Electromagnetism

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

2 πd
Thursday, 16 September 2010

LANGUAGE


THE LANGUAGE OF ELECTRICITY

Term Definition Word list
Charge an electrical quantity based on an excess or deficiency of electrons
static
a form of electricity where charge does not flow continuously
electricity


NOTES

ELECTRO
STATICS

CHARGING OBJECTS

Based on atomic structure
Electron (-ve charge)

Neutron

Proton (+ve charge)

The process Empty space

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.

CHARGE INTERACTION

Oppositely charged objects Objects with the same
attract each other charge repel each other

+ - + + - -

+ - + + - -

Demo: The Van der Graaf Generator


EXAMPLES

Read p52 and 53 (Y10 Pathfinder) and then offer some examples to the class
discussion:

Static electricity
around us

The History of Electricity Generation

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)

CLOUD TO GROUND


CLOUD TO CLOUD


BLUE JETS AND SPRITES


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.
-- ++++
----
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
This symbol object.
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 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

CURRENT
VOLTAGE

IN
TR
O DU
CT
IO
N

WHAT IS ELECTRICITY?


THE ELECTRICAL CIRCUIT -
introducing the idea of the
electron pump


1. What is an electric current?

2. What are the two requirements necessary for an electric
current to exist?

Power Supply

+ -

A conducting path

THE GRAVITY MODEL
http://regentsprep.org/Regents/physics/phys03/bsimplcir/default.htm

A

B

1.Which part of the model represents the power supply?

2. Which part of the model represents the component?

3. What type of current is being modelled?

THE WATER MODEL


2. Which part of the model represents the conducting path?

3. Which part of the model represents charge?

THE BIKE MODEL

C

A

B


2. Which part of the model represents the conducting path?

3. Which part of the model represents charge?

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” ??


ELECTRIC CIRCUITS
Requirements: Power Supply
a power supply
conducting path around which charge
(electrons or ions) can flow.
+ -
components (and sometimes meters)

A component

Current A conducting path
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)
http://regentsprep.org/Regents/
physics/phys03/bsimplcir/default.htm

ELECTRIC CURRENT

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:

Red V Black
Connecting an ammeter + -

I
I
A
Red 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


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)

where V = potential difference or voltage (Volts, V)
V = ∆Ep
q ∆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)

Unit of Voltage: Joule per Coulomb or Volt
(JC-1) or (V)
Example
V
Consider the voltage across a lamp: -
+

A B
1A

If V = 6V then a coulomb of charge has 6J more electrical potential energy at
point A than it does at point B

Notes CIRCUIT SYMBOLS

demo of circuit components ->

+ - V

A •

Two wires joined Two wires crossing

Cell Lamp

Battery (two cells in
Switch
series)
Battery (several
Diode
cells)

Voltmeter Ammeter

Resistor Power supply

variable resistor
fuse
(rheostat)

RE
SI
& ST
OH A N
LA M’ CE
W S


RESISTANCE & ELECTROCUTION - 1


RESISTANCE & ELECTROCUTION - 2


FACTORS AFFECTING RESISTANCE


FACTORS THAT AFFECT RESISTANCE

Also, some materials conduct electricity better than others Eg. Copper is better than iron

CONDUCTORS & INSULATORS

In a conductor, electrons In an insulator,
are free to flow electrons are fixed

_______________

Label the materials that the _______________
arrows are pointing to

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


RESISTANCE
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 Unit of resistance
I is the ohm, Ω
V

I R
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)
24
20
16
12
8
4

0 1 2 3 4 5 6
I (A)

WHATS HAPPENING TO THE RESISTANCE AS THE CURRENT INCREASES?

24 A conductor that retains a constant
1 V (V) 20 temperature as the current is increased:
16
12
8
4

0 1 2 3 4 5 6 I (A)

A conductor that is allowed to heat
24
2 V (V) up as the current is increased
20
16
12
8
4
I (A)
0 1 2 3 4 5 6

A conductor that is cooled progressively
3 V (V) 24
20 as the current is increased
16
12
8
4
I (A)
0 1 2 3 4 5 6

MODELLING TEMPERATURE
INCREASE IN A WIRE

SHAKING


LIMITATIONS OF OHM’S LAW
24
V (V) 20
When a temperature of a lamp increases its
16 resistance increases
12
8
4
I (A)
0 1 2 3 4 5 6

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) 24
The resistance of a thermistor decreases as its
20
16
temperature decreases
12
8
4
I (A)
0 1 2 3 4 5 6


BASIC RESISTANCE PROBLEMS

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?

