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Physics 2135 and the
Shakespearian Tragedy
Faraday’s Law,
Maxwell’s Equations
Electromagnetic Waves
Magnetic
Fields
Circuits
Electric Energy
Electric Fields
Coulomb’s Law
Geometrical
Optics
Wave
Optics
Final Exam
Announcements
 Physics 2135 spreadsheets for all sections, with Exam 2 scores,
will be available today on the Physics 2135 web site. You need
your PIN to find your grade.
 Preliminary exam average is about 71.1% (10 sections out of 13
reporting). OK! Scores ranged from a low of 41 to a high of 200
(1? student). I will fill in the ??’s during the “live” lecture and in its “.ppt” file.
Physics 2135 Exam 2 will be returned in recitation Thursday. When
you get the exam back, please check that points were added
correctly. Review the course handbook and be sure to follow proper
procedures before requesting a regrade. Get your regrade requests
in on time! (They are due by “next” Thursday’s recitation.)
On a separate sheet of paper, briefly explain the reason for your regrade request. This should be based on the work actually shown
on paper, not what was in your head. Attach to the exam and turn it in by the end of your next Thursday’s recitation.
“Next” = first Thursday after Spring Break.
Today’s agenda:
Induced emf.
You must understand how changing magnetic flux can induce an emf, and be able to
determine the direction of the induced emf.
Faraday’s Law.
You must be able to use Faraday’s Law to calculate the emf induced in a circuit.
Lenz’s Law.
You must be able to use Lenz’s Law to determine the direction induced current, and
therefore induced emf.
Generators.
You must understand how generators work, and use Faraday’s Law to calculate numerical
values of parameters associated with generators.
Back emf.
You must be able to use Lenz’s law to explain back emf.
Induced emf and Faraday’s Law
Magnetic Induction
We have found that an electric current can give rise to a
magnetic field…
I wonder if a magnetic field can somehow give rise to an
electric current…
It is observed experimentally that changes in magnetic flux
induce an emf in a conductor.
B
An electric current is induced if there is a closed circuit (e.g.,
loop of wire) in the changing magnetic flux.
I
B
A constant magnetic flux does not induce an emf—it takes a
changing magnetic flux.
Note that “change” may or may not not require observable (to
you) motion.
 A magnet may move through a loop of wire
wire, or a
loop of wire may be moved through a magnetic field.
These involve observable motion.
N
I
S
I B
v
region of
move magnet toward coil magnetic field
change area of loop
inside magnetic field
this part of the loop is
closest to your eyes
N
S
rotate coil in
magnetic field
changing I
induced I
changing B
 A changing current in a loop of wire gives rise to a
changing magnetic field (predicted by Ampere’s
law) which can induce a current in another nearby
loop of wire.
In the this case, nothing observable (to your eye) is moving,
although, of course microscopically, electrons are in motion.
Induced emf is produced by a changing magnetic flux.
Today’s agenda:
Induced emf.
You must understand how changing magnetic flux can induce an emf, and be able to
determine the direction of the induced emf.
Faraday’s Law.
You must be able to use Faraday’s Law to calculate the emf induced in a circuit.
Lenz’s Law.
You must be able to use Lenz’s Law to determine the direction induced current, and
therefore induced emf.
Generators.
You must understand how generators work, and use Faraday’s Law to calculate numerical
values of parameters associated with generators.
Back emf.
You must be able to use Lenz’s law to explain back emf.
We can quantify the induced emf described qualitatively in the
previous section of this lecture by using magnetic flux.
Experimentally, if the flux through N loops of wire changes by
dB in a time dt, the induced emf is
dB
ε = -N
.
dt
*Faraday’s Law of
Magnetic Induction
ε average = - N
B
.
t
Faraday’s law of induction is one of the fundamental laws of
electricity and magnetism.
I wonder why the – sign…
Your text, page 959 of the 14th edition, shows how to determine the direction of the induced emf. Argh!
Lenz’s Law, coming soon, is much easier.
*Well, one expression of Faraday’s Law
In the equation
ε = -N
dB
,
dt
Faraday’s Law of
Magnetic Induction
 B   B  dA is the magnetic flux.
This is another expression of Faraday’s Law:
dB
 E  ds = - dt
We’ll use this version
in the next lecture.
The fine print, put here for me to ponder, and not for students to worry about. A magnetic force does no work on a moving (or stationary) charged particle. Therefore the magnetic force
cannot change a charged particle’s potential energy or electric potential. But electric fields can do work. This equation shows that a changing magnetic flux induces an electric field,
which can change a charged particle’s potential energy. This induced electric field is responsible for induced emf. During this lecture, we are mostly going to examine how a changing
magnetic flux induces emf, without concerning ourselves with the “middleman” induced electric field.
Example: move a magnet towards a coil of wire.
N=5 turns
A=0.002 m2
dB
= 0.4 T/s
dt
I
d B  dA
dB
ε = -N
= -N
dt
dt
d BA 
ε = -N
dt
N
+
S
v
-
(what assumptions did I make here?)
dB
ε = -NA
dt
T
ε = - 5  0.002 m2   0.4  = -0.004 V

