Chapter 21 Electromagnetic Induction Faraday’s Law ac Circuits
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Transcript Chapter 21 Electromagnetic Induction Faraday’s Law ac Circuits
Chapter 21
Electromagnetic Induction
Faraday’s Law
ac Circuits
We found in chapter 20 that an electric current can give
rise to a magnetic field, and that a magnetic field can
exert a force on a moving charge.
I wonder if a magnetic field can somehow give rise to an
electric current…
21.1 Induced emf
It is observed experimentally that changes in magnetic
fields induce an emf in a conductor.
An electric current is induced if there is a closed circuit
(e.g., loop of wire) in the changing magnetic field.
A constant magnetic field does not induce an emf—it
takes a changing magnetic field.
Passing the coil through the
magnet would induce an emf in
the coil.
They need to calibrate
their meter!
Note that “change” does not require observable (to you)
motion.
A magnet may move through a loop of wire, or a
loop of wire may be moved through a magnetic
field (as suggested in the previous slide). These
involve observable motion.
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 latter case, nothing observable (to your eye) is
moving, although, of course microscopically, electrons
are in motion.
As your text puts it: “induced emf is produced by a
changing magnetic field.”
21.2 Faraday’s Law
To quantify the ideas of section 21.1, we define magnetic
flux. In an earlier chapter we briefly touched on electric
flux. This is the magnetic analog.
Because we can’t “see” magnetic fields directly, we draw
magnetic field lines to help us visualize the magnetic
field. Remember that magnetic field lines start at N poles
and end at S poles.
A strong magnetic field is represented by many magnetic
field lines, close together. A weak magnetic field is
represented by few magnetic field lines, far apart.
We could, if we wished, actually “calibrate” by specifying the number of
magnetic field lines passing through some surface that corresponded to a
given magnetic field strength.
Magnetic flux B is proportional to the number of
magnetic field lines passing through a surface.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/fluxmg.html
Mathematically, magnetic flux B
through a surface of area A is
defined by
B
A
B = B A cos
OSE:
B = BA
where B is the component of
field perpendicular to the
surface, A is the area of the
surface, and is the angle
between B and the normal to the
surface.
B
B
A
When B is parallel to the surface,
=90° and B = 0.
When B is perpendicular to the
surface, =0° and B = BA.
B
A
B=0
B=0
=0
A
B
The unit of magnetic flux is the Tm2, called a weber:
1 Wb = 1 T m2 .
In the following discussion, we switch from talking about
surfaces in a magnetic field…
…to talking about loops of wire in a magnetic field.
Now we can quantify the induced emf described
qualitatively in the previous section
Experimentally, if the flux through N loops of wire
changes by an amount B in a time t, the induced emf is
ΔφB
ε = -N
.
Δt
Not an OSE—
not quite yet.
This is called Faraday’s law of induction. It is one of the
fundamental laws of electricity and magnetism, and an
important component of the theory that explains
electricity and magnetism.
I wonder why the – sign…
Experimentally…
…an induced emf always gives rise to a current whose
magnetic field opposes the change in flux—Lenz’s law.*
Think of the current resulting from the induced emf as
“trying” to maintain the status quo—to prevent change.
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 increased the
magnetic field, further increasing the current, making the
magnetic field still bigger…
Among other things, conservation of energy would be
violated.
*We’ll practice with this in a bit.
Ways to induce an emf:
change B
change the area of the loop in the field
Ways to induce an emf (continued):
change the orientation of the loop in the field
Conceptual example 21-1 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.
Conceptual example 21-2 Practice with Lenz’s Law
In which direction is the current induced in the coil for
each situation shown?
(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)
ΔφB
?
Remember ε = - N
Δt
Now that we are experts on the application of Lenz’s law,
lets make our induced emf equation official:
ΔφB
ε = N
.
Δt
This means it is up to you to use Lenz’s law to
figure out the direction of the induced current
(or the direction of whatever the problem wants.
Example 21-3 Pulling a Coil from a Magnetic Field
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
(a) Find the change in flux through the coil.
Initial: Bi = 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.
The current must flow counterclockwise to induce a
downward magnetic field (which replaces the “removed”
magnetic field).
