ICT role in 21st century education and it's challenges.
Electromagnetism
1. ELECTROMAGNETISM
1. Define the following terms & phrases (charge, static electricity, charging by friction/contact/induction,
conductor, insulator, uniform/non-uniform charge distribution, earthing and electrical discharge)
2. Describe the behaviour of like and unlike charges.
3. Name and give symbols for the following: DC power supply, cell, battery, switch, lamp,
resistor, variable resistor, wires joined, wires crossing, ammeter, voltmeter & fuse
4. State the symbol and metric unit for: charge, current, voltage, resistance & power
5. Define the following terms and phrases: Ohm’s law, DC electricity, series, parallel, current
rules and voltage rules.
6. Describe the differences and similarities in the way ammeters and voltmeters are used.
7. Draw and interpret DC circuit diagrams
8. Solve problems using V = IR, P = VI, P = E and Rtotal = R1 + R2
t
9. Describe the magnetic field patterns around permanent magnets, the earth, currents and
coils.
10. State the symbol and unit for magnetic field.
11. Use the right hand grip rule to determine relative field (B) and current (I) directions.
12. Describe how the magnetic field due to a current in a straight wire varies with the size of
current and the distance from the wire.
13. Solve problems using B = µ0I
2πd
Thursday, 4 November 2010
4. THE LANGUAGE OF ELECTRICITY
Term Definition Word list
Charge an electrical quantity based on an excess or deficiency of electrons
static
electricity
a form of electricity where charge does not flow continuously
Thursday, 4 November 2010
7. The process
1. Charge transfer - When two objects are in contact with each other, one object can
transfer electrons to the other object. Protons are not transferred because they are
“locked” into the nucleus.
2. Charge imbalance - When the two objects are moved away from each other the
process of charge transfer is unable to be reversed.
• Positively charged objects have had electrons removed
• Negatively charged objects have gained electrons
Oppositely charged objects attract each other. Those with like charges repel.
Proton (+ve charge)
Neutron
Electron (-ve charge)
Empty space
Based on atomic structure
CHARGING OBJECTS
Thursday, 4 November 2010
9. Static electricity
around us
Read p52 and 53 (Y10 Pathfinder) and then offer some examples to the class
discussion:
The History of Electricity Generation
EXAMPLES
Thursday, 4 November 2010
10. LIGHTNING Warm air currents ascend. Ice
crystals descend removing electrons
from cloud particles in this zone.
A zone of positively charged cloud
particles results
At ground level the air becomes
ionized (by losing electrons) and
these positively charged particles
are attracted to the negatively
charged base of the cloud to give
rise to a lightning bolt (an upstrike)
Thursday, 4 November 2010
14. CHARGING OBJECTS
Methods of charging
Induction - the object being charged is not in contact with the object doing the
charging (usually a rod or a ruler). It involves charge transfer to or from the earth to
generate the charge imbalance in the object being charged. A charge imbalance is a
non-uniform charge distribution
Eg.
++++
- - - -
+
+
+
+
This symbol
represents
a
connection
to the earth
- -
Contact - the object being charged is contacted by the other object and charge is
transferred directly from one object to another.
Friction - the object is rubbed by a material that has a greater or lesser affinity for
electrons and a transfer takes place. It is more the contact than the friction that is
necessary for the charging to take place.
The problem with moisture in the air:
Moisture prevents objects from holding a
charge because it transfers charge to or
from the object resulting in a neutral
object.
Thursday, 4 November 2010
15. The importance of the material
Conductors - are materials that allow charges to flow through them. They do not
hold a static charge because any charge imbalance is easily conducted away.
Insulators - are materials that doe not allow charges to flow through them. They will
hold a static charge because the charge imbalance is not easily conducted away.
Example -
ESA: Ex 15A Q.1 ESA: Ex 15B Q.1, 2 & 3
Thursday, 4 November 2010
16. SHOCKS AND SPARKS
Flow of charge through the body causes the shock
Ionisation or flow of electrical energy from the charged object causes the spark
Examples
Climbing out of a car seat at the gas station. The seat becomes negatively charged
because the electrons are moving from the person’s clothing to the seat. When the
person is out of the car and in contact with the ground the electrons from the car
seat are able to be transferred to the air particles causing these particles to become
ionised. It is these negative ions that can move across a gap between the seat and
the person. This results in the production of a spark.
The shock is caused by the ions losing electrons to the person’s body and these
electrons flowing through the body and into the earth (which is charge flowing into
the person’s body. These charges
Explanation of discharging in terms
of electron transfer to/from an
object to result in neutrality.
Thursday, 4 November 2010
20. THE ELECTRICAL CIRCUIT -
introducing the idea of the
electron pump
Thursday, 4 November 2010
21. + -
A conducting path
Power Supply
1. What is an electric current?
2. What are the two requirements necessary for an electric
current to exist?
Thursday, 4 November 2010
23. 1.Which part of the model represents the power supply?
2. Which part of the model represents the conducting path?
3. Which part of the model represents charge?
THE WATER MODEL
Thursday, 4 November 2010
24. 1.Which part of the model represents the power supply?
2. Which part of the model represents the conducting path?
3. Which part of the model represents charge?
A
B
C
THE BIKE MODEL
Thursday, 4 November 2010
25. THE BIKE MODEL
one link
THINK OF A LINK AS REPRESENTING A COULOMB OF
CHARGE
1. In terms of this model, what do you think is meant by the
term “current” ??
Thursday, 4 November 2010
26. http://regentsprep.org/Regents/
physics/phys03/bsimplcir/default.htm
A conducting path
Power Supply
+ -
Requirements:
a power supply
conducting path around which charge
(electrons or ions) can flow.
components (and sometimes meters)
A component
Current
is a flow of electrons through a circuit
Two types:
AC - Alternating current (electrons vibrate back and forth in wires)
DC - Direct current (electrons flow in wires in one direction only)
Conventional current - the direction in which positive charges would flow in wires
if they could.
Conventional current is from positive to negative in a circuit (see diagram above)
ELECTRIC CIRCUITS
Thursday, 4 November 2010
27. Current - is the rate of flow of electrical charge
- it is the number of coulombs of electrical charge that passes a point in
one second. A coulomb is 6.25 x 1018
charges
- I = electric current (measured in amps, A) by an ammeter:
Connecting an ammeter
I
+ -
A
V
I
Red
Red Black
Black
For charge to flow around an electrical circuit there is a need for a voltage source and
a conducting path that is continuous and connects the positive to the negative
terminal of the power supply.
+ -
Example
charge flows through the
circuit as indicated by the
arrows
ELECTRIC CURRENT
Thursday, 4 November 2010
28. VOLTAGE
Voltage
• Voltage (V) is a measure of the energy lost or gained between two points in a
circuit.
• It is measured in the units volts , (V)
Unit of Voltage: Joule per Coulomb or Volt
(JC-1
) or (V)
V = ∆Ep
q
where V = potential difference or voltage (Volts, V)
∆Ep = change in potential energy that a charge
experiences when it moves from one side to the
other side of a component (Joule, J)
q = the unit of charge (Coulomb, C)
If V = 6V then a coulomb of charge has 6J more electrical potential energy at
point A than it does at point B
1A
V
A B
+
-Consider the voltage across a lamp:
Example
Thursday, 4 November 2010
29. + - V
A •
Two wires joined Two wires crossing
Cell Lamp
Battery (two cells in
series)
Switch
Battery (several
cells)
Diode
Voltmeter Ammeter
Resistor Power supply
fuse
variable resistor
(rheostat)
demo of circuit components ->
Notes CIRCUIT SYMBOLS
Thursday, 4 November 2010
35. FACTORS THAT AFFECT RESISTANCE
Also, some materials conduct electricity better than others Eg. Copper is better than iron
Thursday, 4 November 2010
36. In an insulator,
electrons are fixed
In a conductor, electrons
are free to flow
Label the materials that the
arrows are pointing to
_______________
_______________
CONDUCTORS & INSULATORS
Thursday, 4 November 2010
37. THE VOLTAGE-CURRENT RATIO
1. Consider a lamp in an electrical circuit:
12V
2A
12V represents the energy difference across the lamp. This drives electrons
through the lamp at the rate (or “speed”) of 2A. The voltage:current ratio is
_____
2. Consider a different lamp in an electrical circuit:
12V
1A
This lamp has higher resistance because 12V across this lamp can only drive
electrons through the lamp at a rate of 1A. The voltage:current ratio is _____
This example shows that the greater the voltage:current ratio then the greater the
resistance is. Resistance is the voltage:current ratio
Thursday, 4 November 2010
38. In an insulator,
electrons are fixed
In a conductor, electrons
are free to flow
Label the materials that the
arrows are pointing to
_______________
_______________
STOP OR GO?
Thursday, 4 November 2010
39. Definitions
1. Resistance, R is a measure of the “electrical friction” in a conductor. (the
opposition to the flow of current)
2. It is the ratio of the voltage across a conductor to the current through it.
Resistance = Voltage
Current R = V
I
Unit of resistance
is the ohm, Ω
Resistance is given by the slope or
gradient of a voltage - current graph
Example
In an experiment, the voltage across a lamp is measured and recorded as the current
is increased 1 A at a time. Calculate the resistance of the lamp.
V (V)
I (A)
0 1 2 3 4 5 6
24
20
16
12
8
4
RESISTANCE
V
RI
Thursday, 4 November 2010
40. 1
2
3
A conductor that retains a constant
temperature as the current is increased:
A conductor that is allowed to heat
up as the current is increased
A conductor that is cooled progressively
as the current is increased
V (V)
I (A)0 1 2 3 4 5 6
24
20
16
12
8
4
V (V)
I (A)
0 1 2 3 4 5 6
24
20
16
12
8
4
0 1 2 3 4 5 6
V (V)
I (A)
24
20
16
12
8
4
WHATS HAPPENING TO THE RESISTANCE AS THE CURRENT INCREASES?
Thursday, 4 November 2010
43. For most conductors, as the temperature increases the increased vibration of particles
impedes the flow of electrons. Resistance in the conductor will therefore increase. The
graph slopes upwards.
V (V)
I (A)
0 1 2 3 4 5 6
24
20
16
12
8
4
0 1 2 3 4 5 6
V (V)
I (A)
24
20
16
12
8
4
When a temperature of a lamp increases its
resistance increases
The resistance of a thermistor decreases as its
temperature decreases
LIMITATIONS OF OHM’S LAW
Thursday, 4 November 2010
44. BASIC RESISTANCE PROBLEMS
6. What voltage is needed it a current of 5A is to flow through a resistance of 3Ω?
1. What is the resistance of a bulb it a 240 V supply causes a current of 2 A to flow
through it?
2. What current flows through a heating element of 40Ω resistance when the element
is plugged into a 240 V supply?
5. A current of 2 A flows through a 6Ω resistor. What is the voltage across it?
3. If a current of 3 A is flowing in a resistor across which there is a voltage of 6 V,
what is the resistance?
4. What current must be flowing through a lamp of 0.5Ω resistance if there is a voltage
of 6V across it?
Thursday, 4 November 2010
45. Resistors which are connected end to end are in series with one
another
The total resistance of the series combination, Rs is the sum of the
resistances R1 and R2.
For two or more resistors in series: Rs = R1 + R2 + ...........
Resistors which are connected side by side are in parallel with each other.
The total resistance of the parallel combination, Rp is less than any individual
resistor in the combination.
For two or more resistors in parallel
the total resistance,Rp is given by:
1 1 1 + ....
RP = R1 + R2
RESISTANCE CALCULATIONS
R1 R2
R1
R2
Thursday, 4 November 2010
47. EXERCISES
1. Read p52 to p53 QUIETLY
2. Answer questions 1 to 8 on p54 and 55 in your exercise books. There is no need
to use full sentences
Thursday, 4 November 2010
48. Study the pictures of the appliances shown below and in the
table record the materials in each appliance that are
conductors and insulators.
Material Conductor or Insulator?
Material Conductor or Insulator?
INSULATORS
Thursday, 4 November 2010
50. Draw the following circuit diagrams in the spaces
provided AND when you have finished, assemble them:
1.
2.
3.
Voltmeters are connected _________ components,
ammeters are connected _____ a circuit
CIRCUITS: diagrams & assembly
A
+ -
V
+ -
+ -
Thursday, 4 November 2010
51. CHARACTERISTICS OF SERIES & PARALLEL CIRCUITS
+ -
Series
+ -
Parallel
+ -
+ -
1._
2._
3._
1._
2._
3._
components connected
one other the other.
Single pathway
No junctions
each component has its own
connection with the power
supply
more than one pathway
One or more junctions
Thursday, 4 November 2010
52. http://phet.colorado.edu/simulations/
sims.php?
sim=Circuit_Construction_Kit_DC_Only
CIRCUIT CONSTRUCTION
+ -
1
A
+ -
3
A
+ -
2
A
+ -
1
A
+ -
2
A
+ -
3
A
1. Enter the URL (above) into the address bar of your internet browser.
2. Use the simulation tools to construct each of the following 3 circuits (ensure that you use
identical lamps and an the same power supply for each circuit).
3. Record the current in each circuit and explain your observation.
4. Repeat this exercise for the second set of 3 circuits.
Thursday, 4 November 2010
55. Current in series is constant I1 = I2 = I3
Voltage in series is shared VT = V1 + V2
Note
Voltage is shared in proportion to the size of the resistance
+ -
VT
A1 A3
V1 V2
A2
I1 I3
I2
VOLTAGE & CURRENT IN SERIES CIRCUITS
Thursday, 4 November 2010
56. Current in parallel is shared
IT = I1 + I2
Voltage in parallel is constant
VT = V1 = V2
Note
Current is shared in an inverse proportion to the size of the resistance.
For example:
If R1 = 5 and R2 = 10
and IT = 3
then I1 = 2 and I2 = 1
in other words “charge splits
up as it enters a junction in
a circuit”
+ -
VT
V1
V2
I1
IT
I2
IT
R1
R2
“Double the resistance then halve the current”
PARALLEL CIRCUITS
Thursday, 4 November 2010
57. Example 1
A3 = ________
V1 = ________
V2 = ________
8V+ -
A1
A2
A3
R1
R2
2A
3A V1
V2
8V+ -
R1
V2V1
Example 2
3V
V1 = ________
Rule used ___________________________________________________________
Rules used
____________________________________________________________________
____________________________________________________________________
SIMPLE CIRCUIT CALCULATIONS
Thursday, 4 November 2010
59. For the circuit represented by the circuit diagram above, what is the reading on:
(a) V1
(b) A2
(c) V2
(d) V3
1
+ 9V -
A1
A2
A3
V2 V3
V1
5Ω 10Ω
Examples
ADVANCED CIRCUIT CALCULATIONS
Thursday, 4 November 2010
60. 2
For the circuit represented by the circuit diagram above, what is the reading on:
(a) V3 if V2 = 10 V
(b) A1
(c) A2
(d) A3
(e) What is the value of resistor R?
+ 15V -
A1
A2
A3
V2
V3
V1
5Ω
10ΩR
Thursday, 4 November 2010
61. 3
For the circuit represented by the circuit diagram above, what is the reading on:
(a) V1
(b) V2
(c) V3
(d) A2
+ 12V -
A1
A2
V2
V1
2Ω
3Ω
V34.8Ω1 A
Thursday, 4 November 2010
62. Examples
1
••
100 Ω 100 Ω 100 Ω
Total resistance =
2
Total resistance =
25 Ω 30 Ω 50 Ω
+ -
R1
R2
RT = R1 + R2
RESISTORS IN SERIES
As we add resistors in series the resistance increases and therefore the current
drawn decreases
As we add resistors in parallel the resistance decreases and therefore the current
drawn increases
Thursday, 4 November 2010
69. + -
A
V
IT
9 V
One of these meters has
a very high resistance
The other meter has a
low resistance.
Which is which?
Explain your answer
Thursday, 4 November 2010
70. + -
A
V
IT
9 V
METERS
Ammeter
1. connected in series with
other components
2. has low resistance so
that it doesn’t slow the
current that it is supposed
to be measuring
Voltmeter
1. connected in parallel with other
components
2. has high resistance so that it
doesn’t allow much current to flow
through it. This would reduce the
current and voltage through the
component. It is supposed to be
measuring the voltage
Thursday, 4 November 2010
72. • Power is the rate at which electrical energy is transferred into other forms of
energy.
