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Physics 212
Lecture 12
Today's Concept:
Magnetic Force on moving charges
F  qv  B
Physics 212 Lecture 12, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
The Meters
The Neville Brothers
Trombone Shorty
Michael Franti
Radiators
Tribute to New Orleans and Mardi Gras
Your Comments
“How can a magnetic field point in a direction? What
about the wire example in the prelecture where is
A little at a time!
shown to move in a counter-clockwise/clockwise
direction?”
“This stuff is really neat... It is fun to actually see the calculations for
magnetism. However, since this is the first time I’ve really seen it, it is still a
bit confusing. If you could go through different examples and go over the actual
concepts more, that would be great.”
“Magnets. How do they work?”
Good questions!
“Please explain what causes magnetic fields, and why
But very deep!
they exist. Unless we learn that later or not at all, I
just don’t understand.”
“What are we supposed to do if we have two left
hands?”
“Could you go over this right hand rule again?
There are so many of them...”
RHR confused many
“I refuse to make an attempt at a bad pun.”
05
“We most definitely need examples. Also, every time you post an awful pun, a
puppy dies. Just so you know.”
Physics 212 Lecture 12, Slide
3
Key Concepts:
1) The force on moving charges due to a
magnetic field.
2) The cross product.
Today’s Plan:
1) Review of magnetism
2) Review of cross product
3) Example problem
05
Physics 212 Lecture 12, Slide 4
Magnetic Observations
• Bar Magnets
N
S
N
S
S
N
N
S
• Compass Needles
N
S
Magnetic Charge?
N
07
S
cut in half
N
S
N
S
Physics 212 Lecture 12, Slide 5
Magnetic Observations
• Compass needle deflected by electric current
I
• Magnetic fields created by electric currents
• Direction? Right thumb along I, fingers curl in direction of B
• Magnetic fields exert forces on electric currents (charges in motion)
V
F
F
V
F
I
I
I
I
F
V
12
Physics 212 Lecture 12, Slide 6
Magnetic Observations
P
P
I (out of the screen)
I (into of the screen)
Case I
Case II
The magnetic field at P points
A. Case I: left, Case II: right B. Case I: left, Case II: left
C. Case I: right, Case II: left D. Case I: right, Case II:
right
WHY? Direction of B: right thumb in direction of I,
fingers curl in the direction of B
Physics 212 Lecture 12, Slide 7
Magnetism & Moving Charges
All observations are explained by two equations:
F  qv  B
 0 I ds  rˆ
dB 
2
4 r
14
Today
Next Week
Physics 212 Lecture 12, Slide 8
Cross Product Review
• Cross Product different from Dot Product
– A●B is a scalar; A x B is a vector
– A●B proportional to the component of B parallel to A
– A x B proportional to the component of B perpendicular to A
• Definition of A x B
– Magnitude: ABsinq
– Direction: perpendicular to plane defined by A and B with sense given
by right-hand-rule
16
Physics 212 Lecture 12, Slide 9
Remembering Directions: The Right Hand Rule
y
F  qv  B
x
F
B
qv
17
z
Physics 212 Lecture 12, Slide 10
Checkpoint 1a
Three points are arranged in a uniform magnetic
field. The B field points into the screen.
1) A positively
charged
particle
located
atstationary.
point A and
stationary.
A positively
charged particle
is located
at is
point
A and is
The is
direction
of the
magnetic
force on the
is
The direction
of particle
the magnetic
force on the particle is:
A. right
E. zero
B. left
C. into the screen
D. out of the screen
“all are into the screen”
“Right hand rule says its pointing out of the screen”
“qvXB=0 if v=0.”
18
Physics 212 Lecture 12, Slide 11
Checkpoint 1a
Three points are arranged in a uniform magnetic
field. The B field points into the screen.
1) A positively
charged
particle
located
atstationary.
point A and
stationary.
A positively
charged particle
is located
at is
point
A and is
The is
direction
of the
magnetic
force on the
is
The direction
of particle
the magnetic
force on the particle is:
A. right
E. zero
B. left
C. into the screen
D. out of the screen
F  qv  B
The particle’s velocity is zero.
There can be no magnetic force.
18
Physics 212 Lecture 12, Slide 12
Checkpoint 1b
Three points are arranged in a uniform magnetic
field. The B field points into the screen.
1)positive
A positively
charged
at point
A magnetic
and is stationary.
The
charge moves
fromparticle
A towardisB.located
The direction
of the
force on the
particle
The isdirection of the magnetic force on the particle is:
A. right
E. zero
B. left
C. into the screen
D. out of the screen
“it should be diagonally outward”
“v is up, b is into screen, right hand rule says it should be left.”
“Because the magnetic force goes into the screen, so will the direction”
21
Physics 212 Lecture 12, Slide 13
Checkpoint 1b
Three points are arranged in a uniform magnetic
field. The B field points into the screen.
1)positive
A positively
charged
at point
A magnetic
and is stationary.
