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Chapter 18 Lecture
Chapter 18 Lecture
Electromagnetic
Induction
Electromagnetic induction
 It is possible to have a current in a closed
loop of wire without using a battery.
 This phenomenon of inducing a current
using a changing magnetic field is called
electromagnetic induction.
Inducing an electric current
 At the top and bottom sections of the loop, a
component of the B field is perpendicular to the
velocity of the loop as it moves toward the right.
 The magnetic field exerts a force on each
electron in the wire, which causes the electrons
to accelerate clockwise around the loop.
Discovery of electromagnetic
induction
 In the early 1800s, induced current was
too weak to observe with existing
galvanometers; coils of wire were needed
to amplify the effect but insulation did not
exist.
 American physicist Joseph Henry devised
a method for insulating wires by wrapping
them in silk; he was the first person to
observe a current being induced in a coil.
 In 1831, Faraday used Henry's insulation
method to induce current in coils in his
laboratory.
Inducing an electric current
 When the magnet or coil is moved or rotated with
respect to the magnetic field, or when the area of the
coil is changed, the number of magnetic field lines
going through the area of the coil increases or
decreases.
 Alternatively, when the number of magnetic field lines
through the coil's area changes, a corresponding emf
is produced in the coil, which leads to the induced
current.
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Magnetic Flux
Magnetic flux
• Magnetic flux is a
physical quantity that
measures the number
of magnetic field lines
through a coil's area.
The amount of magnetic
field that is passing
through a given surface.
Φ
Magnetic flux depends
on several factors.
Let’s take a look!
Magnetic flux is the amount of magnetic field
that is passing through a given surface.
Magnetic flux is the amount of magnetic field
that is passing through a given surface.
Therefore, the stronger the field, the greater
the magnetic flux through the loop!
Therefore, the larger the area, the greater
the magnetic flux through the loop!
Small flux
Larger flux
Lastly, the more perpendicular the loop is to the
field, the greater the magnetic flux through the loop!
The more the magnetic field goes through the loop,
the greater the flux will be.
Max flux
Less flux
Zero flux
Small flux
Larger flux
How can we measure the relative direction of a field and a surface?
Answer: With the area vector!
A
A
The area vector points perpendicularly away
from the surface, and is used to compare the
orientation of the surface to the direction of an
external magnetic field.
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Magnetic Flux
B
B
A
B
A
A
θ
F = BAcosq
B: Strength of magnetic field
A: Area of loop
If the angle between the
area vector and the B field
is 0°, then there is a
maximum magnetic flux
through the loop.
θ: Angle between B and A
If the angle between the
area vector and the B field
is 90°, then there is zero
magnetic flux through the
loop.
Rank the loops, in order of increasing Magnetic flux
a)
b)
c)
d)
e)
f)
g)
h)
Units: Tm2
Derived unit: Weber (Wb)
g=e
f
b=d
a=h
c
Zero
BA/4
BA/2
BA
2BA
Michael Faraday
Lenz’s Law and Faraday’s Law
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Faraday’s Discovery
Any induced current in a loop of wire will also have
an induced magnetic field associated with it.
A changing
magnetic flux
through a loop of
wire causes an
induced current in
the wire.
In order to determine the direction of the induced B
field in the center of the loop, we can use RHR #3!
Now On To Lenz’s Law!
Right-Hand Rule #3
If you wrap your fingers
around like the current in a
loop of wire, your thumb
will show the direction of
the magnetic field in the
center of the loop caused
by that current!
The induced current will flow so that
its induced magnetic field opposes
the change in flux through the loop.
Iinduced
I
B
Binduced
Lenz’s Law
The induced B field in a loop of wire will oppose the
change in magnetic flux through the loop.
If you try to increase the flux
through a loop, the induced
field will oppose that increase!
If you try to decrease the flux
through a loop, the induced
field will replace that decrease!
Flux and Change in Flux
Magnetic flux is a vector that always points in
the same direction as the magnetic field.
However, it is really ΔΦ that we are interested in!
If the flux is increasing, then ΔΦ points
in the same direction as Φ.
