Today’s agendum:
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 (part 1).
You must understand how generators work, and use Faraday’s “Law” to calculate numerical
values of parameters associated with generators.
2
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…
3
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.
4
Passing the coil through the
magnet would induce an emf in
the coil.
They need to calibrate
their meter!
5
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
(as suggested in the previous slide). These involve
observable motion.
N
I
S
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
6
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 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.
7
Today’s agendum:
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 (part 1).
You must understand how generators work, and use Faraday’s “Law” to calculate numerical
values of parameters associated with generators.
8
We can quantify the induced emf described qualitatively in the
last few slides 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
Faraday’s “Law” of Induction is one of the fundamental laws of
electricity and magnetism.
I wonder why the – sign…
Your text, pages 997-998, shows how to determine the direction of the induced emf. Argh! Lenz’s Law,
coming soon, is much easier.
9
ε = -N
dB
dt
Faraday’s “Law” of
Magnetic Induction
where the magnetic flux is
 B   B  dA.
This is another version of Faraday’s “Law”:
 
d B We’ll use this version
 E  d s   dt in a later lecture.
 
In a future lecture, we’ll work with
 E  ds
Web page with pictures of a whole bunch of applications:
http://sol.sci.uop.edu/~jfalward/electromagneticinduction/electromagneticinduction.html
10
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 assumption did I make here?)
dB
ε = -NA
dt
T

ε = - 5  0.002 m   0.4  = -0.004 V
s

2
11
Ways to induce an emf:
 change B
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
12
Ways to induce an emf (continued):
 change the orientation of the loop in the field
=90
=45
=0
13
Example: a uniform (but time-varying) magnetic field passes through a
circular coil whose normal is parallel to the magnetic field. The coil’s area is
10-2 m2 and it has a resistance of 1 m. B varies with time as shown in the
graph. Plot the current in the coil as a function of time.
d BA 
dB
dB
ε===- A
dt
dt
dt
ε
A dB
ε = IR  I = = R
R dt
For 0 < t < 3:
.01 T
=1 s
.01 .01

dB B .01
A dB
=
=
 I=== - .0333 A
dt
t
3
R dt
.001 3
For 3 < t < 5:
dB
=0 I=0
dt
14
Example: a uniform (but time-varying) magnetic field passes through a
circular coil whose normal is parallel to the magnetic field. The coil’s area is
10-2 m2 and it has a resistance of 1 m. B varies with time as shown in the
graph. Plot the current in the coil as a function of time.
For 5 < t < 11:
dB B -.01
=
=
dt
t
6
.01 -.01
A dB
 I==R dt
.001 6
I(t)
.01 T
= + .0167 A
=1 s
-.0333 A
15
Today’s agendum:
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 (part 1).
You must understand how generators work, and use Faraday’s “Law” to calculate numerical
values of parameters associated with generators.
16
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.
17
…violating conservation of energy and ripping apart the very
fabric of the universe…
18
Practice with Lenz’s Law.
In which direction is the current induced in the coil for each
situation shown?
(counterclockwise)
(no current)
19
(counterclockwise)
(clockwise)
20
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)
Remember ε = - N
d B
?
dt
Now that you are experts on the application of Lenz’s “Law”,
remember this:
You can use Faraday’s “Law” 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
21
wants).
Today’s agendum:
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.
22
Motional emf: an overview
An emf is induced in a conductor moving in a magnetic field.
Your text introduces four ways of producing motional emf. We
will cover the first two in this lecture.
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
derive these
P = INBA sin t 
23
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
derive these
P = FP  v =I Bv
24
Next time we will look at two more examples of motional emf…
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
You could also solve this using Faraday’s” Law” by constructing a “virtual” circuit using “virtual” conductors.
25
4. Flux change through a conducting loop produces an emf:
moving loop.
         