The induced emf is
ΔφB
ε = N
Δt
ε = 100
-3
-1.5
10
Wb
0.1 s
ε = 1.5 V
The induced current is
I =
ε
R
=
1.5 V
100
= 15 mA .
(c) How much energy is dissipated in the coil?
Current flows “only*” during the time flux changes.
E = Pt = I2Rt = (1.5x10-2 A) (100 ) (0.1 s) = 2.3x10-3 J .
(d) What was the average force required?
The loop had to be “pulled” out of the magnetic field, so
the pulling force did work. It is tempting to try and set up
a free body diagram and use Newton’s laws. Instead,
energy conservation gets the answer with less work.
*If there 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.
No flux change.
No emf.
No current.
No work (why?).
F
D
Flux changes. emf induced. Current flows.
Work done.
No flux change. No emf. No current. (No work.)
The energy calculated in part (c) is the energy dissipated
in the coils while the current is flowing. The amount
calculated in part (c) is also the mechanical energy put
into the system by the force.
Ef – Ei = [ Wother]If
0
See 2 slides
Ef – Ei = F D
back for F and D.
F = Ef / D
F = (2.3x10-3 J) / (0.05 m)
F = 0.046 N
21.3 emf Induced
in a Moving Conductor
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
ℓ
shown.
The rod moves to the right with a
A
speed v for a time t.
vt
The rod moves a distance vt and the area of the loop
inside the magnetic field increases by an amount
A = ℓ v t .
The loop is perpendicular to the magnetic field, so the
magnetic flux through the loop is B = BA. The emf
induced in the conductor can be calculated using
Ampere’s law:
ΔφB
ε = N
Δt
Δ BA
ε = 1
Δt
ε =
B ΔA
Δt
B v Δt
ε =
Δt
ε = B v.
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.
OSE? (not yet)
ε =B v
This “kind” of emf is called “motional emf” because it
took motion to induce it.
The induced emf causes current to flow in the loop.
v
B
Magnetic flux inside the loop
increases (more area).
ℓ
System “wants” to make the flux stay
the same, so the current gives rise to
I
A
a field inside the loop into the plane
vt
of the paper (to counteract the “extra”
flux).
Clockwise current!
The induced emf causes current to flow in the loop.
Giancoli shows an alternate method for getting , by
calculating the work done moving the charges in the wire.
Electrons in the moving rod (only
the rod moves) experience a force
F = q v B. Using the right hand
rule,* you find the the force is “up”
the rod, so electrons move “up.”
“Up” here refers only to the orientation on the page,
and has nothing to do with gravity.
B
I
v
ℓ
A
vt
Because the rod is part of a loop, electrons flow
counterclockwise, and the current is clockwise (whew, we
got that part right!).
*Remember, find the force direction, then reverse it if
the charge is an electron!
The work to move an electron from the bottom of the rod
to the top of the rod is W = (force) (distance) = (q v B) (ℓ).
Going way back to the beginning of
the semester, Wif = q Vif .
But Vif is just the change in
potential along the length ℓ of the
loop, which is the induced emf.
Going way back to the beginning of
the semester, W = (q v B) (ℓ) = (q ).
F=qvB
B
I
v
ℓ
A
vt
Solving (q v B) (ℓ) = (q ) for gives = B ℓ v, as before.
I won’t ask you to reproduce the derivation on an exam,
but a problem could (intentionally or not) ask you to
calculate the work done in moving a charge (or a wire)
through a magnetic field, so be sure to study your text.
“But you haven’t given us the OSE yet!”
Good point! The derivation assumed B, ℓ, and v are all
mutually perpendicular, so we really derived this:
OSE
ε = B v
where B is the component of the magnetic field
perpendicular to ℓ and v, and v is the component of the
velocity perpendicular to B and ℓ.
Example 21-4 An airplane travels 1000 km/h in a region
where the earth’s field is 5x10-5 T and is nearly vertical.
What is the potential difference induced between the wing
tips that are 70 m apart?
The derivation of = B ℓ v on
slides 15 and 16 assumed the area
through which the magnetic field
v
passes increased.
My first reaction is that the magnetic
flux through the wing is not
changing because neither the field
nor the area of the wing is changing.
True, but wrong reaction! The alternate derivation shows
that the electrons in the moving rod (airplane wing in this
case) experience a force, which moves the electrons.