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?

5. A current of 2 A flows through a 6Ω resistor. What is the voltage across it?

6. What voltage is needed it a current of 5A is to flow through a resistance of 3Ω?


RESISTANCE CALCULATIONS
Resistors which are connected end to end are in series with one
another
R1 R2

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.
R1

R2

The total resistance of the parallel combination, Rp is less than any individual
resistor in the combination.

For two or more resistors in parallel 1 1 1 + ....
the total resistance,Rp is given by: RP = R1 + R2

SERIES &
PARALLEL

CIRCUITS: diagrams & assembly

Draw the following circuit diagrams in the spaces
provided AND when you have finished, assemble them:

1.

+ -

2.
+ -

3.
A
V

+ -

Voltmeters are connected _________ components,
ammeters are connected _____ a circuit

CHARACTERISTICS OF SERIES & PARALLEL CIRCUITS

Series Parallel

+ -
+ -

+ -
+ -

components connected each component has its own
1._ 1._ connection with the power
one other the other.
supply
2._ Single pathway 2._ more than one pathway

3._ No junctions 3._ One or more junctions


http://phet.colorado.edu/simulations/ CIRCUIT CONSTRUCTION
sims.php?
sim=Circuit_Construction_Kit_DC_Only

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.

1 2 3
+ - + - + -
A A A

3
+ -
2
A
1 + -

+ - A

A


DRAW THE CIRCUITS AND SET THEM UP


CIRCUIT
RULES

VOLTAGE & CURRENT IN SERIES CIRCUITS

+ -

VT
A1 A3

I1 I3

I2
A2

V1 V2

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

PARALLEL CIRCUITS

+ -
Current in parallel is shared
IT VT IT
R1
IT = I1 + I2
I1 in other words “charge splits
up as it enters a junction in
a circuit”
V1
R2
I2 Voltage in parallel is constant

V2 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
“Double the resistance then halve the current”
then I1 = 2 and I2 = 1

SIMPLE CIRCUIT CALCULATIONS
Example 1

+ 8V -
A3 = ________

3A A1 V1 V1 = ________
2A
A2 R1 V2 = ________

A3 R2

Rules used V2
____________________________________________________________________
____________________________________________________________________
Example 2
+ 8V -
3V
V1 V2 V1 = ________

R1
Rule used ___________________________________________________________

Example 3

+ 8V - A4 A3 = ________
3V
4A A1 V1 V3 A4 = ________
3A
A2 R1 V2 = ________

V3 = ________
A3 R2
Rules used
V2 ______________________
______________________
______________________
Example 4 ______________________
______________________
+ 8V - A4
A1 = ________
A1
3A A4 = ________
A2
Rule used
____________________
4A
____________________
A3 ____________________

ADVANCED CIRCUIT CALCULATIONS
Examples

+ 9V -
1

A1 V1 A3

5Ω 10Ω
A2

V2 V3
For the circuit represented by the circuit diagram above, what is the reading on:
(a) V1

(b) A2

(c) V2

(d) V3


2 + 15V -

V1

5Ω A3
V2 A1

V3
R 10Ω
A2

(a) V3 if V2 = 10 V

(b) A1

(c) A2

(d) A3

(e) What is the value of resistor R?


3 + 12V -

V1

1A 2Ω 4.8Ω V3
A1

V2
3Ω
A2


(a) V1

(b) V2

(c) V3

(d) A2


RESISTORS IN SERIES

+ -

R1

RT = R1 + R2

R2
Examples

1 2

25 Ω 30 Ω 50 Ω
100 Ω 100 Ω 100 Ω
• •

Total resistance = Total resistance =

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

OT
HE
R
ST
UF
F


ESA: 13B Q.5 to 8

INTRODUCING “EMF” & “ELECTRICAL POTENTIAL


METERS


+ 9V -
One of these meters has
IT a very high resistance
The other meter has a
A low resistance.

Which is which?

