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Physics 152physics.valpo.edu/courses/p152/lectures/lecture36notes.pdf · Lecture Ch. 33:...
Transcript of Physics 152physics.valpo.edu/courses/p152/lectures/lecture36notes.pdf · Lecture Ch. 33:...
1
Physics 152Friday,
April 27, 2007
• LLenz s Law Practice
• IInductors
http://www.voltnet.com/ladder/
• Help sessions
• W 9 - 10 pm in NSC 119
• MasteringPhysics
• Hwk #5 due Fri., May 4
• Final Exam available Friday,
May 4 from Mrs. Wellsand.
AnnouncementsFriApr
.27.
Phys
152
Hint: Be able to do the homework(graded AND recommended) andyou’ll do fine on the exam!
Available Friday, May 4Due by our scheduled exam period
Monday, May 14 (10:30 am - 12:30 pm)
You may bring one 8.5”X11” index card (hand-written on both sides), a pencil or pen, and ascientific calculator with you.
I will put any constants and mathematicalformulas that you might need on a singlepage attached to the back of the exam.
AnnouncementsFriApr
.27.
Phys
152
Section 1: graphical problemSection 2: conceptual problem
Sections 3 - 6:One problem on Ch 32One problem on Ch 33
Two problems on material fromExams 1 and 2
REQUIRED Multiple Choice Section:10 questions covering all the above
material, 6 on previous material, 4 onChapters 32 - 34
AnnouncementsFriApr
.27.
Phys
152
Worksheet Problem #1
Ch. 33: Electromagnetic InductionLecture
36Phys
152
In which
direction will
the current
flow?
Two ways to answer this question:
1) The right-hand rule for moving charges
in a uniform magnetic field.
2) Lenz’s Law
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
R Lv
x
Ch. 33: Electromagnetic InductionLecture
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152
2
v to the right. B into the board.
Force on the positive charge carriers is up. Sothe current will go around counterclockwise.
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
R Lv
I
In which
direction will
the current
flow?
1) The right-hand rule for moving charges
in a uniform magnetic field.
Ch. 33: Electromagnetic InductionLecture
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As the bar moves to the right, the flux throughthe loop into the board is increasing.
The current in the loop must create fluxout of the board. So the current goescounterclockwise.
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x x
R Lv
I
In which
direction will
the current
flow?
2) Lenz’s Law
Ch. 33: Electromagnetic InductionLecture
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152
Worksheet Problems
#2, #3, and #4
Ch. 33: Electromagnetic InductionLecture
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A circular loop is oriented with its planeperpendicular to a uniform magnetic fieldwith B = 1.5 T. At an instant when the radiusof the loop = 12.0 cm and is increasing at arate of 3.0 cm/s, what is the magnitude of theEMF induced in the loop?
Worksheet
Problem #5B
1. 0
2. 12 mV
3. 17 mV
4. 25 mV
5. 34 mV
Ch. 33: Electromagnetic InductionLecture
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152
What happenswhen the switchis first closed?
Initially, there is no current in the circuit.
We know the final current in thiscircuit (from Ohm’s Law) will be I = V / R
So, for some period of time, the current ischanging from 0 to V / R.
V R
+
_
Ch. 33: Electromagnetic InductionLecture
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As a result of the changing current, the magneticfield that the current creates changes too.
That magnetic field passes through the planeof this circuit.
There is therefore a changingmagnetic flux through the circuit.
V R
+
_x x x x x x
x x x x x x
What happenswhen the switchis first closed?
Ch. 33: Electromagnetic InductionLecture
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3
The magnetic flux is changingas the current increases.
An EMF is induced in the circuit!
By Lenz’s Law, the induced EMF results in a cur-rent that opposes the changes in magnetic flux.
What happenswhen the switchis first closed?
V R
+
_x x x x x x
x x x x x x
Ch. 33: Electromagnetic InductionLecture
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The very fact that the circuit is a closed
loop made of conducting material has
resulted in an induced EMF in the circuit
which opposes the current!
Of course, once the current attains its final
value, the induced EMF = 0 because the
magnetic field is no longer changing.
V R
+
_x x x x x x
x x x x x x
Ch. 33: Electromagnetic InductionLecture
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152
Unless otherwise stated, we generally ignore
the self-inductance of the circuit itself.
However, you might observe its effect on thecircuits you construct in the laboratory.
When you put a potential differenceacross the circuit, the currentshould ramp up to its final value...
V R
+
_x x x x x x
x x x x x x
Ch. 33: Electromagnetic InductionLecture
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152
We will pay attention to the self-inductance of asolenoid, which as a circuit element is called an
To power supply To power supply
INDUCTOR
Ch. 33: Electromagnetic InductionLecture
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A bunch of current loopsconnected together!
So a magnetic flux exists through the inductor.
As was the case with the electrical circuit of the
last example, an induced EMF will be generated
across the inductor when the current through
the inductor changes.
What is aninductor (solenoid)?