• It is the amount of work done per second
P = E
t
• It can be shown that the electrical power supplied to a device is given by:
P = VI
P
V I
E
P t
P = Power (Watts, W)
E = the amount of energy converted or work done (J)
t = the time taken (s)
P = power (watts, W) (1W = 1 Js-1
)
V = Voltage (volts, V)
I = Current (amps, A)
POWER
Thursday, 4 November 2010
73. POWER & ENERGY
Total energy used by a component/appliance
can be calculated from the equation:
E = P.t
When the power value of the component/appliance is known and this value does not
change over time.
If power changes over time then this change can be graphed.
P (W)
t (s)0 2 4 6 8 10
2
4
6
8
10
12
14 E = Area under the graph
E = 0.5 (9 + 13) = 4.4 J
10
Thursday, 4 November 2010
74. TOTAL POWER USAGE in a parallel circuit
PT = P1 + P2 + P3 + P4
= 12 + 12 + 6 + 6
= 36 W
(ii)
Example
The power usage of the 4 lamps in parallel shown in the circuit below can be
calculated in two ways:
(i) Use the total voltage (supply voltage) and the total current (current drawn from
the supply) to calculate power.
(ii) Add the power usage of each of the components in parallel.
(i)
Lamps 3 & 4: I = P
V
= 12/12
= 1 A
lamps 1 & 2: I = P
V
= 6/12
= 0.5 A
12 V
12 V
12 W
12 V
12 W
12 V
6 W
12 V
6 W
P1
P2
P3
P4
IT = I1 + I2 + I3 + I4 = 0.5 + 0.5 + 1 + 1 = 3 A
P = VTIT = 12 x 3 = 36 A
Thursday, 4 November 2010
75. LAMP BRIGHTNESS IN CIRCUITS
Three main points
(i) The brightness of a lamp depends on its power output since for a lamp, power is
the rate at which electrical energy is converted into light (and heat)
(ii) In a circuit which has values of voltage and current, it is both the voltage and
current that determine brightness.
(iii)The lamp’s resistance will determine that voltage:current ratio that it possesses
Example:
The series circuit (below), shows 2 identical lamps. A third identical lamp is
added to the circuit. Explain how the brightness of the lamps in the circuit
changes
+ -
12 V
Thursday, 4 November 2010
77. TV Stereo Jug Heater Stove torch downlight
Voltage (V) 240 240 240 6
Current (A) 0.3 0.02 8 0.05
Resistance (Ω) 250 450
Power (W) 1000 1200 1800 0.3 140
Running time
(s)
20 min 20 min 10 min 3 h
Energy (J) 1000 125000 300000
POWERING THROUGH THE QUESTIONS
Thursday, 4 November 2010
78. 9V+ -
40Ω
X
0.1 A
0.5 A
10Ω
A 9 V battery is connected in the
circuit shown.
A current of 0.5 A is found to
pass through the 10Ω resistor.
(e) Determine the heat energy generated per second in the whole circuit.
(a) Calculate the voltage across the 10Ω resistor.
(b) Show that the voltage across the parallel combination of resistors is 4.0 V.
(c) If 0.1 A passes through the 40Ω resistor, determine the current through resistor X.
(d) Show that the resistance of resistor X is 10Ω.
“PARALLEL WITHIN SERIES”
Thursday, 4 November 2010
79. PARALLEL CIRCUIT IN ACTION
A car has two tail lights and two brake lights connected as shown in the diagram:
(a) Calculate the resistance of:
(i) a tail light
(ii) a brake light
(b) Calculate the current supplied by the battery when both S1 and S2 are closed.
(c) When the driver takes her foot off the brake S2 is opened state what
happens to the size of the current from the battery and give a reason
for your answer.
Thursday, 4 November 2010
82. (a) 5V
(b) V in series is shared. The combination of the 40Ω resistor and X are in series
with the 10Ω. Therefore V of the combination is 9 - 5 = 4
(c) Current through X = 0.5 - 0.1 = 0.4A because current in parallel is shared. The
total current flowing from the power supply is shared out between the 40Ω
resistor and X.
(d) I through X = 0.4A and V across X = 4V (since voltage in parallel is constant).
R = V/I = 4/0.4 = 10Ω
(e) “Heat energy per second” is the definition of power. For the whole
circuit, I = 0.5A and V = 9V. P = VI = 9 x 0.5 = 4.5W
“PARALLEL WITHIN SERIES”answers
PARALLEL CIRCUIT IN ACTION
(a) The current through a tail light needs to be calculated first:
P = VI => I = P/V = 6/12 = 0.5A For a tail light, R = V/I = 12/0.5 = 24 Ω
For a brake light, P = VI => I = P/V = 12/12 = 1A R = V/I = 12/1 = 12 Ω
(b) I total = 2 x 0.5 + 2 x 1 = 3 A
(c) Less current flows through the battery because there is now more resistance in
the circuit because of the reduction in the number of pathways available for
charge to flow.
Thursday, 4 November 2010
84. THE MAGNETIC FIELD AROUND A BAR MAGNET AND
THE EARTH’S MAGNETIC FIELD
Thursday, 4 November 2010
85. WHAT IS MAGNETISM?
• Magnetism is caused by moving electrons. (The smallest magnetic field is
produced by the motion of 1 electron)
• When electrons move in a common direction, a magnetic field is produced
(sometimes called a magnetic force field)
• A force will be exerted on an iron object placed in a magnetic field.
• A magnetic field is a region in space where a magnetic force can be detected.
The magnetic field around a bar magnet
S
N
The compass needle is
itself a tiny magnet (the
North pole of this magnet
points towards the South
end of the magnet)
Charm compass
Magnetic field lines
Strong magnetic field (high density of lines)
---> Demo c bar magnet & major magnet
Thursday, 4 November 2010
86. • θ changes with time
• Angle of Dip - is the angle that the field lines make with the ground. At the
equator, the angle of dip is zero. Near the poles the angle of dip is close to 90
degrees.
