Transcript Physics 142 Lecture Notes

B-field
into
screen
From these observations alone,
what definite conclusions can be made?
1. as are positively charged, bs negative.
2. as are negatively charged, bs positive.
3. can only say a,b oppositely charged.
B-field
into
screen
From these observations alone,
how can we compare the magnitude of their charges?
1.  qa <qb
2.  qa =qb
3.  qa >qb
4. charges can’t be
directly compared
B-field
into
screen
From these observations alone,
what definite conclusions can be made?
1. as are more massive than bs.
2. as are more massive than gs.
3. bs are more massive than as.
4. The masses cannot be directly compared.
What produces a gravitational field?
A gravitational field exerts a force on?
Mass
Mass
What produces an electric field?
Electric charge
An electric field exerts a force on?
Electric charge
What produces a magnetic field?
Moving electric charge
A magnetic field exerts a force on?
Moving electric charge?
Direction of Magnetic Force
tail
out of
in to
page
page
Drawing vectors in
head
Direction of magnetic force is
“sideways”
 force is perpendicular
to both v and B
 use “right-hand rule”
to find direction
F = q v B sinq
+
A positive charge
enters a magnetic
field region as
shown. The B-field
accelerates it in
what direction?
1) up
2) down
3) right
4) left
5) into the page
6) out of the page
ConcepTest
Magnetic Force
A positively charged beam enters into a magnetic
field region as shown. What is the direction of B?
y
1)+ y
(up)
2) – y (down)
3) + x (right)
4) + z (out of page)
5) – z (into page)
x
Trajectory in a Constant Magnetic Field
Suppose a charge q enters a B field with velocity v.
What will be the path q follows?
Force is always ^ to velocity and B. What is the path?
x x x x x x x x x x x x x x
x x x x x x x x x x x x x x
v
x x x x x x x x x x x x x x
x x x x x x x x x Fx x x x x
+q
R
B
Path will be a circle with
some radius of curvature, R.
Radius of Circular Orbit
magnetic force:
F  qvB
centripetal accel:
v2
a
R
x x x x x x x x x x x x x x
x x x x x x x x x x x x x x
v
x x x x x x x x x x x x x x
x x x x x x x x x Fx x x x x
+q
R
Newton's 2nd Law:
F  ma


mv
R
qB
2
v
qvB  m
R
This has useful
experimental
consequences !
B
ConcepTest
Magnetic Force
x x x x x x x x x x x x
Two particles of the same
charge enter a magnetic field
with the same speed. Which
one has the bigger mass?
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
A
1)
2)
3)
4)
B
A
B
both masses are equal
impossible to tell without weighing the particles
as are ionized Helium (bare Helium nuclei)
2-protons, 2-neutrons (positively charged)
bs are simply electrons(negatively charged)
qa = -2qb
ma=7296mb
     
    
     
v=?
    
     
    
     
R=?
    
     
v= /2Vm/q /B
    
     
    
Velocity Selector
Consider a positively charged ion entering a
region where the electric and magnetic fields
are uniform and perpendicular to each other.
If the particle moves in a straight line, what is
its velocity in terms of E and B?
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
E
For the magnetic force:
direction
magnitude
up
F = qvB
For the electric force?
direction
magnitude
down
F = qE
Sum of the forces on the particle?
v=E/B
Zero (not accelerating)
 |FE| = |FB|  qE = qvB
B
Ratio of charge to mass for an electron
e– “gun”
An electron is accelerated from rest across a
potential difference and then enters a region of
uniform magnetic field, as shown at right. What is
the “charge to mass ratio”, q/m, of the electron?
What is the speed of the electron?
½ mv2 = qV
DV
e–
B
x
x
x
x
x
x
R
x
x
x
x
x
x
(Work-Energy Theorem)
What is the radius of the electron’s orbit?
R = mv / qB
Algebra: determine q/m
q / m = 2V / R2B2
(Earlier today)
(Solve second Eq for v and plug into first)
ConcepTest
I
1
2
I
3
I
I
If all wires carry the same current I, for which of the loops
above is the magnitude of the net force greatest?
A) Loop 1
B) Loop 2
C) Loop 3
D) same for all
ConcepTest
Magnetic Force
A rectangular current loop is in a uniform magnetic field. What
direction is the net force on the loop?
(a) + x
(b) + y
(c) zero
(d) – x
(e) – y
z
B
y
x
ConcepTest
If there is a DC current in the loop in the
direction shown, the loop will
A) move up
B) move down
C) rotate clockwise
D) rotate counterclockwise
E) some combination of moving
and rotating
N
S
Current, I, flows as shown
through a rectangular loop
ab immersed in a uniform
magnetic field B.
a
b
a
b
The force on the top
segment of the
rectangular loop is
1)up.
2)down.
3)into screen.
4)out.
5)left.
6)right
7)zero.
a
b
The force on the
bottom segments of
the rectangular loop is
1)up.
2)down.
3)into screen.
4)out.
5)left.
6)right
7)zero.
a
b
The force on the
left segment of the
rectangular loop is
1)up.
2)down.
3)into screen.
4)out.
5)left.
6)right.
7)zero.
a
b
The force on the
right segment of the
rectangular loop is
1)up.
2)down.
3)into screen.
4)out.
5)left.
6)right.
7)zero.
Viewed from above, looking down

I
q



ANSWERS to CONCEPT QUESTIONS
1. as are positively charged, bs negative.
4. charges can’t be directly compared
R depends not only on the size of the charge, \
but also the mass & velocity of the moving charge!
4. The masses cannot be directly compared.
See above!
4) left
+ charge = Right hand. Fingers down (direction of B-field)
Thumb into page (direction of moving charge).
Palm points left (direction of force on the charge).
5) – z(into page)
1) A
R=mv/qB : the larger the mass, M, the greater R must be
(all other varibles being constant).
ANSWERS to CONCEPT QUESTIONS
A) Loop 1 Everywhere beneath the wire the magnetic field points
into the page. The top segments are therefore pushed
down, the bottom segments up. But not equally!
Near the wire the magnetic field is stronger, so the push is stronger.
(c) zero
C) rotate clockwise
2)down.
Current is at an angle but still crosses field lines…
For the top loop crossing lines as the current moves into the page,
for the bottom, as it comes out of the page.
1)up.
3)into screen.
4)out.
Current runs up through the left (front) segment,
down through the right (back) segment.
It crosses field lines in both cases.
Unlike the pushes up and down on the top/bottom
segments, these do not cancel…they are off-centered pushes
that produce a torque!