Mutual Inductance - Physics & Astronomy
Transcript of Mutual Inductance - Physics & Astronomy
24P21-
Mutual Inductance
1 12 12 2
1 1212
2
N M INM
I
Φ ≡Φ
→ =
212 12
dIdt
Mε ≡ −
12 21M M M= =
A current I2 in coil 2, induces some magnetic flux Φ12 in coil 1. We define the flux in terms of a “mutual inductance” M12:
7P24-
Self InductanceWhat if we forget about coil 2 and ask about putting current into coil 1?There is “self flux”:
1 11 11 1 N M I LINLI
Φ ≡ ≡Φ
→ =
dILdt
ε ≡ −
Figure 30-3
8P24-
Calculating Self Inductance
NLIΦ
= V s1 H = 1 A⋅
Unit: Henry
1. Assume a current I is flowing in your device2. Calculate the B field due to that I3. Calculate the flux due to that B field4. Calculate the self inductance (divide out I)
Figure 30-4
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Group Problem: Solenoid
Calculate the self-inductance L of a solenoid (n turns per meter, length , radius R)
REMEMBER1. Assume a current I is flowing in your device2. Calculate the B field due to that I3. Calculate the flux due to that B field4. Calculate the self inductance (divide out I)
L N I= Φ
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Inductor Behavior
IdILdt
ε = −
L
Inductor with constant current does nothing
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dILdt
ε = −Back EMF
I
0, 0LdIdt
ε> <
I
0, 0LdIdt
ε< >
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Inductors in CircuitsInductor: Circuit element which exhibits self-inductance
Symbol:
When traveling in direction of current:
dILdt
ε = −Inductors hate change, like steady stateThey are the opposite of capacitors!
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LR Circuit
0ii
dIV Ldt
IRε= =− −∑
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LR Circuit
0 dI L dIL Idt R dt R
IR εε ⎛ ⎞= ⇒ = − −⎜ ⎟⎝ ⎠
− −
Solution to this equation when switch is closed at t = 0:
( )/( ) 1 tI t eR
τε −= −
: LR time constantLR
τ =
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LR Circuit
t=0+: Current is trying to change. Inductor works as hard as it needs to to stop it
t=∞: Current is steady. Inductor does nothing.
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LR CircuitReadings on Voltmeter
Inductor (a to b)Resistor (c to a)
c
t=0+: Current is trying to change. Inductor works as hard as it needs to to stop it
t=∞: Current is steady. Inductor does nothing.
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General Comment: LR/RCAll Quantities Either:
( )/FinalValue( ) Value 1 tt e τ−= − /
0Value( ) Value tt e τ−=
τ can be obtained from differential equation (prefactor on d/dt) e.g. τ = L/R or τ = RC
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Group Problem: LR Circuit
1. What direction does the current flow just after turning off the battery (at t=0+)? At t=∞?
2. Write a differential equation for the circuit3. Solve and plot I vs. t and voltmeters vs. t
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Non-Conservative Fields
R=100ΩR=10Ω
Bddd tΦ
⋅ = −∫ E s
I=1A
E is no longer a conservative field –Potential now meaningless
Figure 30-6
Figure 30-8
Figure 30-9
Figure 30-16
Figure 30-19
Figure 30-9
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LR Circuit
t=0+: Current is trying to change. Inductor works as hard as it needs to to stop it
t=∞: Current is steady. Inductor does nothing.
( )/( ) 1 tI t eR
τε −= −
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LR Circuit: AC Output Voltage
0.00 0.01 0.02 0.03 0.00
0.05
0.10
0.15
0
1
2
3
)
)
Current -3 -2 -1 0 1 2 3
Vol
l
I (A
Time (s
Indu
ctor
(V) tmeter across L
Out
put (
V) Output Vo tage
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Mass on a Spring
2
2
2
2 0
d xF kx ma mdt
d xm kxdt
= − = =
+ =
0 0( ) cos( )x t x tω φ= +
(1) (2)
(3) (4)
What is Motion?
0 Angular frequencykm
ω = =
Simple Harmonic Motion
x0: Amplitude of Motionφ: Phase (time offset)
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Mass on a Spring: Energy
0 0( ) cos( )x t x tω φ= +
(1) Spring (2) Mass (3) Spring (4) Mass
Energy has 2 parts: (Mass) Kinetic and (Spring) Potential
22 2
0 0
2 2 20 0
1 1 sin ( )2 21 1 cos ( )2 2s
dxK m kx tdt
U kx kx t
ω φ
ω φ
⎛ ⎞= = +⎜ ⎟⎝ ⎠
= = +
0 0 0'( ) sin( )x t x tω ω φ= − +
Energy sloshes back
and forth
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Simple Harmonic Motion
Amplitude (x0)
0 0( ) cos( )x t x tω φ= −
1Period ( )frequency ( )
2angular frequency ( )
Tfπ
ω
=
=
Phase Shift ( )2πϕ =
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Analog: LC Circuit
Mass doesn’t like to accelerateKinetic energy associated with motion
Inductor doesn’t like to have current changeEnergy associated with current
22
2
1;2
dv d xF ma m m E mvdt dt
= = = =
22
2
1;2
dI d qL L E LIdt dt
ε = − = − =
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Analog: LC Circuit
Spring doesn’t like to be compressed/extendedPotential energy associated with compression
Capacitor doesn’t like to be charged (+ or -)Energy associated with stored charge
