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AMPERE’S LAW
PURPOSE: To verify Ampere’s Law for the magnetic field produced by a solenoid.
APPARATUS: DC power supply, Hall Effect magnetic probe, Fluke high resolution
voltmeter, solenoid and baseboard
BACKGROUND:
Biot-Savart Law: In 1820 while preparing for a lecture
demonstration Danish Professor Hans Christian Oersted made the accidental discovery
that moving electric charge -- an electric current -- produces a magnetic field. This
discovery provided an unexpected link between the phenomena of electricity and
magnetism which had been previously regarded as unrelated. The Biot-Savart law (also,
less frequently, called the differential form of Ampere’s law) tells how to calculate the
magnetic field due to a current i. According to this law the magnetic field dB produced at
a point P separated a distance r from a length ds of a wire carrying a current I is given
by:
dB 
 o Ids  r
.
4  r3
(1)
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Ampere 2/10
 o is a constant called the permeability constant for the vacuum,
o  4  107 T.m.A-1.
The total magnetic field at P is found by integrating Eq. (1) along the wire,
B   dB .
(2)
For example, if the current is flowing in a long straight wire, then substituting Eq. (1) into
Eq. (2) and integrating shows that the magnitude of the magnetic field at a perpendicular
distance R from the wire is
B
o I
.
2 R
(3)
The magnetic field forms concentric circles around the wire in a sense given by a right
hand rule:
Hold the wire with the right hand with the extended thumb pointing in the
direction of the current; then the curled fingers give the sense of the field.
Integral form of Ampere’s Law: Ampere’s Law can also be expressed in an alternative
mathematical form that is sometimes more convenient to use. We will derive the integral
form for the case of the long straight wire, then assert without proof that it holds for all
configurations. Consider a concentric circular path of radius R encircling the long
straight wire. The integral around the close path of the tangential component of the
magnetic field B (given by Eq. (3)) is called a “line integral” and is denoted by  B  dl ,
where the  means dot or scalar product. Note for the case of the long straight wire that
the direction of B is parallel to dl on the circle, so the dot product reduces to Bdl.
Substituting Eq. (3) into the line integral:
o I
o I
2 R   o I
 B  dl 
 dl 
2 R
2 R
 B  dl   o I
(4)
This is the integral form of Ampere’s Law:
The line integral of B around a single closed path is equal
to the permeability of the medium times the current
enclosed.
It holds true for any closed path: non-concentric circle, ellipse, polygon, etc. If there is
no net current passing through the plane of the closed path, the line integral around the
path is zero. This does not mean there is no B field present along the path. Rather it
means that the dot product with the field direction sums to zero, i.e. the dot product can
be positive on one part of the path and negative on another so that the sum adds to zero.
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It is also possible to have currents passing (up and down) through the plane of the path,
and if their directions sum to zero, so will the integral. Think of two adjacent wires with
equal and opposite currents. The closed line integral around them is zero.
If the closed line integral around a closed path is not zero, you know that there is a net
current through the plane of the path.
Solenoids: Equations (1) and (2) can be used to evaluate the magnetic field produced by
a solenoid (coil of wire):
N = number of turns on solenoid (dimensionless)
R = radius of coil (meters)
I = current through solenoid (amp)
L = length of solenoid (meters)
At the center of the solenoid the magnetic field is
BC 
 0 NI
4 R2  L 2
(5)
Another useful measurement is the magnetic field at the end of the solenoid:
BE 
 0 NI
2 R2  L2
(6)
In the following experiment you will test the integral form of Ampere’s Law for a
solenoid.
The Hall Effect: The magnitude of the magnetic field will be measured with a probe
utilizing the Hall Effect, which is illustrated below. When a current I passes through a
conductor, the potential at two points on the side of the conductor, as shown in the figure,
is zero if no magnetic field is applied. But when a magnetic field is applied as shown, the
magnetic force deflects the path of the charges and causes charge to accumulate on the
sides as shown. (In this figure, the convention is used that current is a flow of positive
carriers.) The amount of charge that accumulates depends on the size of the current and
the strength of the magnetic field. The result of the accumulated charge is to induce a
small but measurable potential, called the Hall voltage, that is linearly proportional to the
magnitude of the magnetic field. If the field is reversed, so is the polarity of the induced
voltage. Also note that if the charge carriers were negative (i.e. electrons) then the
polarity of the induced voltage would also be reversed. Thus if you know the direction of
the magnetic field, the Hall Effect allows you to determine the sign of the charge of the
current carriers.