Binduced
Binduced
If the flux is decreasing, then ΔΦ points
in the opposite direction as Φ.
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Φ and ΔΦ Whiteboard
Φ and ΔΦ
State the direction of the flux and the change in flux through
the loop in each of the following cases.
Φ: Into page
ΔΦ: Into page
Φ: Out of page
ΔΦ: Out of page
Φ: Out of page
ΔΦ: Into page
Φ: Out of page
ΔΦ: Zero
Lenz’s Law Challenge Lvl. I
A loop of conducting wire is placed in the region shown
below, in which the magnetic field is increasing in strength.
Which way will the induced current flow in the loop?
Φ: Out of page
ΔΦ: Out of page
Binduced: Into page
Iinduced: Clockwise
Lenz’s Law Challenge Lvl. II
A loop of conducting wire is placed in the region shown below, and
the loop is rotated 90° along the dotted line shown. Which way
will the induced current flow in the loop while it is being rotated?
Φ: Out of page
ΔΦ: Into page
Binduced: Out of page
Iinduced: Counterclockwise
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Lenz’s Law Challenge Lvl. III
The two wire loops shown below are moved at constant velocity
near a current-carrying wire. Which direction, if any, will the
induced current flow in each of the wire loops shown below?
Loop 2
Loop 1
Loop 2
v
v
Loop 1
v
v
I
Loop 1
Φ: Out of page
I
Coils with Multiple Loops!
Each coil acts as its own loop.
If there are N coils,
F = N × BAcosq
Loop 2
Φ: Out of page
ΔΦ: Zero
ΔΦ: Out of page
Binduced: Zero
Binduced: Into page
Iinduced: None
Iinduced: Clockwise
Faraday’s Law
If there is an induced current in a loop
of wire, it must be caused by an
induced emf (voltage)!
The induced emf has the same
magnitude as the rate of change of the
magnetic flux.
Just multiply by N!
Faraday's law of electromagnetic
induction
The average magnitude of the induced eind in a
coil with N loops is the magnitude of the ratio
of the magnetic flux change through the loop
DF to the interval time Dt during which that
flux change occurred multiplied by the
number of loops
eind 
DF
Dt
eind  N
D[BA cos ]
Dt
WHITEBOARD - Faraday’s Law
The loop of wire shown below has a radius of 0.2 m, and is in a
magnetic field that is increasing at a rate of 0.5 T/s.
r
e=
Since the area of the
loop is constant,
Δ(BA) = AΔB
DB
DF D(BA)
=A
=
Dt
Dt
Dt
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I Have Proof!
D(BA) = (B f A f ) - (Bi Ai )
If area doesn’t change, but B does…
WHITEBOARD
The loop of wire shown below has a radius of 0.2 m,
and is in a magnetic field that is increasing at a rate
of 0.5 T/s. The wire contains a 3 Ω resistor.
D(BA) = (B f A)- (Bi A) = A(B f - Bi )
Therefore, when A is unchanged,
D(BA) = ADB
If area changed, but field remained constant,
then you would end up with Δ(BA) = BΔA
Solution
Since the area of the loop is constant,
Δ(BA) = AΔB
e=
DF
D(BA)
DB
=
=A
Dt
Dt
Dt
a) ε = 0.06 Volts
b) I = ε/R = 0.02 A clockwise
Since B and A are constant
DF D(BAcosq )
D(cosq )
e=
=
= BA
Dt
Dt
Dt
r
a) What is the induced emf?
b) What is the induced current, and which way does it flow?
WHITEBOARD
The loop of wire shown below has a radius of 0.2 m, and is in
a 0.8 T magnetic field. The loop is rotated by 90° in 2
seconds along the dotted line shown. The loop contains a 3 Ω
resistor.
r
a) What is the average induced emf?
b) What is the induced current, and which way does it flow?
Reverse Rail Gun!
A metal rod is pulled to the right at constant velocity, along
two conducting rails shown below. What is the direction of the
induced current through the circuit as a result?
cosq f - cosqi = cos(90) - cos(0) = -1
a) ε = 0.05 Volts
b) I = ε/R = 0.016 A counterclockwise
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Whiteboard: What will be εinduced?
