Class 24: Outline
Hour 1:
Inductance & LR Circuits
Hour 2:
Energy in Inductors
P24- 1
Last Time:
Faraday’s Law
Mutual Inductance
P24- 2
Faraday’s Law of Induction
ε
dΦB
= −N
dt
Changing magnetic flux induces an EMF
Lenz: Induction opposes change
P24- 3
Mutual Inductance
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:
N1Φ12 ≡ M 12 I 2
M 12 = M 21 = M
N1Φ12
→ M 12 =
I2
ε12 ≡ −M
dI 2
dt
You need AC
currents!
P24- 4
Demonstration:
Remote Speaker
P24- 5
This Time:
Self Inductance
P24- 6
Self Inductance
What if we forget about coil 2 and
ask about putting current into coil 1?
There is “self flux”:
N1Φ11 ≡ M 11 I1 ≡ LI
NΦ
→L=
I
ε
dI
≡ −L
dt
P24- 7
Calculating Self Inductance
NΦ
L=
I
1.
2.
3.
4.
Unit: Henry
V ⋅s
1H=1
A
Assume a current I is flowing in your device
Calculate the B field due to that I
Calculate the flux due to that B field
Calculate the self inductance (divide out I)
P24- 8
Group Problem: Solenoid
Calculate the self-inductance L of a
solenoid (n turns per meter, length A,
radius R)
REMEMBER
1. Assume a current I is flowing in your device
2. Calculate the B field due to that I
3. Calculate the flux due to that B field
4. Calculate the self inductance (divide out I)
L = NΦ I
P24- 9
Inductor Behavior
L
I
ε
dI
= −L
dt
Inductor with constant current does nothing
P24- 10
Back EMF
ε
dI
= −L
dt
I
dI
> 0,
dt
εL < 0
I
dI
< 0,
dt
εL > 0
P24- 11
Inductors in Circuits
Inductor: Circuit element which exhibits self-inductance
Symbol:
When traveling in
direction of current:
ε
dI
= −L
dt
Inductors hate change, like steady state
They are the opposite of capacitors!
P24- 12
PRS Question:
Closing a Switch
P24- 13
LR Circuit
dI
∑i Vi = ε − IR − L dt = 0
P24- 14
LR Circuit
ε
ε
dI
L dI
⎛
⎞
= −⎜ I − ