PHYSICS
LAB
NOTES
FOR
ELECTRICITY, MAGNETISM
AND
OPTICS
EXPERIMENTS
PHYSICS 7
Los Angeles Harbor College
J. C. FU
R. F. WHITING
© 1992
Thirteenth Edition (F)
February 2004
ELECTRICAL SAFETY
Since we use electrical apparatus daily, we should understand the elements of electrical safety. Electricity can
kill a person in two ways: it can cause the muscles of the heart and lungs (or other vital organs) to malfunction or it
can cause fatal burns.
Even a small electric current can seriously disrupt cell functions in that portion of the body through which it
flows. When the electric current is 0.001A or higher, a person can feel the sensation of shock. At currents 10 times
larger, 0.01A, a person is unable to release the electric wire held in his hand because the current causes his hand
muscles to contract violently. Currents larger than 0.02A passing through the torso paralyze the respiratory muscle
and stop breathing. Unless artificial respiration is started at once, the victim will suffocate. Of course, the victim must
be freed from the voltage source before he can be touched safely; otherwise the rescuer, too, will be in great danger.
A current of about 0.1A passing through the region of the heart will shock the heart muscle into rapid erratic
contractions (ventricular fibrillation) so the heart can no longer function. Finally, currents of 1A and higher through the
body cause serious burns.
The important quantity to control in preventing injury is electric current. Voltage is important only because it
can cause current to flow. Even though your body can be charged to a potential thousands of volts higher than the
metal of an automobile by simply sliding across the car seat, you feel only a harmless shock as you touch the door
handle. Your body cannot hold much charge on itself, and the current flowing through your hand to the door handle is
short-lived and the effect on your body cells is negligible.
In some circumstances, the 120-V house circuit is almost certain to cause death. One of the two wires of the
circuit is almost always attached to the ground, so it is at the same potential as the water pipes in a house. Suppose a
person is soaking in a bathtub; his body is effectively connected to the ground through the water and piping. If his
hand accidentally touches the high-potential wire of the house circuit (by touching an exposed wire on a radio or heater
for example), current will flow through his body to the ground. Because of the large, efficient contact his body makes
with the ground, the resistance of his body circuit is low. Consequently the current flowing through his body is so large
that he will be electrocuted.
Similar situations exist elsewhere. For example, if you accidentally touch an exposed wire while standing on
the ground with wet feet you are in far greater danger than if you are on a dry, insulating surface. The electrical circuit
through your body to the ground has a much higher resistance if your feet are dry. Similarly, if you sustain an electrical
shock by touching a live wire or a faulty appliance, the shock is greater if your other hand is touching the faucet on the
sink or is in the dishwater.
As you can see from these examples, the danger from electrical shock can be eliminated by avoiding a
current path through the body. When the voltage is greater than about 50 V, avoid touching any exposed metal
portion of the circuit. If a high-voltage wire must be touched (for example, in case of a power line accident when help
is not immediately available) use a dry stick or some other substantial piece of insulating material to move it. When in
doubt about safety, avoid all contacts or close approaches to metal or to the wet earth. Above all do not let your body
become the connecting link between two objects that have widely different electric potentials.
Reference: "Physics for Scientists and Engineers" by F. Beuche
MICROSHOCK
Microshock is electrical shock caused by very small amounts of current. As is shown in table 1, currents of
less than 1 milliampere are usually of no consequence. If a shock is delivered directly to the heart, however, even 20
microamperes of current can be dangerous. Current can be delivered directly to the heart through a pacemaker wire.
Wires for use with external (temporary) pacemakers come out to the body through the chest wall or through veins that
lead to an arm, the neck, or elsewhere. If such a wire were touched by a person who was holding onto a light switch,
electric bed frame, television set, or other appliance, many microamperes could be conducted to the pacemaker wire.
Many appliances will supply a good fraction of a milliampere to someone who is grounded. To see that for
yourself, connect an ammeter between the metal parts of an appliance and ground. (Start on a high range to protect
the meter.) Unless there is a very good third wire ground, significant currents will be measured.
EFFECTS OF A 60 Hz ELECTRIC SHOCK
Current
(held one second)
20 mA
1 mA
5 mA
1-10 mA
10-20 mA
30 mA
75-300 mA
5A
Effect
(current applied to skin, unless otherwise noted)
Ventricular fibrillation if applied directly to the heart
Sensation
Maximum harmless current
Mild to moderate pain
May cause muscular contractions, preventing release from shock source
Breathing may stop
Ventricular fribillation may occur
Burns tissues
TABLE OF CONTENTS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
Plotting the Electric Field ................................................. 1
The RC Time Constant .................................................... 3
Resistivity ........................................................................ 6
The Wheatstone Bridge .................................................. 9
Ohm’s Law ...................................................................... 12
Resistors in Series and Parallel....................................... 15
Joule Heat ....................................................................... 19
The Magnetic Field .......................................................... 22
The Tangent Galvanometer ............................................ 24
Capacitive and Inductive Reactance ............................... 27
The Oscilloscope ............................................................. 30
The Visible Spectrum ...................................................... 34
Reflection and Refraction At Plane Surfaces .................. 37
The Thin Lens ................................................................. 40
Atomic Spectra ................................................................ 44
Radioactivity .................................................................... 47
The set of lab experiments that you will be doing this semester will, hopefully elucidate for
you some abstract concepts, enable you to test a few hypotheses or theories using the scientific
method, realize the capabilities or limitations of certain equipment and procedures, and to think
analytically.
Each lab period will begin with a presentation / discussion of the experiment indicated in
your lab notes. Come prepared, having read the references given. An attempt has been made to
have the labs run in parallel with lecture. Share work with your partner so that each person will
have an opportunity to have hands-on experience.
J. C. Fu, Ph.D.
R. F. Whiting, M.S.
Experiment 1: PLOTTING THE ELECTRIC FIELD
PURPOSE:
The purpose of this experiment is to map the electric field around several electrode
configurations.
INTRODUCTION:
In this experiment, we will determine the shape of the electric field arising from three
electrode configurations, namely, the dipole, the capacitor and the lightning rod. The method
used will involve the determination of equipotential lines (lines of constant voltage). Having
established the equipotential lines, the electric field lines can then be drawn, since they are
always at right angles to the equipotential lines.
SUPPLIES & EQUIPMENT:
Electric field plotting apparatus
Leads and connectors
Ruler & French curve
DC Power supply
Digital voltmeter
Plain and carbon paper
PROCEDURE:
1. Set up the apparatus as shown below:
10 V
White paper
Carbon paper
Electrode paper
- + Voltmeter
+
-
2. Locate ten to fifteen points at 2 volts by making a slight impression with the probe. Space the
points about one centimeter apart. Avoid punching holes in the resistive paper.
3. Repeat step two for 4V, 6V, and 8V points.
4. Trace the outline of the electrode configuration. Mark these electrodes as plus and minus to
indicate their polarity. Connect equipotential points of the same voltage marked on the plain
paper with a smooth line using a French curve or flexible ruler.
-1-
5. Draw the electric field lines (about 12 of them) using a French curve. Remember that these
lines have to be at right angles to the equipotential lines. Indicate the direction of the electric
field lines with arrows.
Shown below are representative electric field lines emerging from the positive electrode and
terminating on the surface of the negative electrode in the dipole configuration:
+
-
6. Calculate the charge (q) on the positive electrode of the dipole system by applying the
definition V = kq/r, where r is the (shortest) distance between the center of the positive
electrode and the 2-volt equipotential line and k is the electrostatic constant, 9 X 109 Nm2/C2.
Charge q = ______________________ C
7. Repeat the above procedure for two other electrode configurations.
DATA SHEET:
Your field drawings are your data sheets. Indicate your lab partner(s) name(s)
on the drawings.
-2-
Experiment 2: THE RC TIME CONSTANT
PURPOSE: a) To determine the discharge rate of a capacitor.
b) To determine the time constant  of an RC circuit.
INTRODUCTION:
The rate of charging and discharging a capacitor in an RC circuit is dependent upon the
magnitude of the resistance and capacitance in the circuit.
The time constant () of an RC circuit is the time required to charge a capacitor to 63% of
its maximum capacity, and the time required to reduce the charge to 37% of its maximum value.
In an RC circuit, the time constant  is equal to ReqC (unit is the second).
We will determine the time constant of an RC circuit from its discharge curve.