         
















































































start with this
d B
=dt
derive these
ε=B v
B v
I =
P = I Bv
R
26
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
N
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 motional emf.
27
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.
28
emf, current and power from a generator
ε = NBA  sin t 
ε
NBA
I= =
sin  t 
R
R
P = εI = INBA sin  t 
29
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
N =
N =
εmax
BAω
170 V 
 0.15 T   2×10-2 m2   2 ×60 s -1 
N = 150 (turns)
30
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 rod moves a distance v dt and the area of the loop inside
the magnetic field increases by an amount
dA = ℓ v dt .
31
The loop is perpendicular to the magneticfield,
 so the
magnetic flux through the loop is  B   B  dA  BA .
The emf induced in the conductor can be calculated using
Faraday’s law:
dB
ε = -N
dt
v
B     
d BA 

ε = 1
dt
 ℓ

B dA
ε =

dt
dA
vdt
x
dx
ε = B
B and v are vector magnitudes, so
dt
ε = B v.
they are always +. Wire length is
always +. You use Lenz’s law to
get the direction of the current.
32
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 into
the plane of the paper (to
counteract the “extra” flux).
B 




I
x





v







 ℓ


dA
vdt
Clockwise current!
33
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






F
M




FP

 ℓ


x
34
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










ℓ
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.
35
Today’s agendum:
Motional emf.
You must be able to apply Faraday’s and Lenz’s “Laws” to calculate motional emf, as well
as current and power in circuits powered by motional emf.
Motors and Generators (part 2).
In the previous lecture, we used Faraday’s “Law” to calculate numerical values of
parameters associated with generators. You must also understand conceptually how
motors and generators work.
Back emf.
You must be able to use Lenz’s “Law” to explain back emf.
36
Motional emf
An emf is induced in a conductor moving in a magnetic field.
In the last lecture you learned about two examples of
motional 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
derive these
P = INBA sin t 
37
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
derive these
P = FP  v =I Bv
38
Today we’ll look at two more examples of motional emf.
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
You could also solve this using Faraday’s “Law” by constructing a “virtual” circuit using “virtual” conductors.
39
4. Flux change through a conducting loop produces an emf:
moving loop.
         
         
















































































start with this
d B
=dt
derive these
ε=B v
B v
I =
P = I Bv
R
40
Example 3 of motional emf: moving conductor in B field.
Example 4 of motional emf: flux change through conducting
loop. (Entire loop is moving.)
Let’s work out these two examples now.
Remember, it’s the flux change that produces 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.
41
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.
42
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
“up”
B 




v
–



+










ℓ
 ε=B v
43
Example 4 of motional emf: flux change through conducting
loop. (Entire loop is moving.)
I’ll include some numbers with this example.
44
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
         
         
















































































B = 0.6 T
45
(a) Find the change in flux through the coil.
         
         
















































































 
Initial :  Bi   B  dA  BA
Final: Bf  0
B = Bf - Bi = 0 - BA = - (0.6 T) (0.05 m)2 = - 1.5x10-3 Wb.
46
(b) Find the current and emf induced.
         
         
















































































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).
47
The induced emf is
dB
d(BA)
dA
ε = -N
= -N
= - NB
dt
dt
dt










































d x 
dA
=
=
dt
dt
A
x
v
dx
=
dt
v
48
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 .
49
(c) How much energy is dissipated in the coil?
Current flows “only*” during the time flux changes.
W = 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.
50
The flux change occurs only when the coil is in the process of
leaving the region of magnetic field.
         
         
















































































No
No
No
No
flux change.
emf.
current.
work (why?).
51
         
         
















































































Fapplied
D
Flux changes. emf induced. Current flows. Work
done.
52
         
         
















































































No flux change. No emf. No current. (No work.)
53
(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.
         
         
















































































I
54
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







 
 
 
F2
 
 
 
 





















 
 
 
 
 
L
 1
 














I L4=0
F1
Fpull
Sorry about the “busy” slide!
The forces should be shown
acting at the centers of the
coil sides in the field.
There must be a pulling force to the right to overcome the net
magnetic force to the left.
55
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







 
 
 
F2
 
 
 
 





















 
 
 
 