The
charge moves
fromparticle
A towardisB.located
The direction
of the
force on the
particle
The isdirection of the magnetic force on the particle is:
A. right
E. zero
B. left
C. into the screen
D. out of the screen
F  qv  B
qv
F
21
X
B
Physics 212 Lecture 12, Slide 14
Cross Product Practice
F  qv  B
• protons (positive charge) coming out of screen (+z direction)
• Magnetic field pointing down
• What is direction of force on POSITIVE charge?
A) Left
-x
B) Right
+x
C) UP
+y
D) Down
-y
E) Zero
y
x
B
24
z
Physics 212 Lecture 12, Slide 15
Motion of Charge q in Uniform B Field
• Force is perpendicular to v
– Speed does not change
– Uniform Circular Motion
x vx x x x x x
q
• Solve for R:
x R x x x x vx x
F  qv  B  F  qvB
v
a
R
30
q
F
F
q
x x xF x x x x
2
v2
qvB  m
R
x x x Fx x x x
xv x x x x x x
q
mv
R
qB
x x x x xv x x
Uniform B into page
Physics 212 Lecture 12, Slide 16
R
LHC
mv
p

qB qB
17 miles diameter
Physics 212 Lecture 12, Slide 17
Checkpoint 2a
The drawing below shows the top view
of two interconnected chambers. Each
chamber has a unique magnetic field. A
positively charged particle is fired into
chamber 1, and observed to follow the
dashed path shown in the figure.
What is the direction of the magnetic field in chamber 1?
A. up
B. down
C. into the page
D. out of the page
“Since the direction of the magnetic field has to be perpendicular to the charged
particle’s direction, it will point up..”
“Having a magnetic field going into the screen indicates counterclockwise motion.”
“Thumb points to the left because particle is deflected to the left and fingers point
up because that is velocity vector. Fingers curl out of the screen.”
34
Physics 212 Lecture 12, Slide 18
Checkpoint 2a
The drawing below shows the top view
of two interconnected chambers. Each
chamber has a unique magnetic field. A
positively charged particle is fired into
chamber 1, and observed to follow the
dashed path shown in the figure.
What is the direction of the magnetic field in chamber 1?
A. up
B. down
C. into the page
D. out of the page
F  qv  B
qv
.
B
34
F
Physics 212 Lecture 12, Slide 19
Checkpoint 2b
The drawing below shows the top view
of two interconnected chambers. Each
chamber has a unique magnetic field. A
positively charged particle is fired into
chamber 1, and observed to follow the
dashed path shown in the figure.
Compare the magnitude of the magnetic field in chamber 1 to the magnitude of the
magnetic field in chamber 2
A. |B1| > |B2|
B. |B1| = |B2|
C. |B1| < |B2|
“because the curve is much sharper inside box 1”
“It equals out.”
“Because the particle changed direction more slowly in the second chamber.”
36
Physics 212 Lecture 12, Slide 20
Checkpoint 2b
The drawing below shows the top view
of two interconnected chambers. Each
chamber has a unique magnetic field. A
positively charged particle is fired into
chamber 1, and observed to follow the
dashed path shown in the figure.
Compare the magnitude of the magnetic field in chamber 1 to the magnitude of the
magnetic field in chamber 2
A. |B1| > |B2|
B. |B1| = |B2|
C. |B1| < |B2|
Observation: R2 > R1
R
36
mv
qB
B1  B2
Physics 212 Lecture 12, Slide 21
Calculation
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
x0/2
E
d
enters here
What is B?
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• Conceptual Analysis
–
What do we need to know to solve this problem?
(A) Lorentz
ForceLaw

 
( F  qv  B  qE )
–
–
(B) E field definition
(C) V definition
(D) Conservation of Energy/Newton’s Laws
(E) All of the above
Absolutely ! We need to use the definitions of V and E and either
conservation of energy or Newton’s Laws to understand the motion of
the particle before it enters the B field.
We need to use the Lorentz Force Law (and Newton’s Laws) to
determine what happens in the magnetic field.
Physics 212 Lecture 12, Slide 22
Calculation
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
x0/2
E
d
enters here
What is B?
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• Strategic Analysis
–
–
–
Calculate v, the velocity of the particle as it enters the magnetic
field
Use Lorentz Force equation to determine the path in the field as a
function of B
Apply the entrance-exit information to determine B
Let’s Do It !!
Physics 212 Lecture 12, Slide 23
Calculation
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
E
d
enters here
What is B?
exits here
x0/2
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• What is v0, the speed of the particle as it enters the magnetic field ?
2E
m
(A)
vo 
• Why??
–
–
2qEd
vo 
m
(B)
vo  2 ad
2qE
md
(D)
vo 
(C)
vo 
qEd
m
(E)
Conservation of Energy
•
•
Initial: Energy = U = qV = qEd
Final: Energy = KE = ½ mv02
•
•
a = F/m = qE/m
v02 = 2ad
Newton’s Laws
vo2  2
1 mv 2
o
2
qE
d
m
 qEd
vo 
vo 
2qEd
m
2qEd
m
Physics 212 Lecture 12, Slide 24
Calculation
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
What is B?
x0/2
E
d
enters here
2qEd
vo 
m
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• What is the path of the particle as it moves through the magnetic field?
