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Electricity
N Bronks
Basic ideas…
Electric current is when electrons start to flow around a
circuit. We use an _________ to measure it and it is
measured in ____.
Potential difference (also called _______) is
how big the push on the electrons is. We use a
________ to measure it and it is measured in
______, a unit named after Volta.
Resistance is anything that resists an electric current. It is
measured in _____.
Words: volts, amps, ohms, voltage, ammeter, voltmeter
Electrons are flowing from the
negative to positive side of the
battery through the wires
Note current moves from
positive to negative,
however electrons are
actually are moving in the
opposite direction!
Current
• Flow of electrons
Current 
Charge Passed a point
time
I
Q
t
Current and Charge
Since one Ampere flows when one coulomb of charge passes
a given point in a circuit each second,
Coulomb
Amp =
second
Charge (C)
Current (A) =
time (s)
or
Q
I=
t
Also
Charge (C) =
no. of electrons x charge of one electron
Example 1:
How many electrons are there in 20 Coulombs ?
No. of electrons = total charge / charge of one electron
No. of electrons = 20 / 1.6x10-19
No. of electrons = 1.25 x1020 electrons
Example 2:
The current in a circuit is 5A. What is the charge flowing in :
a. 1 second ?
5 Coulomb
b. 10 seconds ?
50 Coulomb
Summary Question
1. a. The current in a certain wire is 0.35A. Calculate the charge
passing a point in the wire i. in 10 s, ii in 10 min.
b. Calculate the average current in a wire through which a
charge of 15C passes in i. 5s, ii 100s
Ans:
a. i. I = 0.35A
t = 10s
Q = It
Q = 0.35 x 10 = 3.5 C
ii. I = 0.35A
Q = It
t = 10 min = 600s
Q = 0.35 x 600 = 210 C
b. i. Q = 15C
I = Q/t
ii. Q = 15C
I = Q/t
t = 5s
I = 15/5 = 3A
t = 100s
I = 15/100 = 0.15A
Summary Questions
page 47
2. Calculate the number of electrons passing a point in the wire in
10 minutes when the current is a. 1.0µA
b. 5.0A
Ans:
a. I = 1.0µA
t= 10min = 600s
Q = It = 1x10-6 x 600 = 6x10-4 C
No. of electrons = total charge / charge of one electron
No. of electrons = 6x10-4 / 1.6x10-19 = 3.75x1015 electrons
I = 5A
t= 10min = 600s
Q = It = 5 x 600 = 3000 C
No. of electrons = total charge / charge of one electron
No. of electrons = 3000 / 1.6x10-19 = 1.88x1022 electrons
Summary Questions
page 47
3. In an electron beam experiment, the beam current is 1.2mA.
Calculate
a. The charge flowing along the beam each minute
b. The number of electrons that pass along the beam each
minute
Ans:
a. I = 1.2x10-3 A
t= 1min = 60s
Q = It
= 1.2x10-3 x60 = 0.072 C
b. no. of electrons = total charge /charge of one electron
no. of electrons = 0.072/1.6x10-19 = 4.5x1017 electrons
Summary Questions
page 47
A certain type of rechargeable battery is capable of delivering a
current of 0.2A for 4000s before its voltage drops and it
needs to be recharged.
Calculate:
a. The total charge the battery can deliver before it needs to be
recharged.
b. The maximum time it could be used for without being
recharged if the current through it was i. 0.5A, ii. 0.1A
Ans:
a. I = 0.2A
t = 4000s
Q = It = 0.2 x 4000 = 800C
b. i. Q = 800C
I = 0.5A
t = Q/I = 800/0.5 = 1600s
ii. Q = 800C
I = 0.1A
t = Q/I = 800/0.1 = 8000s
Quiz
Current and charge quiz
1. Calculate the charge passing through a lamp
in three minutes when a steady current of 0.4 A
is flowing.