The electrons “pile up” on the left hand wing of the plane,
leaving an excess of + charge on the right hand wing.
Our equation for gives the
potential difference.
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.)
21.4 Changing Magnetic Flux
Produces an Electric Field
From chapter 16, section 6:
E=F/q
OSE:
From chapter 20, section 4:
OSE:
F = q v B sin
For v B, and in magnitude only,
F=qE=qvB
E = v B.
We conclude 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.
Example 21-5 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?
OSE: = B ℓ v
v = / (B ℓ)
In Figure 21-11 (the figure for this example), B is applied
to the blood vessel, so 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
21.5 Electric Generators
Let’s begin by looking at a simple animation of a
generator. http://www.wvic.com/how-gen-works.htm
Here’s a “freeze-frame.”
Normally, many coils of wire
are wrapped around an
armature. The picture shows
only one.
Brushes pressed against a slip ring make continual contact.
The shaft on which the armature is mounted is turned by
some mechanical means.
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.
Alternative right-hand rule for current direction.
B is down. Coil
rotates counterclockwise.
Make an xyz
axes out of your
thumb and first
two fingers.
Thumb along
component of
wire velocity
to B. 1st finger
along B.
2nd finger then points in direction of conventional current.
Hey! The picture got it right!
I know we need to work on that more. Let’s zoom in on
the armature.
v
vB
B
vB
I
Forces on the charges in these parts of the wire are
perpendicular to the length of the wire, so they don’t
contribute to the net current.
For future use, call the
length of wire shown in
green “h” and the other
lengths (where the two red
arrows are) “ℓ”.
One more thing…
This wire…
…connects to
this ring…
…so the current
flows this way.
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!
Again: http://www.wvic.com/how-gen-works.htm
“Dang! That was complicated. Are you going to ask me
to do that on the exam?”
No. Not anything that complicated. But you still need to
understand each step, because each step is test material.
Click here and scroll down to “electrodynamics” to see
some visualizations that might help you!
Understanding how a generator works is “good,” but we
need to quantify our knowledge.
We begin with our OSE = B ℓ v. (ℓ was defined on
slide 16.) In our sample generator on the last 7 slides, we
had only one loop, but two sides of the loop in the
magnetic field. If the generator has N loops, then = 2 N
B ℓ v.
Back to this picture:
v
vB
vB
I
B
This picture is oriented differently than Figure 21-13 in
your text. In your text, is the angle between the
perpendicular to the magnetic field and the plane of the
loop.
v
vB
vB
I
B
The angle is the text is the same as the angle between
vB and the vector v.
Thus, v = v sin .
B is to the wire, so ε = 2 N B v sin θ .
But the coil is rotating, so = t, and v = r = (h/2).
The diameter of the circle of rotation, h, was defined on
slide 16.
h
ε = 2 N B ω sin ωt
2
ε = N B h ω sin ωt
OSE :
ε = N B A ω sin ωt
where A is the area of the loop, f is the frequency of
rotation of the loop, and = 2 f.
Example 21-6 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 0 = 170 V?
ε = N B A ω sin ωt
ε0 = N B A ω
N =
N =
ε0
BAω
170 V
0.15 T 2 10-2 m2 2π 60 s-1
N = 150 (turns)
18-8 Alternating Current
In chapter 21 we learned how a coil of wire rotating in a
magnetic field creates ac current.
A large magnet rotating inside coils of wire would also
produce ac current. (ac current is redundant, isn’t it?)
The voltage produced by the generator is sinusoidal:
The generator web page uses U instead of V as the symbol for voltage.
In chapter 21, we wrote
ε = N B A ω sin ωt .
Using = 2f, V instead of , and grouping constants
together, in chapter 18 Giancoli “decrees” the formula
OSE
V = V0 sin 2π f t
.
V0 is the “peak voltage” and in the US, the frequency of ac
power in your home is f = 60 Hz.
Using Ohm’s law:
OSE
I = I0 sin 2π f t
Where I0 is the peak current.
,
Even though the average current is zero, power is still lost
due to resistance:
P = I2 R
P = I0 R sin 2π f t
2
2
It is easy to show (a bit of calculus) that the average value
of sin2(2 f t) is ½, so the average power developed in a
resistance is
OSE :
I 20 R
V02
P =
=
.
2
2R
The bar above the P denotes “average.” I will write it Pavg
when using text.
