Electromagnetic Induction
Objective:
TSW understand and apply the concept of
magnetic flux in order to explain how
induced emfs are created and calculate
their value and polarity.
You will be responsible for the content contained in
chapter 20, sections 1&2 of the textbook. Take the
time to read these sections.
Chapter 20 Homework Problems:
4, 9, 15, 17, 19, 22, 23
Electromagnetic induction is the process
by which an emf (voltage) is produced in a
wire by a changing magnetic flux.
• Magnetic flux is the product of the
magnetic field and the area through which
the magnetic field passes.
• Electromagnetic induction is the principle
behind the electric generator.
• The direction of the induced current due to
the induced emf is governed by Lenz’s
Law.
Here is a visual model of what we
did in chapter 19:
Loop of wire
Input
Current
and an
External
magnetic field
Electric Motor
Output
Force
(wire moves)
Here is a visual model of what we
will do in chapter 20:
Input
Force
(move wire)
Loop of wire
and an
Output
External
Current
magnetic field
Generator
The next two slides contain
vocabulary and equations, you
should commit them to
memory
Important Terms
alternating current - electric current that rapidly reverses its direction
electric generator - a device that uses electromagnetic induction to
convert mechanical energy into electrical energy
electromagnetic induction - inducing a voltage in a conductor by
changing the magnetic field around the conductor
induced current - the current produced by electromagnetic induction
induced emf - the voltage produced by electromagnetic induction
Faraday’s law of induction - law which states that a voltage can be
induced in a conductor by changing the magnetic field around the
conductor
Lenz’s law - the induced emf or current in a wire produces a magnetic
flux which opposes the change in flux that produced it by
electromagnetic induction
magnetic flux - the product of the magnetic field and the area through
which the magnetic field lines pass.
motional emf - emf or voltage induced in a wire due to relative motion
between the wire and a magnetic field
Equations, Symbols, and Units
  BLv

I
R
  BA cos 

  N
t
P  IV
where
ε = emf (voltage) induced by
electromagnetic induction (V)
v = relative speed between a conductor
and a magnetic field (m/s)
B = magnetic field (T)
L = length of a conductor in a magnetic
field (m)
I = current (A)
R = resistance (Ω)
Φ = magnetic flux (Tm2 = Weber=Wb)
A = area through which the flux is
passing (m2)
= angle between the direction of the
magnetic field and the area through
which it passes
Magnetic Flux
Consider a rectangular loop of wire of height L and width x
which sits in a region of magnetic field of strength B. The
magnetic field is directed into the page, as shown below:
w
L
The magnetic flux is given by the following equation:
  BA cos 
Φ = The magnetic flux (Tm2 = Wb)
B = The magnetic field (T)
A = area of loop (m2)
Φ = angle between the field and the normal
to the area
Faraday’s law states that an induced emf is
produced by changing the flux, but how
could the flux be changed?
• Turn the field off or on.
• Move the loop of wire out of the field
• Rotate the loop to change the angle
between the field and the area of the loop.
Here is Faraday’s Law in equation
form:

N
t
Where
Є = The induced emf (voltage) (V)
ΔΦ = The change in flux (Wb)
Δt = The change in time (s)
N = number of loops.
*Note that an emf is only produced if the flux changes. The quicker the
flux changes the larger the induced emf.
The induced emf in the wire will produce a current in the wire. The
magnitude of the induced current is found using Ohm’s Law:
V  IR
  IR

I
R
The direction of the induced current is found using Lenz’s Law (conservation of
energy).
Lenz’s Law – The induced current will flow in a
direction such that the magnetic field produced by it
opposes the original change in flux. In simple terms
the wire resists the change in flux and wants to go back
to the way things were.
It is helpful to use RHR#2 when using Lenz’s Law.
Example 1: A circular loop of wire with a resistance of 0.5Ω and
radius 30cm is placed in an external magnetic field of 0.2T. The
magnetic field is turned off in .02 seconds.
a) Calculate the original flux of the loop.
b) Calculate the induced emf.
c) Calculate the current induced in the wire.
d) What direction does the induced current have?
e) What other way could the same emf be induced without turning the
field off?
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Example 1: A circular loop of wire with a resistance of 0.5Ω and
radius 30cm is placed in an external magnetic field of 0.2T. The
magnetic field is turned off in .02 seconds.
a) Calculate the original flux of the loop.
The flux (Φ) is the magnetic field times the cross sectional area
  BA cos 
  (0.2T)( )(.30m) 2 cos0
  .057Wb
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Example 1: A circular loop of wire with a resistance of 0.5Ω and
radius 30cm is placed in an external magnetic field of 0.2T. The
magnetic field is turned off in .02 seconds.
b) Calculate the induced emf.

