WJP X, XXXX.XX (20XX)
Wabash Journal of Physics
1
Measuring the Magnetic Moment of a Magnetic Dipole using Two Methods
Daniel Brown and Micah Milliman
Department of Physics, Wabash College, Crawfordsville, IN 47933
(Dated: October 2, 2008)
Applying a uniform external field to a magnet dipole one can easily calculate the dipoles magnetic
moment. There are multiple approaches available that will lead to this value. We first approached
this problem by applying a small angle change to the dipole in a field which created a harmonic
oscillation and led to a magnetic moment value of µ = 0.43 ± 0.03 A m2 (95% CI). We also measured
the moment by balancing the gravitational and magnetic torques on the magnet, giving the value
µ = 0.43 ± 0.01 A m2 (95% CI). This value is within the uncertainty of the values reported by other
research groups using this apparatus.
The magnetic moment of a magnetic dipole is a interesting property that describes the dipoles behavior.
Permanent magnets with north and south ends and induced magnets formed by current flowing through loops
have magnetic moments. Using the right-hand rule you
can determine the direction of the magnetic moment of
the system you are testing. The magnetic moment describes the way the dipole will behave when it encounters
an external magnetic field.
For this experiment we used the Magnetic Torque apparatus made by TeachSpin [2]. This apparatus features
two mounted 195 turn Helmholtz coils with separation
y = 0.1227 ± 0.004 m (95% CI) that create a uniform
magnetic field. In the center of the coils is a cylinder that
the cue ball sits in. The permanent magnetic dipole was
inside a cue ball with mass mtotal = 0.14244 ± 0.00001
kg (95% CI) and radius rsphere = 0.02685 ± 0.0005 m
(95% CI). It also features an air pump that allows the
cue ball to sit frictionlessly in the field. We added a
PASCO Low Voltage AC/DC Power Supply for easier
current adjustment and a Keithley 197A multimeter to
the setup to increase the precision of the current measurement. This is shown in Fig. 1 The two different
methods used slightly different setups. In the harmonic
oscillation procedure a photo-gate and a non-magnetic
rod with mass mrod2 = 0.195 ± 0.01 g (95% CI) was
added to the setup. For the static torque procedure a
plastic weight with mass mmass = 0.00138 ± 0.00001 kg
(95% CI) and thin rod with length lrod = 0.14 ± 0.01 m
(95% CI) and mass mrod = 0.00093 ± 0.00001 kg (95%
CI) were added.
Measuring the magnetic moment depends on a accurate value of the magnetic field that is applied. We used
a DC Magnetometer to measure the field. This instrument is a hall probe that is calibrated by NIST Standards by 2%. We placed the probe vertically in the
field and measured the reading and varied the current.
This is show in Fig. 2. We then took the slope of this
line to find the coefficient of the magnetic field to be
CB = 1.34 ± 0.02 × 10−3 T/A (95% CI). This value is
within uncertainty of the value reported in the TeachSpin
manual, CB = 1.36 ± 0.03 × 10−3 T/A [2].
When a magnetic dipole is put in an external magnetic
field that the dipole aligns with the field, putting it in
equilibrium. The magnetic dipole aligns itself so that
Keithley 197A
Low Voltage AC/DC
Power Supply
Upper Coil
To Lamp Board
Lower Coil
Terminal Strip on Magnet
FIG. 1: This figure shows the circuit diagram for the setup
used in this lab. We added the PASCO Low Voltage AC/DC
Power Supply and the Keithley 197A multimeter to the setup.
The extra devices added an easier way to vary the current
and more accuracy. We wired the devices to the bottom of
the existing apparatus.
its magnetic field lines match the field created by the
external magnetic field. The dipole is then in equilibrium
and at its lowest energy state. If the dipole is offset
by some angle, θ as shown in Fig. 3, the force of the
magnetic field, FB and the force of gravity, Fg will cause
the dipole oscillate. From this oscillation we are able to
calculate the magnetic moment. This can be expressed
by the equation
T2 =
4π 2 IT otal 1
,
µ
B
(1)
where T is period, IT otal is the moment of inertia of the
entire ball configuration, µ is the magnetic moment and
B is the magnetic field. In this situation the moment of
inertia is defined as
WJP X, XXXX.XX (20XX)
Wabash Journal of Physics
of the cylinder, hcyl is the height of the cylinder, and rrod2
is the radius of the added rod. The cylinder is attached
to the ball and measuring the mass of this was tricky.