THE EARTH’S MAGNETIC FIELD
compass
S
N
θ (angle of declination) = 11o
Geographic
North
Earth’s axis
Magnetic
South
Thursday, 4 November 2010
87. Home to millions of species including humans, Earth is currently the only place in the
universe where life is known to exist. The planet formed 4.54 billion years ago, and life
appeared on its surface within a billion years. Since then, Earth's biosphere has
significantly altered the atmosphere and other abiotic conditions on the planet,
enabling the proliferation of aerobic organisms as well as the formation of the ozone
layer which, together with Earth's magnetic field, blocks harmful solar
radiation, permitting life on land.
THE EARTH’S MAGNETIC FIELD IS ESSENTIAL FOR
LIFE ON THE PLANET.
The physical properties of the Earth, as
well as its geological history and orbit,
have allowed life to persist during this
period. The planet is expected to continue
supporting life for at least another 500
million years.
Thursday, 4 November 2010
90. MAGNETIC THEORY
Ferromagnetic materials (Iron, Cobalt and Nickel) can be permanently magnetised.
Electrons spinning in atoms have magnetic fields around them. They set up tiny North
and South poles. Such an arrangement for an electron is called a dipole moment.
[Illustrate a mag. dipole and mention Exchange Coupling]
For most elements:
magnetic fields cancel.
N
S
e
N
S
Iron, Cobalt and Nickel:
Electron structure is such that there is a resultant magnetic field produced by each
atom. These atoms are sometimes called atomic magnets.)
Thursday, 4 November 2010
91. Regions in a metal where the orientation of the magnetic dipoles is the same are
called domains.
Fully magnetised
=> the orientation of
the domains is the
same
Unmagnetised Iron
=> Domains are
scrambled
A domain
Here, a large number
of iron atoms
(magnetic dipoles)
are aligned.
Partially
magnetised
DOMAINS
N
NS
S
Thursday, 4 November 2010
93. “Lining up” the domains - Magnetising
• Stroke the object end to end with a permanent magnet , in the same
direction, using the same pole of the magnet.
• Hold the object inside an D.C or A.C solenoid (Domains line up in the
direction of the magnetic field)
“Scrambling” the domains - Demagnetising
Heat or hammer the magnet (This disturbs the alignment of the domains)
[CAN ALSO BE DEMONSTRATED WITH THE SOLENOID]
MAGNETISING AND DEMAGNETISING [Solenoid Demo]
“Domains are induced into alignment”
- Picking up iron objects
Thursday, 4 November 2010
99. A circular magnetic field is formed around a straight current - carrying conductor:
View from above
•
3D View
The direction of the magnetic
field lines is given by the
Right-hand Thumb Rule
The right hand thumb rule:
Thumb = direction of the electric current
Curled fingers = direction of the circular magnetic field
( “•” represents current directed out of the page)
I
WIRES
Thursday, 4 November 2010
101. PARALLEL WIRES
Piece of card
field lines bunch and this
leads to the wires
repelling each other
Current in opposite directions
I I
Thursday, 4 November 2010
102. B = Magnetic field strength (in Tesla,T)
I = Electric current in the wire (in Amps,A)
µo= the permittivity of free space (the ability of a
material to support a magnetic field (TmA-1
)
d = Distance from the wire (in metres, m)
B = µ0I
2πd
1. Reversing the direction of the current reverses the
direction of the magnetic field.
2. Magnetic field strength (symbol, B) is measured in
NA-1m-1 or Tesla,T.
3. As the current in the wire, I increases the strength of the magnetic field
increases
B α I i.e. B is proportional to I
4. As the distance,d from the wire increases the strength of the magnetic field
decreases.
B α 1/d i.e. B is inversely proportional to d
Note
Thursday, 4 November 2010
103. Example
A special meter able to measure the magnetic field strength at any given point in the
vicinity of a wire is shown below (taking a reading). It measures the magnetic field
strength as 8 x 10-4
T at a distance of 0.01m from the centre of the wire. The current
through the wire is 5 A.
Calculate the value of the constant µo.
Exercises
B I d µo
5 x 10-5
T 2 A 20 mm 3.14 x 10-6
6 x 10-5
T 3A 0.159 m 2 x 10-5
TmA-1
7.2 x 10-5
T 3.11 A 2.2 cm 3.2 x 10-6
TmA-1
1.05 x 10-3
3A 20 mm 4.4 x 10-5
TmA-1
Thursday, 4 November 2010
106. The magnetic field of a solenoid is
similar in shape to that of a bar
magnet:
If the current is known, the poles of the solenoid can be determined using the right
hand thumb rule applied earlier to the straight wire:
Draw the field lines
Complete this:
Field lines are parallel in the core of the solenoid which --> the magnetic field in the
core is uniform.
The density of magnetic field lines is greatest in the core --> the magnetic field
strength is greatest in the core.
THE SOLENOID
Thursday, 4 November 2010
107. Predicting North and South poles:
Thumb points to North
pole of the solenoid
from inside the coil
Curled fingers indicate
the direction of the
current
Factors affecting the strength of the magnetic field:
1. Increasing the current increases the magnetic field strength.
2. Increasing the number of turns of wire per given length of the
electromagnet increases the magnetic field strength
STRENGTH RULZ
Thursday, 4 November 2010
108. Uses of electromagnets
1. Electromagnets in relays are able to open and close electrical circuits (eg. starter
motor circuit in a car).
