Transcript Example 21-5

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21.3-21.5
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21.4-21.5
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review
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Exam 3
Chap. 20-21
Nov. 21
21.6-21.7
18.8
Nov. 3
20.8, 21.121.2
Physics-Speak: when we say this…
“It is clear that much additional work
will be required before a complete
understanding is reached.”
we really mean…
“I don’t understand any
of this.”
Homework Problem 21.7 (a) If the resistance is slowly
increased, what is the direction of the current induced in the
small circular loop?
The current flowing as shown
gives rise to a magnetic field
pointing out of the screen (but
only for the area inside the
larger loop).
If the current remains
constant, the magnetic field,
and therefore the magnetic
flux, through the inner loop
remains constant.


No current would flow in the inner loop.



If the resistance is increased, the current in the outer loop
decreases, and the flux through the inner loop decreases.
Less flux through the inner loop
(indicated by the smaller
symbols), so it “wants” to
restore the flux to its “before”
status.
Current flows in the inner loop
to make the magnetic field
inside the loop point out of
the plane of the screen with its
original magnitude.








increase R
Counterclockwise current, just like the book says!


(b) What if the small loop were to the left, outside the larger
one?
Outside the larger loop, the
magnetic field points into the
screen.
Increasing resistance  less
current  less flux pointing
into the screen  small loop
makes current flow to make
more flux point into the screen
(inside itself)  clockwise
current.





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.
I can see it for this part of the loop, but have great difficulty
for this part of the loop.
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
vB
B
vB
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
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.
Back to this picture:
v
vB
vB
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
vB
vB
I
B
The angle  in the text is the same as the angle between vB
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
17.
 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)
You should read about dc motors and alternators in section
21.5, but I won’t test you over that material.