Transcript What Now??? - UCF Physics

Back to Circuits for a bit ….
11-11-06
Induction - Fall 2006
1
What the heck are we doing?

Today




Monday


More of the same,
Wednesday


7:30 AM we had our problem review session
Continue on with Induction & Inductors
Watch those piling up WebAssigns!
EXAMINATION #3
After Holiday

11-11-06
Complete the remaining six chapters in the
syllabus.
Induction - Fall 2006
2
The loop is pushed into a region
where the magnetic field is into
the page. The motion creates an
induced current in the loop
which in turn produces a
magnetic field at A and B.
25% 25% 25% 25%
The field at A & B are
the same.
2.
The field at A is
stronger than the one
at B
3.
The field at B is
stronger than the one
at A.
4.
None of these.
11-11-06
Induction - Fall 2006
e.
th
es
s.
..
on
e
of
is
N
at
B
ld
Th
e
fie
fie
e
Th
Th
e
fie
ld
ld
at
A
at
A
&
is
B
...
s.
..
1.
3
Today we will consider a Coil
11-11-06
Induction - Fall 2006
4
We will base our discussion on
Faraday’s Law
d
emf  V   E  ds  
dt
Lentz
11-11-06
Induction - Fall 2006
5
Consider the following:




There are N turns in the solenoid.
There is a current flowing in the direction shown.
The power supply is set to 20 volts and the resistor
is 10 ohms.
The coil wire has negligible resistance.
11-11-06
Induction - Fall 2006
6
The steady state current (20 volts, 10
ohms) in the circuit is
11-11-06
Induction - Fall 2006
in
fo
.
es
m
or
e
pe
r
am
d
ee
N
th
a
or
e
M
Le
ss
th
an
2
2
am
pe
am
pe
re
re
s
4.
2
3.
25% 25% 25% 25%
n
2.
2 amperes
Less than 2
amperes
More than 2
amperes
Need more info.
s
1.
7
The DIRECTION of the magnetic field
in the coil is
A
B
Induction - Fall 2006
N
11-11-06
ot
en
o
ug
h
Fr
om
B
in
fo
...
to
B
Fr
om
3.
to
2.
From A to B
From B to A
Not enough
information is
given.
A
1.
A
33% 33% 33%
8
Back to the coil diagram …
B



Recall from the last
discussion that the
magnetic field in the coil is
given by:
n = #turns per unit length
Coil is infinitely long

11-11-06
B  0 ni
Sort of
Induction - Fall 2006
9
Back to the coil diagram …
B  0 nI
B



The Flux through a single turn of the coil is
BA or 0niA.
Now, let’s increase the applied voltage
linearly at a rate of V/t.
The current will change at a rate I/t.
11-11-06
Induction - Fall 2006
10
Back to the coil diagram …
B  0 nI
B
For the single coil (as t0) FARADAY Says:
d
d
dI
emf 
 ( 0 nIA)   0 nA
dt
dt
dt
11-11-06
Induction - Fall 2006
11
Looking into the coil from the end with the red arrow, the
emf around the coil induced current will be
B
50 50
% %
11-11-06
Induction - Fall 2006
co
a
In
In
a
cl
oc
.
..
2.
In a clockwise direction
In a counterclockwise
direction
u.
..
1.
12
So … the induced emf
B


Will create a current that will oppose the
change in the current.
The induced emf will therefore oppose the
applied voltage (also an emf) from the power
supply.
11-11-06
Induction - Fall 2006
13
So for the single coil
emf 
d single
dt
d
dI
 (  0 nIA)   0 nA
dt
dt
becomes
dI
dt
for a coil of N turns, " back emf" is
dI
dI
emf  -  0 nNA   L
dt
dt
I N 0 nIA N sin gle
L   0 nNA  N 0 nA 

I
I
I
emf1coil    0 nA
11-11-06
Induction - Fall 2006
14
Definition of Inductance L
N B
L
i
UNIT of Inductance = 1 henry = 1 T- m2/A
B is the flux near the center of one of the coils
making the inductor
11-11-06
Induction - Fall 2006
15
An inductor in the form of a solenoid
contains 420 turns, is 16.0 cm in
length, and has a cross-sectional area
of 3.00 cm2. What uniform rate of
decrease of current through the
inductor induces an emf of 175 μV?
11-11-06
Induction - Fall 2006
16
N B nl   0 nIA
2
L

  0 n lA
I
I
11-11-06
Induction - Fall 2006
17
Look at the following circuit:



