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Electromagnetic Induction


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Title: Electromagnetic Induction

Electromagnetic Induction
Electromagnetic Induction
Faraday was working with two coils mounted on a
wooden spool. Whenever he put a large amount of
current into the primary coil, a small amount of
current was produced in the secondary coil. This
flow of current in the secondary coil was only
present for a very short time.
Michael Faraday
Using a galvanometer he was also able to detect a
small amount of current when the current to the
primary was turned off.
After a lot of further investigations he
determined that a changing magnetic field induced
(created) a current flow in another completely
separate solenoid.
Electromagnetic Induction
Demo Pass a Magnet quickly through a coil of
wire attached to a galvanometer.
The galvanometer will show a current flow when
the magnet moves in and out of the coil.
The strength of the magnetic field.
The amount of current that is induce depends on
The rate of change of the magnetic field.
The number of coils in the solenoid.
Electromagnetic Induction
This is known as Faraday's Law.
It describes the amount of emf (voltage) induced
when a coil of wire is subjected to a changing
magnetic field.
? emf (voltage) V
N number of coils in the solenoid
?F change in magnetic flux (Webber's Wb
?t change in time (s)?
Electromagnetic Induction
What is magnetic flux?
It is a combination of the magnetic field
intensity and the area in which the magnetic
field lines are in.
?F ?BA
The combination of BA must be changed in order to
generate an induced current.
B magnetic field intensity (Tesla)?
A Area of the coil or magnetic field (m2)?
Electromagnetic Induction
Examine each of these two side-by-side magnetic
flux situations. Do you see why the left side has
the greatest magnetic flux?
The Magnetic flux is increasing as the ring is
moving upwards. Why?
Electromagnetic Induction
Example 1
A 200 turn circular coil of radius 0.15 m is
quickly rotated in a time of 0.12 s from being
perpendicular to being parallel to a 0.56 T
magnetic field as shown below.
?t 0.12 s
What is the average induced emf (voltage)?
Electromagnetic Induction
66 V
(The negative sign is just there to remind us in
which direction the induced current is going.)?
Electromagnetic Induction
Example 2
A coil having 45 loops and as area of 0.35 m2 is
initially placed perpendicular to 0.75 T magnetic
field. The field is reversed in direction to a
magnitude of 0.62 T in a time of 0.25 s. The coil
is connected to a 40 ohm resistor. What is the
magnitude of the current through the resistor?
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Electromagnetic Induction
86.31 V
Electromagnetic Induction
Remember Ohm's Law!
But what about the direction of the current?
This will take a little explaining!
Electromagnetic Induction
Lenz's Law
The induced emf always gives rise to a current
whose own magnetic field opposes the original
change in flux.
The coil is squished together reducing the area
to zero. Which direction will the induced current
flow in the coil during the collapse?
Electromagnetic Induction
What is the direction of the induced current in
the coil as the magnetic field is reversed?
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Electromagnetic Induction
Another Example
A bar magnet is quickly brought near a single
coil of copper wire.
What direction will the current flow in the
copper coil when the magnet is brought closer to
the coil?
Electromagnetic Induction
One last example
What direction will the current flow through the
meter as the bar magnet is pulled out?
Answer Current will Downwards through the meter
Electromagnetic Induction
EMF Induced in a Moving Conductor
v x ?t
But A L x(v x ?t) N 1
Electromagnetic Induction
B x L x(v x ?t)?
This equation will determine the voltage placed
across a conductor as it moves in a magnetic
x x x x x x x x
x x x x x x x x
v 5.6 m/s
1.9 V
L 0.45 m
x x x x x x x x
(The top of the conductor will become positively
charged! right hand rule)?
x x x x x x x x
x x x x x x x x
B 0.75 T
Electromagnetic Induction
The back (or Counter) Emf in a Motor
A motor simply a set of coils with current
running through them placed in a magnetic field.
The split-ring commutator allows the current to
change direction every ½ turn keeping the motor
coil turning.
Motor Demo
But the turning coil crossing the field lines
generate a induced current that opposes the
original applied current that makes the motor
Electromagnetic Induction
This induced current produces an opposing voltage
called back emf. It acts against the voltage
causing the motor to spin in the first place.
As the motor spins up to speed the back emf
increases in value until most of the applied
voltage is cancelled out.
At start up the motor only sees the applied
voltage. Since the resistance of the motor is
relatively constant the motor draws a lot of
current when starting up.
Most heavy duty motors found in refrigerators or
table saws have a start capacitor which supplies
some of this current at start up so that it
doesn't trip the circuit breaker when the motor
is first turned on. (In older homes you may have
noticed that some lights in a kitchen dim when
the refrigerator motor (compressor) turns on.)?
When the motor is running at full speed there is
lots of back emf. The effective voltage across
the motor is much less than at start up.
Electromagnetic Induction
Since the effective voltage is reduced when the
motor is running at full speed, so then is the
current required. A motor running at full speed
requires much less current than at start up.
(Once the refrigerator motor is running at full
speed the lights go back to their usual
A very similar effect can cause an electric drill
to burn out its motor. If the drill is being
used to drill a hole and the drill bit is grabbed
and held stationary, the back emf through the
drill's motor is reduced to zero. The current
through the motor goes very high. High enough in
some cases to melt the wires in the motor
windings and burn outthe drill's motor.
Electromagnetic Induction
Vb Emf - IR
Vb back emf
Emf voltage applied to the motor
I current applied to the motor
R resistance of the motor (constant)?
Electromagnetic Induction
The armature windings of a motor have a
resistance of 8.0 ohms. The motor is connected to
120 V. The motor draws 2.0 A while running at
full speed.
a) What current is drawn by the motor at start up?
120 V