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The Earth’s Magnetic Field: You will use the Earth’s magnetic field to calibrate the
Hall probe. To a good approximation the Earth can be thought of as a large bar magnet.
The magnetic field lines leave the south geomagnetic pole and enter the north
geomagnetic pole. At our latitude the Earth’s magnetic field is about 0.58 G and is
pointed down about 70o to the horizontal. Thus the horizontal component is about 0.20
G. You will use this number for calibration. [The gauss (G) is an older unit of magnetic
field that is related to the tesla by: 1 tesla = 10,000 gauss. The gauss is still frequently
used because its size is more convenient for many cases.]
PROCEDURE: You will use a Hall Effect magnetic field probe to measure the
magnetic field. The detector is mounted on a clear plastic block that can be oriented in
the magnetic field. The output is amplified and recorded by a highly sensitive voltmeter.
With this device you will experimentally evaluate the line integral of the B-field around a
large wire coil that has a strong field. The experimental technique is very simple. The
closed path is broken up into a series of equal length steps, l . For each step you will
use the Hall probe to measure the component of B that is parallel to  l , Bp  Bcos
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where  is the angle between the magnetic field direction and the path direction. The
line integral is then the sum of these components as you move the Hall probe around the
closed path. That is
 B  dl   Bl cos  B l
p
(7)
Note the Hall probe measures the component of the magnetic field perpendicular to its
plane; if the field is oriented parallel to the current through the probe, there will be no
Hall voltage. So the plane of the probe must be oriented perpendicular to  l .
Warnings: Keep magnetic material (steel watch bands, bracelets, etc.) away from the
experiment. Also, current carrying wires, power supplies, computers, etc. have magnetic
fields. DO NOT CONNECT OR DISCONNECT THE SOLENOID WHEN THE
POWER SUPPLY IS “ON.”
1. Calibration Procedure for Hall Probe(Vernier magnetic field
sensor)
Open Logger Pro software on pc.
The Hall probe measures the magnetic field parallel to the axis of the probe at the probe
tip.
Your task is to zero the sensor before you can take any measurements with it. This
paragraph explains the zeroing procedure conceptually. It is important to remember that
the Hall probe detects ONLY the component of the magnetic field vector perpendicular
to plane of the slab of the Hall probe. In your case the Hall probe is inside the pen-like
sensor parallel to its cross-sectional area. Zeroing means that the probe reads zero when
the component of the magnetic field perpendicular to the probe is zero. As we are on
Earth, which has its own magnetic field and there are lost of magnetic materials around
us, the zero reading can only be achieved when there is no perpendicular component of
the combined field of Earth and environment to the plane of the probe. This
is achieved when the probe is pointed to the direction that is between the
directions that yield the minimum and maximum field reading. The
following steps are required to zero the probe.
a) Make sure your power supply is turned OFF. Open “Ampere’s Law.cmbl, in
your Lab 8 experiment folder; there is no need to Collect data. You should
see a small window in the lower-left hand corner of the window, labeled
“Magnetic Field” (B). Take the magnetic field probe (long pen-like sensor
connected via cable to the LabPro interface) and make sure the tip is straight
(not bent, see photo on the right). Set the probe to the range of 6.4 mT by
sliding the switch on the holder.
b) Place the probe on the large rectangular solenoid board, on the graphic labeled “Zero
Field Compass”, such that the sensor is over the center of the “clock”. IMPORTANT:
Make sure that the arrow and plus symbol (which form more or less a cross) at the tip of
the sensor is always oriented UP, whenever the sensor is held in a horizontal orientation.
Notice that B will vary as you sweep it around the clock
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c) Find the orientation (hour on “clock”) that yields the highest B reading and note it down;
also find the orientation that yields the lowest B reading (which should be 180 degrees
from the highest). To zero the sensor, simply point it in the orientation in the orientation
of 90 degrees from either (this orientation should yield approximately the average of
those values) For example:
-0.053 @ 5 o’clock
-0.006 @ 11 o’clock
—————————————
-0.030 @ 2 o’clock or 8 o’clock (average)
When it reads your average value, hold it at that position, go up to Experiment—>Zero in
the Menu bar of the Logger Pro, and record your reading.
