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Electromagnetism
Moving Charges In
Magnetic and Electric
Fields
Magnetic Forces – Charged
Particles
I. Charged
particles in
external
magnetic fields
 a charged particle
in motion, induces
a magnetic field
around the particle
which is
perpendicular to
the motion of the
particle


induced magnetic field around a particle
interacts with the external magnetic field
the induced magnetic field arrow is
attracted toward the south and north poles
of the magnets resulting in a downward
force.


Below the particle, the induced and external
magnetic fields repel one another to create a
downward force.
The result of a charged particle going through a
magnetic field: particle will be deflected by a
force which is perpendicular to both the
original direction of the particle's motion
and the external magnetic field.
Magnitude of the deflecting force

The deflecting force
on a charged particle
moving through an
external magnetic
field is calculated
using:
| Fm| = q v B sin θ






where:
Fm = deflecting force
from the magnetic field
(N)
B =magnetic flux
density or magnetic
field strength (Tesla)
(T)
q = charge of moving
particle (C)
v = speed of particle
(m/s)
θ = angle between v
and B Note: The
maximum deflecting
force will occur
when θ= 90o. Thus
sin 90o= 1 and Fm=
qvB.
Example:

A 20 g particle with a charge of +2.0 C
enters 0.20 T a magnetic field at 90o to
the field. If the speed of the particle is 40
m/s, what is the acceleration that is
experienced by the particle in the diagram
Fm  qv B
below?
Fm  (2.0C )(40m / s)(0.20T )
Fm  16 N
Fm  Fnet
Fm
16 N
a

m 0.020kg
a  800  8.0  102 m / s 2 out of the page
 An
alpha particle enters a 50 T field at
30°to the field at a speed of 500 m/s.
What is the magnitude of the
deflecting force experienced by the
alpha particle? (An α+2 particle has a
charge of 2 x 1.60 x 10-19C = 3.2 x
10-19C.)
Fm  qv B sin 
Fm  (3.20  10
Fm  4.0  10
19
15
N
C )(500m / s)sin 30(50T )
Applications of Magnetic Forces
a. Mass spectrometer
sin 30
www.edumedia-sciences.com/en/a105-mass-spectrometer
http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/MassSpec/masspec1.htm
b. Van Allen radiation belts
c. Black and white television

http://www.colorado.edu/physics/2
000/tv/big_picture.html
The Movement of Charges Through
Electric and Magnetic Fields
Simultaneously


-when a charge
Fe
passes through a
E
magnetic field which is
q
perpendicular to an
Fe  Eq
electric field, it can
pass through
undeflected
Eq  qvB
when no deflection
of the charge occurs, E  vB
the magnetic force is
E
equal to the electrical
v
B
force
Fm  qvB
Fe  Fm
When the forces are equal, (Fm = Fe) the
speed of the charge can be determined
Fm is down by 3rd
LHR when e is in B
 Fe is up, e is
attracted to the
positive plate
 When Fe up = Fm
down, e passes E
and B undeflected
 Speed of electron
can be determined

Example

Eg) An electron enters a magnetic field of 2.00 x
10 -3 T at 90 degrees. An electric field of 1000
N/C is perpendicular to the magnetic field.
Determine the kinetic energy of the electron as it
passes undeflected between the two fields.
Fe  Fm
Eq  qvB
E  vB
E
v
B
1000 N / C
v
2.00 103 Ns / Cm
v  5.00 105 m / s
1 2
Ek  mv
2
1
Ek  (9.11 1031 kg )(5.00  105 m / s )
2
Ek  1.14 1019 J (kgm 2 / s 2 )
Current Balance
 -when a current carrying wire is positioned
perpendicular to B, the current in the wire can be
adjusted to equal the force of gravity downwards
with the magnetic force upwards.
This is
referred to
as current
balance
where Fg
downwards
equals Fm
upwards.
Fg (downwards)  Fm (upwards)
mg  IlB
Example:

Eg) A 6.00 m length of wire is to a magnetic field
of If the mass of the wire in the magnetic field
is 15.0 g, what current must pass through the
wire to suspend it?
Fg (downwards )  Fm (upwards )
mg  IlB
0.0150kg (9.81m / s)  I (0.600m)(4.00  10 2 T )
I  6.13 A