PHY 114 A General Physics II 11 AM-12:15 PM TR Olin 101...
Transcript of PHY 114 A General Physics II 11 AM-12:15 PM TR Olin 101...
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 1
PHY 114 A General Physics II 11 AM-12:15 PM TR Olin 101
Plan for Lecture 14 (Chapter 32):
Inductance
1.Inductors as a circuit element
2.RL, LC, and RLC circuits
3.Energy stored in an inductor
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 2
Remember to send in your chapter reading questions…
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 3
Web assign question -- hint
1000 turns/m
dtdN B
cΦ
−=E
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 4
Web assign question -- hint
A ( ) ( )tRBRBB ωπθπ cos
2cos
2
22
==⋅=Φ ABθ = ωt
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 5
Electric generator:
θ = ωt
( )( )
( )tNBAdt
tBAdN
ωω
ω
sin
cos
=
−=E
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 6
( )( )
( )
( ) ( )tItR
NBAR
I
RtNBA
dttBAdN
ωωω
ωω
ω
sinsin
:bygiven is coil in thecurrent , is coilgenerator in the resistance theIf
sin
cos
max≡==
=
−=
E
E
Electric generator:
I t
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 7
JB
sB
0
0
:form alDifferenti
:form Integral
:law sAmpere'
µ
µ
=×∇
=⋅∫ inId
Summary:
( )
0
0
:form alDifferenti
:form Integral
:law sGauss'
ερ
ε
=⋅∇
=⋅∫
E
ArE inQd ( )0
0
=⋅∇
=⋅∫B
ArB d
( ) ( )
t
dAr dtddE
∂∂
−=×∇
⋅−=⋅ ∫∫BE
Bsr
:law sFaraday'
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 8
Faraday’s law for EMF induced in a current loop:
dtd BΦ
−=E: of Examples
dtd BΦ
dtdIN
dtdB
I
NIB
0
0
: with timechanging is If
:solenoid inside Field
µ
µ
=
=
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 9
“Mutual” inductance
=Nc
turns/m1000=
sNA
dtdIANN
dtdN s
cB
cc
0µ−=
Φ−=E
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 10
“Self” inductance
dtdIAN
dtdIANN
dtdN ss
sB
ss
200 µµ
−=−=Φ
−=E
L inductance
Amp/s 1Volt 1 Henry 1
:inductance of Units
=
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 11
Inductors in a circuit Example: LR circuit
dtdIL−=E
0
:closedswitch With
=−−dtdILRIEMFE
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 12
Example: LR circuit
0
:closedswitch With
=−−dtdILRIEMFE
( )
( )
−=
−=
−−
=−
−=−=
−− 01)(
ln
:equation aldifferenti Solve
0
ttLR
EMF
EMF
EMF
EMFEMF
eR
tI
ttLRI
R
dtLR
IR
dI
IRL
RILR
LdtdI
E
E
E
EE
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 13
Example: LR circuit
t
I
( )
( )( )
Volts/Amp/s)Volts/(Amp
/ 1
1)(
:closedswitch With
/0
0
=
=−=
−=
−−
−−
τ
ττ RLeR
eR
tI
ttEMF
ttLR
EMF
E
E
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 14
The LR circuit reminds us of:
A. Why physics class is so beautiful. B. Why physics class is so terrible. C. RC circuit. D. Money in the bank.
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 15
Energy storage in inductor:
dtdIL−=E
2
2
2
21
21
21
LIU
LIddU
dtLIdtddtI
dtdILdt
dtdQVVdQdU
=⇒
=⇒
====
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 16
Example: LR circuit continued
( ) ( )11
1)(
0
:placed wireandopen switch With
ttLR
EMFttLR
eR
eItI
dtLR
IdI
dtdILRI
−−−−==
−=
=−−
E
I
t
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 17
LC circuit This circuit has the following time dependence after the switch is closed: A. Charge decays with time B. Charge increases with time C. Charge remains constant
with time D. Charge increases and
decreases with time
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 18
LC circuit
01
0
0
:closedswitch With
2
2
2
2
=+
=−−
=−−
QLCdt
Qddt
QdLCQ
dtdIL
CQ
( )
LC
tQtQQtQ
1 where
cos)(:)0( assumingSolution
max
max
≡
===
ω
ω
s
LC1
VoltsCoulombs
Amp/sVolts
1
FaradHenry11
:unitsCheck
=⋅
=
⋅=≡ω
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 19
LC circuit For L=1 H, C=1 F; ω=1 rad/s
Q/Qmax
t
ω=2πf
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 20
The LC circuit is mathematically analogous to: A. Nothing seen previously seen in class B. An elastic bouncing ball C. A swinging pendulum D. A mass attached to a spring E. All above, except A
The LC circuit is useful:
A. Because of its mathematical analogies B. Physicists like simple formulae like cos(ωt) C. It could be useful in science experiments D. It could be useful for toys E. It could be useful for everyday appliances
3/20/2012 PHY 114 A Spring 2012 -- Lecture 14 21
LRC circuit:
01
0
2
2
=−++
=−−−
E
E
QCdt
dQRdt
QdL
RIdtdIL
CQ
( )2
2/
21'
)'cos(1)(:0)0( assumingSolution
−=
−=
==−
LR
LC
teCtQtQ
LRt
ω
ωE
Q/CE
t