetic field , which near
the surface has a magnitude 5x10-5 T and points north.
Assume that your house is located near the equator, where
the B field produced by Earth is approximately parallel to
Earth’s surface. The clothesline is 10 m long. The clothes
and the line have a mass of 2 kg.
𝐸𝑎𝑟𝑡ℎ
𝐹𝐵 = 1.6 𝑁
𝐵 = 0.133 𝑇
What direction should you orient the clothesline?
What current is needed to support it?
Is this a promising way to support the clothesline?
𝐵 = 0.146 𝑇
13
2/13/2017
WHITEBOARD - SOLUTION
𝐹𝐵 = 𝐿𝐵𝐼𝑠𝑖𝑛(𝜃)
𝐼=
What will happen when you place an
ECCLoop inside the magnetic field?
𝐹𝐵
𝐿𝐵𝑠𝑖𝑛(𝜃)
𝐹𝐵 = 𝐹𝐸 𝑜𝑛 𝐶𝐿
𝐼=
𝑚∙𝑔
𝐿𝐵𝑠𝑖𝑛(𝜃)
𝐼 = 39,200 𝐴
Torque exerted on a magnetic loop
 A magnetic field exerts forces of the current
passing through the wires of a coil, resulting in a
torque on the coil.
𝜏𝐵𝑜𝑛𝐶 = 𝑁𝐵𝐴𝐼𝑠𝑖𝑛(𝜃)
14
2/13/2017
WHITEBOARD
 A 500 turn coil of wire is hinged to the top of a
table. Each side of the movable coil has a length
of 0.5 m.
 In which direction should a magnetic field point to
help lift the free end of the coil off the table?
 Determine the torque caused by a 0.7 T field
pointing in the direction described above, when
there is a 0.80 A current through the wire.
𝜏𝐵𝑜𝑛𝐶 = 𝑁𝐵𝐴𝐼𝑠𝑖𝑛(𝜃)
Summary of the differences between
gravitational, electric, and magnetic forces
 The gravitational and electric forces exerted on
objects do not depend on the direction of
motion of those objects, whereas the magnetic
force does.
 The forces exerted by the gravitational and
electric fields are always in the direction of the
g or E field, but the force exerted by the
magnetic field on a current-carrying wire is
perpendicular to both the B field and the
electric current.
𝜏𝐵𝑜𝑛𝐶 = 70 𝑁𝑚
Magnetic force exerted on a single moving
charged particle
MAGNETIC FORCE ON
A CHARGED
PARTICLE
Direction of the force that the magnetic field
exerts on a moving charged particle
 We can use the right-hand rule for
the magnetic force to predict the
direction in which electrons in the
oscilloscope will be deflected.
 The magnetic field exerts a force on a
current-carrying wire, which is made of
moving electrons.
 The magnetic field also exerts a force on
each individual electron.
 The magnetic field also exerts a force on
other moving charged particles, such as
protons and helium nuclei.
RIGHT HAND RULE #2
B
v
FB
© 2014 Pearson Education, Inc.
15
2/13/2017
MAGNETIC FORCE ON A
CHARGED PARTICLE
MAGNETIC FORCE ON A
CHARGED PARTICLE
B
B
+
V
- V
V
+
V
+
-
V
- V
+
V
-
V
B out of the paper
+
V
- V
MAGNETIC FORCE ON A
CHARGED PARTICLE
B
V
- V
V
+
V
+
-
V
- V
B
-
V
- V
V
V
+
-
V
- V
V
+
-
V
- V
-
+
-
+
V
V
V
V
-
+
-
+
V
V
+
V
V
B
-
V
+
V
-
V
B
V
-
+
-
V
V
-
V
+
MAGNETIC FORCE ON A
CHARGED PARTICLE
V
V
V
+
+
V
+
-
+
B into the paper
+
MAGNETIC FORCE ON A
CHARGED PARTICLE
V
MAGNETIC FORCE ON A
CHARGED PARTICLE
B
+
V
+
V
V
B
16
2/13/2017
Magnetic force exerted by the magnetic field
on an individual charged particle
 The magnitude of the magnet force that the
magnetic field of magnitude B exerts on a particle
with electric charge q moving at speed v is
WHITEBOARD
 Each of the lettered dots shown in the figure represents
a small object with an electric charge +2.0 x 10 –6 C
moving at a speed 3.0 x 107 m/s in the directions shown.