R
VR
CHARGING A CAPACITOR
VR, I
C
Q
VC
Vo, Io
Vo
0
0
t (sec)
Voltage and Current Across R
Qo
0
t (sec)
t (sec)
Voltage Across C
Charge On C
R
Switch
C
DISCHARGING A CAPACITOR
VR, I
Vo, Io
Vo
0
0
t (sec)
Voltage and Current Across R
Q
VC
Qo
0
t (sec)
t (sec)
Voltage Across C
-3-
Charge On C
SUPPLIES & EQUIPMENT:
DC power supply 6 V = Vo. Timer
Capacitor (10 F or 6 F)
DVM (H/P) digital voltmeter: Rinternal = 11 X 106 )
Composition resistor (20 M)
Conducting leads
Breadboard
PROCEDURE:
A. CHARGING:
1. Set up apparatus as in Figure I. Check source voltage, V o _______________________
2. Close the switch.
3. Monitor the voltage drop across the resistor on the DVM at "DCV" setting.
4. When the reading on the DVM has leveled off to 0.00 V, open the switch and disconnect from
the power supply.
5. The capacitor is now fully charged. Vc = Vo
B. DISCHARGING:
1. With the capacitor fully-charged, set up the apparatus as in Figure 2.
2. Reset timer.
3. Close switch. Record VR ` every 10 sec. until readings level off.
4. Tabulate data. Calculate Q from Q = CVR.
5. Plot graph (Q vs. Time).
6. Estimate the time constant  from the graph. Indicate on the graph the time at which the
discharging capacitor has only 37% of the initial charge left.
7. Calculate the percent error of  (from graph) when compared with (calculated).
Charging:
Discharging:
C
V
R
C
V
R
Vo
Fig. 1
Fig. 2
-4-
DATA SHEET: The RC Time Constant
R = 20 X 106 (composition resistor)
Rint = Internal resistance of voltmeter
1
Req =

1  1
C = __________ X 10-6 F
Q = CVc
Vc = V R
R

Vc(max) = Vo ________________

Rint
= ReqC = time constant
Data and Calculations Table:
Time (s)
VC (V)
Q (C)
VC(max) = Vo





 calc. = _________
 graph = ________
% error = ________
-5-
Experiment 3: RESISTIVITY
PURPOSE:
The purpose of this experiment is to determine the resistivity of copper and German silver
wires.
INTRODUCTION:
There are three factors that influence the resistance of an ohmic resistor (one which
follows Ohm's law). These factors are the physical shape of the resistor, the material
characteristics of the resistor and the temperature. In this laboratory exercise we shall investigate
how the material makeup of the resistor affects its resistance.
If a uniform metal conductor is considered, it is found that its resistance is directly
proportional to its length, that is, the impeding effects of the resistor increase as the length of the
resistor increases. It is also found that resistance is inversely proportional to the cross-sectional
area of the resistor. A good analogy of these effects is water flowing through a pipe. As the
length of the pipe increases, the frictional effects of the pipe wall and the friction generated by the
water itself increase as the length of the pipe increases, thereby decreasing water flow, which is
analogous to the current in a circuit. Also, it is seen that the volume flow of water increases as
the radius of the pipe increases, suggesting that the resistance to current flow has decreased.
This dependence of the resistance on the length and cross-sectional area of the resistor
can be stated as follows:
L
R =  
A
where , the constant of proportionality, called the resistivity, is characteristic of the material of
which the conductor is made.
In this experiment, the value of the resistivity for copper and German silver shall be
determined. The literature value of the resistivity for these is:
German Silver:  = 33 X 10-6 -cm
Copper:  = 1.70 X 10-6 -cm
SUPPLIES & EQUIPMENT:
Rheostat, 20 
Leads & connectors
Voltmeter
Ammeter
Power supply, 3 VDC
Wire spools
PROCEDURE:
1. Set up the apparatus as shown in the diagram below.
A
3 VDC
Rheostat
Wire Spool
-6-
V
2. For the nickel silver, determine the voltage drop across it and the current through it. Use the
rheostat to keep the current in the wire below 0.30 ampere. From these data, compute the