V Explain your answer


METERS

Ammeter
+ 9V -
1. connected in series with IT
other components
A
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 V
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


POWER
&
ENERGY

POWER

• Power is the rate at which electrical energy is transferred into other forms of
energy.
• It is the amount of work done per second

E E = the amount of energy converted or work done (J)
P = E
t P t t = the time taken (s)

• It can be shown that the electrical power supplied to a device is given by:

P P = power (watts, W) (1W = 1 Js-1)

P = VI V I V = Voltage (volts, V)
I = Current (amps, A)
P = Power (Watts, W)


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)

14 E = Area under the graph
12

10

8
E = 0.5 (9 + 13) = 4.4 J
6 10

4

2

0 2 4 6 8 10 t (s)

TOTAL POWER USAGE in a parallel circuit

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 1 & 2: I = P
12 V V
= 6/12
= 0.5 A
12 V
P1 6W Lamps 3 & 4: I = P
V
12 V = 12/12
P2 6W
=1A

12 V IT = I1 + I2 + I3 + I4 = 0.5 + 0.5 + 1 + 1 = 3A
P3 12 W P = VTIT = 12 x 3 = 36 A

12 V
P4 12 W (ii) PT = P1 + P2 + P3 + P4
= 12 + 12 + 6 + 6
= 36 W

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

+ -


EXERCISES


POWERING THROUGH THE QUESTIONS

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
20 min 20 min 10 min 3h
(s)

Energy (J) 1000 125000 300000


“PARALLEL WITHIN SERIES”
0.5 A
+ 9V -
A 9 V battery is connected in the
circuit shown. 10Ω
A current of 0.5 A is found to 0.1 A
pass through the 10Ω resistor. 40Ω

X

(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.

(e) Determine the heat energy generated per second in the whole circuit.


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.


Y11 Sci - W & W


answers “PARALLEL WITHIN SERIES”
(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 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.

MAGNETIC
FIELDS

THE MAGNETIC FIELD AROUND A BAR MAGNET AND
THE EARTH’S MAGNETIC FIELD


---> Demo c bar magnet & major magnet
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

Charm compass
S

Magnetic field lines The compass needle is
itself a tiny magnet (the
N North pole of this magnet
points towards the South
end of the magnet)

Strong magnetic field (high density of lines)

THE EARTH’S MAGNETIC FIELD

θ (angle of declination) = 11o
Geographic
North Magnetic
South

compass
S

N

Earth’s axis

• 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 IS ESSENTIAL FOR
LIFE ON THE PLANET.

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 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.


MAGNETIC
DOMAINS

MAGNETIC MAGIC -->
MAGNETIC DOMAINS


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]

N
N

e
S S

For most elements:
magnetic fields cancel.

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.)

DOMAINS

Regions in a metal where the orientation of the magnetic dipoles is the same are
called domains.

A domain
Unmagnetised Iron Here, a large number
of iron atoms
=> Domains are (magnetic dipoles)
scrambled are aligned.

Partially
magnetised

S N
Fully magnetised

=> the orientation of
the domains is the S N
same

BREAKING A MAGNET


MAGNETISING AND DEMAGNETISING [Solenoid Demo]

“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]

“Domains are induced into alignment”

- Picking up iron objects


WIRES


WIRES

A circular magnetic field is formed around a straight current - carrying conductor:

3D View
I View from above

The direction of the magnetic
field lines is given by the
Right-hand Thumb Rule

•

( “•” represents current directed out of the page)

The right hand thumb rule:
Thumb = direction of the electric current
Curled fingers = direction of the circular magnetic field


“View” from above”

I

•

Magnetic field
lines


B = Magnetic field strength (in Tesla,T)
I = Electric current in the wire (in Amps,A)
B = µ0I
2 d π 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)

Note

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
BαI 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


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 2A 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

CURRENT CARRYING CONDUCTOR AND
MAGNETIC FIELD -----> COIL


COILS


THE SOLENOID

The magnetic field of a solenoid is
similar in shape to that of a bar
magnet:
Draw the field lines
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:

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.

STRENGTH RULZ

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

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


ELECTROMAGNETS

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.

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.

THE COIL GUN
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.