2. Used in scrap yards to lift car bodies.
3. Create the ringing sound in electric bells.
4. Electromagnets in the recording heads of tape recorders are used to magnetise
the audio tape during recording.
A solenoid which contains an iron core is called an electromagnet.
Adding an iron core increases the strength of the magnetic field because
the iron core itself becomes magnetised and adds to the magnetic field of
the solenoid.
ELECTROMAGNETS
Thursday, 4 November 2010
109. Induced magnetism
An unmagnetised object will have have its
domains aligned and therefore develop a north
and south pole. The object can be picked up
by the magnet because opposite poles attract.
x x x x x x x x
- - - - - - - -
North end of coil South end
S NN
Attraction to South of coil
Attraction to
North of coil
The dipoles in the object change along the rod as the rod is drawn into the coil and it
is this dipole change which pulls the rod into the coil
Cross-section of coil
THE COIL GUN
Thursday, 4 November 2010
115. plastic rod (charged by rubbing with a cloth)
small pieces
of torn paper
Observation:
Explanation:
1
2 Balloon rubbed against hair
Removed from head and then brought back to hair
PLAYING AROUND WITH STATIC ELECTRICITY
Observation:
Explanation:
Thursday, 4 November 2010
116. 3
Balloon rubbed against jersey
Release
Observation:
Explanation:
4
cotton
(a) Each balloon charged separately by rubbing against
the sleeve of a jersey
(b) Holding the balloons by the cotton, release them,
allowing them to come close to each other.
Observation:
Explanation:
Thursday, 4 November 2010
117. Charged plastic rod is held near a thin stream of water
http://phet.colorado.edu/new/simulations/sims.php?
sim=Balloons_and_Static_Electricity
Observation:
Explanation:
6
5
(a) Straw, charged at both
ends (using a woollen
cloth)
(b) Straw, also charged using a
woollen cloth held horizontally
and brought close
(c) Repeat (b) using a silk cloth.
Observation:
Explanation:
Thursday, 4 November 2010
118. Cap
Insulating material
Body of electroscope
Leaf
Base
Aim
to charge an electroscope by both induction and by contact and to draw charge distribution
diagrams
Method
1. Follow the instructions below
2. Write observations as you perform each step
3. Complete the diagrams only after recording the observations (you may need some help with
these)
Part 1 - Charging by induction
Equipment
dry cloth/jersey
perspex rod
ebonite rod
electroscope
1. Charge the rod by rubbing it with a dry cloth/jersey and hold the rod
near the cap of the electroscope
Observation:
If the rod was positively charged the charge distribution diagram would look like this:
++++
- - - -
+
+
+
+
CHARGING OBJECTS
Thursday, 4 November 2010
119. Complete the diagram to show how charges would distribute on the electroscope should the rod
be negatively charged.
- - - -
2. With the rod in this position, earth the cap with your finger.
Observation: ++++
- - - -
+
+
+
+
This symbol
represents
a
connection
to the earth
Complete the diagram to show how charge moves
when the cap of the electroscope is earthed
Thursday, 4 November 2010
120. 3. Unearth the cap of the electroscope without removing the
charged rod
Observation:
Draw the charge distribution diagram (by adding to
the existing diagram on the right) showing the
situation once this charge movement has finished.
++++
- - - -
+
+
+
+
Draw the resultant charge distribution and the new
position of the leaf on the diagram (right).
4. Remove the charged rod
Observation:
Finally, complete the diagram (right).
Thursday, 4 November 2010
121. Part 2 - Charging by contact
Method
1. Follow the instructions below
2. Write observations as you perform each step
3. Complete the diagrams only after recording the observations (you may need some help with
these)
1. A positively charged rod is held near the cap of the electroscope.
2. The rod makes contact with the cap.
3. The rod is removed.
++++
Thursday, 4 November 2010
122. Cap
Insulating material
Body of electroscope
Leaf
Base
Aim
to charge an electroscope by both induction and by contact and to draw charge distribution
diagrams
Method
1. Follow the instructions below
2. Write observations as you perform each step
3. Complete the diagrams only after recording the observations (you may need some help with
these)
Part 1 - Charging by induction
Equipment
dry cloth/jersey
perspex rod
ebonite rod
electroscope
ALL CHARGED UPLab 12
1. Charge the rod by rubbing it with a dry cloth/jersey and hold the rod
near the cap of the electroscope
Observation:
If the rod was positively charged the charge distribution diagram would look like this:
++++
- - - -
+
+
+
+
The leaf of the electroscope springs up.
Thursday, 4 November 2010
123. Complete the diagram to show how charges would distribute on the electroscope should the rod
be negatively charged.
- - - -
2. With the rod in this position, earth the cap with your finger.
Observation: ++++
- - - -
+
+
+
+
This symbol
represents
a
connection
to the earth
Complete the diagram to show how charge moves
when the cap of the electroscope is earthed
-
-
-
-
++++
- -
Electrons at the cap are
repelled by the negatively
charged rod.
Leaf of the electroscope drops
Electrons at the cap are held in
position by the positively
charged rod. The earth supplies
electrons to the positively
charged leaf and lower stem.
Thursday, 4 November 2010
124. -
-
-
-
3. Unearth the cap of the electroscope without removing the
charged rod
Observation:
Draw the charge distribution diagram (by adding to
the existing diagram on the right) showing the
situation once this charge movement has finished.
++++
- - - -
+
+
+
+
Draw the resultant charge distribution and the new
position of the leaf on the diagram (right).
4. Remove the charged rod
Observation:
Finally, complete the diagram (right).
The cap and leaf now have
no overall charge.
Electrons on the cap are
still held in position.
