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Physics 12 Electromagnetic Induction
In 1819, when Oersted demonstrated the ability of a steady current to
produce a steady magnetic field, scientists assumed that a steady magnetic field
would produce a steady current. It seems logical and yet it doesn’t work. It wasn’t
until 1831 that Michael Faraday discovered the basic principle of electromagnetic
induction. Faraday went on to develop the first electric generator, modern versions
of which are now used to provide electricity to homes and industries all over the
world.
Faraday’s Law of Electromagnetic Induction
Faraday made his discovery by experimenting with conductors in the vicinity of
magnetic fields. His investigations involved three situations:
i) moving a wire through the jaws of a horseshoe magnet
ii) plunging a bar magnet into and out of the core of a coil
iii) touching the iron core of a coil with a bar magnet, then removing the magnet
In the first case, Faraday found that electric current only flowed while the
conductor was cutting across the magnetic field. In the second case, electric current
only began to flow when the bar magnet was moving into or out of the coil. And in
the third situation, electric current was observed in the coil when the iron cylinder
was being magnetized or demagnetized.
Faraday was able to combine these three conditions into one general
statement, now known as the law of electromagnetic induction.
Law of Electromagnetic Induction
An electric current is induced in a conductor whenever the magnetic
field in the region of the conductor changes.
Induction can be demonstrated with a device that Faraday constructed and
used himself in his early studies of the induction effect. Known as Faraday’s iron ring, it consists
of a doughnut-shaped ring of soft iron with two separate coils of wire wound around it, as in
Figure 4.
The primary coil is connected through a switch to a voltage source. The secondary
coil is connected directly to a galvanometer (a device for measuring current). When the switch is
closed there is a current in the primary circuit, causing the entire iron ring to become magnetized.
This sudden increase in magnetic field strength causes a current to be induced momentarily in the
secondary coil. Once the current in the primary coil is steady and the magnetic field in the iron ring
is established, the induced current no longer exists.
If the switch is then opened, the iron ring becomes demagnetized and the consequent
decreasing magnetic field strength once again induces a momentary current in the secondary coil,
this time in the opposite direction. This effect is known as mutual induction. In fact, the two coils
do not have to be coupled with an iron ring; the iron ring merely acts to strengthen an effect that
would be present in any case.
By observing what happens when a bar magnet is plunged into the core of a coil that
is connected to a galvanometer, it is possible to conclude that the factors affecting
the magnitude of the induced current are
• the number of turns on the induction coil
• the rate of change of the inducing magnetic field
• the strength of the inducing magnetic field
Faraday’s discovery of induced currents opened the door to readily accessible and
cheap sources of current. This current could be used to power electric motors and other devices,
saving time and money for many people and industries. However, there were still some questions
to be answered. What determines the direction of the induced current? Where does
the energy associated with the induced current come from?
A few years after Faraday’s discovery of induction, a German physicist working in
Russia, Heinrich Lenz (1804–1865), applied the law of conservation of energy and succeeded
in stating this relationship.
Lenz (pronounced “Lents”) reasoned that when a current is induced through a
conductor, the induced current itself sets up a magnetic field. This magnetic field, which we will
call the induced field, then interacts with the inducing field, either attracting it or repelling it. In
determining which of these interactions is the more likely one, it will help to consider an example.
When the N-pole of a bar magnet enters a coil, a galvanometer will indicate an
induced current through the coil. When the N-pole is removed, the galvanometer will indicate a
current in the opposite direction. Using the S-pole of the bar magnet causes induced currents in
directions opposite to the above. Evidently there is some simple relationship between the action of
the inducing field and the direction of the induced current.
Lanz’s Law is really an application of the Law of Conservation of Energy. Energy cannot
be created nor destroyed. You cannot get electrical energy from nothing. It must come from
somewhere. As indicated in diagram (a), in order to see work done, the right hand side of the core
must be the same pole as the approaching pole, otherwise “S” will attract the incoming N pole, and
“energy” of plunging in the bar magnet will not be needed.
Similarly, when the bar magnet is being pulled away, in order to see “work done”, the
right hand side of the core must be south pole, so that work is needed to overcome the attraction.
Lenz’s Law
For a current induced in a coil by a changing magnetic field, the
electric current is in such a direction that its own magnetic field
opposes the change that produced it.
It should be noted that Lenz’s law is really just a rule to determine the direction of the
induced current and contains nothing that wasn’t already implied by Faraday. Also, the changing
magnetic field really produces an electric field, and the induced current is just a way that it can be
detected.
Example 1) Determine the direction of the electric current for the case in Figure below.