Cross-section of coil

Attraction to
x x x x x x x x
North of coil
Attraction to South of coil
N S N

- - - - - - - -

North end of coil South end

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

ELECTROMAGNETS IN RELAYS


PRACTICAL


STATIC


PLAYING AROUND WITH STATIC ELECTRICITY

1 plastic rod (charged by rubbing with a cloth)

small pieces
of torn paper

Observation:

Explanation:

2 Balloon rubbed against hair
Removed from head and then brought back to hair

Observation:

Explanation:


3
Balloon rubbed against jersey
Release

Observation:

Explanation:

cotton
(a) Each balloon charged separately by rubbing against
4
the sleeve of a jersey

(b) Holding the balloons by the cotton, release them,
allowing them to come close to each other.

Observation:

Explanation:

5

(a) Straw, charged at both
(b) Straw, also charged using a
ends (using a woollen
woollen cloth held horizontally
cloth)
and brought close

(c) Repeat (b) using a silk cloth.

Observation:

Explanation:

Charged plastic rod is held near a thin stream of water

6 Observation:

Explanation:

http://phet.colorado.edu/new/simulations/sims.php?
sim=Balloons_and_Static_Electricity

Equipment
CHARGING OBJECTS dry cloth/jersey
Cap
perspex rod
Insulating material
ebonite rod
electroscope
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

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:

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
Complete the diagram to show how charge moves represents
a
when the cap of the electroscope is earthed connection
to the earth

Draw the charge distribution diagram (by adding to ++++
the existing diagram on the right) showing the ----
situation once this charge movement has finished.

+ +
+ +

3. Unearth the cap of the electroscope without removing the
charged rod

Observation:

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).

Part 2 - Charging by contact

Method
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.


Lab 12 ALL CHARGED UP Equipment
dry cloth/jersey
Cap perspex rod
Insulating material ebonite rod
electroscope
Body of electroscope
Leaf

Base

Aim
to charge an electroscope by both induction and by contact and to draw charge distribution
diagrams
Method
these)
Part 1 - Charging by induction

1. Charge the rod by rubbing it with a dry cloth/jersey and hold the rod
near the cap of the electroscope ++++
----
Observation:

The leaf of the electroscope springs up.
+ +
+ +

If the rod was positively charged the charge distribution diagram would look like this:

Complete the diagram to show how charges would distribute on the electroscope should the rod
be negatively charged.

Electrons at the cap are
repelled by the negatively ----
++++
charged rod.

- -
- -
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.

2. With the rod in this position, earth the cap with your finger.

Observation: -- ++++
----
Leaf of the electroscope drops

+ +
+ +
This symbol
Complete the diagram to show how charge moves represents
a
when the cap of the electroscope is earthed connection
to the earth

Draw the charge distribution diagram (by adding to ++++
the existing diagram on the right) showing the - - - -
situation once this charge movement has finished.

-
The cap and leaf now have +
- +-
+ +
no overall charge. -
Electrons on the cap are
still held in position.

3. Unearth the cap of the electroscope without removing the ++++
charged rod - - - -

Observation:
Leaf of the electroscope remains in the “dropped”
position. The charge distribution has not changed -
+ +
- -
+
- +
Draw the resultant charge distribution and the new
position of the leaf on the diagram (right).
Negative charge redistributes itself
4. Remove the charged rod around the metal parts of the
electroscope leaving the stem and - -
Observation: leaf with an overall negative charge

The leaf of the electroscope springs up. - -+
+
- -
+ + -
-
Finally, complete the diagram (right).

Part 2 - Charging by contact

Method
these)

1. A positively charged rod is held near the cap of the electroscope. ++++
- - - -
Charge separation occurs. Positive
repels positive at the stem/leaf
+ +
+ +

2. The rod makes contact with the cap. -

Electrons migrate up
+ + into the rod
+ +

3. The rod is removed. + + +

The electroscope is now left with
+ + an overall positive charge.
+ +


SPARKS

12 Physics > resources > electricity > DC
electricity > videos


THE VAN DER GRAAF
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

______________

• 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.