I=
Q
t
Q=It
Q = 0.4 x 3 x 60 = 72 Coulomb
Current and charge quiz
2. Calculate the number of electrons flowing
through a resistor when a current of 2.3 flows
for 5 minutes
Q
I=
t
Q=Ixt
Q = 2.3 x 5 x 60 = 690 Coulomb
no. of electrons = 690 /1.6x10-19 = 4.31x1021
Current and charge quiz
3. What is the current in a circuit if 2.5x1020
electrons pass a given point every 8 seconds
Charge (C) =
no. of electrons x charge of one electrons
Charge (C) =
2.5x1020 x 1.6x10-19 = 40 Coulombs
I=
Q
t
Current = 40/8 =
5 Amps
Current and charge quiz
4. How long does it take for a current of 0.3A to
supply a charge of 48C?
Q
I=
t
t = Q/I
t = 48/0.3 = 160 seconds
Current and charge quiz
5. How many electrons pass a point when a
current of 0.4A flows for 900 seconds?
Q
I=
t
Q=Ixt
= 0.4 x 900
= 360 Coulomb
no. of electrons = total charge / charge of one electron
no. of electrons = 360 / 1.6 x 10-19 = 2.25 x 1021
Current and charge quiz
6. A torch bulb passes a current of 120 mA.
How many coulombs of charge flow through the lamp in 1
minute?
Q=Ixt
= 120x10-3 x 60
= 7.2 Coulomb
Current and charge quiz
7. A car battery is rated as 36 A h.
In principle this means it could pass a current of 1 A for 36 h
before it runs down. How much charge passes through the
battery if it is completely run down?
Q=Ixt
= 1 x 36 x 60 x 60
= 129600 Coulomb
H/W
• 2004 HL Q4
More basic ideas…
Another battery
means more current
as there is a
greater push on the
electrons
The extra
resistance from the
extra bulb means
less current
Current in a series circuit
If the current
here is 2
amps…
The current
here will
be…
2A
The
current
here will
be… 2A
And the
current
here will
be…
2A
In other words, the current in a series
circuit is THE SAME at any point
Current in a parallel circuit
A PARALLEL circuit is one where the current has a “choice
of routes”
Here comes the current…
Half of the current
will go down here
(assuming the bulbs
are the same)…
And the rest will
go down here…
Current in a parallel circuit
If the
current
here is 6
amps
And the
current here
will be… 6A
The current
here will be…
2A
The current
here will be…
2A
The current
here will be…
2A
Voltage in a series circuit
If the voltage
across the
battery is 6V…
V
Voltmeter
always in
parallel
…and these
bulbs are all
identical…
…what will the
voltage across
each bulb be?
V
V
2V
Voltage in a series circuit
If the voltage
across the
battery is 6V…
…what will the
voltage across
two bulbs be?
V
V
4V
Voltage in a parallel circuit
If the voltage across
the batteries is 4V…
What is the
voltage here?
4V
V
And here?
V
4V
Summary
In a SERIES circuit:
Current is THE SAME at any point
Voltage SPLITS UP over each component
In a PARALLEL circuit:
Current SPLITS UP down each “strand”
Voltage is THE SAME across each”strand”
An example question:
6V
A3
3A
A1
V1
A2
V2
V3
Advantages of parallel circuits…
There are two main reasons why parallel circuits are used
more commonly than series circuits:
1) Extra appliances (like bulbs) can be added without
affecting the output of the others
2) When one
breaks they
don’t all fail
Resistance
Resistance is anything that will
RESIST a current. It is measured
in Ohms, a unit named after me.
That makes me so happy
Georg Simon Ohm
1789-1854
The resistance of a component can be
calculated using Ohm’s Law:
Resistance
(in )
=
V
Voltage (in V)
Current (in A)
I
R
An example question:
Ammeter
reads 2A
A
What is the resistance across this bulb?
V
As R = volts / current = 10/2 = 5
Assuming all the bulbs are the same what is
the total resistance in this circuit?
Voltmeter
reads 10V
Total R = 5 + 5 + 5 = 15 
More examples…
3A
6V
12V
3A
2A
4V
2V
1A
What is the
resistance of
these bulbs?