s
Ways to induce an emf:
 change B
Possible homework hint:  B   d B   B  dA   B(t) dA if B varies but loop  B.
 change the area of the loop in the field
Possible homework hint: for a circular loop, C=2R, so A=r2=(C/2)2=C2/4, so you can express d(BA)/dt in terms of dC/dt.
Ways to induce an emf (continued):
 changing current changes B through conducting loop
a
I
B
b
Possible Homework Hint.
The magnetic field is not uniform through the loop, so you
can’t use BA to calculate the flux. Take an infinitesimally thin
strip. Then the flux is d = BdAstrip. Integrate from a to b to
get the flux through the strip.
Ways to induce an emf (continued):
 change the orientation of the loop in the field
=90
=45
=0
Today’s agenda:
Induced emf.
You must understand how changing magnetic flux can induce an emf, and be able to
determine the direction of the induced emf.
Faraday’s Law.
You must be able to use Faraday’s Law to calculate the emf induced in a circuit.
Lenz’s Law.
You must be able to use Lenz’s Law to determine the direction induced current, and
therefore induced emf.
Generators.
You must understand how generators work, and use Faraday’s Law to calculate numerical
values of parameters associated with generators.
Back emf.
You must be able to use Lenz’s law to explain back emf.
Experimentally…
Lenz’s law—An induced emf always gives rise to a current
whose magnetic field opposes the change in flux.*
N
I
+
S
v
-
If Lenz’s law were not true—if there were a + sign in
Faraday’s law—then a changing magnetic field would produce
a current, which would further increase the magnetic field,
further increasing the current, making the magnetic field still
bigger…
*Think of the current resulting from the induced emf as “trying” to maintain the status quo—
to prevent change.
…violating conservation of energy and ripping apart the very
fabric of the universe…
Practice with Lenz’s Law.
In which direction is the current induced in the coil for each
situation shown? Practice on your own. In lecture, skip to here.
(counterclockwise)
(no current)
(counterclockwise)
(clockwise)
Rotating the coil about the vertical diameter
by pulling the left side toward the reader
and pushing the right side away from the
reader in a magnetic field that points from
right to left in the plane of the page.
(counterclockwise)
Faraday’s Law
ε = -N
dB
dt
You can use Faraday’s Law (as written above) to calculate the
magnitude of the emf (or whatever the problem wants). Then
use Lenz’s Law to figure out the direction of the induced current
(or the direction of whatever the problem wants).
The direction of the induced emf is in the direction of the
current that flows in response to the flux change. We usually
ask you to calculate the magnitude of the induced emf ( || )
and separately specify its direction.
Magnetic flux is not a vector. Like electrical current, it is a scalar. Just as we talk about current direction
(even though it is not a vector), we often talk about flux direction (even though it is not a vector). Keep
this in mind if your recitation instructor talks about the direction of magnetic flux.
Today’s agenda:
Induced emf.
You must understand how changing magnetic flux can induce an emf, and be able to
determine the direction of the induced emf.
Faraday’s Law.
You must be able to use Faraday’s Law to calculate the emf induced in a circuit.
Lenz’s Law.
You must be able to use Lenz’s Law to determine the direction induced current, and
therefore induced emf.
Generators.
You must understand how generators work, and use Faraday’s Law to calculate numerical
values of parameters associated with generators.
Skip to slide 27 (the skipped slides
don’t add anything new).
Back emf.
You must be able to use Lenz’s law to explain back emf.
Induced emf: an overview
An emf is induced in a conductor moving in a magnetic field.
Your text introduces four ways of producing induced emf.
1. Flux change through a conducting loop produces an emf:
rotating loop.
A