 
L
 1
 














I
Fpull
F1
This calculation assumes the coil is pulled out “uniformly;” i.e.,
56
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
     










































x
I
Fpull
D
The work done by the pulling force is equal to the electrical
energy provided to (and dissipated in) the coil.
57
Handy Hint: Debugging Your Homemade Generator
If you build a generator and it
doesn’t seem to be working…
European model shown.
US model also available.
…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.
58
Today’s agendum:
Motional emf.
You must be able to apply Faraday’s and Lenz’s “Laws” to calculation motional emf, as well
as current and power in circuits powered by motional emf.
Motors and Generators (part 2).
In the previous lecture, we used Faraday’s “Law” to calculate numerical values of
parameters associated with generators. You must also understand conceptually how
motors and generators work.
Back emf.
You must be able to use Lenz’s “Law” to explain back emf.
59
electric motors and web applets
Generator: source of mechanical energy rotates a current loop
in a magnetic field, and mechanical energy is converted into
electrical energy.
Electric motor: a generator “in reverse.” Current in loop in
magnetic field gives rise to torque on loop.
A dc motor animation is here.
Details about ac and dc motors at hyperphysics.
Other useful animations here.
True Fact you didn’t know: all electrical motors operate on smoke. Every
motor has the correct amount of smoke sealed inside it at the factory. If
this smoke ever gets out, the motor is no longer functional. I can even provide
the source of this true information. http://www.xs4all.nl/~jcdverha/scijokes/2_16.html#subindex
60
detailed look at a generator (if time permits)
Let’s begin by looking at a simple animation of a generator.
http://www.wvic.com/how-gen-works.htm
brush
Here’s a “freeze-frame.”
Normally, many coils of wire are
wrapped around an armature.
The picture shows only one.
slip ring
Brushes pressed against a slip ring make continual contact.
The shaft on which the armature is mounted
is turned by some mechanical means.
Disclaimer: there are some oversimplifications in this analysis which an expert would consider
“errors.” Anyone who is an expert at generators is invited to help me correct these slides!
61
Let’s look at the current direction in this particular freezeframe.
B is down. Coil
rotates counterclockwise.
Put your fingers
along the
direction of
movement. Stick
out your thumb.
Bend your fingers 90°. Rotate your hand until the fingers
point in the direction of B. Your thumb points in the direction
of conventional current.
62
One more thing…
This wire…
…connects to this
ring…
…so the current
flows this way.
Another way to generate electricity with hamsters: give them little
magnetic collars, and run them through a maze of coiled wires.
http://www.xs4all.nl/~jcdverha/scijokes/2_16.html#subindex
63
Later in the cycle, the current still flows clockwise in the
loop…
…but now this*
wire…
…connects to this
ring…
…so the current
flows this way.
Alternating current! ac!
Without commutator—“dc.”
64
*Same wire as before, in different position.
Today’s agendum:
Motional emf.
You must be able to apply Faraday’s and Lenz’s “Laws” to calculation motional emf, as well
as current and power in circuits powered by motional emf.
Motors and Generators (part 2).
In the previous lecture, we used Faraday’s “Law” to calculate numerical values of
parameters associated with generators. You must also understand conceptually how
motors and generators work.
Back emf.
You must be able to use Lenz’s “Law” to explain back emf.
65
back emf (also known as “counter emf”) (if time permits)
A changing magnetic field in wire produces a current. A
constant magnetic field does not.
We saw how changing the magnetic field experienced by a
coil of wire produces ac current.
But the electrical current produces a magnetic field, which by
Lenz’s law, opposes the change in flux which produced the
current in the first place.
http://campus.murraystate.edu/tsm/tsm118/Ch7/Ch7_4/Ch7_4.htm
66
The effect is “like” that of friction.
The counter emf is “like” friction that opposes the original
change of current.
Motors have many coils of wire, and thus generate a large
counter emf when they are running.
Good—keeps the motor from “running away.” Bad—”robs”
you of energy.
67
If your house lights dim when an appliance starts up, that’s
because the appliance is drawing lots of current and not
producing a counter emf.
When the appliance reaches operating speed, the counter emf
reduces the current flow and the lights “undim.”
Motors have design speeds their engineers expect them to run
at. If the motor runs at a lower speed, there is less-thanexpected counter emf, and the motor can draw more-thanexpected current.
If a motor is jammed or overloaded and slows or stops, it can
draw enough current to melt the windings and burn out. Or
even burn up.
68
Two brief examples (if time permits):
Induced emf on an airplane wing.
Blood flow measurement.
69
Example: An airplane travels 1000 km/h in a region where the
earth’s field is 5x10-5 T and is vertical. What is the potential
difference induced between the wing tips that are 70 m apart?

































