Intro Physics II
Administrivia
Intro Physics II Physics 11b
! !
HW #8 due Friday at 4 PM HW #9 out today
Lecture 17 Applications of Induction Inductance RL Circuits
Summary !
Ampere’s Law !
!
Restatement of Biot-Savart
Magnetic Flux Redux ! !
!
What We Will Learn Today
Magnetic Induction !
!
ΦB =
=0 on closed surface ≠0 on open surface
ΦB
∫
∫∫
r r B d l = µ0 I
r r B ⋅ da = 0
Closed Surface
∫∫
r B nˆ da
Changing magnetic field → electric field ! Field is nonconservative ! No longer the gradient of a scalar potential Faraday’s Law
E =
r
r
∫ E ⋅ dl = −
Closed Path
∫∫
r r B da
!
Applications of Induction
!
Inductance
!
R-L Circuits
dΦ B dt
1
DC Magnetic Motor
DC Magnetic Motor
Magnetic motor consists of three major parts
!
! ! !
!
!
! !
At top, brushes disconnect Angular momentum carries loop through small angle
!Now opposite commutator brushes make contact
Loops of area A starts || to Bfield
!
!
Rotates under torque !
Source of voltage Region of large B-field Rotating current-carrying coil ! Commutator brushes
Current I creates dipole ⊥ to B-field ! Into page, in this picture Torque on dipole loop: BIA Begins to rotate under torque
!
!
Induction Generator Construction is inverse of motor !You force the coil to rotate inside the field by providing mechanical source of power !
!
!
Current reverses direction Dipole moment switches direction
Induction Generator !
For constant rate of rotation ! !
!
As loop turns !
Usually, it is steam turning a turbine
!
As loop turns ! ! !
Dot product of B.da changes Flux changes Voltage induced around coil ! Current flow in coil
!
θ=ωt B.da=BAcosθ=BAcosωt F=NBacoswt E=-(-NBAωsinωt)= NBAωsinωt
AC electrical voltage from ! ! !
A coil of wire Permanent magnets A steam turbine
2
AC Transformer
Quiz: AC Transformer
Take output of rotating coil Vin Pass through an iron-core solenoid (windings / length = N1/L=n1) !Wrap another set of N2 windings around the solenoid (second solenoid) ! !
Vin (from generator)
!
!
Field in first solenoid ! B=µn I=µn I sinwt 1 1 0 Flux through second solenoid ! Φ=N BA=µn N AI sinwt 2 1 2 0 ! dΦ/dt=µn N A dI/dt 1 2 ! !
!
!Suppose the secondary winding was not directly over the solenoid
!Would there still be an output voltage?
E=µn1N2 dI/dt Output voltage of second set of windings Not necessarily the same as input voltage!
Quiz: AC Transformer Vin (from generator)
Vin (from generator)
Suppose the secondary winding was not directly over the solenoid !
Mutual Inductance Given two objects As current flow in one object changes, voltage in other is induced
! !
!
!
Would there still be an output voltage? !
!
Yes ! Some small fringe fields outside would still be present and would alternate flux
Some constant of proportionality (calculable in many cases) called M12 V2=-M21dI1/dt
Example: two solenoidal coils (AC transformer)
!
! !
!
It works the other way, too ! ! !
!
M21=µn1N2A General case ! M =N Φ/I 21 2 1
Changes in I2 induce voltage in V1 M21=M12: “Mutual inductance”=M E=-MdI/dt
Units of M: henry !
1 V / (A/s) = 1 s V/A = 1 Ωs
3
Step-Up Transformers and Electrical Distribution A city consumes all of its electrical power at 120 V !Total power consumption is P !
!
Total current at 120 V is P/V
Self-Inductance The phenomenon even happens within one object! !Iron-core solenoidal coil !
! !
As a power company, you would like to supply sufficient current for the city’s power
!
!
!
You’d like to lose as little on the way to the city, in the power lines, as possible
!
!
Self-inductance !
If you supply at a voltage V’ ! !
!
I’/I=V/V’ > I’=P/V’ Power loss in transmission lines: I2R=I’2R=P2R/V’2 ! Want V’ as large as possible Must step down V’ > 120 V with transformer
!
A circuit element with large L is known as an inductor
Inductor in DC Circuit
!
Kirchoff’s Current Law
!
Kirchoff’s Voltage Law
IL=IR=I
!
Vbatt-VL-VR=0 V-LdI/dt-IR=0 V-IR=LdI/dt dt=LdI/(V-IR)
! ! ! !
V
R
I
∫
1
L
V0
dt = IR ∫0 L I
ln(V − IR )0 =
−
1
I
t
ln (V − IR )0 =
Exponentiate
V
V − IR ln
R
I
V − IR
t
dI
I 0
−
Mself=NΦ/I ! Φ=µnAI ! n=N/L 2 ! M self=L=µN A/L E=-LdI/dt
!
