section 4 - electricity and magnetism - with notes page
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ElectricityTRANSCRIPT
PAL (IGCSE) – PHYSICS Section 4 Electricity and Magnetism
Electricity and Magnetism
PAL (IGCSE) Single ScienceRevision Book - Section 4
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PAL (IGCSE) – PHYSICS Section 4 Electricity and Magnetism
Syllabus Content_______________________________
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Syllabus Details________________________________
4. Electricity and magnetism4.1 Simple phenomena of magnetismCore• State the properties of magnets
Attract
Attract
Repel
Repel
• Give an account of induced magnetism
INDUCED MAGNETISM: If a piece of iron is brought close to a magnet it becomes magnetic and is attracted to the magnet.
• Distinguish between ferrous and non-ferrous materials
FERROUS: Containing a large proportion of Iron (e.g. Iron, steel)NON-FERROUS: Containing no iron
• Describe methods of magnetization and of demagnetization
MAGNETISATION: Stroking a magnet across a material (e.g. iron)
DEMAGNETISATION: Hitting the material
• Describe an experiment to identify the pattern of field lines round a bar magnet
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Compass
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• Distinguish between the magnetic properties of iron and steel
Material Type PropertiesIron Soft magnetic material Magnetism is temporarySteel Hard magnetic material Magnetism is permanent
• Distinguish between the design and use of permanent magnets and electromagnets
Type of magnet Design UsePermanent Hard magnetic material For applications where
magnetism is needed over long periods – fridge doors
Electromagnet Uses a solenoid to create magnetic field
For applications where the magnetic field needs to be turned on and off – Scrap metal moving
4.2 Electrical quantities4.2 (a) Electric chargeCore• Describe simple experiments to show the production and detection of electrostatic charges
Perspex Rod
Rubbed
+ + + + + + + + + + + + + + + + + +
-
Cloth
Polythene Rod
Rubbed
- - - - - - - - - - - - - - - - - - - - -
-Cloth
Electron transfer
Electron transfer
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Positive Rod
Negative charge attracted up to cap
Gold leaf diverges Positive charges repelled down to leaf
Negative charge attracted to earth
+ + + + + + + + +
+ + + + + + + + +
- - - - - - - - - - - - - - - - - -
+
+ +
+
+
+
+
+
+
+
++
++
----
----
A Gold Leaf Electroscope
GOLD LEAF ELECTROSCOPE
If a charge is moved close to the cap the gold leaf rises
• State that there are positive and negative charges
There are two type of charges; POSITIVE and NEGATIVE
• State that unlike charges attract and that like charges repel
LIKE CHARGES – RepelUNLIKE CHARGES - Attract
• Describe an electric field as a region in which an electric charge experiences a force
Electric Fields
+
+ The electric field shows the region in which a (positive) charge feels a force
Field lines
• Distinguish between electrical conductors and insulators and give typical examples
ELECTRICAL CONDUCTOR: Charges are able to flow through the material (metals)
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ELECTRICAL INSULATOR: Charges are unable to flow through the material (plastics)
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Supplement• State that charge is measured in coulombs
Charge (Q) is measured in coulombs [C]
• State the direction of lines of force and describe simple field patterns, including the field around a point charge and the field between two parallel plates
+
- - - - - - - - - - - -
+ + + + + + + + +
Electric Fields
Point +ve Charge
Parallel Plates
-
Point -ve Charge
• Give an account of charging by induction
+ + + + + + + + +
+ + + + + + + + + - - - - - - - - - - - - -
-
-
-
-
-
-
+
+
+
++
Neutral Electroscope Charged rod brought close
Charged rod brought close to top of electroscope The positive rod attracts electrons to the top of the electroscope
(induced charge)
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• Recall and use the simple electron model to distinguish between conductors and insulators
Electrical Conduction
Metal Wire
Metal Ions
Electrons
In a conductor the charges are free to move (the electrons in a metal) In an insulator the charges are not free to move
4.2 (b) CurrentCore• State that current is related to the flow of charge
CURRENT: Related to the flow of charge
• Use and describe the use of an ammeter
A
Ammeters…• measure the Current flowing in AMPS (A)• are always placed in series
A
A
1
2
32
1
Ammeter A1 measures the total current through bulb 1 and 2
Ammeter A2 measures current through bulb 1
Ammeter A3 measures current through bulb 2
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Supplement• Show understanding that a current is a rate of flow of charge and recall and use the equation I = Q /t
Current [A] = Charge flowed [C]
Time [s]
I = dQ
dt
1 amp = 1 Coulomb
1 second
• Distinguish between the direction of flow of electrons and conventional current
Direction of
electron flow
Direction of
conventional current
4.2 (c) Electro-motive forceCore• State that the e.m.f. of a source of electrical energy is measured in volts
EMF is measured in Volts [V]
Supplement• Show understanding that e.m.f. is defined in terms of energy supplied by a source in driving charge round a complete circuit
Electromotive force (e.m.f.): The total energy difference per unit charge around a circuit