Determine the magnetic force (magnitude and direction)
that a 0.10-T magnetic field exerts on each object.
FBonq = Bvq sin
𝐹𝐵 = 6 𝑁 (𝑖𝑛𝑡𝑜)
 Where  is the angle between the direction of the
velocity of the particle and the direction of the B
field. The direction of this force is determined by
the right-hand rule for the magnetic force. If the
particle is negatively charged, the force points
opposite the direction given by the right-hand
rule.
𝐹𝐵 = 4.8 𝑁 (𝑜𝑢𝑡)
𝐹𝐵 = 6 𝑁 (left)
WHITEBOARD
WHITEBOARD
𝐹𝐵 = 3.2𝑥10−13 𝑁
𝐹𝐵 = 1.728𝑥10−13 𝑁
𝐹𝐵 = 8.64𝑥10−16 𝑁
𝐹𝐵 = 6.4𝑥10−22 𝑁
Circular motion in a magnetic field
 The force exerted by the
magnetic field always
points perpendicular to
the particle's velocity,
toward the center of the
particle's path.
 The particle will
move along a circular
path in a plane
perpendicular to the
field.
𝐹𝐵 = 0 𝑁
𝑚𝑣 2
𝐹𝑐 =
𝑟
In a B-field, the magnetic force exerted on a moving
particle will always be perpendicular to its velocity vector.
v
F
This means that
magnetic force can
never speed up a
particle, and can never
slow down a particle.
v
F
It can only change the
particle’s direction!
F
v
17
2/13/2017
Another way of saying this is…
Magnetic force cannot do work!
A B-field can never add or remove kinetic energy from
a system. It can only change the system’s direction of
motion while maintaining a constant speed.
Since kinetic energy is a scalar quantity, this will leave
the system’s kinetic energy unchanged.
WHITEBOARD
 A magnetic field of 0.2 T forces a proton
beam of 1.5 mA to move in a circle of
radius 0.1 m. The plane of the circle is
perpendicular to the magnetic field.
 What is the speed of the proton?
Always 90°
W = FΔxcosθ = 0 J
𝑣 = 1916168
𝐹𝐶 = 𝐹𝐵
WHITEBOARD
𝑚𝑣 2
= 𝐵𝑣𝑞
𝑟
𝐵𝑞𝑟
𝑣=
𝑚
0.2 ∙ 1.6𝑥10−19 ∙ 0.1
𝑣=
1.67𝑥10−27
𝑣 = 1916168
𝑚
𝑠
 Of the following, which is the best estimate
of the work done by the magnetic field on
the protons during one complete orbit of
the circle?
(A) 0 J
(B) 10-22 J
(D) 102 J
(C) 10-5 J
(E) 1020 J
𝑚
𝑠
Cosmic rays
 Cosmic rays are electrons, protons, and other
elementary particles produced by various astrophysical
processes, including those occurring in the Sun and
sources outside the solar system.
 Earth's magnetic field serves as a shield against harmful
cosmic rays, causing them to deflect from their original
trajectory toward Earth.
The auroras
 Charged particles moving in Earth's magnetic
field follow helical paths around the magnetic
field lines.
18
2/13/2017
INTENSITY MODULATED RADIATION
THERAPY
 An IMRT machine
accelerates electrons to
the desired kinetic
energy, then uses a
magnetic field to bend
them into a target,
resulting in the
production of X-rays.
Movable metal leaves
then shape the X-ray
beam to match the
shape of the tumor.
INTENSITY MODULATED RADIATION
THERAPY
WHITEBOARD
 Estimate the magnitude of
the magnetic field needed
for an IMRT machine. For
the estimate, assume that
the electrons are moving
at a speed of 2 x 108 m/s,
the mass of the electrons
is 9 x 10–31 kg, and the
radius of the turn is
5 cm.