Inductance in DC Circuit
L
Throw switch Large dI/dt, large flux change in solenoid’s interior Large “back-EMF” induced in solenoid ! Lenz’s Law
V/R
1−
R
=−
=e
I =e
R
R − t L
R
t L/R
t
I=
t
R − t L
− L V 1 − e R
t
4
Charge Buildup in Capacitor (Lecture 13) I = I 0e
−t RC
=V
R
e
−t τ
! !
τ=RC dQ/dt=I
Q = ∫ Idt
C
Power in resistor
I R
∞
0
V/R
RI
2R t 2 − L 0
Total heat energy dissipated
∫
2 0
dt RI e
−2 R Lt
L = RI 2R
1 2 U = LI 0
2 0
Where did the energy come from? !
L/2R
t
t
L/R
I = eC e
−R t L
= I 0e
−R t L
B-field Energy 2
R
dI − IR = 0 dt
dI = −I R L dt dI R =− dt L I ln I = − R t + C L
V/R
P
P
R
t
L
Initial current I0 Kirchoff Voltage
−L
I
Dissipated Power
!
!
VR
τV τV − t / τ τV (1 − e − t / τ ) = CV (1 − e −t / τ ) e − = R R R
!
L
CV
Q=−
Q=
!
Q
τV − t / τ e +C R Initial τV (1) + C condition 0 = Q ( 0 ) = − Q=0 R
Discharging Inductor
!Resistor can’t store energy !Must be stored in inductor somewhere !With I0, there is magnetic field inside inductor !
!
!
2
dU = !
B=µnI0=µNI0/L ! I =BL/µN 0 L=µN2A/L 2
2
Energy density 1B 2
For arbitrary B-fields 2
U=
∫
d x
1 B( x ) 2 µ
2
2
1 2 1 µN A BL 1 B LI = AL = 2 2 L N 2
3 S
U=
1B V 2
5
Inductors and Capacitors
V I = 1 − e R
t
!
V − I = e RC R
Capacitors ! !
!
Initially let current flow Eventually block current
!
! !
!
E-field buildup Energy stored in E-field !
!
Inductors
−t /τ ) Charge buildup Q = CV (1 − e !
!
!
U=1/2 CV2
Upon discharge (RC circuit) ! Initial current decays exponentially as stored energy dissipated as heat through resistor Characteristic time: RC
!
LC Circuit !
Initially block current Eventually let flow freely Current buildup ! B-field buildup ! Energy stored in B-field !
!
R − t L
Characteristic time: L/R
!
No battery No initial current
Kirchoff current rule
!
Kirchoff voltage rule
!
! ! !
IC=IL
Q/C-LdI/dt=0 Take one derivative; dQ/dt=-I -I/C-Ld2I/dt2=0
!
Same as harmonic motion!
!
Q is integral of I
ω0 = 1
!
!
! ! !
Q0=ImaxSqrt(LC) Imax=Q0/Sqrt(LC)
Q=
I max
1 I = I max sin t −ϕ LC ω
cos(ω 0 t − ϕ ) = Q0 cos(ω 0 t − ϕ )
Current is flowing without battery! Current is sinusoidal in time Voltage across the circuit elements sinusoidal in time ! But out of phase (sin vs. cos)
Energy in capacitor !
LC
Note
LC
+Q -Q
Energy and LC Circuit
Q = Q0 cos(ωt − ϕ ) I = Q0 sin(ωt − ϕ ) !
C
!
!
The Solution
L
Let capacitor start with charge Q0 !
U=1/2 LI2
Upon discharge (RL circuit) ! Initial current decays exponentially as stored energy dissipated as heat through resistor
I
!
Energy in inductor !
!
½ C V2 = ½ Q2/C 2 2 ! ½ Q 0 /C cos ωt
½ LI2 2 2 2 2 2 ! ½ LI = ½ L (Q 0 /LC) sin ωt= ½ (Q0 /C) sin ωt
Sum: Constant ½ Q2/C (=initial energy) ! !
There’s no resistor to dissipate power Power goes from being stored in capacitor’s E-field to inductor’s B-field
6
Time Behavior EC
Summary !
EL
Applications of Induction ! !
VC(t)
!
Mutual Inductance !
t
!
Microphones, Generators Transformers
Current through one conductor induces current through another ! M depends on geometry and NΦ M= materials
Self Induction and Inductance
!
Q(t)
I(t) !
Energy in B-field
dU =
dI 2 dt
I
“Back-EMF” induced in conductor itself R ! Called L, measured in “Henrys” − t V I = 1 − e L Inductor Circuit R ! Opposes initial energy flow 1 B2 !
