Download Magnetic Mass Balance

Magnetic Mass Balance
The magnetic mass balance can measure masses down to hundredths of a gram –
that’s 10-5kg. It achieves this by using a seesaw balance to compare the weight of a
mass with a force magnetic generated force. Because magnetic forces can be quite
small and well controlled, then a delicate balance can be constructed. Subsequently,
this sensitive measuring device can be used for experiments.
Seesaw Balance:
First let’s revise how a seesaw works. A parent and a child with very different masses
can balance on a seesaw provided that they set their distances from the fulcrum in the
ratio of their masses. It sounds complicated, but everyone just does it automatically.
Referring to the diagram below, a seesaw is in balance when
W1 x1 = W2 x2,
or
m1 g x1= m2 g x2.
The same relation holds even if one of the forces is not a weight, but results from a
magnetic field interacting with a current carrying wire say. In the second diagram, the
seesaw is balanced when
W1 x1 = F x2,
or
x1
m1 g x 1 = F x 2 .
x2
W2 = m2 g
W1 = m1 g
x1
x2
F
W1 = m1 g
This second relation can be rearranged to provide a delicate way to measure an
unknown force F.
F = m1 g x 1 / x2 .
This is exactly how a beam-balance scale works. A set of small known masses is
counterbalanced with an unknown force F. Obviously, if the seesaw is designed so
that x1 = x2, then we have
F = m1 g
so the unknown force F is exactly equal to the weight.
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Magnetic Mass Balance:
The seesaw balance principle can now be used to design a magnetic mass balance.
Referring to the diagram below, a square shaped wire carrying a current I is
suspended from strings and balanced. The square coil is arranged so that one set of
sides points North-South. One of the North-South sides is placed in a magnetic field
B pointing East-West using, say a horseshoe magnet. The length of the side is L and
we assume that the entire side is within the same magnetic field B. A mass m is
suspended on the side opposite the horseshoe magnet.
Strings
North
Horseshoe
magnet
I
Magnetic
Field
Insulating
straw pivot
Mass
Current loop
On the diagram above, show the balanced forces of the weight of the mass m and the
magnetic force acting on the wire between the horseshow magnet.
Write the magnitude of the weight force W.
(W = m g)
Write the magnitude of the magnetic force F.
(F = B I L sin 90o = B I L)
Write a relation between the magnetic force and the weight force assuming the seesaw
arms are of the same length.
(W = F, m g = B I L)
For typical values of B = 5 BEarth = 25 × 10-5T, I = 2A, L = 0.05m, estimate the value
of the mass required to balance the seesaw.
(m = B I L / g = 2.55
× 10-6kg
= 0.003g)
Why is this device orientated North-South and East-West as it is?
You should find that this setup can create a very sensitive mass balance.
A very important skill is to be able to design and test experimental ideas before you
get in the lab. In the next section, we are going to design and test some experiments.
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The Practical Investigation
You will be involved in using a sensitive “current balance” to determine the Magnetic
Field Strength B of a solenoid by finding the force on a current-carrying conductor
which is placed in the field of the solenoid. You are expected to collect sufficient data
to show any relationships between the variables investigated.
Ammeter 1
I1
+
_A
Coil
Power Supply 1
Rheostat 1
Loop
I2
Ammeter 2
A
Power Supply 2
Rheostat 2
+
The diagram above show a sensitive current balance that can be used to measure
the force on a short length of current–carrying wire in a magnetic field that is
produced by an electric current flowing in a coil (solenoid). If the balance is so
aligned that the end of the U-shaped metal loop is perpendicular to the magnetic field
while the sides are parallel to it, only the end will be subject to a force from this
magnetic field.
The force on the end of the loop can be measured by balancing it with a known mass
hung from the other end of the balance.
Part 1:
Finding the relationship between the force on a current-carrying
conductor and the electric current flowing through it.
1.
Balance the loop without any current flowing through the loop and
solenoid.
2.
Move the nut on the bolt in or out until the loop is horizontal.
3.
Adjust the rheostat in the solenoid circuit until a current of approximately
3A is flowing.
4.
Add one 0.01g mass (one ring) and adjust the current flowing in the
loop until the loop is balanced.
5.
Repeat this step by changing the mass (0.01g at a time) and adjusting
the current in the loop each time to balance the loop. Record your data
for 10 different masses.
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Part 2:
Finding the relationship between the force on a current-carrying
conductor and the magnetic field strength across it
1.
Balance the loop without any current flowing through the loop and
solenoid.
2.
Move the nut on the bolt in or out until the loop is horizontal.
3.
Adjust the rheostat in the loop circuit until a constant current of
approximately 2A is flowing.
4.
Add one 0.01g mass (one ring) and adjust the current flowing in the
solenoid circuit until the loop is balanced.
5.
Repeat this step by changing the mass (adding 0.01g at a time) and
adjusting the current in the solenoid each time to balance the loop.
Record your data for 10 different masses.
Detailed calculations, tables and graphs are essential.
Notes
1. You will be given the number of turns of the solenoid and its length.
2. Electric currents through the solenoid and loop should not exceed 4
amperes. This experiment may be conducted with smaller values of
current through the solenoid.
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Experimental Design:
Every good experiment has
 one independent variable (which you vary), and
 one dependent variable (that changes because you changed the independent
variable).
 All the other variables must be controlled (which means they are not changed).
IMPORTANT: The independent and dependent variables will be plotted on a graph
and a line of best fit will be used to determine the quantity of interest.
This is because every measurement has errors. Some of the errors are in the positive
direction and others are in the negative direction. Plotting all your measurements on a
graph and taking the line of best fit will average over all the positive and negative
errors to leave an error close to zero. This is the basis of statistics and data analysis.
For our experiments, we will always plot the dependent variables on the vertical axes,
and the independent variable on the horizontal axes. The regression line of best fit
will then be used to determine the quantity of interest. Usually, the regression line
will be linear, but will sometimes be a polonomial and sometimes a power law.
Exp: Measurement of Magnetic Field of Horseshoe Magnet:
Design an experiment to determine the strength of the magnetic field of the horseshoe
magnet B. Complete a rough indicative table of expected values and a graph of
expected shape. (This planning step is best done before you try to do any experiment.)
Theory: m = (B L / g) I
Independent variable:
Dependent variable:
Gradient:
DEP =
DEP vs IND
DEP (units)
IND =
IND (units)
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(Several options: m = (B L / g) I, IND=I, DEP=m, Grad=(BL/g)
(m = (B I / g) L, IND=L, DEP=m, Grad=(BI/g)
(1 / I = (B L / g) 1 / m, IND=1 / m, DEP= 1 / I, Grad=(BL/g)
(Using different planets: m = (B L I) 1 / g, IND=1 / g, DEP=m, Grad=(BLI)
Examples provided for top option of (m = (B L / g) I, IND=I, DEP=m, Grad=(BL/g)
m
0
0.00E+00
1
1.51E-06
2
2.94E-06
3
4.90E-06
4
6.03E-06
5
9.13E-06
6
9.17E-06
7
1.15E-05
Gradient = 1.6×10-6
= B L / g.
This gives
B = 1.6×10-6 g / L
= 3.14×10-4T
m vs I
1.40E-05
1.20E-05
1.00E-05
m (kg)
I
8.00E-06
6.00E-06
y = 1.6309E-06x
2
4.00E-06
R = 9.8974E-01
2.00E-06
0.00E+00
0
2
4
6
I (A)
6
8
10
This compares with my
assumed value of
2.5×10-4T. The
difference results from
the random noise added.