Switch is open
NO current flows in the circuit.
All is at peace!
11-11-06
Induction - Fall 2006
18
At the INSTANT that
the switch is closed,
the current through
the resistor is:
Induction - Fall 2006
’t
te
ll
ro
Ze
11-11-06
an
Can’t tell
C
3.
33%
R
2.
Zero
E/R
1.
33%
E/
33%
19
Three years after the
switch is closed, the
current through the
resistor is:
i..
w
e
D
on
’t
c
ar
e
..
w
E/
R
3.
ro
2.
E/R
Zero
Don’t care .. we will
be out by then!
Ze
1.
33% 33% 33%
11-11-06
Induction - Fall 2006
20
Graph?
IR
E/R
Probably looks something
Like this.
time
11-11-06
Induction - Fall 2006
21
Close the
circuit…



After the circuit has been closed for a long
time, the current settles down.
Since the current is constant, the flux through
the coil is constant and there is no Emf.
Current is simply E/R (Ohm’s Law)
11-11-06
Induction - Fall 2006
22
Close the
circuit…




When switch is first closed, current begins to flow
rapidly.
The flux through the inductor changes rapidly.
An emf is created in the coil that opposes the
increase in current.
The net potential difference across the resistor is the
battery emf opposed by the emf of the coil.
11-11-06
Induction - Fall 2006
23
Looking at the math
dI
emf   L
dt
11-11-06
Ebattery  V (notation)
dI
 V  IR  L  0
dt
Induction - Fall 2006
24
Just as we did with the capacitor, we can
solve this equation and we get:
E
t / 
I  (1  e )
R
L
   time constant
R
11-11-06
Induction - Fall 2006
25
The growth
11-11-06
Induction - Fall 2006
26
Death of the current:
dI
IR  L  0
dt
solution
E t / 
I e
R
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Induction - Fall 2006
27
Graph
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Induction - Fall 2006
28
Consider the Solenoid Again…
l
 B  ds   i
0 enclosed
 Bl   0 nli
n turns per unit length
11-11-06
Induction - Fall 2006
or
B   0 ni
29
Inductance & Geometry
N B nlBA nl 0 niA
L


i
i
i
or
L   0 n 2 Al
or
inductance
2
L/l 
 n A
unit length
Depends only on geometry just like C and
is independent of current.
11-11-06
Induction - Fall 2006
30
Max Current Rate of
increase = max emf
VR=iR
~current
11-11-06
Induction - Fall 2006
31
E
(1  eRt / L )
R
L
  (time constant)
R
i
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Induction - Fall 2006
32
IMPORTANT QUESTION





Switch closes.
No emf
Current flows for a
while
It flows through R
Energy is conserved
(i2R)
WHERE DOES THE ENERGY COME FROM??
11-11-06
Induction - Fall 2006
33
For an answer
Return to the Big C

E=e0A/d


+dq
+q
11-11-06
-q

We move a charge dq
from the (-) plate to the
(+) one.
The (-) plate becomes
more (-)
The (+) plate becomes
more (+).
dW=Fd=dq x E x d
Induction - Fall 2006
34
The calc

q
dW  (dq ) Ed  (dq ) d  (dq )
d
e0
e0 A
d
d q2
W
qdq 

e0 A
e0 A 2
or
2
2


1

Ad
1

1
2
2


W
(A) 
 e 0  2  Ad  e 0 E Ad
2e 0 A
2 e0
2  e0 
2
d
energy
1
2
u
 e0 E
unit volum e 2
11-11-06
Induction - Fall 2006
The energy is in
the FIELD !!!
35
What about POWER??
di
E  L  iR
dt
i :
di 2
iE  Li  i R
dt
power
to
circuit
power
dissipated
by resistor
Must be dWL/dt
11-11-06
Induction - Fall 2006
36
So
dWL
di
 Li
dt
dt
1 2
WL  L  idi  Li
2
1
WC  CV 2
2
11-11-06
Induction - Fall 2006
Energy
stored
in the
Capacitor
37
WHERE is the energy??
l
 B  ds   i
0 enclosed
0l  Bl   0 nil
B   0 ni
or
B
 0 Ni
l
  BA 
11-11-06
Induction - Fall 2006
 0 Ni
l
A
38
Remember the Inductor??
N
L
i
N  Number of turns in inductor
i  current.
Φ  Magnetic flux throu gh one turn.
?????????????
11-11-06
Induction - Fall 2006
39
So …
N
i
N
i
L
1
1 N 1
W  Li 2  i 2
 N i
2
2
i
2
 0 NiA

l
1   0 NiA 
1
2
2 2 A
W  Ni
0 N i

2  l  2 0
l
L
11-11-06
Induction - Fall 2006
40
1
A
W
 N i
2 0
l
2
0
2 2
From before :
 0 Ni
B
l
1 2 2A
1 2
W
Bl

B V (volume)
2 0
l 2 0
or
W
1 2
u

B
V 2 0
11-11-06
ENERGY IN THE
FIELD TOO!
Induction - Fall 2006
41
IMPORTANT CONCLUSION


A region of space that contains either a
magnetic or an electric field contains
electromagnetic energy.
The energy density of either is proportional to
the square of the field strength.
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Induction - Fall 2006
42
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Induction - Fall 2006
43