v


70
The electrons “pile up” on the left hand wing of the plane,
leaving an excess of + charge on the right hand wing.
The equation for  at the bottom of
slide 10 gives the potential difference.
(You’d have to derive this on a test.)
ε =B v
ε = 5×10-5 T   70 m   280 m/s 
ε = 1V

































v


+
No danger to passengers! (But I would want my airplane
designers to be aware of this.)
71
Example: Blood contains charged ions, so blood flow can be
measured by applying a magnetic field and measuring the
induced emf. If a blood vessel is 2 mm in diameter and a 0.08
T magnetic field causes an induced emf of 0.1 mv, what is the
flow velocity of the blood?
=Bℓv
v =  / (B ℓ)
If B is applied  to the blood vessel, then B is  to v. The
ions flow along the blood vessel, but the emf is induced across
the blood vessel, so ℓ is the diameter of the blood vessel.
v = (0.1x10-3 V) / ( (0.08 T)(0.2x10-3 m) )
v = 0.63 m/s
72
Today’s agendum:
Induced Electric Fields.
You must understand how a changing magnetic flux induces an electric field, and be able
to calculate induced electric fields.
Eddy Currents.
You must understand how induced electric fields give rise to circulating currents called
“eddy currents.”
Displacement Current and Maxwell’s Equations.
Displacement currents explain how current can flow “through” a capacitor, and how a timevarying electric field can induce a magnetic field.
Leftovers.
Any assorted topics I need to “cover” before finishing magnetism.
73
Time-Varying Magnetic Fields
and Induced Electric Fields
A Changing Magnetic Flux Produces an Electric Field?
dB
ε = -N
dt
V = Vb - Va = - Ed
ε = V = - Ed
-N
d B
= - Ed
dt
This suggests that a changing magnetic flux produces an
electric field. This is true not just in conductors, but anywhere in space where there is a changing magnetic field.
74
The previous slide uses an equation
derived specifically for the electric
field between two large parallel
plates. Let’s see what a more
general analysis gives us.
Consider the conducting loop of
radius r around (but not in) a region
where the magnetic field is into the
page and increasing.

 
 


 
r
 

This could be a wire loop around
the outside of a solenoid.
The charged particles in the conductor are not in a magnetic
field, so they experience no magnetic force.
But the changing magnetic flux induces an emf around the
loop.
dB
ε = dt
75
The induced emf causes a counterclockwise current (charges move).
But the magnetic field did not
accelerate the charged particles
(they aren’t in it). Therefore, there
must be a tangential electric field
around the loop.
E

 
 

I
E
E

 
r
 

E
The work done moving a charged particle once around the
loop is.
W = qV = qε
The sign is positive because the particle’s kinetic energy
increases.
76
We can look at work from a different
point of view.
ds
E

 
 

I
The electric field exerts a force qE
on the charged particle. The
instantaneous displacement is
always parallel to this force.
E
E

 
r
 

E
Thus, the work done by the electric field in moving a charged
particle once around the loop is.
W  2r qE  qE  ds
The sign is positive because the particle’s displacement and
the force are always parallel.
77
Summarizing…
d B

dt
W  q
W  2r qE  qE  ds
ds
E

 
 

I
E
E

 
r
 

E
Combining and generalizing…
 
q  qE  ds  q  E  d s
 
d B
   E  ds  
dt
 
d B
 E  d s   dt
78
Generalizing still further…
The loop of wire was just a
convenient way for us to visualize
the effect of the changing magnetic
field.
The electric field exists whether or
not the loop is present.
ds
E

 
 

I
E
E

 
r
 

E
 
d B
 E  d s   dt
A changing magnetic flux gives rise to an electric field.
Was there anything in this discussion that bothered you?
79
q
E = k 2 , away from +
r
ds
E

 
 

I
-
+
E
This should bother you: where are
the + and – charges in this picture?
E

 
r
 

E
Answer: there are no + and – charges. Instead, there are
electric field lines that form continuous, closed loops.
Huh?
80
But wait…there’s more!
A potential energy can be defined
only for a conservative force.*
A potential energy is a single-valued
function.