resistance using Ohm's law. V = IR
3. Record the length and the wire gauge of the wire.
4. Determine the experimental resistivity from  = RA/L. Compare this to the literature value.
5. Repeat the procedure for two other voltage and current settings.
6. Repeat steps 1 through 4 for the copper wire.
-7-
DATA SHEET: Resistivity
Data and Calculations Tables:
Type of Wire
German Silver
Gauge of wire
Radius of Wire
Area of Wire
Length of Wire
(cm)
(cm2)
(cm)
Voltage Across Wire
(V)
Current Through Wire
(A)
Experimental Resistivity
(-cm)
Literature Value of Resistivity
(-cm)
Percent Difference
Type of Wire
Copper
Gauge of wire
Radius of Wire
Area of Wire
Length of Wire
(cm)
(cm2)
(cm)
Voltage Across Wire
(V)
Current Through Wire
(A)
Experimental Resistivity
(-cm)
Literature Value of Resistivity
(-cm)
Percent Difference
-8-
Experiment 4: THE WHEATSTONE BRIDGE
PURPOSE: To measure resistance using a null (zeroing) method.
INTRODUCTION:
The Wheatstone bridge is a circuit designed to measure an unknown resistance by
comparison with other known resistances. It is called a "bridge" circuit because a galvanometer is
bridged across two parallel branches. (abd and acd in Figure 2.) In the balanced condition, there
is zero deflection on the galvanometer. By making use of the ratio of resistance on the two
branches, the value of the unknown resistance can be determined.
Furthermore, referring to the diagrams below, we have:

Vab = Vac
When
V b = Vc
and no current goes through
Wheatstone Bridge
G
b
I1 Rs = I2 RL
Rs
I R
I1 = 2 L
Rs
When Vb = Vc
Rx
a
LL
LR
Fig. 1
I1Rx = I2RR
Since:

d
Vbd = Vcd
Then

G
c
I R
Rx = 2 R
I1
I 2R R
Rx =
I2 R L / R s
R
Rx = R s R
RL
L
RR =  R 
A
b
Rs
a
I1
I2
Rx
G
c
RL
L
and RL =  L 
A
Fig. 2
L
 Rx = R s R
LL
Eq. 1
-9-
d
RR
SUPPLIES & EQUIPMENT:
Galvanometer
Wheatstone bridge
Standard Resistor 50 &100 
Leads & connectors
DC supply 3V
"Unknown resistors"
Composition 47
Decade box 20 - 80 
Rheostat, 20 
Spool of wire #30 copper
PROCEDURE:
1. Set up apparatus as in Figure 1.
2. Start with the composition resistor as Rx.
3. Balance the bridge by zeroing/nulling so that the galvanometer reads zero. current.
4. Record Rs, LL, LR in the data table.
5. Calculate Rx using Eq. 1.
6. Repeat steps 1 through 5 for the other unknowns.
Note that: Cu = 1.72 X10-6 -cm
Diameter of #30 Wire = 0.02548 cm
Spool length = 2000 cm
- 10 -
DATA SHEET: Wheatstone Bridge
Resistor
Rs
()
LL
(m)
LR
Rx
Rx
From Eq. 1
Nominal Value
(m )
()
()
Composition
Decade
Spool
Rheostat
- 11 -
% difference
Experiment 5: OHM'S LAW
PURPOSE: 1. To learn about simple DC circuit elements and Ohm's Law (V = IR)
2. To determine the power consumption of a resistor.
INTRODUCTION:
Ohm's law states the relation between the electrical potential difference, the current and
the resistance in a circuit. This is one of the most frequently used relationships in practical
electricity and electronics.
In this experiment, we will generate a set of current readings corresponding to a set of
voltage readings. The functional relationship between the current and the voltage will then be
determined from a graph of the data. The slope of the graph is then related to Ohm's law.
SUPPLIES & EQUIPMENT:
Voltmeter
DC power supply, 3 Volts
Ammeter
Rheostat, 20 
Leads & connectors
Decade resistance box
PROCEDURE:
1. Set up apparatus as shown in Figure 1.
V
R
V
= DVM (digital voltmeter to measure VR
(parallel wiring).
A
= VOM (volt-ohmmeter) to measure I
(series wiring).
A
Rheostat
VR = DC power supply = 3 Volts.