Practice with Ohm’s Law
Ohms
4
15
2
9
6
Volts
100
150
30
45
48
Amps
25
10
15
5
8
VARIATION OF CURRENT (I) WITH P.D.
(V)
A
+
6V
-
V
Nichrome
wire
Method
1.
Set up the circuit as shown and set
the voltage supply at 6 V d.c.
2.Adjust the by moving the slider of the
potential divider to obtain different
values for the voltage V and hence for
the current I.
3.Obtain at least six values for V and I
using the voltmeter and the ammeter.
4.Plot a graph of V against I
Variations
(a) A METALLIC CONDUCTOR
With a wire
(b) A FILAMENT BULB
(c) COPPER SULFATE SOLUTION
WITH COPPER ELECTRODES
(d) SEMICONDUCTOR DIODE
Done both ways with a milli-Ammeter and the
a micro Ammeter
Current-voltage graphs
I
I
I
V
V
V
1. Resistor
Current
increases in
proportion
to voltage
2. Bulb
As voltage increases
the bulb gets hotter
and resistance
increases
3. Diode
A diode only
lets current go
in one direction
Factors affecting
Resistance of a conductor
• Resistance depends on
– Temperature
– Material of conductor
– Length
– Cross-sectional area
Temperature
The resistance of a metallic
conductor increases as the
temperature increases e.g. copper
The resistance of a
semiconductor/insulator decreases
as the temperature increases e.g.
thermistor.
VARIATION OF THE RESISTANCE OF A
METALLIC CONDUCTOR WITH TEMPERATURE
10º C
10ºC
Ω
Digital
thermometer
Wire wound
on frame
Water
Glycerol
Heat source
Method
1.
Set up as shown.
2.
Use the thermometer to note the temperature
of the glycerol, which is also the temperature of the
coil.
3.
Record the resistance of the coil of wire using
the ohmmeter.
4.
Heat the beaker.
5.
For each 10 C rise in temperature record the
resistance and temperature using the ohmmeter and
the thermometer.
6.
Plot a graph of resistance against temperature.
Graph and Precautions
R

Precautions
- Heat the water slowly so temperature does not
rise at end of experiment
-Wait until glycerol is the same temperature as
water before taking a reading.
Factors affecting Resistance of
a conductor
Length
Resistance of a uniform
conductor is directly
proportional to its length.
i.e. R  L
Cross-sectional area
Resistance of a uniform
conductor is inversely
proportional to its crosssectional area.
i.e.
R1
A
Factors affecting
Resistance of a conductor
• Material
The material also affects the resistance of a conductor
by a fixed amount for different materials. This is known
as resistivity ().
R = L
A
 = Resistivity
Unit: ohm meter  m
RESISTIVITY OF THE MATERIAL OF A
WIRE
Micrometer
Nichrome
wire
Crocodile clips
l

Metre stick
Bench
clamp
Stand
Method
1. Note the resistance of the leads when the crocodile clips are
connected together. Could also be precaution.
2. Stretch the wire enough to remove any kinks or ‘slack’ in the
wire.
3.Read the resistance of the leads plus the resistance of wire
between the crocodile clips from the ohmmeter. Subtract the
resistance of the leads to get R.
4.Measure the length l of the wire between the crocodile clips,
with the metre stick.
5.Increase the distance between the crocodile clips. Measure the
new values of R and l and tabulate the results.
6.Make a note of the zero error on the micrometer. Find the
average value of the diameter d.
 R
1.
Calculate the resistivity ρñ    A,
2

d
l
where A =
4
2.
Calculate the average value.
Precautions Ensure wire is straight
and has no kinks like ....