Example 2) Determine the pole of the bar magnet that is being inserted into the induction coil in
Figure.
In the same way, if a conducting rod is moved through a magnetic field, the magnetic
field of an induced current opposite the magnetic field that produced the induced current (Lenz’s
law).
Direction of current:
 Palm opposite velocity (v)
 Fingers into the page
 Thumb will point up in direction of current.
Lenz’s law tells us that the magnetic force on the conducting rod must be in the
opposite direction to its motion (opposite to V). If we use our right hand (opened) and point the
palm opposite to V and our fingers in the direction of the induced current. In the diagram above,
the current flows counter-clockwise through the circuit.
Now it is your turn to determine how the induced current flow in the given diagram
below.
What is the magnitude of the induced current?
The magnitude of the induced current is found using Ohm’s law.
and
V=IR
V = B┴ l v
where V = induced voltage
B┴ = magnetic field component perpendicular to………
l = length of the conducting rod
v = speed of the conducting rod
Derivation of V = B┴ l v
(Method I)
When a conducting rod is moved through a magnetic field, the + charges in the rod experience a
magnetic force.
F = q v B┴
These positive charges would move to the upper end of the rod, leaving the bottom end
negative. Because of the difference in charge between the ends of this rod, there is a potential
difference (induced voltage). When the positive charges were forced through this rod, work is done.
W=Fd
or
∆E = F l (l = length of the conductor)
but
F = Fm = q v B┴ and ∆E = q V
therefore q V = q v B┴ l
to simplify
V (ε) = v B┴ l
Derivation of V = B┴ l v
(Method II)
To derive this equation we can use the concept of magnetic flux (Φ).
Faraday’s law of induction: The induced voltage is proportional to the rate of change in
the magnetic flux (Φ).
Now let us return to our circuit:
Change in magnetic flux:
∆ Φ = B┴ l ∆ x
Rate of change in magnetic flux:
Since
x
= v , then
t
V = B┴ l v
Example 1) A conducting rod 25.0 cm long moves perpendicular to a magnetic field, B = 0.20 T at
a speed of 1.0 m/s. Calculate the induced voltage in the rod?
Example 2) The conducting rod in the diagram below is 15 cm long, and is moving at a speed of
2.0 m/s perpendicular to a 0.30 T magnetic field.
If the resistance in the circuit is 4.0 Ω, what is the
magnitude and direction of the current (electron
flow) through the circuit?
Example 3) A rectangular coil of wire containing 5 loops is moved at a speed of 4.5 m/s
perpendicular to a 0.85 T magnetic field as shown below. If the length of the side of the
coil moving perpendicular to the field is 0.28 m and the resistance in the circuit is 3.0 Ω.
a) What is the induced current?
b) What is the direction of the current (electron flow)?
Example 4) A solid conductor travels 15 m/s across a uniform 0.086 T magnetic field. Which side
(X or Y) is positively charged and what is the emf across this conductor?
Example 5) A 600 turn circular coil with an area of 2.64 x 10–2 m2 is perpendicular to a 0.08 T
field. The magnetic field changes to 0.01 T in the opposite direction in 0.25s.
PHY 12
Electromagnetism
Lesson Notes: Motors and Back emf
Motors are devices that convert electrical energy to mechanical energy. Essentially,
a motor is a generator run in reverse: instead of a current being generated by a rotating loop, a
current is supplied to the loop by a source of emf, and the magnetic torque on the current-carrying
loop causes it to rotate.
A motor can perform useful mechanical work when a shaft connected to its rotating
coil is attached to some external device. As the coil in the motor rotates, however, the changing
magnetic flux through it induces an emf which acts to reduce the current in the coil. If it increased
the current, Lenz’s law would be violated. The phrase back emf is used for an emf that tends to
reduce the applied current. The back emf increases in magnitude as the rotational speed of the coil
increases. We can picture this state of affairs as the equivalent
circuit in Figure 20.23. For illustrative purposes, assume that the
external power source supplying current in the coil of the motor
has a voltage of 120 V, that the coil has a resistance of 10 Ω, and
that the back emf induced in the coil at this instant is 70 V. The
voltage available to supply current equals the difference between
the applied voltage and the back emf, 50 V in this case. The
current is always reduced by the back emf.
When a motor is turned on, there is no back
emf initially, and the current is very large because it’s limited
only by the resistance of the coil. As the coil begins to rotate, the
induced back emf opposes the applied voltage and the current in
the coil is reduced. If the mechanical load increases, the motor slows down, which decreases the
back emf. This reduction in the back emf increases the current in the coil and therefore also
increases the power needed from the external voltage source. As a result, the power requirements
for starting a motor and for running it under heavy loads are greater than those for running the
motor under average loads. If the motor is allowed to run under no mechanical load, the back emf
reduces the current to a value just large enough to balance energy losses by heat and friction.