Leaf of the electroscope remains in the “dropped”
position. The charge distribution has not changed
- - - -
-
- -
-+
+
+
++
+
- -
-
- -
-
-
-
Negative charge redistributes itself
around the metal parts of the
electroscope leaving the stem and
leaf with an overall negative charge
The leaf of the electroscope springs up.
++++
+
+
Thursday, 4 November 2010
125. Part 2 - Charging by contact
Method
1. Follow the instructions below
2. Write observations as you perform each step
3. Complete the diagrams only after recording the observations (you may need some help with
these)
1. A positively charged rod is held near the cap of the electroscope.
2. The rod makes contact with the cap.
3. The rod is removed.
++++
- - - -
+
+
+
+
+
+
+
+
-
Electrons migrate up
into the rod
+ + +
+
+
+
+
The electroscope is now left with
an overall positive charge.
Charge separation occurs. Positive
repels positive at the stem/leaf
Thursday, 4 November 2010
126. 12 Physics > resources > electricity > DC
electricity > videos
SPARKS
Thursday, 4 November 2010
127. • When the generator is turned on, the electric motor begins turning the belt.
• The belt is made of rubber and the lower roller is covered in silicon tape. Silicon has
a greater affinity for electrons than rubber and so it captures electrons from the
belt. The belt in turn must capture electrons from the dome, leaving the dome
positively charged.
Label the picture of the Van der Graaf
(left) using the labels in the box below:
Lower roller
Belt - A piece of surgical tubing
Output terminal - an aluminium or steel sphere
Upper roller - A piece of nylon
Motor
Upper brush - A piece of fine metal wire
Lower Brush
______________
______________
______________
______________
______________
______________
______________
Reference: http://science.howstuffworks.com/vdg3.htm
THE VAN DER GRAAF - HOW IT WORKS
Thursday, 4 November 2010
128. THE VAN DER GRAAF - OBSERVATIONS & EXPLANATIONS
1. Small dome held close to generator dome
2. Hair stands on end when contact is made with the generator
dome
3. Aluminium foil plates flying of the top of the generator dome
Drawn observation
Drawn observation
Drawn observation
Explanation
Explanation
Explanation
Thursday, 4 November 2010
129. 4. Sparking - a result of ionisation --> thorough step by step
explanation
5. The shock that is felt ------> Charge travelling from/into the
earth, through the body
Thursday, 4 November 2010
132. ADDING BULBS IN PARALLEL
1. Set up each of the following circuits, one after the other (making a mental note of
the brightness of the lamps in the circuit.
2. For each circuit read the ammeter and record the current in the space provided.
+ -
1
A
+ -
2
A
+ -
3
A
Current = ______ A
Current = ______ A
Current = ______ A
Observation
Explanation
8 V 8 V 8 V
Thursday, 4 November 2010
133. ADDING BULBS IN SERIES
1. Set up each of the following circuits, one after the other (making a mental note of
the brightness of the lamps in the circuit.
2. For each circuit read the ammeter and record the current in the space provided.
+ -
1
A
Current = ______ A
+ -
3
A
Current = ______ A
Observation
+ -
2
A
Current = ______ A
Explanation
Thursday, 4 November 2010
134. CURRENT IN THE SERIES CIRCUIT IS CONSTANT
Aim
to look for a pattern in the current through bulbs and resistors in a series circuit.
1. Use ONE ammeter in the three different places shown in the circuit diagram.
2. Without changing the setting on the power pack or the variable resistor write the
current readings in the spaces provided (below):
Equation
8V+ -
A2
A1
A3
A1 = ______ A
A2 = ______ A
A3 = ______ A
Thursday, 4 November 2010
135. CURRENT IN THE PARALLEL CIRCUIT IS SHARED
Aim
to look for a pattern in the current through bulbs and resistors in a parallel circuit.
1. Use ONE ammeter in the each of the four places shown in the circuit diagram.
2. Without changing the setting on the power pack record your results below:
Conclusion
Current in a parallel circuit is ____________ .
Equation relating the currents
Results
A1 = ____ A
A2 = ____ A
A3 = ____ A
A4 = ____ A
8V+ -
A1
A2
A3
A4
Thursday, 4 November 2010
136. VOLTAGE IN THE SERIES CIRCUIT IS SHARED
Aim
to look for a pattern in the voltages across bulbs and resistors in a series circuit.
Conclusion
The power supply voltage is _____________ between the components in the circuit
8V+ -
V1
V2 V4
V3
A
Equation relating the voltages
1. Set up the circuit (below)
2. Use ONE voltmeter in the three different places shown in the circuit diagram.
3. Without changing the setting on the power pack record your results below:
Results
V1 (power supply) = __ V
V2 (variable resistor) = __ V
V3 (bulb) = __ V
V3 (ammeter) = __ V
Thursday, 4 November 2010
137. VOLTAGE IN THE PARALLEL CIRCUIT IS CONSTANT
Aim
to look for a pattern in the voltages across bulbs and resistors in a series circuit.
Conclusion
The voltage across components connected in parallel is __________
Equation relating the voltages
Results
V1 = ___ V
V2 = ___ V
V3 = ___ V
8V+ -
V1
V2
V3
1. Set up the circuit (below)
2. Use ONE voltmeter in the three different places shown in the circuit diagram.
3. Without changing the setting on the power pack record your results below:
Thursday, 4 November 2010
138. VOLTAGES AND CURRENTS IN SERIES AND PARALLEL
Aim
to investigate voltage and current in a series circuit that has a parallel portion in it.
1. Set up the circuit (below)
2. Use ONE voltmeter in the four different places shown in the circuit diagram
and ONE ammeter in the four different places shown.