N
t
.057Wb
  (1)
.02s
  2.85V
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Example 1: A circular loop of wire with a resistance of 0.5Ω and
radius 30cm is placed in an external magnetic field of 0.2T. The
magnetic field is turned off in .02 seconds.
c) Calculate the current induced in the wire.
  IR

I
R
2.85V
I
0.5
I  5.7A
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Example 1: A circular loop of wire with a resistance of 0.5Ω and
radius 30cm is placed in an external magnetic field of 0.2T. The
magnetic field is turned off in .02 seconds.
d) What direction does the induced current have?
Use lenz’s law – The induced current will
flow in a direction such that the magnetic
field produced by it opposes the change in
flux.
Since the magnetic field is turned off it
will be decreasing out of the page, so the
induced current will produce a field out of
the page in order to restore it.
Use the right hand rule – fingers out of
the page inside the loop and grab the
wire. Your thumb gives the direction of
the current.
The current is counter clockwise
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e) Rotate the loop
Let’s do some examples
predicting the induced current
direction using Lenz’s Law
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•decreasing
• flux •
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Bout
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CCW
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Bin decreasing flux
CW
Bout increasing flux
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Bin increasing flux
CCW
If you are really struggling applying Lenz’s
Law, then memorize the following table:
decreasing
Increasing
B out
CCW
CW
B in
CW
CCW
Example 2: A circuit with a total resistance of R is made using a set
of metal wires and a copper bar. The magnetic field (B) is directed
into the page as shown in the diagram. The bar of length L starts
on the left and is pulled to the right at a constant velocity.
a) Calculate the induced emf in the circuit.
b) Calculate the current induced in the wire.
c) What direction does the induced current have?
d) What is the magnitude and direction of the magnetic force that
opposes the motion of the bar?
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v
a) Calculate the induced emf in the circuit.
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Δx
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
N
t
(B)(x)(L)
  (1)
t


  BLv

v
x
v
t
b) Calculate the current induced in the wire.
V  IR
  IR

BLv
I 
R
R
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
v
c) What direction does the induced current have?
Use Lenz’s Law – The flux is increasing into the page, so the induced current
will produce a field that opposes this. So the induced magnetic field is out of
the page. Using RHR put your fingers out of the page, inside the loop and
grab the wire. Your thumb gives the direction of the induced current, which is
CCW (Shown in green)
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v
d) What is the magnitude and direction of the magnetic force that
opposes the motion of the bar?
FB  BIL
BLv 
FB  B
L
 R 
To find the direction of the magnetic force use the
RHR on the copper bar. Fingers in the direction of
the field, thumb in the direction of the current and
the palm of your hand gives the direction of the
magnetic force, which is to the left
B 2 L2v
FB 
R
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FB X
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v
The last example led to the equation for the motional emf.
The motional emf is the voltage induced in a wire as it
moves in an external magnetic field. The induced emf will
produce a current in the wire, which will in turn result in a
force that opposes the motion of the wire. You don’t get
something for nothing.
Motional emf
Where
Є = The induced emf (V)
  BLv
B = The external magnetic field (T)
L = The length of the wire (m)
v = The velocity of the wire (m/s)
Example 3: A conducting rod of length 0.30 m and resistance 10.0 Ω
moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is
directed out of the page.
v
L
B (out of the
page)
a) Find the emf induced in the rod.
b) Find the current in the rod and the direction it flows.
c) Find the power dissipated in the rod.
d) Find the magnetic force opposing the motion of the rod.
Example 3: A conducting rod of length 0.30 m and resistance 10.0 Ω
moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is
directed out of the page.
v
L
B (out of the
page)
a) Find the emf induced in the
rod.
  BLv
  (0.20T)(0.30m)(2.0m /s)
  .12v