 
 


 
 

E
If this electric field E is due to a
conservative force, then the
potential energy of a charged
particle must be unchanged when it
goes once around the loop.
*The work done by the force is independent of path.
81
But the work done is
 
W  q  E  d s  qE 2r 
Work depends on the path!
I and F

 
 


 
r
 

E
If we tried to define a potential
energy, it would not be singlevalued:
 
U F  U I  q  E  d s
 
d B
U F  U I  q  E  d s  
 0 even if I  F
dt
U is not single-valued! We can’t define a U for this E!
(*%&^#!)
82
Induced Electric Fields: a summary of the key ideas
A changing magnetic flux induces an electric field, as given by
Faraday’s Law:


d B
 E  d s   dt
This is a different kind of electric field than the one you are
familiar with; it is not the electrostatic field caused by the
presence of charged particles.
Unlike the electrostatic electric field, this “new” electric field is
 

nonconservative.
E  EC  E NC
“conservative,” or “Coulomb”
“nonconservative”
83
Stated slightly differently: we have “discovered” two different
ways to generate an electric field.
Coulomb Electric Field
Ek
“Faraday” Electric Field
q
r
2
, away from
 
d B
 E  d s   dt
Both “kinds” of electric fields are part of Maxwell’s
Equations.
Both “kinds” of electric fields exert forces on charged particles.
The Coulomb force is conservative, the “Faraday” force is not.
84
Direction of Induced Electric Fields
The direction of E is in the direction a positively charged
particle would be accelerated by the changing flux.
 
d B
 E  d s   dt
Use Lenz’s Law to determine the direction the changing
flux would cause a current to flow. That is the direction
of E.
85
Example
A long thin solenoid has 900 turns per meter and a radius of
2.50 cm. The current is increasing at a steady rate of 60 A/s.
What is the magnitude of the induced electric field near the
center of the solenoid 0.5 cm from the axis of the solenoid?
86
Some Revolutionary Applications of Faraday’s Law
 Magnetic Tape Readers 
 Phonograph Cartridges 
 Electric Guitar Pickup Coils
 Ground Fault Interruptors
 Alternators
 Generators
 Transformers
 Electric Motors
87
The Big Picture
You have now learned Gauss’s Law for both electricity and
magnetism… and Faraday’s Law of Induction:
  q enclosed
 E  dA  o
 
d B
 E  d s   dt
 
 B  dA  0
These equations can also be written in differential form:

E 
0
B
×E=t
B  0
88
Today’s agendum:
Induced Electric Fields.
You must understand how a changing magnetic flux induces an electric field, and be able
to calculate induced electric fields.
Eddy Currents.
You must understand how induced electric fields give rise to circulating currents called
“eddy currents.”
Displacement Current and Maxwell’s Equations.
Displacement currents explain how current can flow “through” a capacitor, and how a timevarying electric field can induce a magnetic field.
Leftovers.
Any assorted topics I need to “cover” before Spring Break.
89
Eddy Currents
You have seen how a changing magnetic field can induce a
“swirling” current in a conductor (the beginning of this lecture).
If a conductor and a magnetic field are in relative motion, the
magnetic force on charged particles in the conductor causes
circulating currents.
These currents are called “eddy currents.”
Eddy currents give rise to magnetic fields that oppose any
external change in the magnetic field.
90
Eddy Currents
Eddy currents are useful in generators, microphones, and roller
coaster brakes (among other things).
However, the I2R heating from eddy currents causes energy
loss, so if you don’t want energy loss, you probably think eddy
currents are “bad.”
91
Conceptual Example: Induction Stove
An ac current in a coil in the stove
top produces a changing
magnetic field at the bottom of a
metal pan.
The changing magnetic field gives
rise to a current in the bottom of
the pan.
Because the pan has resistance, the current heats the pan. If
the coil in the stove has low resistance it doesn’t get hot but
the pan does.
An insulator won’t heat up on an induction stove.
Remember the controversy about cancer from power lines a few years back? Careful studies
showed no harmful effect. Nevertheless, some believe induction stoves are hazardous. 92
Today’s agendum:
Induced Electric Fields.
You must understand how a changing magnetic flux induces an electric field, and be able
to calculate induced electric fields.
Eddy Currents.
You must understand how induced electric fields give rise to circulating currents called
“eddy currents.”
Displacement Current and Maxwell’s Equations.
Displacement currents explain how current can flow “through” a capacitor, and how a timevarying electric field can induce a magnetic field.
Leftovers.
Any assorted topics I need to “cover” before Spring Break.
93
Displacement Current
Apply Ampere’s “Law” to a
charging capacitor.
ds
 