4.2 (d) Potential differenceCore• State that the potential difference across a circuit component is measured in volts
Potential difference (pd) is measured in Volt [V]
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• Use and describe the use of a voltmeter
V Voltmeter in parallel
Voltmeter measures the potential difference (‘voltage’) across a component
4.2 (e) ResistanceCore• State that resistance = p.d./current and understand qualitatively how changes in p.d. or resistance affect current
V
Ohms law
Material
A
Cu
rre
nt (
A)
Potential Difference (V)
If a material obeys OHMS LAW the current will increase in PROPORTION to the potential difference applied
• Recall and use the equation R = V/I
Resistance [W] =Potential Difference [V]
Current [A]
R = VI
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• Describe an experiment to determine resistance using a voltmeter and an ammeter
V
A
Measuring resistance
Ammeter in series
Voltmeter in parallel
R = V / I
• Relate (without calculation) the resistance of a wire to its length and to its diameterSupplement• Recall and use quantitatively the proportionality between resistance and length, and the inverse proportionality between resistance and cross-sectional area of a wire
rLR =
A
R = Resistancer = Resistivity of materialL = Length of conductorA = Area
Summary
Increasing length = increasing resistance Increasing cross-sectional area = decreasing resistance
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4.2 (f) Electrical energySupplement• Recall and use the equationsP =IV and E = IVt
P = VIPower [W] = Voltage [V] x Current [A]
Energy [J] = Voltage [V] x Current [A] x Time [s]
E = VIt4.3 Electric circuits4.3 (a) Circuit diagramsCore• Draw and interpret circuit diagrams containing sources, switches, resistors (fixed and variable), lamps, ammeters, voltmeters, magnetizing coils, transformers, bells, fuses and relays
V
A
Cell
Variable resistor
Resistor
Lamp
Voltmeter
Ammeter
Switch
Bell
Fuse
Transformer
Magnetizing Coils
Relay
Supplement• Draw and interpret circuit diagrams containing diodes and transistors
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Diode
Transistor
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4.3 (b) Series and parallel circuitsCore• Understand that the current at every point in a series circuit is the same
Series Circuit
I = I = I = IA A
A
1 3
4
1 2 3 4A2
• Give the combined resistance of two or more resistors in series
Resistors in Series
R1 R2
Rtotal = R1 + R2
• State that, for a parallel circuit, the current from the source is larger than the current in each branch
Parallel (Branched) Circuit
I = I + I
A1
1 2 3
A2
A3
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• State that the combined resistance of two resistors in parallel is less than that of either resistor by itself
Resistors in Parallel
R1
R2
1/Rtotal = 1/R1 + 1/R2
• State the advantages of connecting lamps in parallel in a lighting circuit
Parallel Circuit advantage: If one lamp fails the other lamps in parallel continue to function
Supplement• Recall and use the fact that the sum of the p.d.s across the components in a series circuit is equal to the total p.d. across the supply
V
V1 = V2 + V3
V
V
1
2 3
A
A A
1
2 3R1 R2
Rtotal = R1 + R2I1 = I2 = I3
Series Circuits
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• Recall and use the fact that the current from the source is the sum of the currents in the separate branches of a parallel circuit• Calculate the effective resistance of two resistors in parallel
V
V1 = V2 = V3
V
V
1
2
3
A
A
A
1
2
3
R1
R2
1/Rtotal = 1/R1 + 1/R2
I1 = I2 + I3
Parallel Circuits
4.3 (c) Action and use of circuit componentsCore• Describe the action of a variable potential divider (potentiometer)
Output voltage
Potential Divider
The output voltage can be varied by changing the position of the potential divider
The voltage dropped across that section of the resistor will vary
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• Describe the action of thermistors and light dependent resistors and show understanding of their use as input transducers
LDR : Light Dependent Resistor
Thermistor
• High light intensity = Low resistance
• Low light intensity = High resistance
• High temperature = Low resistance
• Low temperature = High resistance
Light Dependent Resistor (LDR)
LDR
10 kW
When light shines on the LDR the resistance decreases
When light shines on the LDR the p.d. across the fixed resistor increases
Thermistor
Thermistor
10 kW
When the temperature of the Thermistor increases the resistance decreases
When the temperature of the Thermistor increases the p.d. across the fixed resistor increases
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• Describe the action of a capacitor as an energy store and show understanding of its use in time delay circuits
V
Capacitors
• When the switch is closed the capacitor starts to charge up (store electrical energy)• Voltmeter reading will slowly increase as resistance of the capacitor increases
Time delay circuit
• When the switch is closed the capacitor starts to charge up• The lamp will slow get brighter as the resistance of the capacitor increases• Is the switch is opened the capacitor will discharge and the lamp will slowly get dimmer
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• Describe the action of a relay and show understanding of its use in switching circuits
Low Current Circuit
High Current Circuit
Air- Con
Using Relay Circuits