Example 3: A conducting rod of length 0.30 m and resistance 10.0 Ω
moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is
directed out of the page.
v
b) Find the current in the rod and the
direction it flows.
  IR

I
R
.12V
I
10
I  .012A
L
B (out of the
page)
The flux is increasing outward so the
induced current will produce a
magnetic field inward. The current will
flow CW
Example 3: A conducting rod of length 0.30 m and resistance 10.0 Ω
moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is
directed out of the page.
v
L
c) Find the power dissipated in the rod.
P  IV
P  (.012A)(.12V )
P  1.44  10 3 W
B (out of the
page)
Example 3: A conducting rod of length 0.30 m and resistance 10.0 Ω
moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is
directed out of the page.
v
d) Find the magnetic force opposing the
motion of the rod.
FB  BIL
FB  (0.20T)(.012A)(0.30m)
FB  7.2  10 4 N
Opposing the direction of the external force.
L
B (out of the
page)
Example 4: A square loop of sides a = 0.4 m, mass m =
1.5 kg, and resistance 5.0 Ω falls from rest from a
height h = 1.0 m toward a uniform magnetic field B
which is directed into the page as shown.
(a) Determine the speed of the loop just before it enters
the magnetic field.
As the loop enters the magnetic field, an emf ε and a
current I is induced in the loop.
(b) Is the direction of the induced current in the loop
clockwise or counterclockwise? Briefly explain how
you arrived at your answer.
When the loop enters the magnetic field, it falls through
with a constant velocity.
(c) Calculate the magnetic force necessary to keep the
loop falling at a constant velocity.
(d) What is the magnitude of the magnetic field B
necessary to keep the loop falling at a
constant velocity?
(e) Calculate the induced emf in the loop as it enters and
exits the magnetic field.
a
a
h
B
Example 4: A square loop of sides a = 0.4 m, mass m = 1.5
kg, and resistance 5.0 Ω falls from rest from a height h =
1.0 m toward a uniform magnetic field B which is directed
into the page as shown.
(a) Determine the speed of the loop just before it enters the
magnetic field.
a
a
Ug  K
h
1 2
mgh  mv
2
v  2gh
v  (2)(10m /s2 )(1.0m)
m
v  4.5
s
B
Example 4: A square loop of sides a = 0.4 m, mass m =
1.5 kg, and resistance 5.0 Ω falls from rest from a
height h = 1.0 m toward a uniform magnetic field B
which is directed into the page as shown.
As the loop enters the magnetic field, an emf ε and a
current I is induced in the loop.
(b) Is the direction of the induced current in the loop
clockwise or counterclockwise? Briefly explain how
you arrived at your answer.
The flux is increasing into the page, so the
induced current will produce a magnetic field
out of the page. So the current will be CCW
a
a
h
B
Example 4: A square loop of sides a = 0.4 m, mass m =
1.5 kg, and resistance 5.0 Ω falls from rest from a
height h = 1.0 m toward a uniform magnetic field B
which is directed into the page as shown.
When the loop enters the magnetic field, it falls through
with a constant velocity.
(c) Calculate the magnetic force necessary to keep the
loop falling at a constant velocity.
a
a
h
FB  Fg
FB  mg
m
FB  (1.5kg)(10 2 )
s
FB  15N
B
Example 4: A square loop of sides a = 0.4 m, mass m =
1.5 kg, and resistance 5.0 Ω falls from rest from a
height h = 1.0 m toward a uniform magnetic field B
which is directed into the page as shown.
(d) What is the magnitude of the magnetic field B
necessary to keep the loop falling at a
constant velocity?
FB  BIL
15N  BIL
BLv 
15N  B
L
 R 
B 2 L2v
15N 
R
(15N)R
B

2
Lv
  BLv
IR  BLv
BLv
I
R

(15N)(5.0)
 10.2T
m
2
(0.4m) (4.5 )
s
a
a
h
B

Example 4: A square loop of sides a = 0.4 m, mass m =
1.5 kg, and resistance 5.0 Ω falls from rest from a
height h = 1.0 m toward a uniform magnetic field B
which is directed into the page as shown.
(e) Calculate the induced emf in the loop as it enters and
exits the magnetic field.
  BLv
m
  (10.2V )(0.4m)(4.5 )
s
  18.4V
a
a
h
B