B
dB
=dt
start with this
ε = NBA  sin  t 
side view
I=
NBA
sin  t 
R
P = INBA sin  t 
derive these
2. Flux change through a conducting loop produces an emf:
expanding loop.
B 









v







 ℓ


dA
x=vdt
=-
dB
dt
start with these
FM = I  B
ε =B v
I =
ε B v
=
R
R
P = FP  v = I Bv
derive these
3. Conductor moving in a magnetic field experiences an emf:
magnetic force on charged particles.
B
–




+





v





start with these











F = q E+ v  B
ℓ
 =E

(Mr. Ed)
derive this
ε =B v
This is called “motional” emf, because the emf results from the motion of the conductor in the magnetic field. The other
three examples in this lecture are “induced” emf because they involve the change of magnetic flux through a current loop,
as described by Faraday’s Law.
4. Flux change through a conducting loop produces an emf:
moving loop.
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start with this
dB
=dt
derive these
ε =B v
B v
I =
P = I Bv
R
Let’s look in detail at each of these four ways
of using flux change or motion to produce an
emf.
Method 1…
Generators and Motors: a basic introduction
Take a loop of wire in a magnetic field
and rotate it with an angular speed .
A


side view
B
S
B =B  A = BA cos   
Choose 0=0. Then
 = 0  t = t .
B = BA cos  t 
dB
ε = dt
Generators are an application of induced emf.
N
A

side view
B
If there are N loops in the coil
dB
ε = -N
dt
ε = -N
d BA cos  t  
dt
ε = NBA  sin  t 
The NBA equation!
|| is maximum when  = t = 90° or 270°; i.e., when B is
zero. The rate at which the magnetic flux is changing is then
maximum. On the other hand,  is zero when the magnetic flux
is maximum.
emf, current and power from a generator
ε = NBA  sin  t 
ε
NBA
I= =
sin  t 
R
R
P = εI = INBA sin  t 
None of these are on your starting equation sheet!
Example: the armature of a 60 Hz ac generator rotates in a
0.15 T magnetic field. If the area of the coil is 2x10-2 m2, how
many loops must the coil contain if the peak output is to be
max = 170 V?
ε = N B A ω sin  ωt 
εmax = N B A ω
I will not work this in
lecture. Please study it on
your own! Know where
the 2 comes from.
N =
N =
“Legal” for me,
because I just
derived it (but
not for you)!
εmax
BAω
170 V 
 0.15 T   2×10-2 m2   2 ×60 s -1 
N = 150 (turns)
Let’s look in detail at each of these four ways
of using flux change or motion to produce an
emf.
Method 2…
Another Kind of Generator: A Slidewire Generator
Recall that one of the ways to induce an emf is to change the
area of the loop in the magnetic field. Let’s see how this
works.
v
B
A U-shaped conductor and a

moveable conducting rod are placed

in a magnetic field, as shown.
 ℓ

The rod moves to the right with a

dA
constant speed v for a time dt.
vdt
x
The area through which flux passes
increases by dA.
The loop is perpendicular to the magnetic field, so the
magnetic flux through the loop is B =  B  dA = BA. Calculate
the induced emf using Faraday’s law:
dB
ε = -N
dt
d BA 
ε = 1
dt
ε =
B dA
dt
ε = B
dx
dt
ε = B v.
B 




x





v







 ℓ


dA
vdt
B and v are vector magnitudes, so
they are always +. Wire length is
always +. You use Lenz’s law to
get the direction of the current.
Direction of current?
The induced emf causes current to flow in the loop.
Magnetic flux inside the loop
increases (more area).
System “wants” to make the flux
stay the same, so the current gives
rise to a field inside the loop
directed into the plane of the paper
(to counteract the “extra” flux).
B 




I
x





v





vdt
Clockwise current!
What would happen if the bar were moved to the left?


