Reference:Nelson
PhysicsVCE Units 3 & 4Chapters
5 8Page 128
Revision of Unit 2: ELECTRICITY
Household Electricity
Wiring
·the
voltage at the active pin varies sinusoidally with time
·the
voltage at the neutral pin is the voltage of the mid-point of the
sine curve and should be earthed
·the
bottom earth pin connects the metal frame of the appliance to earth
so that it cannot become live if a fault develops
·Some
modern appliances dont use the earth pin.
·Instead
they are double insulated. The
live wires and parts are insulated and the appliance itself is made of
insulating material.
·Household
appliances are connected in parallel so that they can be operated separately.
Safety Devices
·A
fuse is a short piece of wire which melts if the current is too
large.
·A
circuit breaker is an electromagnetic device which opens a switch
if the current is too large.It
acts more quickly and can be reset more easily.
·Earth
leakage protection opens
a switch if the current between active and neutral exceeds about 50 mA.
Electric Shock
·High
voltage alone will not cause harm.
·A
current greater than 0.001 A gives the sensation of electric shock.
·At
currents greater than0.01 Athe
hand muscles contract so violently the wire cannot be released.
·If
a current greater than 0.1 A passes through the heart it will cease to
function and currents greater than 1 A cause serious burns to the tissue.
·The
household circuit will cause death whenever a person touches a live wire
and the body completes the circuit to earth.
·This
can occur for a person standing or lying in water.
Energy Consumption
·Energy
used is the power rating of the appliance by time it is on
·The
unit of energy used is the Kilowatt Hour where power is measured in kilowatts
and time in hours.One kilowatt hour
equals 3,600,000 Joules
Electric Circuits
Electric Field and Potential Difference
·A
positive charge Q (symbol Q, unit Coulomb) released in an electric field
accelerates in the direction of the field due to an electricforce
FE
·The
voltage at a circuit point tells you the amount of electrical potential
energy per Coulomb of charge available to do work in the remaining circuit
elements.
·The
magnitude of a constant field between two point a distance d apart and
with potential difference V (unit Volt = Joule per Coulomb) is
·The
energy gained or work done (symbol W, unit Joule) by charge Q as it moves
through the potential difference is
W
= VQ = EdQ
Electric Current
·If
an amount of charge Q passes a circuit point in t seconds the current is
in the direction the positive charges move and is given by:
Power
P
= IV
·the
work done W in t seconds is:
W
= IVt
Ohmic and Non-ohmic Resistors
·For
Ohmic devices, the resistance is constant and a graph of V vs I is a straight
line through the origin with gradient R. They obey Ohms Law.
·Common
examples of Ohmic resistors are conductors at constant temperature and
the commercially available ceramic resistors which are colour coded in
bands to show their resistance.
·Many
electronic devices are non-ohmic e.g. diodes, capacitors
Series and Parallel Circuits
·The
equivalent resistance RES for R1 , R2,
R3, ... in series is:
RES
= R1 + R2 + R3 + ....
·Circuit
elements in parallel all have the same potential difference.
·The
equivalent resistance REP for R1 , R2,
R3, ... in parallel is:
E.1MAGNETIC
FIELDS
A
magnetic field is the region around a magnet (or other magnetic device)
where a magnetic force is felt.In
magnets the direction of the magnetic field is from North to South.
E.1.1Magnetic
Field Around a Current Carrying Wire
A
magnetic field can also be produced when a current passes through a wire.
Demo:Wire
& compass on OHP
The
magnetic field is in fact 3 dimensional and is circular around the wire.
The
direction of the magnetic field depends on the direction of the current.
SYNNOT'SSTUPENDOUS
RIGHT-HAND RULE
This
rule works in the following way.If
you point the thumb of your right hand in the direction of the current
then curl your fingers around the wire.Your
fingers give the direction of the magnetic field.
E.1.2Magnetic
Field of a Coil
If
we have a single coil of wire, a magnetic field is set up.The
direction of this magnetic field can be worked out using the Right Hand
rule.
E.1.3Magnetic
Field of a Solenoid
A
solenoid is simply a coil of wire that has many turns in the coil.The
direction of the magnetic field can be worked out using Synnot's right
hand rule for solenoids which says wrap your fingers around the coil in
the direction of the current and your thumb points to North.In
fact the magnetic field pattern around a solenoid is the same as that around
a bar magnet.
E.1.4Magnetic
Flux Density
The
fancy name for magnetic field is magnetic flux density.It
is a vector, has the symbol
B and is measured in a unit called Tesla
(T).
E.1.5Drawing
Magnetic Fields
When
drawing magnetic fields the following convention is used.If
the magnetic field is out of the page a dot is drawn 8.If
the magnetic field is into the page a cross is drawn U.