Take the diameter of the wire at
different angles
Resistors in series and Parallel
I
IT
R1
V1
R2
R3
VT  V1  V2  V3
V
I2
R1
I1
R2
IT  I1  I 2  I 3
Resistors in series and Parallel
I
IT
R1
V1
R2
R3
VT  V1  V2  V3
IRT  IR1  IR2  IR3
RT  R1  R2  R3
V
I2
R1
I1
R2
Resistors in series and Parallel
I
IT
R1
V1
R2
R3
IT  I1  I 2  I 3
V
V V
V



R T R1 R 2 R 3
1
1
1
1
 

RT R1 R2 R3
V
I2
R1
I1
R2
H/W
• 2005 HL Q9
Wheatstone Bridge
Uses
– Temperature control
– Fail-Safe Device (switch
circuit off)
– Measure an unknown resistance
B
A
I
C
D
– R1 = R3 (When it’s balanced
R2 R4 Galvanometer reads
zero)
Metre Bridge
R1 = R2 (|AB|)
|BC|
Effects of an Electric Current
• Heat
• Chemical
• Magnetic
Chemical Effects of an
Electric Current
• Electrolysis is the chemical effect of an
electric current
• Voltameter consists of electrodes, an
electrolyte and a container
• Inactive electrodes are electrodes that
don’t take part in the chemical reaction e.g.
platinum in H2SO4
• Active electrodes are electrodes that take
part in the chemical reaction e.g. copper in
CuSO4
Chemical Effects
• Ion is an atom or molecule that
has lost or gained 1 or more
electrons
• Charge Carriers in an electrolyte
are + and – ions
Uses
Electroplating to make metal
look better, prevent corrosion
Purifying metals
Making electrolytic capacitors
Current-voltage graphs
I
I
V
V
1. Active
Electrodes
2. Inert Electrodes
e.g. Copper in
Copper Sulphate
e.g. Platinum in
Water
Current Carriers
Medium
Carrier
Solid (Metal)
Electrons
Liquid (Electrolyte)
Ions
Gas
Electrons and Ions
Resistance in Semiconductors
1) Normal conductor like
metal resistance increases
as vibrating atoms slow
the flow of electrons
Resistance
2) Thermistor – resistance
DECREASES when
temperature INCREASES –
Due to more charge carriers
being liberated by heat
Resistance
Temperature
Temperature
Fuse – Safety device
Fuses are designed to melt
when too large a current
tries to pass through them
to protect devices.
Prevent Fires
Modern fuse boxes contain
MCB (Miniature circuit
breakers) that trip when
too much current flows to
protect the circuit
2A
5A
Which Fuse
• A i-pod charge uses 200W and is
plugged into the mains at 230v. What
fuse is in the plug?
• P=I.V
• 200=I.230
• I = 200/230 = 0.87A is current used
• So the most the fuse should be is a 1A
Other safety devices…
1)
Insulation and double insulation
In some parts of Europe they have
no earth wire just two layer of
insulating material the sign is
2)
Residual Current Circuit Breaker
An RCCB (RCB) detects any difference in
current between the live and neutral
connectors and the earth it switches off
the current when needed. They can also
be easily reset.
Electrical Safety
Fuse on live wire !!
• A combination of fuse and Earth
The casing touches the bare wire and it becomes live
That
Hurts!
A.C. Supply
The fuse will melt to
prevent electrocution
and the electricity is
carried to earth
Wiring a plug
1. Earth
wire
4. Live
wire
5. Fuse
2. Neutral
wire
3. Insulation
6. Cable
grip
Capacitors
• A device for storing charge.
• A pair of metal plates are
separated by a narrow gap
-
+
+
+
+
+
+
electrons
-
-
-
-
capacitor charge
charged capacitor
capacitor discharge
electrons
Charge & Discharge
Capacitor Construction
• Two metal plates
• Separated by insulating
material
• ‘Sandwich’ construction
• ‘Swiss roll’ structure
• Capacitance set by...
A
C 
d
Uses of
Capacitors
• Storing charge for quick
release – Camera Flash
• Charging and discharging
at fixed intervals –
Hazard Lights
• Smoothing rectified
current – See
Semiconductors
Smoothing
• Add capacitor
variable capacitor
smoothing capacitors
Parallel Plate Capacitors
•
1.
