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CHARGED PARTICLES IN ELECTRIC AND MAGNETIC
FIELDS
F  q v B sin 
E
F
q
E
V
d
q
charge on particle [coulomb C]
v
velocity of charged particle [m.s-1]
B
magnetic flux density [tesla T]

angle between directions of magnetic field and motion of charged particle
[degrees]
E
electric field strength [V.m-1 N.C-1]
F
force acting on particle [newton N]
FB
force acting on particle in magnetic field [N]
FE
force acting on particle in electric field [N]
V
potential difference between a pair of charged parallel plates [volt V]
d
distance between parallel plates [m]
Electric
Field
large electric field intensity
small electric field intensity
+Q
+ + + +
-
+
d
-Q
-
POSITIVE charge
Equation Mindmap eq17:
+
NEGATIVE charge
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-
-
-
V
-
uniform electric field
1
E
Parallel plates – uniform electric field
V
 constant
d
F
FE  q E
E
F
Electron gun – motion of a charged
particle in an electric field
v
+Q
V
F
+Q
-Q
E
E
F
v
-Q
d
straight
line path
parabolic path
straight
line path
Work = change in kinetic energy
V
W  F d  q E d  q
d
E K  12 m v 2
1
2

 d  qV

m v 2  qV
Equation Mindmap eq17:
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Motion of a charged particle in a magnetic field
+q
FB  q v B sin 
v
+I
Direction of force on moving charged particle
in a magnetic field is given by the right hand
palm rule
F
out of page
F
palm face
B fingers
B
motion of a
positive charge in a
magnetic field

v
v  q 
thumb
v   q  thumb
v (+q)
B fingers

v  q 
motion of a
negative charge in
a magnetic field
F palm face
Equation Mindmap eq17:
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e
S
N
B
v  q 
right hand palm rule
v  q 
thumb
palm
facing down
fingers
B
F
The direction of the force on an electron beam in a cathode ray tube is
given by the right hand palm rule.
Motion of a charged particle in a uniform magnetic field
FB  q v B
uniform magnetic field B into page
v
negative charge q
enters uniform
B-field
FB causes a change
in direction, no
change in speed
since FB acts at
right angle to v
F
r
force F on charge always directed to centre of circle
Circular motion: Magnetic force = Centripetal force
m v2
qB
FB  q v B Fc 
FB  Fc v 
r
m
Equation Mindmap eq17:
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J.J. Thompson’s e/me Experiment
Motion of charged particle in a cross magnetic and electric fields
J. J. Thompson - measured qe / me  conclusive evidence that cathode rays were a stream of
electrons.
_
Electric field E  electrons
deflected up
Magnetic field B 
electrons deflected down
Electron gun  acceleration
of electrons
heater
element
cathode
+
accelerating
voltage VA
eVA  12 mev 2  v 
+
X
d
_
6 Vac
anode
deflection
plates V
Z
I
Y
coils to produce
B field into page
2eVA
me
Electric force FE = magnetic force FB
 zero deflection of electron beam
eV
FE  eE 
d
2 e VA
FB  ev B  eB
me

e
V2

me 2 d 2 VA B
Motion of electron in uniform magnetic field  circular motion of electron as magnetic force also
directed perpendicular to the motion  magnetic force FB = centripetal force FC
FB  FC
v2
ev B  m
R
v
2eVA
me
FE  FB  eE  evB
v
E
B

e
E
V
 2
me RB d RB2
Equation Mindmap eq17:
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Example
An electron beam travels through a cathode ray tube. A south pole of a bar magnet is placed
above the beam causing the beam to be deflected. The magnitude of the B-field at the
location of the electron beam is 0.0875 T. The beam of electrons has been accelerated by a
voltage of 6.65 kV. What is the magnitude of the force acting on the electrons and which
way is the beam deflected?
Solution
How to approach the problem: ISEE
Category: work done on charge by a potential difference
qV  12 mv 2
force on a charged particle in a B-field
FB  q v B sin 
Diagram:
direction of force – right hand palm rule
Sketch the physical situation
Choose a [3D] set of axes for the direction of B, v and F
Summary of given information
Summary of unknown information
Solve the problem
Evaluate your answers
Equation Mindmap eq17:
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N
+y
thumb
v  q 
fingers B
S
B
e
-
+x
FB
palm face
+z
v  q 
beam will be deflected
to the right
right hand palm rule
F
B
magnitude: electron charge q = 1.602x10-19 C
electron mass: me = 9.101x10-31 kg
accelerating voltage: V = 6.65 kV = 6.65x103 V
B- field: B = 0.875 T
v = ? m.s-1
FB = ? N
From the diagram, using the right hand palm rule the electron beam is deflected towards
the right.
The work done by the accelerating voltage increases the kinetic energy of the electrons,
therefore, we can calculate the speed of the electrons in the beam.
qV  12 me v 2



(2) 1.602  1019 6.65  103
2 qV
v

m.s-1  2.965  107 m.s-1
31
me
9.109  10
A moving charge in a magnetic field experience a force



FB  qv B  1.602  1019 2.965  107  0.0875  T  1.56 1013 T
Equation Mindmap eq17:
Doing Physics on Line
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