Study Guide
Magnetic Force on Moving Charged
-
University:
Kentucky State University -
Course:
PHY 212 | General Physics II Academic year:
2021
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Views:
435
Pages:
26
Author:
diamundgalx45p
Magnetic force on a charge
• magnetic field B creates force acting on moving charged
particle
F = q v B
force on
particle
velocity of
charged particle
magnetic field
vector
Recall slide 4: SI unit for magnetic field: T (Tesla)
1 kg
1 T=
Cs
These units come from the
magnetic force equation, which
appears two slides from now.
Quiz: verify that T (Tesla) is force / (charge.velocity) Brush-up: Vector cross product: Magnitude
C=AxB
C = AB sin θ
F = qvB sin θ
Brush-up: Vector cross product: Direction
C is perpendicular to both A and B
F is perpendicular to both v and B Brush-up: Vector cross product: Direction
Direction:
Right Hand Rule (RHR):
Point thumb in direction of
velocity, fingers in magnetic
field direction. Then palm
direction is direction of
force on charge.
v
x B
= F
thumb × four fingers = palm
Many alternative versions of the RHR exist!
All versions are equivalent!
Recommendation:
Memorize one version and stick to it! Examples: Direction of the magnetic force
F
+
Another version
of RHR:
Curl fingers as if
rotating v into B.
Thumb is in the
direction of F.
B
-
v
F
v
F?
-
B
B
v
Force is “down”
because charge is
negative. More about direction of the magnetic force Brush-up: Cross product in terms of vector components
F = qv B
ˆi
F = q det v x
B
x
ˆj
vy
By
kˆ
vz
Bz
need lots of cross products of unit vectors ˆi, ˆj, kˆ
All of the RHRs are just techniques for determining the direction of
vectors in the cross product without having to do any actual math. Brush-up: cross products of unit vectors
Cyclic property:
“forward”
“backward”
i j k i j k
i j k i j k
ˆi ˆj = kˆ
ˆj ˆi = -kˆ
Example:
ˆi ˆj
ˆk ˆj = det 0 0
0 -1
kˆ
ˆ
1 = i 0 - -1 = ˆi
0
Magnetic Force Vs Electric Force
Electric force acts in the direction of the electric field.
FE = qE
Electric force is nonzero even if v=0.
Magnetic force acts perpendicular to the magnetic field.
FB = qv B
Magnetic force is zero if v=0.
FB v = 0 = q 0 B = 0 Magnetic Force Vs Electric Force
Electric force does work in displacing a charged particle.
FE = qE
E
F
+
D
WF =F D =FD = qED
Electric force can be used to accelerate particles.
Magnetic force does not do work in displacing a charged
particle!
B v
FB = qv B
WF = F ds = 0
+ds
F
is a vector pointing out of the paper/board/screen
(looks like an arrow coming straight for your eye).
is a vector pointing into the paper/board/screen
(looks like the feathers of an arrow going away from eye). Magnetic Force Vs Electric Force
Magnetic force does not do work in displacing a charged
particle!
WF = F ds = 0
Recall work-energy theorem …!
Thus, the change in the kinetic energy of a charged particle
moving through a magnetic field is zero. Hence, when a
charged particle moves with a given velocity through a magnetic
field, the field can alter the direction of the velocity, but not the
speed or the kinetic energy.
Copyright © 2009 Pearson Education, Inc. ^
Example 1: a proton is moving with a velocity v = v0j in a
^
region of uniform magnetic field. The resulting force is F = F0i.
What is the magnetic field (magnitude and direction)?
v = v 0 ˆj
F = F ˆi
0
F = qv B
Blue is +, red is -
ˆi
ˆ
F = F0 i = q det v x
B
x
ˆj
vy
By
kˆ
v z = e
B z
ˆi
det 0
B
x
ˆj kˆ
v 0 0
B y B z
F0 ˆi = e ˆi v 0 B z B y 0 ˆj 0 B z B x 0 kˆ 0 B y B x v 0
^
Example: a proton is moving with a velocity v = v0j in ^a region
of uniform magnetic field. The resulting force is F = F0i.
What is the magnetic field (magnitude and direction)?
F0 ˆi = ev 0Bz ˆi 0jˆ eB x v 0kˆ = ev 0B z ˆi + 0 ˆj+ eB x v 0 kˆ
F0 = ev 0B z
F0
Bz =
ev 0
and
and
0 = eB x v 0
Bx = 0
What is By???? It could be anything. Not enough information is
provided to find By. We can’t find the magnitude and direction
of the magnetic field using only the information given! ^
Example: a proton is moving with a velocity v = v0j in a region
^
of uniform magnetic field. The resulting force is F = F0i. What is
the minimum magnitude magnetic field that can be present?
F0
Bz =
ev 0
By = 0
and
Bx = 0
for a minimum magnitude magnetic field, so...
F0
Bmin =
ev 0
F0
Bmin =ev k
0 Motion of a charged particle
in a uniform magnetic field
Example 2: an electron travels at 2x107 m/s in a
plane perpendicular to a 0.01 T uniform magnetic
field. Describe its path. Example: an electron travels at 2x107 m/s in a plane
perpendicular to a 0.01 T uniform magnetic field.