 ℓ


dA
As the bar moves through the magnetic field, it “feels” a force
FM = I  B
A constant pulling force, equal in
magnitude and opposite in direction,
must be applied to keep the bar
moving with a constant velocity.
FP = FM = I B
v
B 




I
x






F
M




FP
 ℓ


Power and current.
If the loop has resistance R, the current is
ε B v
I = =
.
R
R
And the power is
P = FP  v = I Bv
P = I  IR  = I2R
(as expected).
B 




I





v










x
Mechanical energy (from the pulling force) has been
converted into electrical energy, and the electrical energy is
then dissipated by the resistance of the wire.
ℓ
You might find it useful to look at Dr. Waddill’s lecture on
Faraday’s Law, from several semesters back. Click here to view
the lecture.
If the above link doesn’t work, try copying and pasting this
into your browser address bar:
http://campus.mst.edu/physics/courses/24/Handouts/Lec_18.ppt
Let’s look in detail at each of these four ways
of using flux change or motion to produce an
emf.
Method 3…
Example 3 of motional emf: moving conductor in B field.
Motional emf is the emf induced in a conductor moving in a
magnetic field.
If a conductor (purple bar) moves
with speed v in a magnetic field,
the electrons in the bar experience
a force
FM = qv  B = - ev  B
B 




v
–



+










“up”
ℓ
The force on the electrons is “up,” so the “top” end of the
bar acquires a net – charge and the “bottom” end of the bar
acquires a net + charge. The charges in the bar are
“separated.”
This is a simplified version of reality, but it gives you the right “feel.”
The separated charges in the bar produce an electric field
pointing “up” the bar. The emf across the length of the bar is
ε= E
The electric field exerts a
“downward” force on the electrons:
FE = qE = - eE
An equilibrium condition is reached,
where the magnetic and electric
forces are equal in magnitude and
opposite in direction.
ε
evB = eE = e
 ε =B v
“up”
B 




v
–


E

+










ℓ
Homework Hint!
When the moving conductor is
“tilted” relative to its direction of
motion (when v is not perpendicular to the conductor), you must
use the “effective length” of the
conductor.*
V  E   E
 E
B   v

  

+


–
 ℓ
effective


cos
effective
ℓeffective = ℓ sin  if you use the if you define  as the angle relative to the horizontal
Caution: you do not have to use this technique for homework
problems 29.25 or 29.28 (if assigned this semester). Be sure to understand
why!
*This is because the magnetic force is , and not directed along the conductor. Let’s not worry about showing this.
Let’s look in detail at each of these four ways
of using flux change or motion to produce an
emf.
Method 4…
Example 4 of induced emf: flux change through conducting
loop. (Entire loop is moving.)
I’ll include some numbers with this example.
Remember, it’s the flux change that induces the emf. Flux has no direction associated with
it. However, the presence of flux is due to the presence of a magnetic field, which does have
a direction, and allows us to use Lenz’s law to determine the “direction” of current and emf.
A square coil of side 5 cm contains 100 loops and is positioned
perpendicular to a uniform 0.6 T magnetic field. It is quickly
and uniformly pulled from the field (moving  to B) to a region
where the field drops abruptly to zero. It takes 0.10 s to
remove the coil, whose resistance is 100 .
5 cm
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B = 0.6 T
(a) Find the change in flux through the coil.
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Initial: Bi =
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 B  dA= BA .
Final: Bf = 0 .
B = Bf - Bi = 0 - BA = - (0.6 T) (0.05 m)2 = - 1.5x10-3 Wb.
(b) Find the current and emf induced.
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Current will begin to flow when the coil starts to exit the magnetic
field. Because of the resistance of the coil, the current will eventually
stop flowing after the coil has left the magnetic field.