The size of the capacitor depends on
The Distance the plates are apart d
-
+
-
+
-
+
d
Parallel Plate Capacitors
2 /.The area of overlap A
-
+
+
+
A
Parallel Plate Capacitors
•
3/.The material between ()
-
-
-
+
+
+
+
+
+
+
High  material
Called a
DIELECTRIC
Finding Capacitance
Vs
capacitanc e 
VA
Equations
For the parallel plate capacitor
Capacitance
In Farads
Permitivity in
Fm-1
C = A
d
Distance in
meters
Area
In m2
Example 1
The common area of the plates of an air
capacitor is 400cm2 if the distance between the
plates is 1cm and ε0=8.5x10-12Fm-1.
C = 0 A
d
C=
8.5x10-12Fm-1x 0.04m2 =3.4x10-11F.
0.01m
Capacitance experiment on the
internet
Equations
Capacitance on any conductor
Capacitance
In Farads
C = Q
V
Potential
Difference
in volts
Charge in
Coulombs
Placing a charge of 35μC on a conductor
raises it's potential by 100 V. Calculate
the capacitance of the conductor.
Info Q = 35μC and V = 100V find C=?
Using Q=VC or C = Q/V
= 35 x 10-6/100
= 35 x 10-8 Farads
Equations
Energy stored on a capacitor
Energy
Stored
Capacitance
In Farads
2
C
(V)
Work Done = ½
Voltage
Squared
Example 3
Find the capacitance and energy stored of a parallel
plate capacitor with 2mm between the plates and
150cm2 overlap area and a dielectric of relative
Permittivity of 3. The potential across the plates
is 150V.
A = 150cm2=0.015m2,
d = 2x10-3m,
ε = 3xε0 = 27x10-12Fm-1
As C = ε0A/d = 27x10-12 x 0.015/0.002 = 2.025x10-9 F
Energy stored = ½ C V2 = ½ x 2.025x10-9x (150)2
= 2.28x10-5 Joules
Types of Batteries
Type of Battery
Contains
Uses
Wet cell
rechargeable
Lead and acid
Cars, industry
Dry cell
rechargeable
Nickel, cadmium,
lithium
Mobile phones,
power tools
Dry cell nonrechargeable
Zinc, carbon,
manganese,
lithium
Torches, clocks,
hearing aids
Why use rechargeable batteries?
Why use standard batteries?
• Long long-term expense
• No need for charger
• Can be used many times
• Less expensive
• Less energy to produce
• Rechargeables contain carcinogens
There are 2 types of currents:
• Direct Current (DC) – Where
electrons flow in the same
direction in a wire.
There are 2 types of currents:
• Alternating Current (AC) –
electrons flow in different
directions in a wire
DC and AC
V
DC stands for “Direct
Current” – the current only
flows in one direction:
AC stands for “Alternating
Current” – the current
changes direction 50 times
every second (frequency =
50Hz)
Time
1/50th s
240V
Find Root Mean Square of
voltage by
T
Vrms= Vpeak/ √2
V
Used to compared
power output of
D.C. with A.C.
The National Grid
Power station
Step up
transformer
Step down
transformer
Homes
If electricity companies transmitted electricity at
240 volts through overhead power lines there
would be too much energy lost by the time
electricity reached our homes.
This is explained by JOULES LAW
The National Grid
Power station
Step up
transformer
Step down
transformer
Homes
Power Transmitted is = P = V.I
JOULES LAW gives us the power turned into heat
Power Lost = I2R
So if we have a high voltage we only need a small
current. We loss much less energy
Power loss in Transmission lines
A power company wants to send 100000w of
power by a line with a resistance of 12
ohms. If it uses 100A as the current
Power transmitted = V . I
100000 = V . 100
So V=1000Volts
But the loss is from Joules law = I2R
= (100)2.12 = 120000watts
Power loss in Transmission lines
If we want the same power but use only 1A
as the current
Power transmitted = V . I
100000 = V . 1
So V=100000Volts
But the loss is from Joules law = I2R
= (1)2.12 = 12watts
10000 times less!