E 1 = −M
E = −L
dI dt
2 µ
7
Intro Physics II Physics 11b
! !
HW #8 due Friday at 4 PM HW #9 out today
Lecture 17 Applications of Induction Inductance RL Circuits
Summary !
Ampere’s Law !
!
Restatement of Biot-Savart
Magnetic Flux Redux ! !
!
What We Will Learn Today
Magnetic Induction !
!
ΦB =
=0 on closed surface ≠0 on open surface
ΦB
∫
∫∫
r r B d l = µ0 I
r r B ⋅ da = 0
Closed Surface
∫∫
r B nˆ da
Changing magnetic field → electric field ! Field is nonconservative ! No longer the gradient of a scalar potential Faraday’s Law
E =
r
r
∫ E ⋅ dl = −
Closed Path
∫∫
r r B da
!
Applications of Induction
!
Inductance
!
R-L Circuits
dΦ B dt
1
DC Magnetic Motor
DC Magnetic Motor
Magnetic motor consists of three major parts
!
! ! !
!
!
! !
At top, brushes disconnect Angular momentum carries loop through small angle
!Now opposite commutator brushes make contact
Loops of area A starts || to Bfield
!
!
Rotates under torque !
Source of voltage Region of large B-field Rotating current-carrying coil ! Commutator brushes
Current I creates dipole ⊥ to B-field ! Into page, in this picture Torque on dipole loop: BIA Begins to rotate under torque
!
!
Induction Generator Construction is inverse of motor !You force the coil to rotate inside the field by providing mechanical source of power !
!
!
Current reverses direction Dipole moment switches direction
Induction Generator !
For constant rate of rotation ! !
!
As loop turns !
Usually, it is steam turning a turbine
!
As loop turns ! ! !
Dot product of B.da changes Flux changes Voltage induced around coil ! Current flow in coil
!
θ=ωt B.da=BAcosθ=BAcosωt F=NBacoswt E=-(-NBAωsinωt)= NBAωsinωt
AC electrical voltage from ! ! !
A coil of wire Permanent magnets A steam turbine
2
AC Transformer
Quiz: AC Transformer
Take output of rotating coil Vin Pass through an iron-core solenoid (windings / length = N1/L=n1) !Wrap another set of N2 windings around the solenoid (second solenoid) ! !
Vin (from generator)
!
!
Field in first solenoid ! B=µn I=µn I sinwt 1 1 0 Flux through second solenoid ! Φ=N BA=µn N AI sinwt 2 1 2 0 ! dΦ/dt=µn N A dI/dt 1 2 ! !
!
!Suppose the secondary winding was not directly over the solenoid
!Would there still be an output voltage?
E=µn1N2 dI/dt Output voltage of second set of windings Not necessarily the same as input voltage!
Quiz: AC Transformer Vin (from generator)
Vin (from generator)
Suppose the secondary winding was not directly over the solenoid !
Mutual Inductance Given two objects As current flow in one object changes, voltage in other is induced
! !
!
!
Would there still be an output voltage? !
!
Yes ! Some small fringe fields outside would still be present and would alternate flux
Some constant of proportionality (calculable in many cases) called M12 V2=-M21dI1/dt
Example: two solenoidal coils (AC transformer)
!
! !
!
It works the other way, too ! ! !
!
M21=µn1N2A General case ! M =N Φ/I 21 2 1
Changes in I2 induce voltage in V1 M21=M12: “Mutual inductance”=M E=-MdI/dt
Units of M: henry !
1 V / (A/s) = 1 s V/A = 1 Ωs
3
Step-Up Transformers and Electrical Distribution A city consumes all of its electrical power at 120 V !Total power consumption is P !
!
Total current at 120 V is P/V
Self-Inductance The phenomenon even happens within one object! !Iron-core solenoidal coil !
! !
As a power company, you would like to supply sufficient current for the city’s power
!
!
!
You’d like to lose as little on the way to the city, in the power lines, as possible
!
!
Self-inductance !
If you supply at a voltage V’ ! !
!
I’/I=V/V’ > I’=P/V’ Power loss in transmission lines: I2R=I’2R=P2R/V’2 ! Want V’ as large as possible Must step down V’ > 120 V with transformer
!
A circuit element with large L is known as an inductor
Inductor in DC Circuit
!
Kirchoff’s Current Law
!
Kirchoff’s Voltage Law
IL=IR=I
!
Vbatt-VL-VR=0 V-LdI/dt-IR=0 V-IR=LdI/dt dt=LdI/(V-IR)
! ! ! !
V
R
I
∫
1
L
V0
dt = IR ∫0 L I
ln(V − IR )0 =
−
1
I
t
ln (V − IR )0 =
Exponentiate
V
V − IR ln
R
I
V − IR
t
dI
I 0
−
Mself=NΦ/I ! Φ=µnAI ! n=N/L 2 ! M self=L=µN A/L E=-LdI/dt
!