Note that the equations for Pavg contain
2
2
I
V
I 2 avg = 20 and V2 avg = 20 .
sorry, eqn. editor
won’t let me put bar
over 2 characters
The rms (root mean square) value of a quantity is obtained
by taking the square root of the average of the square of
that quantity. The equations for Pavg contain the mean
square values of current and voltage.
We write
OSE :
I 2
Irms =
avg
=
I0
=
V0
2
= 0.707 I0
and
OSE :
Vrms =
V
2
avg
2
= 0.707 V0 .
2
V
2
Then P = I rms
R = rms .
R
Not sure why I didn’t make this “official” last
year. I think I’ll make it “official” this year…
Un the US, we talk about rms voltage when we refer to
household line voltage. The peak voltage is thus
V0 = 2 Vrms =
2 120 V = 170 V .
Example 18-11 (a) Calculate the resistance and the peak
current in a 1000 W hair dryer connected to a 120 V line.
(b) what happens if it is connected to a 240 V line in
Britain?
(a) Our OSE P = IV = I2R = V2/R works if we replace P by
Pavg and I and V by Irms and Vrms.
P = Irms Vrms
Irms
P
=
Vrms
Irms =
1000 W
120 V
Irms = 8.33 A
I0 =
2 Irms = 11.8 A
Vrms
R=
Irms
R=
120 V
11.8 A
R = 14.4 .
(b) Assume the hair dryer’s resistance does not change
with temperature (in reality, it probably increases).
P =
Vrms
2
R
240 V
14.4
2
P =
P = 4000 W .
You just melted your hair dryer!
Example 18-12 Each channel of a stereo receiver is
capable of an average power output of 100 W into an 8
loudspeaker. What is the rms voltage and rms current fed
to the speaker (a) at the maximum power of 100 W, and (b)
at 1 W?
Vrms
P =
R
Vrms =
2
PR
Vrms = Irms R
Irms
Vrms
=
R
we’ll use
these for
both parts
(a) at 100 W and 8 :
Vrms =
100 W 8 = 28 V
Irms =
28 V = 3.5 A
8
(b) at 1 W and 8 :
Vrms =
Irms =
1 W 8 = 2.8 V
2.8 V = 0.35 A
8
21-7 Transformers
No, no, no…
A transformer is a device for increasing or
decreasing an ac voltage.
Pole-mounted
transformer
Power Substation
ac-dc
converter
A transformer is basically two coils of wire wrapped
around each other, or wrapped around an iron core.
When an ac voltage is applied to the primary coil, it
induces an ac voltage in the secondary coil.
A “step up” transformer increases the output voltage in
the secondary coil; a “step down” transformer reduces it.
The ac voltage in the primary coil causes a magnetic flux
change given by
ΔφB
VP = NP
.
Δt
The changing flux (which is efficiently “carried” in the
transformer core) induces an ac voltage in the secondary
coil given by
ΔφB
VS = NS
.
Δt
Dividing the two equations gives the transformer equation
OSE :
VS
NS
=
.
VP
NP
For a step-up transformer, NS > NP and VS > VP (the voltage
is stepped up).
For a step-down transformer, NS < NP and VS < VP (the
voltage is stepped down).
Transformers only work with ac voltages; a dc voltage
does not produce the necessary changing flux.
A step-up transformer increases the voltage. Is this an
example of “getting something for nothing?”
No, because even though transformers are extremely
efficient, some power (and therefore energy) is lost.
If no power is lost, we can use P = IV to get
OSE :
IS
NP
=
.
IP
NS
flipped!
If transformers only work on ac, how come you showed a
picture of an ac-dc converter a few slides back?
Ever wanted to cut open one of those ac-dc converters and
see what they look like inside?
An ac-dc converter first steps down the 120 volt line
voltage, and then converts the voltage to dc:
fewer turns in the
secondary coil
a diode is a device
that lets current flow
one way only (dc)
Pictures of ac-dc converter came from
http://www.howstuffworks.com/inside-transformer.htm
Example 21-9 A transformer for home use of a portable
radio reduces 120 V ac to 9 V dc. The secondary contains
30 turns and the radio draws 400 mA. Calculate (a) the
number of turns in the primary; (b) the current in the
primary; and (c) the power transformed.