3V
Fig. 1
Rh = rheostat to regulate VR and I.
2. Set R at 50 .
3. Regulate the voltage across R by adjusting the rheostat.
4. Start with the maximum voltage across R. Record the current I (ammeter reading) and VR.
- 12 -
5. Decrease VR by 0.5 V. Record I and VR.
6. Repeat step 5 for 5 more sets of readings.
7. Plot VR vs.  and compute R from the slope of the line.
V
V = IR,
Slope = R
I
8. Compare to known value of R set on decade box. Calculate the percent difference.
9. Repeat steps 3 through 8 for R set at 500  on the decade resistance box.
- 13 -
DATA SHEET: Ohm’s Law
Data Table 1: R = 50 
Reading
Voltage V
(Volts)
Current I
(Amperes)
Power
(Watts)
Current I
(Amperes)
Power
(Watts)
1
2
3
4
5
6
7
R = ___________ (Measured from graph)
Percent difference = _______________
Data Table 2: R = 500 
Reading
Voltage V
(Volts)
1
2
3
4
5
6
7
R = ___________ (Measured from graph)
Percent difference = _______________
- 14 -
Exp. 6: RESISTORS IN SERIES AND PARALLEL
PURPOSE:
To learn to wire simple series and parallel circuits on a breadboard and to verify the rules
pertaining to these circuits.
INTRODUCTION:
Breadboards allow an experimenter to build and modify circuits very easily, without
soldering. The Protoboard.10 that you will be using contains sockets that are connected together
in the following arrangement:
G
GLOBAL SPECIALTIES
proto-board.10
V1
V2
GND
Each vertical line represents a set of five sockets that are permanently wired together
under the breadboard. Each of the eight horizontal lines represents 25 sockets wired together.
Banana wires can be attached to the three terminals in the upper-right of the breadboard, and
short wires can be run from the base of these terminals to the horizontal sockets to provide
sources of voltage and a ground. Short wires can then connect different sets of sockets together,
with circuit devices such as resistors and integrated circuits plugged in as well.
SERIES CIRCUIT GEOMETRY AND EQUATIONS:
R1
R2
I1
R3
I2
Io = I 1 = I 2 = I 3
I3
Vo = V 1 + V 2 + V 3
Req = R1 + R2 + R3
Io
Io
Vo = Io·Req
Vo
Fig. 1a
- 15 -
PARALLEL CIRCUIT GEOMETRY AND EQUATIONS:
Io = I 1 + I 2 + I 3
I1
R1
I2
R2
I3
R3
Io
Vo = V 1 = V 2 = V 3
Req =
Io
1
1
R1

1
R2

1
R3
Vo = Io·Req
Vo
Fig. 1b.
APPARATUS:
Hewlett-Packard multimeter
Carbon resistors (1000, 470, 270)
SPST switch
Breadboard (protoboard.10)
Leads and connectors
Colored pencils
Power Supply 6 VDC
Multimeter
PROCEDURE:
1. Set the DigiTec multimeter to function as an ohmmeter, plug it into the V1 and GND terminals,
insert the exposed end of each of two short wires through the hole at the base of the two
terminals and screw them tightly in place. By inserting these two short wires into various
sockets, convince yourself that the sockets are connected to each other as described in the
Introduction. Unplug the ohmmeter.
Resistor Color Code
2. Determine
the
nominal
resistance of each carbon
resistor by using the color
code bands. For example,
yellow-violet-brown-silver
becomes 4-7-1-10%, so the
resistance of the resistor is
47 X 101 = 470 ± 47.
Tens Digit
Ones Digit
Exponent
Tolerance
Black = 0
Brown = 1
Red = 2
Orange = 3
Yellow = 4
Green = 5
Blue = 6
Violet = 7
Grey = 8
White = 9
Silver 10% tolerance
Gold 5% tolerance
3. Before you plug into the DC power supply, set up the circuit as in Figure 2a with the switch
open. Then plug into the DC power supply and close the switch. Use alligator clip adaptors
over the banana plugs of the patch cords.
4. Repeat step 3 for the other seven circuits.
5. Calculate R = V/I, from Ohm’s law. Disconnect the breadboard from the power supply, and set
one of the multimeters to act as an ohmmeter. Measure the resistance of each resistor and of
the two combinations directly with the ohmmeter, and calculate the percent difference of this
reading from the value calculated from Ohm’s law.
