Transcript The force is always perpendicular to velocity, so it cannot change

The Magnetic Field
The force

F
on a charge q moving with a velocity

  
F  q( E  v  B)

v
The magnitude of the force
F  qvB sin 
[ B]  Newtons /(Coulomb  meter / sec)
1T (tesla)  1w / m  1Newton / C  m / s  10 gauss
2
4
4
BEarth  1Gauss  10 T
left-hand rule
right-hand rule
Motion in magnetic field

1) Uniform B ,

2) Uniform B ,
3) Nonuniform
 
vB
 
vB

B

E 0

 
F  qv  B
 
v || B F  0
 
v  B F  qvB


F  ma

ir
Fr  mar
F  0
2
v
 qvB  mr 2  m
r
mv
r
qB
The angular velocity
v
v
qB
 

r mv
m
qB

The angular velocity (cyclotron frequency f 
)
2
v
v
qB
 

r mv
m
qB
does not depend on velocity!
The force is always perpendicular to velocity, so it cannot
change the magnitude of the velocity, only its direction.
The work done by the magnetic force is zero!
Motion of a charged particle under the action of a
magnetic field alone is always motion with constant
speed.
Exercise 1
An electron, q=1.6 10-19C moves with velocity



5
5
v  6 10 ix  4 10 iy in meters per sec ond


2
B

0
.
1
i
webers
/
m
It enters a magnetic field with
x
What is the force on the electron?
Using Crossed

E
and

B Fields
Velocity selector
qvB  qE  0
E  vB
E
v
B
independent of the mass of the particle!
Mass spectrometer
m1v
R1 
qB
v
E
B
m1 E
R1 
qB 2
m2 E
R2 
qB 2
Thomson’s e/m experiment
1897: Cavendish Laboratory
in Cambridge, England
1 2
2eV
mv  eV  v 
2
m
E
v
B
E
2eV
e
E2

 
B
m
m 2VB 2
Electron motion in a microwave
oven
A magnetron in a microwave oven emits
electromagnetic waves with frequency
f=2450 MHz. What magnetic field strength
is required for electrons to move in circular
paths with this frequency?
Problem 5
Hall effect: The magnetic force on the
charge carries in a wire can be used to
determine their sign. Show that there will
be an electric field, set up inside a wire in
a magnetic field, that is perpendicular to
the direction of the current. You should be
able to show that the sign of the field
depends on whether the mobile charges
are positive or negative.
qE Hall  qvB
i
v
nAq
EHall
iB
jB


nAq nq
You place a slab of copper, 2.0 mm thick and 1.5
cm wide, in a uniform magnetic field with
magnetic field with magnitude 0.40 T. When you
run a 75-A current in the +x direction, you find by
careful measurement that the potential at the left
side of the slab is 0.81V higher than at the right
side of the slab. From this measurement,
determine the concentration of mobile electrons
in copper.
Exercise 3
A wire of length l and mass m is suspended as
shown. A uniform magnetic field of magnitude B
points into the page.
What magnitude and direction would a current,
passing through a wire, have to have so that the
magnetic and gravitational forces would cancel?
Problem 4
A metal wire of mass m can slide without friction
on two parallel, horizontal, conducting rails. The
rails are connected by a generator which delivers a
constant current i to the circuit. There is a constant,
vertical magnetic field, perpendicular to the plane
of the rails. If the wire is initially at rest, find its
velocity as a function of time.
B
l
i
generator
mv
r
qB
The angular velocity
v
v
qB
 

r mv
m
qB
Uniform magnetic field,
 
vB
Uniform
  
B, v  B
When a charged particle has velocity components both
perpendicular and parallel to a uniform magnetic field, the
particle moves in a helical path. The magnetic field does no
work on the particle, so its speed and kinetic energy remain
constant.
Example: A proton ( 1.60 1019 C, m  1.67 1027 kg) is
placed in the uniform magnetic field directed along the
x-axis with magnitude 0.500 T. Only the magnetic
force acts on the proton. At t=0 the proton has velocity
components
vx  1.50 105 m / s, vy  0, vz  2.00 105 m / s.
Find the radius of the helical path, the angular speed
of the proton, and the pitch of the helix (the distance
traveled along the helix axis per revolution).