final
initial
The current must flow clockwise to induce an “inward”
magnetic field (which replaces the “removed” magnetic field).
The induced emf is
dB
d(BA)
dA
ε = -N
= -N
= -NB
dt
dt
dt
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x
d x 
dA
=
=
dt
dt
dA
x
v
dx
=
dt
v
ε = - NB v
“uniformly” pulled
x
5 cm
m
v =
=
= 0.5
t
0.1 s
s
m

ε = 100   0.6 T  0.05 m  0.5 
s

ε = 1.5 V
The induced current is
I =
ε
R
=
1.5 V
100 Ω
= 15 mA .
(c) How much energy is dissipated in the coil?
Current flows “only*” during the time flux changes.
E = P t = I2R t = (1.5x10-2 A)2 (100 ) (0.1 s) = 2.25x10-3 J
(d) Discuss the forces involved in this example.
The loop has to be “pulled” out of the magnetic field, so there
is a pulling force, which does work.
The “pulling” force is opposed by a magnetic force on the
current flowing in the wire. If the loop is pulled “uniformly”
out of the magnetic field (no acceleration) the pulling and
magnetic forces are equal in magnitude and opposite in
direction.
*Remember: if there were no resistance in the loop, the current would flow indefinitely. However, the
resistance quickly halts the flow of current once the magnetic flux stops changing.
The flux change occurs only when the coil is in the process of
leaving the region of magnetic field.
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No
No
No
No
flux change.
emf.
current.
work (why?).
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Fapplied
D
Flux changes. emf induced. Current flows. Work
(equal to FD) is done.
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No flux change. No emf. No current. (No work.)
(e) Calculate the force necessary to pull the coil from the field.
Remember, a force is needed only when the coil is partly in
the field region.
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I
Fmag = N IL  B
Multiply by N because there
are N loops in the coil.
where L is a vector in the direction of I having a magnitude
equal to the length of the wire inside the field region.
         
         
F3
L
          3
L2
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F2
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L
 1
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about the “busy” slide!
I L4Sorry
=0
The forces should be shown
acting at the centers
Fpullof the
coil sides in the field.
F1
There must be a pulling force to the right to overcome the net
magnetic force to the left.
Magnitudes only (direction shown in diagram):
Fmag = NILB = 100  1.5 10-2 5 10-2   0.6  = 4.5 10-2 N = Fpull
         
         
F3
L
          3
L2
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F2
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L
 1
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I
Fpull
F1
This calculation assumes the coil is pulled out “uniformly;” i.e.,
no acceleration, so Fpull = Fmag.
Work done by pulling force:

 
Wpull =Fpull  D = NILB ˆi  L ˆi
Wpull = 100  1.5 10-2 5 10-2   0.6  5 10-2  = 2.25 10-3 J
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x
I
Fpull
D
The work done by the pulling force is equal to the electrical
energy provided to (and dissipated in) the coil.
Concluding Remarks
I will post a PowerPoint illustrating the operation of motors,
discussing back emf, and giving two more examples of
motional emf calculations. You can find the PowerPoint in the
“resources” section for this lecture on the course web site.
These slides (posted under “resources”) are intended to
supplement your reading. They are optional. I will not post a
narration of their content.
Now that you are an expert at designing generators, I thought
it wise to direct you towards some handy gadgets you can use
if you need to debug your non-working generator.
Handy Hint: Debugging Your Homemade Generator
If you build a generator and it
doesn’t seem to be working…
…or if you want to test a wall
socket to see if it is “live…”
…simply purchase a Vilcus Plug
Dactyloadapter from ThinkGeek
( http://www.thinkgeek.com/stuff/41/lebedev.shtml )
and follow the instructions.
European model shown.
US model also available.