Ch. 33: Electromagnetic InductionLecture
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Again, by Lenz’s Law, the
induced EMF will generate
a current to oppose the
changing magnetic flux
through the inductor.
= N B A = Nμ0
N
LIA = μ
0
N2
LAI
What is the magnetic flux through aninductor of length L with N turns anda current I flowing through it?
Ch. 33: Electromagnetic InductionLecture
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4
What is the inducedEMF through such acoil?
=d
dt= μ
0
N2
LA
dI
dt
= LdI
dt
where L is theinductance andis defined to be
L = μ0
2N
AL
Ch. 33: Electromagnetic InductionLecture
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We can derive another definition by combining
AND
That is, equate these two statements of therelationship of EMF to changing fluxes andchanging currents.
TRUE ONLY FORA SOLENOID!!!!
= LdI
dt= N
d
dt
L = μ0
2N
AL
Ch. 33: Electromagnetic InductionLecture
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152
L NI
=
[L] = [N][ ]
[I]= turns
Wb
A
[L] =
Tm2
A=
N
Amm
2
A
[L] =
Nm
A2=
Js
AC=
Vs
A= H
General expression forself-inductance
HENRY
Ch. 33: Electromagnetic InductionLecture
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Worksheet Problem #6
Two solenoids have the same cross-
sectional area. Solenoid B, however, is
twice as long and has twice the number
of turns as Solenoid A. The ratio of the
self-inductance of Solenoid B to that of
Solenoid A is
1. 1 / 4
2. 1 / 2
3. 1 / 1
4. 2 / 1
5. 4 / 1
Ch. 33: Electromagnetic InductionLecture
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Inductors set up a back-EMF in a circuit asthe current in the circuit changes. Themagnitude of that EMF is given by
As a circuit element, therefore, inductorsaffect the rate at which the current in acircuit changes.
In some respects, their effect on a circuitis analogous to the role of a capacitor.
= LdI
dt
Ch. 33: Electromagnetic InductionLecture
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Recall, capacitors are circuit elements
which store energy in an electric field
between a positively charged plate and
a negatively charged plate.
Inductors store energy in a magnetic field.
The amount of energy stored
in the magnetic field of an
inductor is given byU I=
1
2
2L
Ch. 33: Electromagnetic InductionLecture
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5
Plugging in the voltage
across an inductor
The power consumed by a
circuit element is given by
P = I LdI
dt
P = I V
dU
dt= I L
dI
dt
Power is energy per
time, so we have
How do we go about deriving this?
Ch. 33: Electromagnetic InductionLecture
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Now, integrating both sides
dU = L I dI
How do we go about deriving this?
dU
dt= I L
dI
dt
U I=1
2
2L
Ch. 33: Electromagnetic InductionLecture
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V R
+
_
L
Let’s set up Kirchhoff’s loop equation:
V + VL + V
R = 0
V LdI
dtIR = 0
I tV
Ret L R( ) ( )/( / )
= 1
This differential equationhas a known solutiongiven by
Ch. 33: Electromagnetic InductionLecture
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152
I
t
I=V/R
I=0.632V/R
I t I et( ) ( )max
/= 1
I V Rmax
/=
= the time constant = L / R
Ch. 33: Electromagnetic InductionLecture
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V R
+
_
L
We can use the loop rule and Ohm’s law to
determine the potential difference across the
inductor:
V = VL + V
R
V
R(t) = I R = I
max(1 e
t / ) R =V
R(1 e
t / ) R
Ch. 33: Electromagnetic InductionLecture
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V R
+
_
L
V t V eR
t( ) ( )/= 1
V t V V e Vet t
L( ) ( )/ /
= =1
Induced Back - EMF in the Inductor
Ch. 33: Electromagnetic InductionLecture
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V R
+
_
L
6
Worksheet Problem #7
Ch. 33: Electromagnetic InductionLecture
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152
V R
+
_
L
In the circuit below, L = 7.00 H, R = 9.00 ,and V = 120 V. What is the self-inducedemf in the inductor 0.200 s after theswitch is closed?
Worksheet Problem #8
Ch. 33: Electromagnetic InductionLecture
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If we now open switch S1 and simultaneouslyclose switch S2, we once again have a changingcurrent in our circuit, inducing an EMF acrossour inductor.
VL + V
R = 0
LdI
dtIR = 0 I t I e
t L R( ) max
/( / )=
This time, the loop rule says:
V R
+
_
L
S2
S1
withsolution
Ch. 33: Electromagnetic InductionLecture
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I
t
I=V/R
I=0.368V/RI t I e
t( ) max
/=
I V Rmax
/=
= the time constant = L / R
V R
+
_
L
S2
S1
If we allowed ourprevious circuit toreach its stable state…
Ch. 33: Electromagnetic InductionLecture
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= =V V VeL R
t /
Using Ohm’s Lawand the loop rule
we know
VR = I R
VL + V
R = 0
V R
+
_
L
S2
S1
Ch. 33: Electromagnetic InductionLecture
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Worksheet Problem #9
Ch. 33: Electromagnetic InductionLecture
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