• When the Thermistor is hot its resistance will be low• The current will flow in the circuit A (and through the coil in the relay circuit)• The relay will close allowing the current to flow through circuit B and so the Air-conditioner will turn on
A B
A LOW CURRENT CIRCUIT CONTROLS A HIGH CURRENT CIRCUIT
Thermistor
relay
Supplement• Describe the action of a diode and show understanding of its use as a rectifier
V
A
A Diode allows current to flow in one direction only
Cu
rre
nt (
A)
Voltage (V)
A diode allows current to flow in only one one direction
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Input ACCurrent
Output Rectified Current
Rectifier
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• Describe the action of a transistor as an electrically operated switch and show understanding of its use in switching circuits
Transistor circuit
• A small current flow through the base• This allows a large current flow from the collector to the emitter
• Recognise and show understanding of circuits operating as light sensitive switches and temperature-operated alarms (using a relay or a transistor)
LDR Circuits
Thermistor Circuits
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4.3 (d) Digital electronicsSupplement• Explain and use the terms digital and analogue
Analogue Signal Digital Signal
1 0 0 1 0 1 1
• A continuous wave with data carrier in the amplitude and frequency of the wave
• A signal consisting of digital pulses (1’s or 0’s), which is a coded version of the analogue signal
• State that logic gates are circuits containing transistors and other components
LOGIC GATES: Circuits containing transistors and other components
• Describe the action of NOT, AND, OR, NAND and NOR gates• State and use the symbols for logic gates (candidates should use the American ANSI#Y 32.14 symbols)
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• Design and understand simple digital circuits combining several logic gates
Example of Logic Gate Usage (NOT GATE)
LDR
+V
0 V
X
Y
Light Operated Switch
• In GOOD lighting the voltage at point X is high• The output voltage at Y will be low• The Lamp will stay off
• In DIM lighting the voltage at point Y is low• The output voltage at Y will be high• The Lamp will go on
4.4 Dangers of electricityCore• state the hazards of– damaged insulation
DAMAGED INSULATION – risk of electrocution when handling wires etc
– overheating of cables
OVERHEATING OF CABLES – insulation will melt and wires become exposed
– damp conditions
DAMP CONDITIONS – impure water conducts electricity and so risk of electrocution
• Show an understanding of the use of fuses and circuit-breakers
FUSE: A thin piece of wire which melts and breaks if too much current passes through it. Used to cut a circuit if the current is too high.
CIRCUIT BREAKER: Automatic electrical switch which will cut a circuit if the current is too high
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4.5 Electromagnetic effects4.5 (a) Electromagnetic inductionCore• Describe an experiment that shows that a changing magnetic field can induce an e.m.f. in a circuit• Show understanding that the direction of an induced e.m.f. opposes the change causing it
As the magnet is moved in and out of the coil the magnetic field in the coil changes
The direction of the induced EMF and so current opposes the change causing it (i.e. makes a magnetic field opposite to the field of the moving magnet
Supplement• State the factors affecting the magnitude of an induced e.m.f.
FACTORS EFFECTING MAGNITUDE OF INDUCED E.M.F.
Increasing strength of magnet = increased induced E.M.F.Increasing velocity of motion of magnet = increased induced E.M.F.Increasing # of coils in solenoid = increased induced E.M.F.
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4.5 (b) a.c. generatorCore• Describe a rotating-coil generator and the use of slip rings
Rotation
OutputAC Generator
Slip rings
AC GENERATOR
If a coil is rotated in a permanent magnetic field the coil experiences a changing magnetic field
If a wire experiences a changing magnetic field an EMF is induced If the wire is connected into a circuit a current will flow The slip rings allow the current to flow around a complete circuit as
the coil rotates
• Sketch a graph of voltage output against time for a simple a.c. generator
Indu
ced
emf
Time
Sinusoidal output
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4.5 (c) TransformerCore• Describe the construction of a basic iron-cored transformer as used for voltage transformations• Recall and use the equation(Vp /Vs) = (Np /Ns)Supplement• Describe the principle of operation of a transformer• Recall and use the equation Vp Ip = Vs Is (for 100% efficiency)
AC Input AC Output
Vp Np
Vs Ns
=
Vp = Voltage in primary coilVs = Voltage in secondary coilNp = Number of turns in primary coilNs = Number of turns in secondary coil
Magnetic Field Transformer core
Primary Coil Secondary Coil
VpIp = VsIs for 100% efficiency
THE TRANSFORMER…
An AC current is passed through the primary coil A varying magnetic field is induced in the coil and the iron core The secondary coil experiences a varying magnetic field An EMF and so current is induced in the secondary coil The output current is also alternating
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• Describe the use of the transformer in high voltage transmission of electricity
Output from Factory2000 MW23500 V85000 A
Power lines need high voltage so low current400000 V5000A
House needs low voltage230 V
Step-up Transformers
Step-downTransformers
Transformers needed to change the voltage
TRANSFORMERS ARE USED IN ELECTRICTY TRANSMISSION BECAUSE...