Reference: http://science.howstuffworks.com/vdg3.htm

THE VAN DER GRAAF - OBSERVATIONS & EXPLANATIONS

1. Small dome held close to generator dome
Drawn observation Explanation

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



DC


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 2 3
8V 8V 8V
+ - + - + -

A A A

Current = ______ A

Current = ______ A

Current = ______ A

Observation

Explanation


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 2 3

+ - + - + -

A A A

Current = ______ A Current = ______ A Current = ______ A

Observation

Explanation


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):

A1 + 8V -
A1 = ______ A
A3 A2 = ______ A
A3 = ______ A
A2

Equation


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:

+ 8V -
Results
A1 A4 A1 = ____ A

A2 = ____ A
A2
A3 = ____ A
A4 = ____ A
A3

Conclusion
Current in a parallel circuit is ____________ .

Equation relating the currents


VOLTAGE IN THE SERIES CIRCUIT IS SHARED

Aim
to look for a pattern in the voltages across bulbs and resistors in a series circuit.

1. Set up the circuit (below)
2. Use ONE voltmeter in the three different places shown in the circuit diagram.

+ 8V -
Results
V1 V1 (power supply) = __ V

V2 V4 A V2 (variable resistor) = __ V

V3 V3 (bulb) = __ V
V3 (ammeter) = __ V

Conclusion
The power supply voltage is _____________ between the components in the circuit

Equation relating the voltages

VOLTAGE IN THE PARALLEL CIRCUIT IS CONSTANT

Aim
to look for a pattern in the voltages across bulbs and resistors in a series circuit.

2. Use ONE voltmeter in the three different places shown in the circuit diagram.

+ 8V -

V1 Results
V1 = ___ V
V2
V2 = ___ V
V3 = ___ V

V3

Conclusion
The voltage across components connected in parallel is __________

Equation relating the voltages

VOLTAGES AND CURRENTS IN SERIES AND PARALLEL

Aim
to investigate voltage and current in a series circuit that has a parallel portion in it.

2. Use ONE voltmeter in the four different places shown in the circuit diagram
and ONE ammeter in the four different places shown.

Results
A1 + 8V -
A1 = __A
V1 A2 = __A
V2 A4 V4
A3 = __A
V3
A4 = __A
A2
V1 = __V
A3 V2 = __V
V3 = __V
V4
V4 = __V

OHMIC
CONDUCTORS


Lab 14 OHM’S LAW

Method
1. Set up the following circuit using
iced water to cool the immersion + -
coil.
2. Increase the voltage in regular
increments through an A V
appropriate range (widest
possible range)
ice
immersion coil beaker
water

Results

Voltage setting of Power pack
2 4 6 8 10 12
(V)

Voltage, V (V)

Current, I (A)


Draw a graph
of Voltage
against Current
on the grid
provided

Conclusion

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.

Repeat the experiment but this time replace the coil with a lamp (that will
increase in temperature as the current through it increases)

RESISTANCE


RESISTANCE AND POWER in series and parallel

Measurements Resistance
Resistance calculated Calculated
specified Voltage Current
from Power output
measurements

R1

R2

R3

across drawn from
Resistance calculated For any circuit, set power
combination the power
(formula provided for parallel resistors)
of resistors supply supply voltage to 8V

R1 & R2 in series

R 1 & R2 & R3 in series

R1 & R2 in parallel

R 1 & R2 & R3 in parallel


POWER

CALCULATING POWER OUTPUT OF APPLIANCES WITH AN
ELECTRONIC MONITOR

Appliance

UNDER
CONSTRUCTION


ELECTROMAG


HANGING MAGNETS I

1. Cut out the net
2. Suspend the
and fold it at the
magnet in the
dotted lines to
cradle
create a cradle

S N
+

3. Use a short
length of cotton S N
to suspend the
4. Repeat using a magnet from a
second magnet. retort stand


1. Position your two magnets in the
HANGING MAGNETS II orientations shown
2. For each orientation, record your
observations
N
A B
S N
S N N S
S

Observation Observation

C D
S

S N S N S N

N

Observation Observation


STROKING MAGNETS

A. Try magnetising
an iron nail using
a magnet

S N

B. Once you have
finished, check for a
magnetic field using a
charm compass.


PLOTTING MAGNETIC FIELDS

Follow the instructions (below) and draw your observations

A. Place one or more B. Use a pencil to mark the
compasses around a bar North pole of each
magnet magnet using a dot
Charm compasses (moved around in a
variety of positions around the magnet

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.


Electromagnetism

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