Φ: Into page
ΔΦ: Into page
Binduced: Out of page
Iinduced: Counterclockwise
e=
DF D(BA)
D(A)
D(xL)
=
=B
=B
Dt
Dt
Dt
Dt
Motional EMF
e=
DF D(BA)
D(A)
D(xL)
=
=B
=B
Dt
Dt
Dt
Dt
Dx
= BL
!
Dt
= BLv
Using Motional EMF
When the amount of a loop’s area that in a magnetic field is
changing at a constant rate, Faraday’s Law gives the result
e = BLv
• L is the side that is not changing.
• v is the rate of change of the side
that is entering or leaving the field.
a) What is the current through the resistor?
I
I
I
I
In terms of the quantities shown above, write an expression for
a) The current through the resistor, and direction of current.
b) The power output of the resistor in the circuit.
c) The force required to pull the metal bar at constant velocity.
ε = BLv
I = ε/R
I = BLv/R
Φ: Into page
ΔΦ: Into page
Binduced: Out of page
Iinduced: Counterclockwise
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b) What is the power output of the resistor?
c) What is the force required to pull the bar at constant speed?
FB
I
Using RHR #2, you can determine that when there
is a current flowing through the circuit, the moving
metal bar will feel a magnetic force to the left.
P = I2R
P = (BLv)2/R
P = B2L2v2/R
Therefore, to pull the bar at constant velocity, you must
exactly balance out the magnetic force BIL.
c) What is the force required to pull the bar at constant speed?
Let’s analyze the results for a second…
FB
I
FB = BIL
FB = B(BLv/R)L
FB = B2L2v/R
In order to pull the
bar at constant
velocity, you must
exactly match this
force by pulling to
the right
The electric generator
Presistor = B2L2v2/R
F = B2L2v/R
P = Force x velocity!
The power output of the resistor will be exactly equal to the
power delivered to the system by pulling the rod.
The emf of a generator
 Consider a very simple
device consisting of a
loop of wire attached to
a turbine.
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Peak voltage of a generator
e in max = NBAw.
𝜔=2 ∙𝜋∙𝑓
WHITEBOARD
WHITEBOARD
 The label on the Schmidt E6 bicycle dynamo
headlight, indicates that the light has a power
output of 3 Wand a peak emf of 6 V. The
generator (also called a dynamo) for the lightbulb
has a cylindrical hub that rubs against the edge if
the bike tire, causing a coil inside the generator
to rotate. When the bicycle is traveling at a speed
of 5.4 m/s the tire makes 80 revolutions every
second. The B field in the vicinity of the coil is
uniform and has a magnitude of 0.10 T. The coil
is a rectangle with dimensions 1.0cm x 3.0cm.
Without taking the light apart, determine how
many turns there are in the generator coil.
Transformers
 A transformer is a device that increases or
decreases the maximum value of an
alternating emf.
Transformers
 An alternating emf
across the primary
coil is converted
into a larger or
smaller alternating
emf
across
the
secondary coil.
Transformers
 If the transformer is perfectly efficient, then
the rates of change of the magnetic flux
through one turn of each coil are the same.
Using the results from Faraday's law, we find:
 The emf in the secondary coil can be
substantially larger or smaller than the emf in
the primary coil depending on the number of
turns in each coil.
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Magnetic Core
Transformers
Transformers
 Transformers are used in many electronic
devices.
 They are also essential for transmitting
electric energy from a power plant to
customers' homes.
 Due to the electrical resistance of the
power lines, much electric energy is
converted into thermal energy.
 To reduce losses, the transmission of
electric energy is done at high-peak emf.
Transformers
 Some transformers are designed to increase emf,
such as those converting the 12-V potential
difference of the car battery to the 20,000-V
potential difference needed to produce a spark in
the engine's cylinder.
I
QUICK THINKING
PRACTICE
Whiteboard
II
In the scenarios shown above, two identical magnets are held near two
identical loops of conducting wire. In case I, the magnet is held a distance x
from the loop, and in case II, is held a distance 2x from the loop. Which of the
following is true?
(A) In case I, the wire will have twice the induced current as in case II.