The output of a power station is high current The transformer is used to convert this to low current and high voltage
before passing through the overhead cables A second transformer is used to lower the current and voltage before
the supply enters homes
• Give the advantages of high-voltage transmission
HIGH VOLTAGE TRANSMISSION = Low energy losses through heating in the cables
• Explain why energy losses in cables are lower when the voltage is high
High voltage …. Low current …… Less heating of the cable …… Low energy loses
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4.5 (d) The magnetic effect of a currentCore• Describe the pattern of the magnetic field due to currents in straight wires and in solenoidsSupplement• State the qualitative variation of the strength of the magnetic field over salient parts of the pattern• Describe the effect on the magnetic field of changing the magnitude and direction of the current
Magnetic field around a current carrying wire
GENERAL CONVENTION FOR MAGNETIC FIELD LINES...
The arrow points towards the south pole The spacing of the field lines is proportional to the magnetic field
strength If the current changes direction the magnetic field will reverse in
direction
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Solenoid
Coil
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• Describe applications of the magnetic effect of current, including the action of a relay
ACTION OF A RELAY…
With no current through the coil there is no magnetic field in the coil
The contacts will be separate When a current passes through the coil a magnetic field is induced The contacts will become magnetized and so close
4.5 (e) Force on a current-carrying conductorCore• Describe an experiment to show that a force acts on a current-carrying conductor in a magnetic field, including the effect of reversing:(i) the current(ii) the direction of the field
A current carrying wire in a magnetic field
Magnetic Field Current Carrying Wire
Direction of Force
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If the current is reversed – The force will act in the opposite directionIf the field direction is reversed – The force will act in the opposite direction
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Supplement• State and use the relative directions of force, field and current
The Left Hand Rule
• Describe an experiment to show the corresponding force on beams of charged particles
A moving charge in a magnetic field
S
N
Electron
Force
When using the left hand rule for charged particles… For electrons the direction of current is the opposite direction to the
electron velocity For protons the direction of current is the same direction as the
electron velocity
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4.5 (f) d.c. motorCore• State that a current-carrying coil in a magnetic field experiences a turning effect and that the effect is increased by increasing the number of turns on the coil• Relate this turning effect to the action of an electric motorSupplement• Describe the effect of increasing the current
Electric Motor Force
Force
Commutator
Rotation
THE ELECTRIC MOTOR
A current is passed through a coil in a magnetic field If a current carrying wire is placed in a magnetic field it experiences a force The coil feels a force which rotates the coil The commutator ensures that the current flows around the coil to give a force
which always acts to rotate the coil in the same direction
To Increase the rate of rotation...
Increase the number of coils Increase the strength of the magnetic field Increase the current
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4.6 Cathode-ray oscilloscopes4.6 (a) Cathode raysCore• Describe the production and detection of cathode rays
Electrons are given off from a heated Tungsten wire
DETECTION OF CATHODE RAYS
Cathode rays can be detected by a fluorescent screen When the rays are incident on the screen the screen emit’s light
• Describe their deflection in electric fields
Cathode rays
Cathode rays are deflected towards the positive of an electric field This indicates that the cathode rays are negatively charged
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• State that the particles emitted in thermionic emission are electrons
The particles emitted in thermionic emission are electrons
Electrons flow around the circuit Thermionic
Emission
4.6 (b) Simple treatment of cathode-ray oscilloscopeSupplement• Describe (in outline) the basic structure and action of a cathode-ray oscilloscope (detailed circuits are not required)
Hot Filament
Anode
Electron Beam
Screen
Cathode Ray Oscilloscope
Deflection coils
Electrons are generated by the hot element The electrons are accelerated towards the anode and focused into a
beam
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The deflection coils direct the beam at a certain position in the screen
The fluorescent screen indicates the position of the beam
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• Use and describe the use of a cathode-ray oscilloscope to display waveforms
KEY CONTROLS….
Time base: Time taken for beam to pass through one horizontal division [Sec/div]
Vertical amplifier gain: The vertical scale control [Volts/div]
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