15 A
8.0 O
b) What is the back Emf when the motor is running
at full speed?
Vb 120 V - 2.0 x 8.0
104 V
Electromagnetic Induction
Electromagnetic Induction
A transformer is used to either increase
(Step-UP) or decrease (Step-Down) the voltage for
various uses.
Transformers consist of two separate coils of
wire and usually a conducting iron metal support.
It is the arrangement of the two coils that
creates either a step-up or step-down transformer.
Transformers only work with AC (alternating
Electromagnetic Induction
The transformer is very efficient device. It has
no moving parts. The two coils that make up the
transformer are not in contact with each other so
how is it able to change the voltage?
The answer lies in using AC (alternating current)
to create a changing current flow which produces
a changing magnetic field on one side of the
transformer. This induces a changing magnetic
field on the other coil which induces a new
current flow in this coil.
Since the only a changing magnetic field will
induce a new current, the transformer only works
if the current is AC (which is changed 60 times
every second).
Let us examine how to work with the transformer
Electromagnetic Induction
For all ideal Transformers
Energy In Energy Out
And Energy Power x Time
Power x time In Power x time Out
Power In Power Out
And Power V x I
Electromagnetic Induction
Transformer Equation
Vinput x Iinput Voutput x Ioutput
Vi x Ii Vo x Io
Electromagnetic Induction
Let us have another look at Faraday's law and
apply it to transformers
Remember ? voltage
Vi x Ii Vo x Io
Replace the V on each side with Faraday's Law
No?BA x Io
Ni?BA x Ii

The B and A and t are the same on both sides and
cancel each other out.
Electromagnetic Induction
This results in a new equation
NiIi NoIo
Putting this equation with the Power one results

Electromagnetic Induction
A small step-down transformer is plugged into a
wall outlet supplying 120 V. The transformer has
1800 windings on the input side and 40 windings
on the output side. The input current is 0.0050 A.
a) What is the output voltage of this

Vo (Vi x No)/Ni
2.7 V
Vo (120 x 40) / 1800
Electromagnetic Induction
b) What is the output current of this

Io (Vi x Ii) / Vo
Io (120 V x 0.0050 A) / 2.7 V
Io 0.22 A
c) What is the power of this transformer?
P V x I
2.7 V x 0.22 A
0.59 W
Electromagnetic Induction
Large Transformers used in Power and Sub-stations.
These transformers convert high voltage to lower
household voltage.
Small Transformers used in converting 120 V to
6.0 v for small devices.
Electromagnetic Induction
Transformers are used for stepping up voltage to
very high values for efficient transmission over
long distances from power stations and also used
to reduce voltage to more household friendly
voltage of 240 V.
Electromagnetic Induction
But why is electrical energy transported at such
high voltage? Why do we have high voltage power
lines that cross our province?
First of all large scale power production is
usually done far away from populated centers.
Dams are usually created across large rivers at
some distance from a large urban center. Like
wise for large thermal generating plants or
Nuclear power plants.
Second of all no electrical transportation system
is 100 efficient. All electrical wires have some
resistance and hence generate heat when current
is flowing in the wires.
Third the amount of energy lost per second in a
power line called Power loss can be calculated
from this equation
I current in the wires
R resistance of the wires
Electromagnetic Induction
Typically the resistance in power line wires is
quite small at around 5 10 ohms.
The current flow on the other hand can be quite
significant as power (energy) has to be supplied
to all those who need it.
Lets us do a sample calculation as to why high
voltage is used to transport electrical energy
Typically power stations generate power at 5 -
15000 MW (5 million watts to 15 billion watts) at
around 500 V and around 10000 or more Amps
Assuming a resistance of 10 O in the wires what
would be the power loss if the electrical energy
was transported at this voltage and current.
Electromagnetic Induction
Assuming Current is 10000 A and resistance is 10
O then the power loss would be
PL 100002 x 10
PL 1000000000 W
PL 1000 MW
This is 200 x the available energy!!!!
The wires would get very hot and maybe melt. This
would be one huge toaster and no power would
reach those who needed it!
Electromagnetic Induction
Now let us redo the calculation but use a
transformer to increase the voltage up to high
Since the same power has to be delivered by
increasing the voltage the current has to be
Let's increase Voltage to 250000 V
5 x 106 W 250000 x I
I 5 x 106 / 250000
At a high voltage of 250 KV only 20 amps of
current is needed
I 20 A
Electromagnetic Induction
Re-doing the Power Loss calculation at the new
PL 202 x 10
PL 4000 W (A good size electric Heater)
This an insignificant amount of power loss and
would allow a delivery of 99.92 of the energy
Remember transformers themselves are very
efficient as well so very little power is lost in
using them.
The high voltage that power is transmitted at is
too dangerous to use the consumer level so
step-down transformers change it back to a more
manageable level.
Electromagnetic Induction
Have another look at the transmission diagram.
You should now have a good understanding of why
all the changes in voltage have to happen.
Electromagnetic Induction
This the end of electromagnetism and of Physics
I hope you found it both interesting and
Good luck on your final exams and next year!
Electromagnetic Induction
Electromagnetic Induction
Electromagnetic Induction
Electromagnetic Induction
Electromagnetic Induction
Electromagnetic Induction
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