2. Closed Line Integral Enclosing No Currents
Next, turn off the power supply and connect the solenoid red lead to the +12V (black lead
to any black ground plug) on the supply. Now, turn on the power supply. You will do the
line integral for the closed path that lies to the side of the solenoid, but does not enclose
part of the solenoid. Notice that there is a strong field as you move around the path, but
when you do the summation the components (some are positive, others negative) cancel
as predicted by Ampere’s Law.
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3. Closed Line Integral Enclosing Solenoid Current
The second closed path passes inside the solenoid and loops outside and around to return
to the initial position. This integral will not be zero, but will contain information about
the current and number of coil turns in the solenoid. (The total current enclosed by a
closed path through the center is NI.) However, the current and coil turns can be related
to the field strength at the end of the coil, Eq.(6). Substituting Eq. (6) into Eq. (4), we get
 o NI  2BE R2  L2   B  dl   Bpl
(8)
Thus, the line integral can be evaluated in two different ways -- either by summing
around the path as you did in step 2 or by calculating from the dimensions of the coil and
one measurement of the magnetic field, namely at the end of the solenoid. According to
Eq. (8) you will need to measure: R, L, BE.
In Part 3B, after doing the closed line integral, you will evaluate and compare the
magnetic fields at the end and center of the solenoid.
4. Another Closed Loop.
In Part 3 your closed path went through the center of the solenoid. Finally, do the path
around the entire solenoid. What would you expect according to Ampere’s Law? Hint:
what are the directions of ALL the currents passing through the plane of the closed
integral.
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AMPERE’S LAW & MAGNETIC FIELDS
Name: _____________________________________________________Section: ______
Partner: ________________________________________________Date: ____________
To speed up the experiment you may want to directly record your data into an Excel
spreadsheet instead of the tables given below and use it to make the calculations.
1. Calibration
“zero reading” = ______________(
). Enter units.
2. Closed Line Integral Enclosing No Currents.
Turn off the power supply and connect the solenoid leads to the +12 V supply. Then,
turn on the power supply. Set the output voltage to about 11 V. Do the closed line
integral for the path labeled “A.” REMEMBER NOT TO FLIP THE PROBE AROUND
IN DOING THE PATH INTEGRAL - THE PROBE END MUST ALWAYS FACE IN
THE SAME DIRECTION (similar to pushing a toy car around a track.)
l = 1 cm path increment. Enter the magnetic field.
Bp( ) – enter units.
Calculate  Bpl = _________; Is this what is predicted by Ampere’s Law? Examine
this result by calculating the ratio of the sum of positive intervals to the sum of the
negative intervals. What is the expected theoretical ratio of those sums? What does this
tell you about the total sum?
3A. Closed Line Integral Enclosing Currents.
Repeat the above procedure for path “B.” Also, measure the dimensions of the solenoid,
R, L, and put the probe at one end to measure BE. USE CM UNITS FOR R, L, and l.
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Ampere 9/10
Enter the Hall Probes voltages (V) for each l = 2 cm path increment. (The larger
increment will speed up the process.) There are twenty eight 2 cm increments. Again, do
not flip around the probe; it must always face in the same sense as you move around the
closed path.
Bp( ) – enter units.
Calculate
 B l = _________
p
Measure the solenoid dimensions and BE(V) which should be reduced to BE (gauss).
Make sure that you measured BE in the same sense (direction) as you made the line
integration; otherwise, you will have the wrong sign convention.
BE (gauss)
R(cm)
2
2
2BE R  L = ______________
L(cm)
How closely does 2BE R2  L2   Bp l ?
Explain your answer in terms of the prediction from Ampere’s Law.
3B. Center and End Fields
Predict the center field BC from your measured BE by using Equations (5) and (6). Show
your math.
Predicted BC(gauss) = ______________
Now, use the Hall probe and check this. Measure the Hall probe voltage at the center of
the solenoid __________ volts. Now convert this to a magnetic field intensity.
Measured BC (gauss)= ______________
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How close are they? What are the sources of errors?
4. Loop Around Perimeter of Solenoid.
Finally use path “C” which takes you around the entire solenoid (not through it.) First,
predict what the closed path integral will give you according to Ampere’s law.
Predicted  Bpl = _________
Bp(gauss)
Calculate
 B l = _________
p
Discuss the result according to Ampere’s Law.
5. Questions
What is the ratio BE/BC if the solenoid were very short, shaped like a bicycle tire. Show
each step in your math.
What is the ratio BE/BC if the solenoid were long, shaped like a hose? Show each step in
your math.