Joules law
10°C
A
Digital
thermometer
Calorimeter
Heating coil
Lid
Water
Lagging
Method
1.
Put sufficient water in a calorimeter to cover
the heating coil. Set up the circuit as shown.
2.
Note the temperature.
3.
Switch on the power and simultaneously start
the stopwatch. Allow a current of 0.5 A to flow for
five minutes. Make sure the current stays constant
throughout; adjust the rheostat if necessary.
4.
Note the current, using the ammeter.
5.
Note the time for which the current flowed.
6.
Stir and note the highest temperature.
Calculate the change in temperature ∆.
Calculation and Graph
∆
I2
Repeat the above procedure for increasing
values of current I, taking care not to exceed
the current rating marked on the rheostat or
the power supply. Take at least six readings.
Plot a graph of ∆(Y-axis) against I 2 (X-axis).
A straight-line graph through the origin verifies that
∆  I 2 i.e. Joule’s law.
Electrical Power lost as Heat P  I2 is Joules law
The power lost (Rate at which heat is produced) is
proportional to the square of the current.
H/W
• 2006 HL Q 4
Experiment to Show shape of
Electric Field
• The electrodes connected to high voltage
source is placed in the shallow glass dish
containing a mixture of semolina and castor
oil. The semolina aligns itself along the lines
of the electric field.
The Electroscope
-
The
electroscope
detects charge
The Gold leaf
and post repel
each other
- +
+
+
+
H/W
• 2006 HL Q9
Electric and Magnetic Fields
• Electric Field- region of space where a
charged particle feels a electrostatic
force.
• Magnetic field – region where a magnet
feels a force other than gravity.
• Field lines are the path a positive charge
or north pole would travel
Coulomb's Law
• Force between two charged bodies
Q1
Force = f
d
Q2
 Q1.Q2
d2
Put this as a sentence to get a law!
Coulomb Calculations
Force =f
 Q1.Q2
d2
• We replace the proportional with a
equals and a constant to get an
equation
Force =
f
=
Q1.Q2
4d2
 = permitivity as in capacitors
Coulomb's Law Calculations
• Force between these bodies
d=2m
2C
Force
=
f
4mC
=
Q1.Q2
4d2
 = 3.4 x 10-11
Coulomb's Law Calculations
• Force between these bodies
d=2m
2C
Force
=
f
4mC
=
2 x 0.004
4 x3.4 x 10-11x 22
Coulomb's Law Calculations
• Force between these bodies
d=2m
2C
Force
=
f
4mC
= 7.49 x 10-15 N
Coulomb's Law Calculations
• Force between these bodies
2C
d=2m
4mC
Electric Field Strength
Electric Field Strength
=
E = F/q
=
E = 7.49 x 10-15 N /2C
= 3.75 x 10-15 N /C
Precipitator
• Carbon and
ash - can be
removed
from waste
gases with
the use of
electrostatic
precipitators
Precipitator
• Dirt
particles are
charged
then made to
stick to
oppositely
charged
plates
Photocopier
•
•
•
•
•
•
Charging:
Exposure:
Developing:
Transfer:
Fusing:
Cleaning:
Potential Difference (V)
Potential difference is the work done
per unit charge to transfer a charge
from one point to another (also Voltage)
i.e
V=W
Q
Potential Difference (V)
V=W
Q
 Unit Volt V or J C-1
 Volt is the p.d. between two points if one joule of
work is done bringing one coulomb from one point to
the other
 Potential at a point is the p.d. between a point and
the Earth, where the Earth is at zero potential

Current in a Magnetic Field
N
S
N
S
Current in a Magnetic Field
A conductor carrying a current in a
magnetic field will always feel a force
Current
N
S
Magnetic
Field
Force
The force is perpendicular to the current and the
field. – This is THE MOTOR EFFECT
Fleming’s Left Hand Rule
I used my left hand to show
the direction the wire would
move
The Size of the Force
Force = F = B.I.l
Where B = Magnetic Field Density in Tesla (T)
I= Current in Amps (A)……………………………
L = length if the conductor in metres…
Example What is the force acting on a conductor of
length 80cm carrying a current of 3A in a 4.5T
magnetic field?