(B) In case II, the wire will have twice the induced current as in case I.
(C) In case I, the wire will have a greater induced current, but not twice as
great.
(D) In case II, the wire will have a greater induced current, but not twice as
great.
(E) Neither wire will have any current induced in it.
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Whiteboard Warmup!
I
Motional EMF Whiteboard
A rectangular loop of wire is made to move at a constant velocity
into, through, and out of the magnetic field shown below.
II
v
v
(A)
(B)
(C)
(D)
(E)
In case I, the wire will have twice the induced current as in case II.
In case II, the wire will have twice the induced current as in case I.
In case I, the wire will have a greater induced current, but not twice as great.
In case II, the wire will have a greater induced current, but not twice as great.
Neither wire will have any current induced in it.
v
Determine the direction of the induced current in the loop
Current is caused by a change in magnetic flux!
a) as the cart is entering the field
b) as the cart travels through the field
c) as the cart is leaving the field
There is no change in flux in either case!
Round II!
What will be the direction of the net magnetic force
exerted the cart at each of the points shown?
No change in magnetic flux
v
v
v
v
v
v
I
I: CW
I: Zero
I: CCW
I
Hint: You will need to use RHR #2 for each segment
of current-carrying wire that is in the B-field
The net magnetic force will oppose the loop of wire entering
the field, and then oppose the loop of wire leaving the field!
There is zero net magnetic force while the cart is completely within the B field.
v
FB
v
v
FB
You will have to push the cart into the field, then let it glide
through the field, and finally you have to pull it out!
A square loop of wire of resistance R and side a is oriented
with its plane perpendicular to a magnetic field B, as shown
above. What must be the rate of change of the magnetic
field in order to produce a current I in the loop?
(A) IR/a2
(B) Ia2/R
(C) Ia/R
(D) Ra/I
(E) IRa
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I=
e
R
DF
D(BA)
DB
A
Dt
Dt
Dt
=
=
=
R
R
R
a2
I=
DB
Dt
R
DB IR
=
Dt a 2
Since it is a conductor, we know that the electrons are mobile.
Since the electrons are moving to the right in the B-field, we
can use RHR # 2 to determine that…
B
v
A neutral loop of conducting wire is moved through a uniform
magnetic field as shown below. What will happen as a result?
B
v
(A) A clockwise current will flow around the loop
(B) A counterclockwise current will flow around the loop
(C) The top of the loop will become positively charged and the
bottom of the loop will become negatively charged.
(D) The top of the loop will become negatively charged and the
bottom of the loop will become positively charged.
(E) None of the above.
The electrons within the conductor will feel an upward force!
They will flow to the top of the conductor,
giving it a net negative charge.
B
v
The electrons within the conductor will feel an upward force!
This will leave the bottom of the conductor
with a net positive charge.
FUN PPLICATIONS
Solenoid
Solenoids are useful!
Φ = N BA
*
They multiply the magnetic flux, and therefore the
induced emf, by the number of turns that the wire has
e=
DF D(NBA)
=
Dt
Dt
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Ring Launcher!
Ring Launcher!
Bcoil
I
I
This induces a current in the ring that
opposes the field of the coil
The current-carrying coil of wire acts like a magnet, with
the field lines coming out of North and into South.
Bring
N
Bcoil
Iinduced
I
I
The current-carrying ring also acts like a magnet, with the
field lines coming out of North and into South.
S
The net result looks like this!
Bcoil
Bring
Iinduced
S
N
S
N
Bring
Iinduced
I
N
S
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Whiteboard: Copper Tube Drop!
Strong repulsion!!!
S
N
S
N
N
S
Iinduced
I
Iind
b) Draw the “magnet” that each
of these sections acts like.
v
c) What will be the result when
the magnet is dropped down the
tube?
Φ: Downward
ΔΦ: Upward
Binduced: Downward
v
v
Iind
a) What will be the direction of
the induced current in each of
these sections of copper tube?
v
Φ: Downward
ΔΦ: Downward
Binduced: Upward
Attracted by the induced
magnet above
v
Repelled from the
induced magnet below
The magnet will fall slowly!!!
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