XXXXXXXXX
XXXXXXXXX
XXXXXXXXX
(A) X X X X X X X X X
• Why??
–
(B)
XXXXXXXXX
XXXXXXXXX
XXXXXXXXX
XXXXXXXXX
(C)
XXXXXXXXX
XXXXXXXXX
XXXXXXXXX
XXXXXXXXX
Path is circle !
•
•
•
Force is perpendicular to the velocity
Force produces centripetal acceleration
Particle moves with uniform circular motion
Physics 212 Lecture 12, Slide 25
Calculation
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
E
d
enters here
2qEd
vo 
m
What is B?
x0/2
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• What is the radius of path of particle?
R  xo
R  2 xo
(A)
(B)
1
R  xo
2
(C)
mvo
R
qB
(D)
vo2
R
a
(E)
• Why??
R
xo/2
XXXXXXXXX
XXXXXXXXX
XXXXXXXXX
XXXXXXXXX
Physics 212 Lecture 12, Slide 26
Calculation
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
What is B?
2
B
xo
2 mEd
q
(A)
vo 
2qEd
m
E
d
enters here
R  1 x0
2
E
B
v
m
BE
2qEd
(B)
(C)
1
B
xo
exits here
x0/2
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
2 mEd
q
B
(D)
mvo
qxo
(E)
• Why??


F  ma
vo2
qvo B  m
R
m vo
B
q R
m 2
B
q xo
B
2
xo
2 qEd
m
2 mEd
q
Physics 212 Lecture 12, Slide 27
Follow-Up
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
What is B?
2
B
xo
x0/2
E
d
enters here
2 mEd
q
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• Suppose the charge of the particle is doubled (Q = 2q),while keeping the
mass constant. How does the path of the particle change?
(A)
X X X X X X X X XX X X X X X X X XX X X X X X X X X
X X X X X X X X XX X X X X X X X XX X X X X X X X X
X X X X X X X X XX X X X X X X X XX X X X X X X X
X X X X X X X X XX X X X X X X X XX X X X X X X X X
No slam dunk.. As Expected !
Several things going on here
(B)
(C)
1. q changes  v changes
2. q & v change  F changes
3. v & F change  R changes
Physics 212 Lecture 12, Slide 28
Follow-Up
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
What is B?
2
B
xo
x0/2
E
d
enters here
2 mEd
q
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• Suppose the charge of the particle is doubled (Q = 2q),while keeping the
mass constant. How does the path of the particle change?
How does v, the new velocity at the entrance,
compare to the original velocity v0?
v
(A) v  o
2
• Why??
1 mv 2
2
v
(B) v  o
(C) v  vo
2
 QEd  2 qEd  2 1 mvo2
2
(D) v  2vo
v 2  2vo2
(E) v  2 vo
v  2 vo
Physics 212 Lecture 12, Slide 29
Follow-Up
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
What is B?
2
B
xo
x0/2
E
d
enters here
2 mEd
q
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• Suppose the charge of the particle is doubled (Q = 2q),while keeping the
mass constant. How does the path of the particle change?
v  2 vo
How does F, the magnitude of the new force at the entrance,
compare to F0, the magnitude of the original force?
(A) F  Fo
2
(B) F  Fo
(C) F  2 Fo
(D) F  2 Fo
(E) F  2 2 Fo
• Why??
F  QvB  2 q  2 vo  B
F  2 2 Fo
Physics 212 Lecture 12, Slide 30
Follow-Up
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
What is B?
2
B
xo
x0/2
E
d
enters here
2 mEd
q
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• Suppose the charge of the particle is doubled (Q = 2q),while keeping the
mass constant. How does the path of the particle change?
v  2vo F  2 2 Fo
How does R, the radius of curvature of the path, compare to
R0, the radius of curvature of the original path?
(A) R 
Ro
2
(B) R 
Ro
2
(C) R  Ro
(D) R  2 Ro
(E) R  2 Ro
• Why??
v2
F m
R
v2
Rm
F
2vo2
vo2
R
Rm
m
 o
2 2 Fo
2 Fo
2
Physics 212 Lecture 12, Slide 31
Follow-Up
A particle of charge q and mass m is accelerated
from rest by an electric field E through a
distance d and enters and exits a region
q,m
containing a constant magnetic field B at the
points shown. Assume q,m,E,d, and x0 are known.
What is B?
2
B
xo
x0/2
E
d
enters here
2 mEd
q
exits here
XXXXXXXXX
X X X X X X X X X x0
XXXXXXXXX
XXXXXXXXX
B
B
• Suppose the charge of the particle is doubled (Q = 2q),while keeping the
mass constant. How does the path of the particle change?
(A)
X X X X X X X X XX X X X X X X X XX X X X X X X X X
X X X X X X X X XX X X X X X X X XX X X X X X X X X
X X X X X X X X XX X X X X X X X XX X X X X X X X
X X X X X X X X XX X X X X X X X XX X X X X X X X X
(B)
Ro
R
2
(C)
A Check: (Exercise for Student)
Given our result for B (above), can you show:
R
1
B
2mEd
Q
R
R o
2
Physics 212 Lecture 12, Slide 32