Describe its path.
• Magnetic force on electron is
always perpendicular to the
velocity
• force does not change speed of
electron
• force only changes direction of
electron
• Thus magnetic force plays the
role of centripetal force i.e. it
is always directed toward the
center of the circular path i.e.
radial (along the radius).
B
F
v
F
- v
Electron will move in a circular path with a constant speed
and centripetal acceleration = v2/r, where r is the radius of
the circle. This is a “uniform” circular motion. Example 3: a proton of speed v travels in a plane perpendicular
to a uniform magnetic field. Find the radius of its orbit.
+
v
Bout
FB r
v
+
+
FB
FB
v
The period T of orbit is:
2πr 2π 2πm
T
v
ω
qB
• force is in radial direction
• magnitude F=|q|vB
• for circular motion, a = v2/r
mv 2
F = q vB =
r
mv
r
qB
The angular speed (rotational frequency),
which is also referred to as the cyclotron
frequency is
v qB
ω
r
m
Quiz: A proton of speed v has period T. A proton of speed 2v has period of ..?
Quiz: An electron of radius r has cyclotron frequency . An electron of radius 3r has frequency of ..?
Quiz: A charged particle of momentum p has radius R. If the momentum is tripled, the radius will be ..? Helical motion in a uniform magnetic field
If v and B are perpendicular, a
charged particle travels in a circular
path. v remains constant but the
direction of v constantly changes.
If v has a component along B, then v
remains constant, and the charged
particle moves in a helical path.
There won’t be any test problems on
helical motion. Cheers!
B
v
+
v Application I: Velocity Selector or Velocity Filter
Bout
----------------
v
E
+
If the particles enter with different velocities,
this device can “select” a particular velocity v.
In many experiments
involving charged
particles, it is important
that all particles move
with essentially the same
velocity. This can be
achieved by applying a
combination of a uniform
electric field E & a
uniform magnetic field
Bout oriented as shown.
Lorentz Force Law: If both electric and magnetic fields are present,
F = q E+ v B .
What velocity will allow the proton to
pass undeflected through region of
uniform electric field E and uniform
magnetic field Bout ? Application I: Velocity Selector
Bout
FE = qE
----------------
v
E
+
v
+
FB = qv B
When the electric and magnetic forces balance
(in other words when Lorentz force is zero),
then the charge will pass straight through.
FE = FB
q E = q vB
or
E
v=
B Application II: Mass Spectrometer
Mass spectrometers separate charges of different mass
according to their mass-to-charge ratio, (m/q).
.When ions of fixed energy enter a region of constant magnetic
field that is perpendicular to the ion path, they follow a circular
path (semicircle).
Therefore, we can
determine (m/q) by
measuring the radius of
S
the path and knowing
the field magnitudes E
and B.
mv
x = 2r and r =
qB
V
B
+q
A variation of this technique was used
by Thomson in 1897 to measure the
ration (m/q) for electrons!
R
x
Example 4: ions from source S enter a region of uniform magnetic
field B that is perpendicular to the ion path. The ions follow a
semicircle and strike the detector plate at x = 1.7558 m from the point
where they entered the field. If the ions have a charge of 1.6022 x 10-19 C,
the magnetic field has a magnitude of B = 80.0 mT, and the
accelerating potential is V = 1000.0 V, what is the mass of the ion?
V
S
B
+q
R
x
Step 1: find the final speed of the ions if they have a charge of
1.6022 x 10-19 C and the accelerating potential is V = 1000.0 V?
Let’s focus on the part of
the diagram where the
ions from the source are
being accelerated.
Let’s assume the ions
start at rest.
V
S
+q
vi = 0
Recall work-energy theorem …!
Don’t get your v’s and V’s mixed
up! Use a script v for speed!
+q
v=? Step 2: find the radius of the circular path followed by the ions in the
region of uniform magnetic field of magnitude B.
Let’s focus on the
magnetic field region of
the diagram.
+q B
v
R
x
Step 2: find the radius of the circular path followed by the ions in the
region of uniform magnetic field of magnitude B.
For motion in a circular path...
B
+q
v2
FB ma m
R
FB qv B qvBsin 90
2
v
q vB m
R
mv
R
qB
x
v
FB R
Step 3: solve for m, using q=1.6022x10-19 C, V=1000.0 V,
B = 80.0 mT, and R=x/2=1.7558/2=0.8779 m.
V
Now we can put it all
together.
S
mv
R
qB
B
+q
R
x
We know q, R, B, and V. We have two
unknowns, v and m, and two equations,
so we can solve for m.
Mrs. Sahar ,
m 3.9515 1025 kg
4th
grade!
1 atomic mass unit (u) equals
1.66x10-27 kg, so m=238.04 u.
Uranium-238! Last lecture: Brush up!
Magnetic force on a charge
F = q v B
Uniform circular motion of a charged particle
in a uniform magnetic field
mv
r
qB
v qB
ω
r
m
2πr 2π 2πm
T
v
ω
qB
Lorentz Force Law: If both electric and magnetic fields are
present,
F = q E+ v B .
Application I:
Velocity Selector
Magnetic Force on Moving Charged
of 26
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