Upcoming Schedule

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Upcoming Schedule
Nov. 3 20.8, 21.1-21.2 Nov. 5 boardwork Nov. 7 21.3-21.5 Quiz 7
Nov. 10 boardwork Nov. 12 21.5-21.7 Nov. 14 boardwork Quiz 8
Nov. 17 review Nov. 19 Exam 3 Chap. 20-21 Nov. 21 18.8 Chapter 22
Must-know facts for all students taking Physics
classes (1) physicists don't like not knowing
things, and (2) never point this fact out to a
physicist.
2
according to CERN web page (worlds biggest
particle accelerator) physicists don't like not
knowing things, and never point this fact out
to a physicist.
Original "homework3.pdf" showed lecture on
November 12 covering 21.6. Corrected file shows
lecture on November 12 covering 21.5-21.7. I
dont expect to finish 21.5 today, so I will
finish 21.5 on the 12th. I also intended all
along to lecture on transformers (21.7). The
homework problem assignments are unchanged.
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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. Lets see how this works.
v
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 speed v for a
time ?t.
?A
v?t
The rod moves a distance v?t and the area of the
loop inside the magnetic field increases by an
amount ?A l v ?t .
got to here, lect 15 fs2002
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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 Amperes law
B and v are vector magnitudes, so they are always
. Wire length is always . You use Lenzs law
to get the direction of the current.
5
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
Magnetic flux inside the loop increases (more
area).
l
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).
I
?A
v?t
Clockwise current!
6
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.
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!
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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,
Wi?f q ?Vi?f .
F q v B
But ?Vi?f 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 ?).
Solving (q v B) (l) (q ?) for ? gives ? B l
v, as before.
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I wont 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 havent given us the OSE yet!
Good point! The derivation assumed B, l, and v
are all mutually perpendicular, so we really
derived this
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.
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Example 21-4 An airplane travels 1000 km/h in a
region where the earths 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
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 calculation I did
3 slides back showed that the electrons in the
moving rod (or airplane wing in this case)
experience a force, which moves the electrons.
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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.
No danger to passengers! (But I would want my
airplane designers to be aware of this.)
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21.4 Changing Magnetic Flux Produces an Electric
Field
From chapter 16, section 6
OSE E F / q
From chapter 20, section 4
OSE F q v B sin?
For v ? B, and in magnitude only,
F q E q v B
E v B.
We conclude that a changing magnetic flux
produces an electric field. This is true not
just in conductors, but any-where in space where
there is a changing magnetic field.
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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
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21.5 Electric Generators
Lets begin by looking at a simple animation of a
generator. http//www.wvic.com/how-gen-works.htm
Heres 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.
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Lets look at the current direction in this
particular freeze-frame.
B is down. Coil rotates counter-clockwise.
B is down.
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.
I can see it for this part of the loop, but have
great difficulty for this part of the loop.
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Alternative right-hand rule for current direction.
B is down. Coil rotates counter-clockwise.
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!
16
I know we need to work on that more. Lets zoom
in on the armature.
v
v??B
v?B
I
B
17
Forces on the charges in these parts of the wire
are perpendicular to the length of the wire, so
they dont 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.
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One more thing
This wire
connects to this ring
so the current flows this way.
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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
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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 17.) 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?.
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Back to this picture
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.
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?
The angle ? in the text is the same as the angle
between v??B and the vector v.
Thus, v? v sin ?.
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B is ? to the wire, so
But the coil is rotating, so ? ? t, and v ? r
? (h/2). The diameter of the circle of
rotation, h, was defined on slide 17.
where A is the area of the loop, f is the
frequency of rotation of the loop, and ? 2 ? f.

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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?
You should read about dc motors and alternators
in section 21.5, but I wont test you over that
material.