VS NS
=
VP
NP
VP
NP
=
VS NS
VP
NP = NS
VS
120 V
NP = 30 turns
9 V
NP = 400 turns
IS
NP
=
IP
NS
NS
IP
=
IS
NP
NS
IP = IS
NP
30 turns
IP = 0.400 A
400 turns
IP = 0.030 A
The power output to the secondary coil is
PS = IS VS
PS = 0.400 A 9 V
PS = 3.6 W .
This is the same as the power input to the primary coil
because our transformer equation derivation assumed
100% efficient transformation of power.
Example 21-10 An average of 120 kW of electrical power is
sent to a small town from a power plant 10 km away. The
transmission lines have a total resistance of 0.40 .
Calculate the power loss if power is transmitted at (a) 240
V and (b) 24,000 V.
This problem does not use the transformer equation, but it
shows why transformers are useful.
P = IV
P
I=
V
PLOST = I2 R
2
PLOST
P
= R
V
(a) at 240 V
120 10 W
=
240
V
3
PLOST
2
0.4
PLOST = 100 103 W = 100 kW .
(a) at 24000 V
120 10 W
=
24000 V
3
PLOST
2
0.4
PLOST = 10 W .
More than 80% of the power would be wasted if it were
transmitted at 240 V, but less than 0.01 % is wasted if the
power is transmitted at 24000 V.
http://www.howstuffworks.com/power.htm
We skipped section 21.6 on counter emf. Read it to see why motors burn out when they
can’t turn, and why your house lights might dim when the fridge comes on.
Chapter 22
Electromagnetic Waves
Every student of E&M should be “exposed” to
electromagnetic waves. Here is your exposure.
22.1 Maxwell’s Equations
Maxwell’s equations involve calculus. They represent the
fundamental laws of electricity and magnetism. We have
seen simpler forms of some of them.
In words, Maxwell’s equations are:
1—a generalized form of Coulomb’s law, relating electric
fields to their sources (charges)
2—a law relating magnetic fields to magnetic poles
3—an equation describing how an electric field is
produced by a changing magnetic field (Faraday’s Law)
4—an equation describing how a magnetic field is
produced by an electric current or changing electric field
(Ampere’s Law)
That a changing electric field can produce a magnetic field
is not one of the predictions of Ampere’s law; it was
hypothesized by Maxwell and verified after his death.
Every student should be exposed to Maxwell’s equations,
so here they are in their integral form…
Maxwell’s equations are to E&M as Newton’s laws are to
mechanics, except Maxwell’s equations are relativistically
correct, and Newton’s laws are not.
22.3 Production of
Electromagnetic Waves
Your text shows how electromagnetic waves can be
produced by oscillating charges on conductors.
These waves travel through space even long after they are
far away from the charges that produced them.
E and B in the radiation fields drop off as 1/r, and the
intensity of the waves drops off as 1/r2.
The electric and magnetic fields are perpendicular to each
other and to the direction of propagation of the wave.
They are also in phase. See figure 22.7, page 666.
From Maxwell’s equations you can show that
electromagnetic waves travel with a speed v = 1/(00)1/2
which is equal to 3x108 m/s, the speed of light.
22.5 Light as an E&M Wave
The Electromagnetic Spectrum
Light was known to behave like a wave long before
Maxwell showed that the speed of E&M waves is the same
as the speed of light. Eventually, it came to be recognized
that light is just an example of an E&M wave.
The frequencies of visible light lie between about 4x10-7
and 7.5x10-7 m, or 400 to 750 nm. The frequency,
wavelength, and speed of a wave are related by v = f, so
for E&M waves, and for light, c = f.
Visible light represents only a minute portion of the
electromagnetic spectrum, see figure 22.10. E&M waves
are typically produced by acceleration of charged
particles.
The sun emits large amounts of IR, visible, and UV
radiation. We detect the IR as heat, the visible as light, and
the UV through skin damage. Note from figure 22.10 that
x-rays, gamma rays, and radio waves are “just” E&M
waves, like light, only of different frequencies.
(sorry, rushed scan)
Symbols for cut and paste
ℓ °
http://www.howstuffworks.com/power.htm
http://www.howstuffworks.com/inside-transformer.htm
http://www.sweethaven.com/acee/forms/frm0502.htm