𝐹𝐶 = 𝐹𝐵
𝑚𝑣 2
𝑟
A proton is launched into a
uniform magnetic field
= 𝐵𝑣𝑞
Sketch the resulting path of the proton.
𝑚𝑣
𝐵=
𝑞𝑟
9𝑥10−31 ∙ 2𝑥108
𝐵=
1.6𝑥10−19 ∙ 0.05
+q
𝐵 = 0.0225 𝑇
The proton will take follow a circular path until it
hits one of the walls of the
Suppose you wanted to hook up the plates to
opposite terminals of a battery, so that the proton
travels straight through the plates, undeflected.
+q
+q
Which way would you need to hook up
the battery? (charges of the plates)
19
2/13/2017
Combining electric and magnetic forces!
FB
+q
E
v
+q
B
E
v
Fq
B
The forces will only be balanced if qvB = Eq.
𝐸
If the strengths of the fields are fine-tuned so that the particle travels
straight through, derive an expression for the velocity of the particle.
FB = Bvq
Fq = Eq
Therefore, only particles with the exact velocity 𝑣 = will
𝐵
make it through.
This device is called a velocity selector and is used in particle
accelerators to hand-pick the right particles for a collision!
Too fast: Magnetic force too strong.
Magnetic force dominates.
E
+q
v
B
Juuuust
right!
What will happen if the particle is not moving fast enough?
(v < E/B) ?
What about if the particle is moving too fast?
(v > E/B) ?
𝑣=
𝐸
𝐵
Too slow: Magnetic force too weak.
Electric force dominates.
FB = Bvq
Fq = Eq
Whiteboard Challenge: Capture the Protons!
mp = 1.67 x 10-27 kg
e = 1.6 x 10-19 C
B = 0.2 T
Beam of protons
with randomly
distributed speeds
Too fast: Magnetic force too strong.
Magnetic force dominates.
2r = 20 cm
E = 400 kV/m
a) Sketch the complete path of the protons that will
make it through the velocity selector undeflected.
Too slow: Magnetic force too weak.
Electric force dominates.
b) Where should a detector be placed along the orange
wall (quantitatively) to measure the number of protons
per second that made it through the velocity selector?
20
2/13/2017
The only protons that make it through the crossed
𝐸
E and B fields must have a speed 𝑣 =
𝐵
𝐸
𝑣=
𝐵
400,000
𝑣=
0.2
𝑣 = 2𝑥106
Magnetohydrodynamic (MHD) generator
𝑚
𝑠
 An MHD generator converts the random kinetic energy
of high-temperature charged particles into electric
potential energy.
Once they are in the region of only B-field, they will immediately
move in uniform circular motion with a radius given by
r=
mv
qB
r = 10.44 cm
And you can use RHR #2 to determine which way they will curve!
Magnetic flow meter
 MHD generators are used at some older coal-fired power
plants to improve the efficiency of power generation.
MAGNETIC FLOW METER
WHITEBOARD
 A magnetic flow meter works only for fluids with
moving ions, which includes most fluids.
 A magnetic field is oriented perpendicular to the
vessel through which the fluid flows. Oppositely
charged ions in the fluid are pushed by the
magnetic field to opposite walls, producing a
potential difference across the walls of the
vessel.
 With this information, we can determine the
fluid's volume flow rate.
 Is the general magnetic flow meter idea feasible for
measuring blood speed in an artery? Estimate the
potential difference you would expect to measure as
blood in an artery passes through a 0.10-T magnetic
field region. Assume the heart pumps 80 cm3 of
blood each second (the approximate volume for
each heartbeat) and the diameter of the artery is 1.0
cm.
