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THE INTEGRAL FORM OF AMPERES LAW

By Greg Hamilton,2014-01-20 04:17
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THE INTEGRAL FORM OF AMPERES LAW

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THE INTEGRAL FORM OF 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: Differential form of Ampere’s 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 produced at dB

r ds a point P separated a distance from a length of a wire carrying a current I is given by:

?Ids ?r o . (1) dB ?3r4?

? is a constant called the permeability constant for the vacuum, o

-1?7.? T.m.A ?4??10o

The total magnetic field at P is found by integrating Eq. (1) along the wire,

. (2) B ?dB ?

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

?Io . (3) B?2?R

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

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field (given by Eq. (3)) is called a “line integral” and is denoted by , where the B B ?dl ?

means dot or scalar product. Note for the case of the long straight wire that the ?

direction of is parallel to on the circle, so the dot product reduces to Bdl. B dl

Substituting Eq. (3) into the line integral:

??IIoo? B ?dl ??R?Idl?2??o22?R?R

?B ?dl ?I (4) ?oThis 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.

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

?NI0? (5) BC22?4RLAnother useful measurement is the flux density at the end of the solenoid:

?NI0? (6) BE22?2RLIn the following experiment you will test the integral form of Ampere’s Law for a solenoid.

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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

o70 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 detector to measure the magnetic field. The detector is mounted on a clear plastic block that can be oriented in the

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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

?l. For each step you will use the path is broken up into a series of equal length steps,

?Hall probe to measure the component of that is parallel to , where B ?l B?Bcos?pis 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

(7) B ?dl ?B?lcos??B?l??p?

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

First, you will calibrate the Hall probe by using the Earth’s magnetic field. The probe

amplifier will give an output voltage (offset voltage) somewhere between 1 and 5 volts

even when there is no applied magnetic field. To determine the zero magnetic field

oreading for the probe you will rotate the probe horizontally through 360. (Use the

inscribed line on the bottom of the clear plastic holder in order to align the probe.) As you

rotate the probe horizontally through the Earth’s magnetic field, you will see a range of

Hall voltages, say, 2.409V to 2.391V, which averages to 2.400V. Thus the probe has a

zero offset, V of 2.400V. Readings greater than 2.400V indicate a positive field, while o

those less are negative fields. You will then use the amplitude of the variation to convert

the probe voltage reading to the magnetic field in gauss. The full peak-to-peak amplitude

variation in the probe voltage corresponds to a change in magnetic field 0.40 G [i.e. the

horizontal component of the Earth’s field changes from +0.20 G to -0.20 G as the probe is

orotated through 360]. The conversion factor b from probe voltage reading to magnetic

field can then be calculated from B = b(V- V). In this example b=0.40 /(2.409-2.391) = o

22.2 G/V

Do the calibration on the “compass” diagram that positions the probe for 12 directions.

Before turning on the 5V supply for the Hall probe, disconnect the +12 V lead going to the solenoid. Turn on the supply and set the Fluke multimeter to (volts DC). Take Vtwelve compass readings and determine V and b from the voltage range. o

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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.

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

22 (8) ?NI?2BR?L?B ?dl ?B?l?oEp?

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, B. E

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

Disconnect the +12V to the solenoid, then turn on the power +5V to the Hall Effect probe.

Place the probe on the “compass” board and record the voltage output for each angle.

Determine the mean value, which is the effective zero offset V, and set the total 0

amplitude equal to 0.40 G to get the conversion factor, b.

V = _________ volt b = ___________ gauss/volt 0

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. 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 uncorrected Hall probe voltages, V.

Hall Probe Voltage (V)

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Now calculate the magnetic field intensities:

B(gauss) p

B?lCalculate = _________; Is this what is predicted by Ampere’s Law? Examine ?p

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 B. USE CM UNITS FOR R, L, and ?l. E

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.

Hall Probe Voltage (V)

Now calculate the magnetic field intensities:

B(gauss) p

B?lCalculate = _________ ?p

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(V) which should be reduced to B (gauss). Make Measure the solenoid dimensions and BEEsure that you measured B in the same sense (direction) as you made the line integration; E

otherwise, you will have the wrong sign convention.

V (volts) B (gauss) R(cm) L(cm) EE

22222BR?L = ______________ How closely does ? 2BR?L?B?l?EEp

3B. Center and End Fields

Predict the center field B from your measured B by using Equations (5) and (6). Show CE

Predicted B(gauss) = ______________ C

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 B(gauss)= ______________ C

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.

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B?l = _________ Predicted ?p

Enter each Hall voltagefor each l = 2 cm path increment.

Hall Probe Voltage (V)

Now calculate the magnetic field intensities:

B(gauss) p

B?lCalculate = _________ ?p

Discuss the result according to Ampere’s Law.

5. Questions

What is the ratio B/B if the solenoid were very short, shaped like a bicycle tire. Show EC