- 16 -
I1
A
I2
R1
V
R2
R3
A
R1
R2
V
Fig. 2a
R3
Fig. 2b
I3
R1
A
R2
R1
R3
V
R2
R3
V
Io
A
Fig. 2d
Fig. 2c
I1
A
R1
I2
V
R2
A
V
Fig. 3a
I3
A
Fig. 3b
R3
V
Io
A
V
Fig. 3c
Fig. 3d
- 17 -
DATA:
Data and Calculations Table 1: Series Circuit
Fig.
Across
2a
R1
2b
R2
2c
R3
2d
Req
Nominal
Resistance
()
Voltage
(Volts)
Current
(Amperes)
R = V/I
()
Resistance
From
Ohmmeter
()
R = V/I
()
Resistance
From
Ohmmeter
()
% difference
Data and Calculations Table 2: Parallel Circuit
Fig.
Across
3a
R1
3b
R2
3c
R3
3d
Req
Nominal
Resistance
()
Voltage
(Volts)
Current
(Amperes)
- 18 -
% difference
Experiment 7:
JOULE
HEAT
PURPOSE:
The object of this experiment is to verify that the electrical equivalent of heat is 4.186 joules
per calorie.
INTRODUCTION:
When an electric current flows in a conductor with a finite resistance, there is always some
loss of energy due to free electron interaction with the lattice ions in the conductor. This so-called
joule heat can be a desired effect such as in an electric toaster or electric heater or it can be an
undesired side effect such as the heat generated in the windings of an electric motor.
If it is assumed that the resistance of the heating coil which we are using has a constant
resistance over a small temperature range, then for a constant voltage, V, and a constant current,
I, the energy expended per unit time or the power, P, is given by: P = V I. Since we have a
constant power, multiplication of the power by the time interval gives the energy expended in the
heating coil in joules:
Electrical energy (J) = P X t = V X I X t
Eq. 1
Since the medium to which this electrical energy flows is, in our case, an insulated
calorimeter cup containing water, we can apply the conservation of energy to this isolated system
and say that the energy dissipated in the conductor is equal to the energy gained by the water,
calorimeter cup, and stirrer-coil-lid combination.
Electrical energy (joules)  Heat energy (calories)
Eq. 2
The heat energy in calories, gained by a substance is given by the product of its mass, m, specific
heat, c , and the temperature difference, T. Stated in equation form:
Heat energy (cal) = mwcwT + (W.E.)cw T + mcccT
Eq. 3
Combining Eq. 1 and Eq. 3, we obtain:
V I t = mwcwT + (W.E.)cw T + mcccT
______ joules = ______ calories
Eq. 4
Where the subscripts w and c stand for water and cup respectively and W.E. stands for the water
equivalent of the lid assembly in grams and is indicated on the lid. From Eq. 4, the electrical
equivalent of heat in joules per calorie can be calculated.
SUPPLIES & EQUIPMENT:
Double wall calorimeter
Ammeter & voltmeter
Battery charger
Thermometer & glycerine
Leads & connectors
Electronic balance
- 19 -
Ice cubes
10 V DC
Stop clock
PROCEDURE:
1. Weigh the inner cup (without insulating ring) of the calorimeter and record the result. Record
the water equivalent of the lid.
2. Fill the inner cup of the calorimeter about ⅔ full with chilled water about 8oC below ambient
temperature. Weigh and record the mass of the cup and water.
3. Calculate the mass of the water.
4. Set up the apparatus as shown below with the rheostat at its maximum resistance. Make
sure that the heating coil is totally immersed.
Heating
V
Coil (Calorimeter)
A
A
A
S
S
Switch
+
+
-
Switch
-
6 Volts
5. Measure and record the initial temperature of the water just before closing the switch.
6. Activate the circuit and start the timer. Stir occasionally throughout the run.
7. Record the current and voltage used during the run. When the temperature of the contents of
the cup is about 8oC above ambient temperature, disconnect the power, stop the timer, stir
the water, then record the final temperature of the contents of the cup.
