Transcript Ch. 33 slides

Chapter 33
Electromagnetic Induction
Phys 133 -- Chapter 33
1
Demo Faraday
Phys 133 -- Chapter 33
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Some experiments
Phys 133 -- Chapter 33
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Motional emf
Phys 133 -- Chapter 33
4
Motional emf (cont.)
Reach equilibrium
Phys 133 -- Chapter 33
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Problem 33.1
A potential difference of 0.050 V is developed across a
10-cm long wire as it moves through a magnetic field at
5.0 m/s. The magnetic field is perpendicular to the wire
axis. What are strength and direction of field?
Phys 133 -- Chapter 33
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Problem 33.1 (ans)
At equilibrium
åF
x
Fm
=0
Fe
Fe = Fm
qE = qvB
E
ÞB=
v
0.05V
= 0.1mm = 0.1T
5.0 s
Out of page
Phys 133 -- Chapter 33
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Induced current
--moving charges in B field, force --> current
--current-carrying wire in field, force
Phys 133 -- Chapter 33
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Do workbook 33.1
Phys 133 -- Chapter 33
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Eddy currents
Phys 133 -- Chapter 33
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Do workbook 33.2 & 4
Phys 133 -- Chapter 33
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Eddy currents (cont.)
Phys 133 -- Chapter 33
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Demo eddy currents
Phys 133 -- Chapter 33
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Do workbook 33.5 & 7
33.5 & 7
Phys 133 -- Chapter 33
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Magnetic flux
Fm = B × A = A B cos qA ,B
ò B ×dA
Fm =
area of loop
Phys 133 -- Chapter 33
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Do workbook 33.9 & 11
Phys 133 -- Chapter 33
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Lenz’s law
There is an induced current in a closed, conducting loop if
and only if the magnetic flux through the loop is changing.
The induced current creates an induced magnetic field which
opposes the change in flux.
--changing flux produces induced current
--induced current creates induced field
--induced field opposes the change
Phys 133 -- Chapter 33
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Do workbook 33.12, 13 & 19
Phys 133 -- Chapter 33
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Problem 33.9
The current in the solenoid is decreasing. The solenoid
is surrounded by a conducting loop. Is there a current
in the loop? If so, is the current in the loop cw or ccw?
Phys 133 -- Chapter 33
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Problem 33.9 (ans)
• Find change in flux
Initially


B1

∆
Change in flux to left
• Find induced field
later
Induced field points to right
B2

• Find induced current
Induced current is
clockwise from behind
Phys 133 -- Chapter 33
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Induced fields
field is there independent of wire
Phys 133 -- Chapter 33
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Faraday’s law
--law of nature
--usually helpful
e
dFm
=
, and Lenz's law
dt
- field can change
-area can change
-both can change
Phys 133 -- Chapter 33
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Do workbook 33.21
Phys 133 -- Chapter 33
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Problem 33.12b
The loop is being pushed out of the 0.20 T magnetic field at
50 m/s. The resistance of the loop is 0.20 . What are the
direction and magnitude of current in the loop?
Phys 133 -- Chapter 33
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Problem 33.12b (ans)
f = BAwith field
Awith field = Lw
= BLw
L
w
df
e=
dt
d
dL
= ( BLw) = Bw
= Bwv
dt
dt
I=
e
R
Counter-clockwise
Phys 133 -- Chapter 33
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Problem 33.35
Two 20-turn coils are tightly wrapped on the same 2.0-cm-diameter
cylinder with 1.0-mm-diameter wire. The current through coil 1 is
shown. A positive current is into the page at the top of the loop.
Determine the current in coil 2 as current vs. time t = 0 s to 0.4 s.
(Assume field of coil 1 passes completely through coil 2.
0.1
Phys 133 -- Chapter 33
0.3
26
Problem 33.35 (ans)
(+)
df 2 dB1
=
N 2 A2
dt
dt
dI
N1m 0 1 N 2 A2
dt
=
L1
I
2.0 1
d2/dt
t (s)
0.1
0.2
0.3 0.4
t (s)
0.1
-2.0
Nm I
B1 = 1 0 1
L1
B1
0.1
f 2 = B1N 2 A2
t (s)
0.1
0.2
0.3 0.4
e=
t (s)
0.2 0.3 0.4
2
0.3 0.4
2
t (s)
0.1
0.2
æ N1m 0 I1 ö
=ç
÷ N 2 A2
L
è 1 ø
0.2
=
0.3 0.4
I=
I2
0.1
Phys 133 -- Chapter 33
df 2
dt
N1m 0
e
R
dI1
N A
dt 2 2
L1
27
R
N1m 0
t (s)
0.2
0.3 0.4
=
dI1
N 2 A2
dt
L1
Problem 33.35 (ans)
(+)
I1 2.0
t (s)
0.1
0.2
0.3 0.4
-2.0
I2
t (s)
0.1
0.2
0.3 0.4
I=
e
R
=
N1m 0
dI1
N A
dt 2 2
RL1
= 79mA
Right to left through resistor
Phys 133 -- Chapter 33
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Generator
Phys 133 -- Chapter 33
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Transformer
N2
V2 =
V1
N1
Phys 133 -- Chapter 33
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Coulomb and non-Coulomb electric fields
Phys 133 -- Chapter 33
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Maxwell and induced magnetic field
-symmetric in E and B
Phys 133 -- Chapter 33
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E & M equations
ò E × dA =
e0
no name
ò B × dA = 0
dF m
d
ò E × ds = - dt = - dt
ò B × ds = m I
0 through
(
Gauss' Law
Qin
[ò B × dA]
Faraday' s Law
Ampere' s Law
)
F =q E +v ´B
Phys 133 -- Chapter 33
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Maxwell’s equations
ò E × dA =
Qin
Gauss' Law
e0
ò B × dA = 0
ò E × ds = -
no name
dF m
d
=dt
dt
ò B × ds = m I
0 through
[ò B × dA]
dF e
+ e 0m 0
dt
Faraday's Law
Ampere/Maxwell's Law
Phys 133 -- Chapter 33
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Maxwell-no charges or current
ò E × dA = 0
ò B × dA = 0
ò
ò
dFm
E × ds = dt
dF e
B × ds =e0m0
dt
+math
¶2E 1 ¶2E
= 2 2
2
¶x
v ¶t
v=
1
e0m0
= 3.00 ´10 8 m/s
Light is an E&M wave!!!
Phys 133 -- Chapter 33
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EM waves
Phys 133 -- Chapter 33
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Problem 33.49
A U-shaped conducting rail that is oriented
vertically in a horizontal magnetic field.
The rail has no electrical resistance and
does not move. A slide wire with mass m
and resistance R can slide up and down
without friction while maintaining electrical
contact with the rail. The slide wire is
release from rest.
a) Describe the motion of the slide wire.
b) Find the terminal velocity of the slide wire
Phys 133 -- Chapter 33
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