Physics 201- Fall 2007
Magnetism Solutions
For a summer internship, you are finishing up your design of a desk-top sized magnetic spectrometer for the purpose of measuring the ratio of C 12 to C 14 atoms in a sample in order to determine its age. The idea is to develop an apparatus that is sufficiently portable that it could be taken into the field for measurements.
Your initial design is much like the spectrometer shown in figure 29-24; although, yours does not have the velocity selector. The plan is that by burning (vaporizing) a sample you will create a gas of carbon atoms.
These atoms will then pass through an “ionizer” that, on average, strips one electron from each atom. By putting the ions through an electrostatic accelerator- two capacitor plates with small holes that permit the ions to enter and leave, these ions are accelerated. The two plates are charged so that they are at a voltage difference of 1000 volts. From here, the ions enter a nearly constant, vertical magnetic field. Your magnetic field needs to be adjusted to have the C 12 and C 14 ions separated by at least 2mm when they strike the detector array.
In my design, I have a magnetic field directed into the page and the ions enter from the left.
(I’m not showing the accelerator or ionizer for simplicity.)
Because of their different masses, the two species of ions will travel along different circular paths. The paths are circular because of the Lorentz force that acts on each charged particle.
Detector array
C 12
X
X
X
X
X
X
X
X
X
X
X
X
X
C 14
X
X
X
X limit the radii to under 0.25m.
X
X
X the magnetic field. Rather, they will have the same energy as all ions traverse the same potential difference. (Assuming that no energy is lost through collisions or other interactions.) r
12
, radius of the C 12 path r
14
, radius of the C 14 path m
12
= 12 u=1.99 x 10 -26 kg, mass of C 12 m
14
= 14 u=2.32 x 10 u= 1.66 x 10 -27
-26 kg, mass of C kg, atomic mass unit
14

!
!
!
!
v
12 v
14
, speed of a C
, speed of a C
12
14
ion
ion
B, strength of the magnetic field e= 1.6 x 10 -19 C, charge of each ion
Δ V= 1000V, potential difference
To relate the radii to the magnetic field, we use the Lorentz force, centripetal acceleration, and Newton’s Second law.
F = ma v 2 m
12 v
12 m
14 v
14 evB = m
12
" r
12
= r
14
= r eB eB
Next, we can use conservation of energy to find the speed of the ions in terms of the potential difference. (Assuming that the initial speed of the ions out of the ionizer is zero.)
" U = K f e " V = 1
2 mv 2
# v
12
=
2 e " V v
14
=
2 e " V m
12 m
14
Now, we can substitute this speed expression into the previous one to find the radii in terms of the known quantities. r
12
= m
12 eB
2 e " m
12
V
=
1
B
2 m
12
" V e r
14 m
14 eB
2 m
14
" V e
=
2 e " m
14
V
=
1
B
To match the necessary condition that the radii differ by at least 1mm, we can subtract these two expressions from each other and set equal to 0.001m. r
14
" r
12
=
1
B
(
2 m
14 e
# V
"
14 " 12
)
1
B
2 m
12 e
# V
= 0.001
m
=
1 2 u # V
B e
This can be solved for the necessary magnetic field.

!
1
B =
0.001m
So, B= 1.26T
2u " V e
(
14 # 12
)
That’s a hefty magnetic field, but one that’s not completely out of the realm of possibility.
But,… what are the radii? Is this still a tabletop apparatus?
Plug in the magnetic field into the radii expressions: r
12
= 0.0125m r
14
= 0.0135m
Well, this certainly classifies as “tabletop.” The radii are just over a cm.
This tells us that we can decrease the magnetic field (making for a less expensive device) and still fit with the 0.25m radius constraint.
Let, r
14
= 0.25m, so B= 0.0682T= 682 gauss
With this field, r
12
= 0.23m
This would more than work for our detector array.
Besides the cost (magnetic field strength) - size balance, there is also the question of flexibility. If we design it so the maximum radius is for C 14 , we will not be able to study heavier ions. For the described project, this sounds okay, but it is something to keep in mind.

!
You want to construct an electromagnet by wrapping wire around a hollow plastic tube (20cm long and 3cm in diameter), then connecting the wire to a power supply. The goal is to have an interior field strength of 0.005 T. Due to other constraints, the coil has to be wound as a single layer of wire. You have two spools of wire from which to choose. #18 wire has a diameter of about 1.02 mm and can carry a current of 6.0A before overheating. #26 wire has a diameter of 0.41mm and can carry up to 1.0A.
We need to find out which, if either, wire would allow us to create the desired magnetic field. Which one? Well, the
#18 wire can carry large currents, so that will help to create large magnetic fields, but… it is fat. This means that we will not be able to make many turns around the 20cm tube.
The #26 wire is thinner and therefore can have more turns, but… it isn’t capable of carrying large currents. Each wire has its strengths and weaknesses.
How do each of these factors, maximum current and thickness, effect the magnetic field?
Let’s assume that our solenoid can be treated as an infinite solenoid. (20cm is more than
3cm, so our solenoid is long and narrow.) This means that we can use the result from
Ampere’s Law:
B = µ o
N l
I
N is the total number of turns l = 20cm
I
!
How many turns can be placed on the tube? That depends on the wire thickness, d .
N = l d
So,
!
B =
µ o
I d
For the #18 wire:
6 A
B =
µ o = 7.4
x 10 " 3
0.00102
m
T
!

!
For the #26 wire:
B =
µ o
1 A
0.00041
m
= 3.1
x 10 " 3 T
It looks like the #18 wire will allow us to create a field of 0.005T. (We can turn down the current to make the field exactly equal to 5 x 10 -3 T)