We determined that the cue ball and the cylinder were
made of the same material, therefore they had the same
density. Using this principle we solved for the mass of
the cylinder. We found
6
Magnetic Field (mT)
5
4
mcyl = mtotal
3
2
1
2
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Current (A)
FIG. 2: The graph shown in this figure shows the uniform
magnetic field produced by the 195 turn Helmholtz coils with
separation y = 0.1227 ± 0.004 m (95% CI). We measured
this value by using a DC Magnetometer which was already
calibrated. We varied the current, which varied the field and
recorded the reading. The slope of the pictured graph gave us
the Tesla/ Ampere value we needed to calculate the magnetic
moment. We found this value to be CB = 1.34 ± 0.02 × 10−3
T/A (95% CI), which is in agreement with the manufacturer
reported value of CB = 1.36 ± 0.03 × 10−3 T/A.
(A)
(B)
B
T =q
1
4π 2 Iball
µCB
Icurrent ,
(4)
where Icurrent is the measured current. This equation is
now in a form that we can use. The most important part
of equation in order to calculate the magnetic moment
is the radical, which is equal to the slope of the fit, m1 .
Solving the radical from Eq. 4 for µ we find
B
µ=
θ
(3)
where Vcyl is the volume of the cylinder and Vtot is the
volume of the cylinder. This provided a better model for
our set-up and more accurate reading of the moment of
inertia.
We will be collecting period and current data and plotting them together. We must manipulate equation 1 to
define the function used in the fit within the areas we are
measuring. Doing this we find that
FB
Non-magnetic rod
Vcyl
,
Vtot
4π 2 IT otal
,
m21 CB
(5)
Fg
θ
r
Rod and plastic mass
FB
μ
μ
Fg
FIG. 3: This figure shows a brief set-up of the apparatus
and the forces that are acting on the cue ball. The cue ball
sits on a cushion of air in a holder with a uniform magnetic
field acting on it. The two different methods require slightly
different setups but in both instances the cue ball apparatus
feels the force of the field, FB , and the force of gravity, Fg .
In the static torque lab we need to make θ = 0 and in the
magnetic oscillation lab we change θ a small amount so that
it will oscillate in the field.
1
2
2
2
IT otal = mrod rrod2
+ msph rsph
(2)
2
5
6
rcyl
hcyl
rsph + hcyl 2
+ mcyl [
+
+(
) ],
4
12
2
where mcyl is the mass of the cylinder, rcyl is the radius
where m1 is the slope of the fit shown in Fig. 4. To
find this we first turned on the air cushion on the TeachSpin apparatus and randomly varied the current through
the coils using the Keithley 197A multimeter, shown in
figure. For this setup a thin non-magnetic rod was inserted into the cue ball. The data used ranged from from
I = 1.000±0.001 A (95% CI) to I = 4.000±0.001 A (95%
CI). We then made sure the rod was pointing vertical and
changed the angle a very small amount, making sure that
the rod traveled perpendicular to the beam of the photogate. Using LoggerPro we recorded the average period,
T , and the standard deviation.
We then plotted the period and current data and applied a fit that we derived from equation 1, as shown
in Fig. 4. From this√graph we see that the value for
m1 = 1.657 ± 0.004 s* A (95% CI). We found using Eq.
5 that the magnetic moment was µ = 0.43 ± 0.03 A m2
(95% CI).
The magnetic moment can also be found by putting
the dipole, rod, and mass in equilibrium between the
force of gravity and the force of the magnetic field. Then
calculate the static torque to determine the magnetic
moment. In this procedure, a rod with a plastic mass
on it is added to the cue ball. We then position the
WJP X, XXXX.XX (20XX)
Wabash Journal of Physics
4
1.8
1.6
3.5
Current (A)
Period (s)
3
1.4
3
1.2
2.5
1
0.8
2
0.02
1
1.5
2
2.5
3
3.5
4
0.04
0.06
4.5
mass somewhere on the rod and adjust the current until
the rod is perpendicular to the field. We recorded between from xmeasured = 0.0345 ± 0.0001m (95% CI) to
xmeasured = 0.1135 ± 0.0001 m (95% CI). We graphed
this data, shown in Fig. 5, and applied a linear fit.