8. Calculate the electrical energy for heating in joules.
9. Calculate the heat energy absorbed in calories.
10. Calculate the electrical equivalent of heat and compare this value to the literature value of
4.186 J/cal.
- 20 -
DATA SHEET: Joule Heat
Mass of inner calorimeter cup _________________
Water equivalent (W. E.) of lid assembly _______________
Data and Calculations Table
Mass of water
Initial temperature, (Ti)
(g)
(oC)
Current, (I)
(A)
Voltage, (V)
(V)
Time
(s)
Final temperature, (Tf)
Electrical energy
Heat energy
Experimental value of the
electrical equivalent of heat
Known value of the electrical
equivalent of heat
(oC)
(J)
(cal)
(J/cal)
(J/cal)
Percent difference
Electrical Energy = Pt = (VI) t (Joules)
Heat Energy = mwcwT + [mLcL] T + mcccT (calories)
cw = 1.000 cal/gramoC
cc = 0.22 cal/gramoC
[mLcL] = mL(W.E.) cw = _______________ cal/oC
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Experiment 8: THE MAGNETIC FIELD
PURPOSE: 1) To determine the shape of magnetic fields around magnetic poles.
2) To map the magnetic field around a bar magnet.
INTRODUCTION:
The shape of a magnetic field is determined the lines of magnetic flux. Since the needle of
a magnetic compass will align itself along such a flux line. It is a simple matter to construct a
magnetic field line by marking the positions of alignment of a compass needle. Another way of
determining the field shape of a magnet is to sprinkle iron filings in the region where one wishes
to see the shape of the field. The filings will then be aligned with the flux lines, thereby
delineating the shape of the field.
In this experiment we shall use both methods to determine the shapes of various fields.
Since the magnetic field is a vector field, there should be a point where the Earth's
magnetic field vector just cancels the magnetic field vector of the magnet. This balance point is
called the neutral point. We shall have an opportunity to find this point in this experiment.
SUPPLIES & EQUIPMENT:
2 Bar magnets
8 ½ X 11 paper
Plexiglas sheet
2 Horse shoe magnets
22 X 34 paper
French curve
Iron filings
Small compass
Plywood board
18 Ruler
12 Ruler
PROCEDURE: Part I: SHAPE OF THE MAGNETIC FIELD USING IRON FILINGS
1. Place a sheet of 8½ X 11 paper on
the Plexiglas sheet away from all
magnets.
N
2. Sprinkle a very thin layer of iron
filings evenly over the paper.
10 cm
S
N
S
3. Put the Plexiglas sheet over the first
magnet arrangement in Figure 1.
S
N
4. Tap sides of paper lightly until the
field lines gradually show up.
N
N
S
S
5. Repeat steps 1 - 4 for the other two
pole arrangements in Figure 1.
6. Sketch the magnetic poles and field
lines.
Fig. 1. Magnet Pole Arrangements
- 22 -
Part II: SHAPE OF THE MAGNETIC FIELD USING A MAGNETIC COMPASS
1. Tape the large paper on the wooden board and align the edge of the paper along the NS axis,
with a compass. Tape the bar magnet on the paper as shown in Figure 2.
2. Draw lines AA, BB and CC as shown below:
North
A
A
B
B
C
C
N
S
Fig. 2: Experimental Setup For Part II
3. Start with line BB approximately 10 cm away from the bar magnet and mark the position of
the tip and tail of the compass needle.
4. Move the tail of the needle to coincide with the tip of the previous compass position. Continue
the process until the field line terminates at each pole of the magnet.
5. Join the dots that mark the changing positions of the compass needle tip with a smooth line.
Repeat the process for positions of 20 and 40 cm.
6. Repeat steps 4 and 5 for lines AA and CC.
7. Explain any irregularities from the expected dipole field in your conclusion.
Part III: DETERMINATION OF THE NEUTRAL POINT
1. Make sure the North pole of the magnet is still aligned with the Earth's magnetic South pole
(geographic North). Find the neutral point by sliding the compass slowly along the BB line. At
the neutral point, the magnetic force due to Earth's magnetic poles are balanced by the
magnetic force of the bar magnet. Sketch a vector diagram indicating why there should be a
neutral point.
DATA SHEET: Your data consists of the drawings you have made.
- 23 -