E.2Magnetic
Forces
E.2.1Magnetic
forces on Currents
Demo:Wire
in magnetic field turn current on and off
Because
a current produces a magnetic field, if a current carrying wire is placed
in a magnetic field there will be a force produced.
The
direction of this force is worked out using Synnot's right hand rule for
forces on currents which states curl your fingers from the current to the
field and your thumb gives the direction of the force.
The
strength of the force depends on four things
i)the
size of the current
ii)the
strength of the magnetic field
iii)the
length of the wire
iv)the
number of wires
The
Strength of the force can be calculated using the following formula
F
= n I L B sinØ
whereØ
= angle between the magnetic field and the current
n
= the number of wires
I
= the current
B
= the magnetic field
L
= the length of the conductor
Example:A
wire carrying a current of 3.4 A is placed at right angles to a magnetic
field of strength 0.20 T, as shown.Calculate
the magnitude and direction of the force on a wire of length 0.80 m
F
= n I L B sin Ø
=
1 ´
3.4 ´
0.80´
0.20´
sin 90°
=
0.54 N
DirectionInto
page
Problem
Set #1:TextPage
146Questions 15 24, 26 39
Prac
#3.9:Investigating
an Electric Motor
E.2.2Force
on Moving Charges
In
the previous section we saw the force on a current carrying wire.But
a current is just a stream of charged particles.From
this we can calculate the force on a single charge moving at speed v.
The
force on a current is given by:
F
= I L B sinø
But
and
\
Substituting
these we get
\
The
direction is worked out using Synnot's right-hand rule for moving charges
which says:For a positive charge
curl your fingers from v to B and your thumb gives the direction
of the force.For a negative charge
the direction is opposite to your thumb.
The
direction of the force is perpendicular to the direction of the velocity.This
causes the particle to move in a curved path.If
the Magnetic field is large enough the particle will move in a circle.
We
haveFmagnetic=vB
q sin 90°
=v
B q
Fcircular
motion=
The
force that is causing the circular motion is the magnetic force.
Sov
B q=
r
=
Where r
is the radius of the circular motion
m
is the mass of the particle
Example:
An
alpha-particle of charge 3.2 ´
10-19
C and mass 6.7 ´
10-27
Kg is fired into a uniform magnetic field of magnitude 0.24 T north with
a uniform velocity of 8.0 ´
106
m/s east.Calculate:
a)the
force on the alpha-particle
b)the
radius of the circular motion
a)F
=
= 
=
N
DirectionUpwards
b)
=0.70
m
E.3Magnetic
Flux
Magnetic
flux is the concentration of the magnetic field.The
flux through an area A, at an angle to a field of flux density B
, is given by
Where
is
the angle between the magnetic field and the normal to the surface of the
area
The
unit of magnetic flux is the Weber (Wb)
Note1.0
Tesla equals 1.0 Weber per square meter,
i.e.
1.0 T = 1.0 Wb/m2
Problem
Set #2:TextPage
148Questions 25, 40 65
E.4Electromagnetic
Induction
Prac
#3.10:Electromagnetic
Induction
E.4.1Induced
Current
Demo:Wire
in magnet, Magnet in coil
In
the previous section we saw that a charge moving through a magnetic field
experiences a force whose magnitude is given by
F
= vBq sinø
If
the charge is moving at right angles to the field then the magnitude of
the force is
F
= vBq
Consider
a conductor moving at right angles to a magnetic field.A
conductor is filled with positive and negative charges each of these will
experience a force (direction given by the right-hand rule).This
is shown in the diagram below.
Thus
an electric current will be produced. However
part of the circuit must be outside the magnetic field or there will be
no flow of electrons.
In
a circuit contained inside the magnetic field two currents will be induced,
each of these opposes each other and no over all current results.Although
a potential difference will result because of the separation of charge.
The
direction of the induced current is given by Synnot's right-hand rule.Curl
our fingers from v to B and your thumb gives the direction
of the current.
E.2.4Induced
EMF
When
a current is induced an EMF is also induced.One
of the scientists to work in this area was Michael Faraday, from his experiments
he concluded that the strength of the induced EMF depended on three things:
1.the
speed of thecoil or magnet
2.the
strength of the magnetic field
3.the
number of turns in the coil.
These
facts lead him to state a law which is summarised in the following formula
(Faraday's Law)
WhereN
= number of turns of wire
= change
in magnetic flux (i.e. B2A2 - B1A1)
=
time taken
Consider
a conductor moving with constant velocity v then
Fapplied
= -Fmagnetic
=
- I L B
If
we then assume that there is no energy loss, we have
mechanical
energy supplied=electrical
energy generated
Hence
E.2.5Lenz's
Law
Lenz's
law helps us work out the direction of the induced current and states that
the induced current is in a direction that opposes the change producing
it.
i.e.The
induced current produces a magnetic field opposing the one producing the
current.