Example 1) A motor has coils with a resistance of 10.0 Ω and is supplied by a voltage of
V = 1.20 x 102 V. When the motor is running at its maximum speed, the back emf is 70.0 V.
Find the current in the coils (a) when the motor is first turned on and
(b) when the motor has reached its maximum rotation rate.
Example 2) A motor is designed to operate on 130 V and draws a current of 13.8 A when it first starts
up. At its normal operating speed, the motor draws a current of 4.40 A.
(a) What is the resistance of the armature coil?
(b) What is the back emf developed at normal speed?
(c) What is the current drawn by the motor at one-third normal speed?
Practice Questions.
1) A dc motor has a resistance of 2.0 ohms. When connected to a 12 V source, with the motor
rotating at its operational speed, a back emf of 5.5 V is generated. What is the current in the
motor at operational speed? 3.3 A
2) An electric motor rotates at various speeds and the current through the armature changes
accordingly. Which pair of conditions occurs when the motor generates the greatest back emf?
a)
b)
c)
d)
Speed: fastest; Current: largest
Speed: slowest; Current: largest
Speed: slowest; Current: smallest
Speed: fastest; Current: smallest
3) A motor is connected to a 12 V dc supply and draws 5.0 A when it first starts up. What will be
the back emf when the motor is operating at full speed and drawing 1.2 A. 9.1 V
4) An electric motor is connected to a 12.0 V power supply. When the armature is prevented from
rotating, the current is 8.0 A. When the motor is running at normal speed, the current is 2.0 A.
What is the back emf in each case? 0V, 9V
5) As a carpenter drills into a beam, friction on the drill bit causes the armature of the drill to slow
down. How will the back emf and the current through the armature change as the drill slows
down?
a) Back emf: decrease; Current: decrease
b) Back emf: increase; Current: increase
c) Back emf: increase; Current: decrease
d) Back emf: decrease; Current: increase
Transformer
A transformer is a device used to increase or decrease AC (alternate current) voltage.
Transformers are widely used because they change voltage with relatively little loss of energy.
The transformer is based on two principles: first, that an electric current can produce a
magnetic field (electromagnetism), and, second that a changing magnetic field within a coil of wire
induces a voltage across the ends of the coil (electromagnetic induction). Changing the current in
the primary coil changes the magnetic flux that is developed. The changing magnetic flux induces
a voltage in the secondary coil.
Ideally, the transformer is perfectly efficient; all the incoming energy is
transformed from the primary circuit to the magnetic field and into the secondary circuit. If this
condition is met, the incoming electric power must equal the outgoing power:
giving the ideal transformer equation
Transformers normally have high efficiency, so this formula is a reasonable approximation.
Example 1) A computer adapter contains a transformer that converts 120 V ac across its primary
windings to 24 V ac across its secondary windings. The primary current is 2.4 A. What is the
secondary current and what is the type of transformer?
Example 2) An ideal transformer has potential difference of 130 V ac across the primary windings
and a potential difference of 520 V ac across the secondary windings. There are 390 turns in
the secondary. The secondary current is
a)
b)
c)
d)
one-sixth the primary current.
twice the primary current.
one half the primary current.
One quarter the primary current.
Example 3) A transformer connected to a 120 V ac supply has 1440 primary and 600 secondary
windings. It delivers a secondary current of 8.0 A. Find the primary current and secondary voltage.
a)
b)
c)
d)
Current: 3.33 A; Voltage: 80 V
Current: 3.33 A; Voltage: 50 V
Current: 16 A; Voltage: 50 V
Current: 16 A; Voltage: 80 V
Example 4) Calculate the load current and load voltage in this transformer circuit:
Iload =
Vload =
Understanding concept.
1) Look at the following transformer. The source of electricity is on the left. Therefore, the primary
coil is on the left, and the secondary is on the right.
Complete the table. Would this be a step-up or step-down(circle one)?
Step-up or step-down?
Voltage
Primary
Secondary
(V)
Current (I)
Number of coils (n)
Power (P)
For the above transformer, what is the output voltage if it is only 75% efficient?
2) A transformer has a primary coil with 200 turns and a secondary coil with 2000
turns. The input voltage is 120 VAC, and runs at 1800 W. Complete the table.
Step-up or step-down?
Voltage
(V)
Current (I)
Number of coils (n)
Power (P)
Primary
Secondary