Flow rate
(chapter 11)
𝑄 =𝐴∙𝑣
𝑄 =𝐴∙𝑣
𝑄 = 𝜋𝑟 2 𝑣
𝑑2
𝑄=𝜋 𝑣
4
4𝑄
𝑣= 2
𝜋𝑑
4 ∙ 8𝑥10−5
𝑣=
𝜋 ∙ 1𝑥10−4
𝑚
𝑣 = 1.02
𝑠
𝐸=
𝑣=
𝐸
𝐵
Electric Field
(chapter 15)
𝐸=
∆𝑉
𝑑
∆𝑉
𝑑
𝐸 = 𝐵𝑣
𝐵𝑣 =
∆𝑉
𝑑
∆𝑉 = 𝐵𝑣𝑑
MASS
SPECTROMETER
∆𝑉 = 0.1 ∙ 1.02 ∙ 0.01
∆𝑉 = 1.02𝑥10−3 𝑉
21
2/13/2017
Mass Spectrometers
Mass spectrometer
 Mass spectrometry is an analytical technique that identifies
the chemical composition of a compound or sample based on
the mass-to-charge ratio of charged particles. A sample
undergoes chemical fragmentation, thereby forming charged
particles (ions). The ratio of charge to mass of the particles is
calculated by passing them through ELECTRIC and
MAGNETIC fields in a mass spectrometer.
 A mass spectrometer helps determine the masses of ions,
molecules, and even elementary particles such as protons
and electrons.
 It can also determine the relative concentrations of atoms of
the same chemical element that have slightly different
masses.
WHITEBOARD
FORCE DIAGRAM - The Velocity Selector
When you inject the sample you want it
to go STRAIGHT through the plates.
Since you have an electric field you also
need a magnetic field to exert a force in
such a way as to BALANCE OUT the
electric force caused by the electric
field.
MASS SPECTROMETER
WHITEBOARD (next slide picture)
1. Draw a force diagram (velocity selector)
2. Write an expression for the velocity v of
the charged particle through the velocity
selector.
3. Write an expression for the electric
potential difference V between plates (in
terms of v, d, and B)
4. Write an expression for the of radius of
the path created by the particle inside the
deflection chamber. (in terms of v, m, q,
and B)
WHITEBOARD
Ratio (q/m)
After leaving the velocity selector in a straight
line, it enters the deflection chamber, which
ONLY has a magnetic field. This field causes
the ions to move in a circle separating them
by mass. This is also where the charge to
mass ratio can then by calculated. From that
point, analyzing the data can lead to
identifying unknown samples.
𝐹𝐶 = 𝐹𝐵
𝑞
𝑣
=
𝑚 𝑟𝐵
𝑚𝑣 2
= 𝐵𝑣𝑞
𝑟
MASS SPECTROMETER
DEFLECTION
CHAMBER
VELOCITY SELECTOR
p
+++++++++++
q
v
----------------
22
2/13/2017
WHITEBOARD
(Expressions for v, V & r)
MASS SPECTROMETER
WHITEBOARD
 An atom or molecule with a single
electron removed is traveling at 1.0
x 106 m/s when it enters a mass
spectrometer's 0.50-T uniform
magnetic field region. Its electric
charge is +1.6 x 10–19 C. It moves
in a circle of radius 0.20 m until it
hits the detector.
1. Determine the magnitude of
the magnetic force that the
magnetic field exerts on the
ion.
2. Determine the magnitude of
the Electric Field between the
plates of the velocity selector.
3. Determine the mass of the
ion.
MASS SPECTROMETER
WHITEBOARD - solution
𝐹𝐵 = 𝐵𝑣𝑞 ∙ 𝑠𝑖𝑛(𝜗)
𝐹𝐵 = 8𝑥10−14 𝑁
MASS SPECTROMETER
𝐸 = 𝐵𝑣
𝐸 = 5𝑥105
𝑉
𝑚
q
𝐹𝐶 = 𝐹𝐵
𝑚=
𝐵𝑞𝑟 ∙ sin(𝜃)
𝑣
𝑚 = 1.6𝑥1−26 𝑘𝑔
MASS SPECTROMETER
MASS SPECTROMETER
q
q
23
2/13/2017
MASS SPECTROMETER
q
24