Example:
Consider
a conductor of length 0.10 m and resistance 0.20 W
pushed on a rail at a speed of 1.0 ms-1.If
it is placed at right angles to a magnetic field of 2.0 T and connected
to an external resistance of 1.0 W,
calculate:
a)the
magnitude of the EMF produced
b)the
current through and the potential difference across the external load
a)EMF=-BLv
=
=-0.20
V
Magnitude
is 0.20 V
b)From
Ohm's law
V
= I Rtotal
I
= 0.17 A
Voltage
across external
V
= I R
= 
=
0.17 V
Note:EMF
stands for Electromotive force and is measured in Volts.
E.2.6Electricity
Generation
We
already know that if a coil of wire is in a magnetic field whose strength
is constant, but we change the area of the coil an EMF will also be produced
(from Faraday's Law).This can be
easily done if the coil is rotating.
The
magnetic flux is given by:
where
ø is the angle between the normal to the coil and the magnetic field.
If
we consider four different positions of the coil then the flux will vary
as shown below:
From
Faraday's Law the EMF induced in N coils is:
which
in this case is:
Mathematically
this can be shown to be:
at
any time t.
Thus
an alternating EMF is set up and it varies as a sine wave.The
maximum value would occur when 
equals
one.Thus:
Example
If
a coil of 0.01 m2 and 400 turns is rotated at 10 Hz in a magnetic
field of flux density 0.10 T, calculate the maximum EMF produced.
=
=25
V
E.2.6.1Generation
for the Home
The
Power companies produce electricity by rotating a magnet inside a wire
coil.
Such
a set up produces a current which alternates.An
alternating current has two main advantages over DC (direct current) power.
1.Less
power lost in transmission
2.Its
voltage is easily altered using a transformer.
E.2.6.2Alternating
Current
Alternating
current is where the direction of flow of the current continually changes.The
number of times the direction changes every second is called frequency.The
frequency is governed by the speed of rotation of the generator.In
Australia the standard frequency is 50 Hz.
A
graph of our domestic power supply, 240 V, 50 Hz AC; varies as follows:
Note:The
peak voltage is 340 volts, but the electricity is given the value that
corresponds to the power transfer of a 240 V DC battery.This
is also known as the root mean square (RMS) value and is calculated as
follows:
Problem
Set #3:TextPage
168Questions 1 51
Read:Nelson
PhysicsVCE Units 3 & 4Pages
180 187
Problem
Set #4:TextPage
191Questions 1 35
Video:Electromagnetism
1 (tape no. 1141)
E.2.7Transformers
A
transformer is used to change (transform) the voltage.A
transformer is basically two coils of wire "coupled" by an iron loop.
If
an alternating EMF is applied across the primary coil, the changing current
will produce a changing magnetic field.This
magnetic field is "channeled" via the iron core to the secondary coil.This
will induce an EMF in the second coil.
It
turns out that:
In
a good transformer energy is conserved and:
Power
In= Power
Out
I.e.
Problem
Set #5:TextPage
176Questions 52 83
Video:Electric
Power (tape no. 2046)
E.2.8 Power
Transmission
The
major problem with the transmission of power is that some of it is lost
during transmission.
The
power loss is kept to a minimum by keeping both I and R as low as possible.
R
is kept lowby using thick wires.
In
order to keep the current low and transmit large amounts of power, it is
necessary to use high voltage.The
highest voltage used for transmission in Australia is 500 kV.
To
insulate the line from the supporting structure either glass or porcelain
insulators are used.
Example
In
a power line, a 240 V AC supply of negligible resistance is connected by
wires of total resistance 4.0 W
to a motor of resistance 116 W.Calculate:
a)the
current flowing through the wires
b)the
power loss in the wires
c)the
voltage drop across the wires
d)the
voltage or potential difference across the motor
e)the
power converted in the motor
a)V
= I R
240
= I (4 + 116)
I=2.0
A
b)P
= I2
R
P
= (2.0)2´
4
P
= 16 W
c)V
= I R
=
2.0 ´
4.0
=
8.0 V
d)V
= I R
=
2.0 ´
116
=
232 V or 2.3 ´
102
V
e)P
= I2
R
=
(2.0)2´
116
=
464 W or 4.6 ´
102
W
E.2.9Load
Curves
The
load curves shown above indicate the demand for electricity during the
day and during different seasons. Each of the curves has similar features
and peak times during the day.
The
electric power is the amount of electrical energy supplied every second.
Rearranging
this we get:
Electrical
Energy (joules)=Electrical
Power (watts) ´
Time (seconds)
This
means that the area under the load curve is the total energy supplied.
Problem
Set #6:TextPage
211Questions 1 48