F
 q v B sin

E

F q
E

V d q v
B

E
F
F
B charge on particle [coulomb C] velocity of charged particle [m.s
-1 ] magnetic flux density [tesla T] angle between directions of magnetic field and motion of charged particle
[degrees] electric field strength [V.m
-1 N.C
-1 ] force acting on particle [newton N] force acting on particle in magnetic field [N]
F
V d
E force acting on particle in electric field [N] potential difference between a pair of charged parallel plates [volt V] distance between parallel plates [m]
Electric
Field large electric field intensity
POSITIVE charge
-small electric field intensity
NEGATIVE charge
-Q
d V uniform electric field
Equation Mindmap eq17 : Doing Physics on Line 1

Parallel plates – uniform electric field E

V
 d constant
F
E
 q E
Electron gun – motion of a charged particle in an electric field
F

E
F

+Q
V
F
-Q +Q
E
E
F v
-Q d straight line path parabolic path
Work = change in kinetic energy
W

F d
 q E d
 q

E
K

1
2 m v 2 d
 qV
1
2 m v
2  qV v straight line path
Equation Mindmap eq17 : Doing Physics on Line 2

Motion of a charged particle in a magnetic field
F
B
 q v B sin

+ q v
+ I
Direction of force on moving charged particle in a magnetic field is given by the right hand palm rule
F
F palm face v
 thumb
B
 v
  fingers motion of a positive charge in a magnetic field
B
v (+q) v

  thumb
B fingers
 v
  motion of a negative charge in a magnetic field
F palm face
Equation Mindmap eq17 : Doing Physics on Line 3

e

S N
B v
  v
  right hand palm rule thumb fingers palm facing down
F
B
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
F
B
 q v B uniform magnetic field B into page v negative charge q enters uniform
B-field
F
B causes a change in direction, no change in speed since F
B acts at right angle to v r
F force F on charge always directed to centre of circle
Circular motion: Magnetic force = Centripetal force
2
F
B
 q v B F c
 m v r
F
B

F c v
 q B m
Equation Mindmap eq17 : Doing Physics on Line 4

e
J. J. Thompson - measured q e
/ m e
 conclusive evidence that cathode rays were a stream of electrons .
Electric field E  electrons deflected up
Magnetic field B  electrons deflected down heater element cathode anode d
_ deflection plates V
+
Electron gun  acceleration of electrons
6 Vac _
+ accelerating voltage V
A
I coils to produce
B field into page eV
A

1
2 m v
2  v

2 eV
A m e
Electric force F
E
= magnetic force F
B
 zero deflection of electron beam
F
E
 eE
 eV d
F
B
 
2 e V
A m e
 e V
2
 m e
2 d V 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
F
B

F
C
 v 2
R v

2 eV
A m e
F
E

F
B
 eE evB v

E

B
X
Y
Z e e

E
2

V m RB d RB 2
Equation Mindmap eq17 : Doing Physics on Line 5

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

1
2 m v
2
force on a charged particle in a B-field
F
B
 q v B sin

direction of force – right hand palm rule
Diagram: 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 : Doing Physics on Line 6

N
+y v thumb
  fingers
B B
S e
-
F
B
+x palm face
+z v
  beam will be deflected to the right right hand palm rule
F
B magnitude: electron charge q = 1.602x10
-19 C electron mass: m e
= 9.101x10
-31 kg accelerating voltage: V = 6.65 kV = 6.65x10
3 V
B- field: B = 0.875 T
F
B
= ? N v = ? m.s
-1
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. v
 qV

1
2 m v
2
2 qV m e


 
19

 3

31
 m.s
-1   -1
A moving charge in a magnetic field experience a force
F
B
 q v B


 
19

 7
     
13
T
Equation Mindmap eq17 : Doing Physics on Line 7