Inductance in DC Circuit
L
Throw switch Large dI/dt, large flux change in solenoid’s interior Large “back-EMF” induced in solenoid ! Lenz’s Law
V/R
1−
R
=−
=e
I =e
R
R − t L
R
t L/R
t
I=
t
R − t L
− L V 1 − e R
t
4
Charge Buildup in Capacitor (Lecture 13) I = I 0e
−t RC
=V
R
e
−t τ
! !
τ=RC dQ/dt=I
Q = ∫ Idt
C
Power in resistor
I R
∞
0
V/R
RI
2R t 2 − L 0
Total heat energy dissipated
∫
2 0
dt RI e
−2 R Lt
L = RI 2R
1 2 U = LI 0
2 0
Where did the energy come from? !
L/2R
t
t
L/R
I = eC e
−R t L
= I 0e
−R t L
B-field Energy 2
R
dI − IR = 0 dt
dI = −I R L dt dI R =− dt L I ln I = − R t + C L
V/R
P
P
R
t
L
Initial current I0 Kirchoff Voltage
−L
I
Dissipated Power
!
!
VR
τV τV − t / τ τV (1 − e − t / τ ) = CV (1 − e −t / τ ) e − = R R R
!
L
CV
Q=−
Q=
!
Q
τV − t / τ e +C R Initial τV (1) + C condition 0 = Q ( 0 ) = − Q=0 R
Discharging Inductor
!Resistor can’t store energy !Must be stored in inductor somewhere !With I0, there is magnetic field inside inductor !
!
!
2
dU = !
B=µnI0=µNI0/L ! I =BL/µN 0 L=µN2A/L 2
2
Energy density 1B 2
For arbitrary B-fields 2
U=
∫
d x
1 B( x ) 2 µ
2
2
1 2 1 µN A BL 1 B LI = AL = 2 2 L N 2
3 S
U=
1B V 2
5
Inductors and Capacitors
V I = 1 − e R
t
!
V − I = e RC R
Capacitors ! !
!
Initially let current flow Eventually block current
!
! !
!
E-field buildup Energy stored in E-field !
!
Inductors
−t /τ ) Charge buildup Q = CV (1 − e !
!
!
U=1/2 CV2
Upon discharge (RC circuit) ! Initial current decays exponentially as stored energy dissipated as heat through resistor Characteristic time: RC
!
LC Circuit !
Initially block current Eventually let flow freely Current buildup ! B-field buildup ! Energy stored in B-field !
!
R − t L
Characteristic time: L/R
!
No battery No initial current
Kirchoff current rule
!
Kirchoff voltage rule
!
! ! !
IC=IL
Q/C-LdI/dt=0 Take one derivative; dQ/dt=-I -I/C-Ld2I/dt2=0
!
Same as harmonic motion!
!
Q is integral of I
ω0 = 1
!
!
! ! !
Q0=ImaxSqrt(LC) Imax=Q0/Sqrt(LC)
Q=
I max
1 I = I max sin t −ϕ LC ω
cos(ω 0 t − ϕ ) = Q0 cos(ω 0 t − ϕ )
Current is flowing without battery! Current is sinusoidal in time Voltage across the circuit elements sinusoidal in time ! But out of phase (sin vs. cos)
Energy in capacitor !
LC
Note
LC
+Q -Q
Energy and LC Circuit
Q = Q0 cos(ωt − ϕ ) I = Q0 sin(ωt − ϕ ) !
C
!
!
The Solution
L
Let capacitor start with charge Q0 !
U=1/2 LI2
Upon discharge (RL circuit) ! Initial current decays exponentially as stored energy dissipated as heat through resistor
I
!
Energy in inductor !
!
½ C V2 = ½ Q2/C 2 2 ! ½ Q 0 /C cos ωt
½ LI2 2 2 2 2 2 ! ½ LI = ½ L (Q 0 /LC) sin ωt= ½ (Q0 /C) sin ωt
Sum: Constant ½ Q2/C (=initial energy) ! !
There’s no resistor to dissipate power Power goes from being stored in capacitor’s E-field to inductor’s B-field
6
Time Behavior EC
Summary !
EL
Applications of Induction ! !
VC(t)
!
Mutual Inductance !
t
!
Microphones, Generators Transformers
Current through one conductor induces current through another ! M depends on geometry and NΦ M= materials
Self Induction and Inductance
!
Q(t)
I(t) !
Energy in B-field
dU =
dI 2 dt
I
“Back-EMF” induced in conductor itself R ! Called L, measured in “Henrys” − t V I = 1 − e L Inductor Circuit R ! Opposes initial energy flow 1 B2 !
E 1 = −M
E = −L
dI dt
2 µ
7
