Download Motion in a magnetic field

Advanced Higher Physics
Unit 2
Motion in a magnetic field
Force
A moving charge in a magnetic field experiences a force equal to:
F  BIl sin 
In data booklet
Now if:
q
I
t
Therefore
and
Since a moving charge is
a current.
l  d  vt
q
F  B vt sin 
t
F  qvB sin 
You need to be able to derive this!
Notes:
•This shows that a charge must be moving to experience a force and
that the movement must not be parallel to the field.
•The most common case is when a charge is moving perpendicular to
the field (θ=90˚, sin90=1). In this case the equation reduces to:
You need to be able
to derive this!
F  qvB
In data booklet
With: F magnitude of the force in N.
q magnitude of the charge in C.
v velocity of the charge in msˉ¹.
B magnetic induction in T (B is perpendicular to v).
Direction of the force
The direction of the force is perpendicular to v and B.
B
F
For a negative charge, use the right-hand
rule.
F
v
electron
The force is out of the screen.
B
v
Direction of the force
The direction of the force is perpendicular to v and B.
B
F
For a positive charge, use the left-hand
rule.
v
Positive charge
The force is into the screen.
v
Direction of the force
B
vsinθ
v
θ
When the velocity is NOT perpendicular
to the magnetic field, then the component
of the velocity perpendicular to the
magnetic field must be used:
vsinθ
Where θ is the angle between v and B.
Example
Calculate the magnitude of the force on an electron travelling at
4.5  10 6 ms 1, in a direction perpendicular to a magnetic field of 75 mT.
B=75mT
v
Electron moving with a velocity of
4.5  10 6 ms 1
Path of a charged particle entering
the field perpendicularly
B
-q
F
r
F
v
v
The particle experiences a force
perpendicular to both B and v
(right-hand rule if q negative).
This force changes the direction
of the particle which in turn
changes the direction of the
force and so on.
The particle moves in a circle, and
so the force is centripetal.
mv2
Hence F  qvB and F 
Equating an cancelling gives:
qB 
mv
r
r
Example
A proton travelling at 2.5  10 6 ms 1, enters a uniform magnetic
field, acting over circular area, along a diameter as shown.
The proton leaves the magnetic field.
Describe the velocity:
(a) Inside the magnetic field
(b)After leaving the magnetic field.
Path of a charged particle entering the
field at an angle.
B
v
θ
vcosθ
vsinθ +q
The velocity of the particle has a
component parallel to the field (vcosθ)
and a component perpendicular to the
field (vsinθ).
The perpendicular component (vsinθ) causes circular motion as
described before.
The parallel component is unaffected by the field and stay
constant.
Combining these two components produce helical motion.
Applications of Electromagnetism
•Read green notes page 24-26.
•Read slide velocity selector and mass spectrometer from
Virtual AH Physics.
JJ Thomson’s experiment to measure
the Charge to Mass ratio for electrons
•Read green notes page 26.
•Read slide charge to mass ratio for an electron from
Virtual AH Physics.
•Follow instructions from Virtual Experiment 5: e/m for an
electron.
Example
A proton travelling at 3.2  10 6 ms 1 enters a uniform magnetic
field 65 mT. The magnetic field is perpendicular to the velocity
direction, into the screen, as shown in the diagram.
a) Calculate the force acting on the proton inside the magnetic field.
b) Calculate the radius of curvature of the proton path in the magnetic field.
c) Describe and draw a sketch to show the path of the proton in and beyond
the magnetic field.
d) A uniform electric field is applied and adjusted so that the path of the
proton is undeflected. Show on a sketch how this field is applied showing the
polarity of any electrodes.
e) Calculate the electric field strength required to produce the
proton path in part d).