21;2
F kx E kx= − =
21 1 1;2
q E qC C
ε = =
1; ; ; ;F x q v I m L k Cε −→ → → → →
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LC Circuit
resistor, and battery. 1. Set up the circuit above with capacitor, inductor,
2. Let the capacitor become fully charged. 3. Throw the switch from a to b 4. What happens?
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LC Circuit It undergoes simple harmonic motion, just like a mass on a spring, with trade-off between charge on capacitor (Spring) and current in inductor (Mass)
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LC Circuit
0 ; Q dI dQL IC dt dt
− = = −
2
2
1 0 d Q Qdt LC
+ =
0 0( ) cos( )Q t Q tω φ= + 01LC
ω =
Q0: Amplitude of Charge Oscillationφ: Phase (time offset)
Simple Harmonic Motion
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LC Oscillations: Energy
222 01
2 2 2E BQQU U U LI
C C= + = + =
2220
0cos2 2E
QQU tC C
ω⎛ ⎞
= = ⎜ ⎟⎝ ⎠
22 2 2 20
0 0 01 1 sin sin2 2 2B
QU LI LI t tC
ω ω⎛ ⎞
= = = ⎜ ⎟⎝ ⎠
Total energy is conserved !!
Notice relative phases
22P25
Damped LC Oscillations
Resistor dissipates energy and system rings down over time
Also, frequency decreases: 2
2 0 '
2 R L
ω ω ⎛ ⎞ = − ⎜ ⎟⎝ ⎠
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This concept (& next 3 slides) are complicated.
Bare with me and try not to get confused
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Kirchhoff’s Modified 2nd RuleB
ii
dV d Nd tΦ
∆ = − ⋅ = +∑ ∫ E s
0Bi
i
dV Nd tΦ
⇒ ∆ − =∑If all inductance is ‘localized’ in inductors then our problems go away – we just have:
0ii
d IV Ld t
∆ − =∑
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Ideal Inductor• BUT, EMF generated
in an inductor is not a voltage drop across the inductor!
d ILd t
ε = −
i n d u c t o r 0V d∆ ≡ − ⋅ =∫E sBecause resistance is 0, E must be 0!
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Energy Stored in InductordIIR Ldt
ε = + +
2 dII I R L Idt
ε = +
( )2 212
dI I R L Idt
ε = +Battery
SuppliesResistor
DissipatesInductorStores
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Energy Stored in Inductor
212LU L I=
But where is energy stored?
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Example: Solenoid Ideal solenoid, length l, radius R, n turns/length, current I:
0B nIµ= 2 2oL n R lµ π=
( )2 2 2 21 12 2B oU LI n R l Iµ π= =
22
2Bo
BU R lπµ
⎛ ⎞= ⎜ ⎟⎝ ⎠
EnergyDensity
Volume
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Energy Density
Energy is stored in the magnetic field!2
2Bo
Buµ
= : Magnetic Energy Density
2
2o
EEu ε
= : Electric Energy Density
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Group Problem: Coaxial Cable
XI I Inner wire: r=aOuter wire: r=b
1. How much energy is stored per unit length? 2. What is inductance per unit length?
HINTS: This does require an integralThe EASIEST way to do (2) is to use (1)
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TransformerStep-up transformer
Bs s
dNdt
ε Φ=
Bp p
dNdt
ε Φ=
s s
p p
NN
εε =
Ns > Np: step-up transformerNs < Np: step-down transformer
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Transmission of Electric Power
Power loss can be greatly reduced if transmitted at high voltage
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Example: Transmission linesAn average of 120 kW of electric power is sent from a power plant. The transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the power is sent at (a) 240 V, and (b) 24,000 V.
5
2
1.2 10 5002.4 10
P WI AV V
×= = =
×(a)
83% loss!!2 2(500 ) (0.40 ) 100LP I R A kW= = Ω =
5
4
1.2 10 5.02.4 10
P WI AV V
×= = =
×(b)
0.0083% loss2 2(5.0 ) (0.40 ) 10LP I R A W= = Ω =
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Recall Polar Dielectrics
Dielectric polarization decreases Electric Field!
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Para/Ferromagnetism
Applied external field B0 tends to align the atomic magnetic moments
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Para/Ferromagnetism
The aligned moments tend to increase the B field
0mκ=B BEκ
= 0EECompare to:
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Para/Ferromagnetism
Paramagnet: Turn off B0, everything disordersFerromagnet: Turn off B0, remains (partially) ordered
This is why some items you can pick up with a magnet even though they don’t pick up other items
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Magnetization Vector
M=0 M>0Useful to define “Magnetization” of material:
1
1 N
iiV V=
= =∑ µM µ 0 0µ= +B B M
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Hysteresis in FerromagnetsThe magnetization M of a ferromagnetic material depends on the history of the substance
Magnetization remains even with B0 off !!!