3. Without changing the setting on the power pack record your results below:
Results
A1 = __A
A2 = __A
A3 = __A
A4 = __A
V1 = __V
V2 = __V
V3 = __V
V4 = __V
8V+ -
V1
V3
V4
V2
A1
A2
A3
A4 V4
Thursday, 4 November 2010
140. OHM’S LAW
Results
Voltage setting of Power pack
(V)
2 4 6 8 10 12
Voltage, V (V)
Current, I (A)
+ -
A V
ice
beaker
water
immersion coil
Method
1. Set up the following circuit using
iced water to cool the immersion
coil.
2. Increase the voltage in regular
increments through an
appropriate range (widest
possible range)
Lab 14
Thursday, 4 November 2010
141. Notes
•The iced water was used to keep the temperature of the coil constant .
•If the iced water was forgotten and the coil was allowed to heat up then the graph would curve up.
Draw a graph
of Voltage
against Current
on the grid
provided
Repeat the experiment but this time replace the coil with a lamp (that will
increase in temperature as the current through it increases)
Conclusion
Thursday, 4 November 2010
143. Resistance
specified
MeasurementsMeasurements Resistance
calculated
from
measurements
Calculated
Power output
Resistance
specified Voltage Current
Resistance
calculated
from
measurements
Calculated
Power output
R1
R2
R3
Resistance calculated
(formula provided for parallel resistors)
Resistance calculated
(formula provided for parallel resistors)
across
combination
of resistors
drawn from
the power
supply
For any circuit, set power
supply voltage to 8V
For any circuit, set power
supply voltage to 8V
R1 & R2 in series
R1 & R2 & R3 in series
R1 & R2 in parallel
R1 & R2 & R3 in parallel
RESISTANCE AND POWER in series and parallel
Thursday, 4 November 2010
147. HANGING MAGNETS I
NS
1. Cut out the net
and fold it at the
dotted lines to
create a cradle
+
NS
2. Suspend the
magnet in the
cradle
3. Use a short
length of cotton
to suspend the
magnet from a
retort stand
4. Repeat using a
second magnet.
Thursday, 4 November 2010
148. HANGING MAGNETS II
1. Position your two magnets in the
orientations shown
2. For each orientation, record your
observations
NS
N
S
NS N S
NS NS NS
N
S
A B
C D
ObservationObservation
ObservationObservation
Thursday, 4 November 2010
149. STROKING MAGNETS
NS
A. Try magnetising
an iron nail using
a magnet
B. Once you have
finished, check for a
magnetic field using a
charm compass.
Thursday, 4 November 2010
150. PLOTTING MAGNETIC FIELDS
A. Place one or more
compasses around a bar
magnet
Charm compasses (moved around in a
variety of positions around the magnet
B. Use a pencil to mark the
North pole of each
magnet using a dot
C. Connect the dots using a
smooth curve
D. Plot several field lines
and mark the North and
South poles of the magnet.
E. Wrap your magnet in glad
wrap and spring iron filings
over it.
Follow the instructions (below) and draw your observations
Thursday, 4 November 2010
235. Main points:
1(b) & 1(c) - need further elaboration. See revised notes.
3(a) & (b)....... confusion re. magnetic and geographic poles being opposite
3(c) .......... Need to appreciate that coil becomes magnetised which causes
dipole and domain alignment in the piston
3 (d) ...... problems creeping in with the algebra
Thursday, 4 November 2010
252. 1. Resistance is ______________ __________ .
2. It is responsible for slowing down the ___________ ___ __________ .
3. When there is resistance present in a conductor or an appliance ____________ energy is
transformed into ________ energy.
4. Three everyday household appliances that have high resistance are:
5. __________________ , ___________________ , _____________________
6. Complete the circuit diagram which shows what happens when a third identical lamp is
added in parallel to a circuit:
+ -
+ -
7. If we continued to add lamps in this fashion explain why the power supply would blow a
fuse.
__________________________________________________________________
__________________________________________________________________
QUESTIONS ON RESISTANCE
Thursday, 4 November 2010
253. APPLYING THE WATER MODEL TO Q.6
1. Pool of water fed by a rut
a rut
channel
2. Dig a channel and water can flow out of the pool
3. Dig another channel. What effect does this have on the flow through the rut?
3. Dig a third channel ........ What effect does this have on the flow through the rut?
Thursday, 4 November 2010
255. + -
Draw using arrows the direction of charge flow in each circuit
For each circuit, highlight the lamps that will glow.
1 2
+ -
3
+ -
4
+ -
NEED FOR A VOLTAGE SUPPLY AND A CONDUCTING PATH
Thursday, 4 November 2010
256. CONDUCTORS & METERS
1. Which of the following substances are conductors of
electricity:
wood, tap water, copper, iron, glass, sodium chloride solution, rubber
2. Draw a circuit diagram that you could use to test
for conductivity using a lamp, dry cell and wires.
3. Explain how a fuse protects a circuit.
4. Voltage is ..........
5. Current is ..........
6.
Starter
Thursday, 4 November 2010
257. In an an Ohm’s law experiment a water-cooled resistor was connected in series with a
power supply and an ammeter. A voltmeter was connected to measure the voltage
drop across the resistor.
The readings on the two meters were recorded. Voltage (V) Current (I)
2 0.15
4 0.31
6 0.45
8 0.59
10 0.75
12 0.92
Draw a graph of V vs I.
What is the meaning of its
slope and what is its unit?
OHM’S LAW
Explain the shape of the graph that would be produced should the resistor be allowed
to heat up.
Thursday, 4 November 2010