 B  ds  o IC
IC
+
+q
-q
IC
94
The shape of the surface used for Ampere’s “Law” shouldn’t
matter, as long as the “path” is the same.
Imagine a soup bowl
surface, with the + plate
resting near the bottom of
the bowl.
Apply Ampere’s “Law” to
the charging capacitor.
ds
IC
+
+q
-q
IC
 
 B  ds  0
The integral is zero because no current passes through the
“bowl.”
95
 
 B  ds  o IC
 
 B  ds  0
Hold it right there pal! You can’t have it both ways. Which is it
that you want?
Evidently, our understanding is incomplete.
96
As the capacitor charges, the electric field between the plates
changes.
A
q = CV = ε Ed 
d
= εEA = εE
ds
IC
As the electric field changes,
the electric flux changes.
E
+
+q
-q
IC
dq d
d
=  ε E  = ε   E 
dt dt
dt
This term has units of current.
Note:  on this slide is 0, not an emf.
97
We define the displacement current to be
d
ID = ε  E  .
dt
The changing electric flux
through the “bowl” surface
is equivalent to the current
IC through the flat surface.
ds
IC
E
+
+q
-q
IC
The generalized (“always correct”) form of Ampere’s “Law” is
then
 
d E
 B  ds  o IC  I D encl  o Iencl  o dt
Magnetic fields are produced by both conduction currents and
time varying electric fields.
98
The “stuff” inside the gray boxes is a new starting equation
and serves as your official starting equation for the
displacement current ID.
 
 B  ds  o IC  I D 
d E
  o I encl   o 
encl
dt
In a vacuum, replace  with 0.
99
The Big Picture
You have now learned Gauss’s “Law” for both electricity and
magnetism, Faraday’s “Law” of Induction, and the generalized
form of Ampere’s “Law”:
  q enclosed
 E  dA  o
 
d B
 E  d s   dt
 
 B  dA  0
 
d E
 B  ds  o Iencl  o dt
These equations can also be written in differential form:

E 
0
B  0
dB
×E=dt
1 dE
  B= 2
+μ 0 J
c dt
100
Today’s agendum:
Induced Electric Fields.
You must understand how a changing magnetic flux induces an electric field, and be able
to calculate induced electric fields.
Eddy Currents.
You must understand how induced electric fields give rise to circulating currents called
“eddy currents.”
Displacement Current and Maxwell’s Equations.
Displacement currents explain how current can flow “through” a capacitor, and how a timevarying electric field can induce a magnetic field.
Leftovers.
Any assorted topics I need to “cover” before Spring Break.
101
leftover: torque on generator coils
In lecture 15, I derived an expression for the torque on a
current-carrying coil in a magnetic field.
net = R  L = WILB sin  =IBA sin 
W
sin 
2
The source of external mechanical
energy provides this torque.
Recall that for linear motion,
P =FV
FR

IL
FL
 IR
B

W
2
A
For rotational motion,
P = 
102
The torque and angular velocity vectors are parallel, so
P = 
P = IBA  sin  t  
FR
 IR
B
For a coil of N loops:
P = INBA  sin  t  
If there are no frictional torques or
other dissipative forces present, the
generator’s power output is equal to
the external power input.
IL
FL
A
103