Transcript Physics_A2_36_ChargedParticlesInCircularOrbits

Book Reference : Pages 113-115
1.
To understand that the path of a charged
particle in a magnetic field is circular
2.
To equate the force due to the magnetic field
to the centripetal force
3.
To examine practical applications of circular
particle displacement
In the previous lesson we have seen that a moving
charged particle is deflected in a magnetic field in
accordance with Fleming’s left hand rule
Magnetic field
coming out of the
page
Initial Path of the
Electron
Conventional
Current
Conventional
Current
Conventional
Current
Force
Force
Force
Electron Gun
Electron Gun
Electron Gun
The force acts perpendicular to the velocity causing the path to
change.... The force acts perpendicular to the velocity causing the
path to change.... The force acts perpendicular to the velocity
causing the path to change.... Circular motion is achieved
The force on the moving charged particle is always at
right angles to the current velocity... Circular motion!
As always with circular motion problems we are looking
a the force to “equate” to the centripetal force
From the last lesson we saw that the force on a charged
particle is F= BQv
BQv
BQv = mv2 / r
Velocity
r = mv/BQ
The path becomes more curved (r reduced) if the flux
density increases, the velocity is decreased or if particles
with a larger specific charge (Q/m) are used
A beam of electrons with a velocity of 3.2x107
m/s is fired into a uniform magnetic field which
has a flux density of 8.5mT. The initial velocity is
perpendicular to the field.
Explain why the electrons move in a circular orbit
Calculate the radius of the orbit
What must the flux density be adjusted to if the
radius of the orbit is desired to be 65mm
[21mm, 2.8mT]
The magic A2 crib sheet quotes the following :
Charge on an electron (e) = -1.6 x 10-19 C
Electron rest mass (me)= 9.11 x 10-31 kg
However it also quotes...
Electron Charge/Mass ratio(e/me) = 1.76x1011 C/kg
Two things.... Do not be phased by this since they could
quote this for another particle... Simply inverting it (m/Q)
for this sort of calculation saves one step in the calculator!
Last lesson we saw the CRT (Cathode Ray Tube)
Such devices can also
be called
Electron guns
Thermionic devices
The cathode is a heated filament with a negative potential
which emits electrons, a nearby positive anode attracts
these electrons which pass through a hole in the anode to
form a beam. This is called Thermionic emission. The
potential difference between the anode and cathode
controls the speed of the electrons.
These are machines which can
be used to analyse the types of
atoms, (and isotopes) present
in a sample
r = mv/BQ
The key to how it works is the
effect that the mass has on the
radius of the circular motion while
keeping the velocity and flux
density constant
First we need to ionise
the atoms in the
sample so that they
become charged...
Electrons are removed
yielding a positive ion
A component known as a “velocity selector” is key
to obtaining a constant velocity
Collimator with
slit
The +ve ions are acted upon
by both an electric field & a
magnetic field.
Only when they are equal &
opposite do the ions pass
through the slit
The electric field is given by F = QV/d & the magnetic field given by
F = BQv Only ions with a particular velocity will allow QV/d = BQv
& hence only ions with that particular velocity make it through the
slit. Note that it is also independent of charge since the Qs cancel
Cyclotrons are a method of producing high energy beams
used for nuclear physics & radiation therapy
An alternating electric field is
used to accelerate the
particles while a magnetic
field causes the particles to
move in a circle, (actually a
spiral since the velocity is
increasing)
Compared to a linear accelerator, this arrangement allows
a greater amount of acceleration in a more compact space
Two hollow D shaped
electrodes exist in a
vacuum. A uniform
magnetic field is applied
perpendicular to the
plane of the “Dees”
Charged particles are injected into a D, the magnetic field
sets the particle on a circular path causing it to emerge
from the other side of this D & to enter the next.
As the particle crosses the gap between the Ds the
supplied current changes direction (high frequency AC) &
the particle is accelerated, (causing a larger radius)
Assuming Newtonian rather than relativistic velocities
apply...
The particle leaves the cyclotron when the velocity causes
the path radius to equal the radius R of the D
v = BQR/m
The period for one cycle of the AC must approximate to
the time for one complete circle (2R) using s=d/t for t
T = 2Rm/BQR
The frequency of the AC will be f=1/T
f=BQ/2m