Using
Force = F = B.I.l
= 4.5x3x0.8
= 10.8N
Two Parallel Wires
• Wires also produce magnetic fields when a
current flows
Attraction
Two Parallel Wires
• The fields act like magnets when the
current flows
Repulsion
The Ampere
• Basic unit of electricity
1m
F=2x10-7N/m
The current flowing is 1A when the force between
two infinitely long conductors 1m apart in a vacuum
is 2x10-7N Per metre of length.
Demo
• OHP and coils and compass
Moving Charge
• When any charged particle moves it is like a small
current of electricity
• It feels the same force
• The crosses show a magnetic field into the screen
e-
Force
Velocity
e-
e-
Moving Charge
• A positive will move the other way
+
e-
Force
Velocity
All charged
particles
moving in
magnetic fields
always have a
force at right
angles to their
velocity so
follow a circular
path due to
FLH Rule
See particles motion
Force 0n a Particle
Force = F = B.q.v
Where B = Magnetic Field Density in Tesla (T)
q=charge on the particle (C)
v=velocity of the particle…
Example What is the force acting on a particle
travelling at 80m/s carrying a charge of 0.1C in a
10T magnetic field?
Using
Force = F = B.q.v
= 10x.1x80
= 80N
Use to define the tesla
Defining Units
• Find a formula
• Say 1 thingys is when all other
thingys are 1
• F = B.q.v
• 1 Tesla in when 1N of force is on a
charge of 1C moving at 1m/s
Demo
• CRT and magnet
Induction
is where changes in the current flow in a
circuit are caused by changes in an
external field.
Moving Magnet
N
Circuit
turning
off and
on
Electromagnetic
induction
The direction of the induced current is
reversed if…
1) The magnet is moved in the opposite
direction
2) The other pole is inserted first
The size of the induced current can be
increased by:
1) Increasing the speed of movement
2) Increasing the magnet strength
3) Increasing the number of turns on
the coil
Demo
• Coils and spot galvo
• Internethttp://phet.colorado.edu/en/
simulation/faraday
Generators (dynamos)
Induced current can be
increased in 4 ways:
1) Increasing the speed of
movement
2) Increasing the magnetic
field strength
3) Increasing the number of
turns on the coil
4) Increasing the area of the
coil
Electric motor
Faraday’s Law
Basically
1. More turns (N) more EMF
2. Faster movement more EMF
Rate of change of FLUX DENSITY is
proportional to induced EMF
Induced EMF = E = - Nd ( =B.A)
dt
Lenz’s Law
The induced EMF always opposes the current/Motion
You get ought for nought
A version of Newton III and of energy conversion
The induction always tries to stop the motion or
change in the field.
The ring
moves away
as the
induced
current is
Aluminum
preventing
Ring
more
induction
Mutual induction
• Induction in a second circuit
caused by changes in a first
circuit
• Main use in a transformer
• As the current changes the field
changes giving a EMF in the
second circuit.
Transformers
This how A.C.
changes
voltage up or
down
V In
V Out
=
Turns 2
Turns 1
Self Induction
• property whereby an electromotive
force (EMF) is induced in a circuit by
a variation of current in the circuit
its self
Current
Back
EMF
D.C. Source
Another example on LENZ’S LAW
Flux Density
• Magnetic flux, represented by the Greek
letter Φ (phi), total magnetism produced by
an object. The SI unit of magnetic flux is
the Weber
• Magnetic field (B) is the flux through a
square meter (the unit of magnetic field is
the Weber per square meter, or Tesla.)
As the flux
expands the
density through
any square meter
decreases