When the rod is perpendicular to the field the force of
gravity on the mass and rod is equal to the force from
the magnetic field on the dipole. This relationship is
described as
µ×B =
X
τ,
(6)
P
where
τ is the sum of the torques on the system. The
magnetic moment and the external magnetic field are
perpendicular so we can rewrite this as
lrod
+ xof f set )
(7)
2
lrod
+ xof f set + xmeasured ),
+ mmass g(
2
where xmeasured is the position of the plastic mass, and
xof f set is the distance that the end of the rod is from the
center of the ball.
We can now see the linear relationship that we will use
for the fit of the data. We now have
µCB Icurrent =mrod g(
Icurrent =
mmass g
+ b,
µCB
(8)
0.1
0.12
Mass Position (m)
Current (A)
FIG. 4: The graph shown in this figure shows the results we
found using the Harmonic Oscillation procedure. We plotted
the current, I in Amperes, and the period of oscillation, T in
seconds. We defined a fit for the data using Eq. 4. We then
used the value of m1 in Eq. 5 to find the magnetic √
moment,
µ. We found this value to be m1 = 1.657 ± 0.004 (s* A (95%
CI).
0.08
FIG. 5: The graph shown in this figure shows the results
we found using the Static Torque procedure. We plotted the
position of the mass, xmass in meters, and the current, I
in Amperes. We added a linear fit to the graph and used
the slope of the fit line, m, in Eq. 8 to find our value for
the magnetic moment, µ. We found this value to be m =
22.70 ± .02 A/m (95% CI).
where b is some constant. To calculate the magnetic moment we need to solve the slope of the fit. Solving Eq. 8
for µ we find
µ=
mmass × g
,
mCB
(9)
where g is gravity, and m is the slope of the linear fit.
From Fig. 5 we see that m = 22.70 ± 0.02 A/m (95%
CI). We found using Eq. 9 that magnetic moment is
µ = 0.43 ± 0.01 A m2 (95% CI).
We have compiled a chart, Fig. 6, that shows the values of µ that other research groups using the same apparatus have measured. The values that we measured,
BM, are in the middle of the two groups measurements.
We believe that the improvements we made in our experiment give us a better value of the magnetic moment for
this specific apparatus.
Although we made many improvements to this experiment, there are still some changes that could be made
if this were repeated. Overall more trials would result in
better data. We did about 20 trials for each method. If
this were doubled or even tripled there would be a better
picture of what is going on. Another area of focus would
be voltage dropping. We had a hard time keeping the
current at a steady value. Fixing this would increase the
accuracy of this experiment. Finally, the levelness of the
apparatus, especially where the cue ball sits, should be
addressed. The apparatus features a level gauge but we
discovered that this is not very accurate. Fixing these
WJP X, XXXX.XX (20XX)
Wabash Journal of Physics
4
few problems should improve this measurement.
TS Feild Graident
TS Far-feild
TS Precessional
TS Oscillation
TS Torque
FP Torque
FP Average
FP Oscillation
BM Torque
BM Oscillation
0.34
0.36
0.38
0.4
0.42
0.44
0.46
0.48
μ (A-m^2)
FIG. 6: The chart shown in this figure presents the values that
our lab and other sources have measured for the magnetic
moment using an apparatus similar to the one used in this
lab. It gives values by Brown and Milliman, BM, Fritsch and
Pizarek [1], FP, and by TeachSpin [2], TS. From this data we
can see that the values we measured are in the middle of the
values provided by the different parties.
[1] Adam Frisch, Tom Pizarek,Two Methods for Determining the Moment of a Magnet Inside a Cue Ball, Wabash
Journal of Physics (2008), Crawfordsville, IN.
[2] Douglas LaFountain, Magnetic Torque* Mτ 1-A, (1997),
Buffalo, NY.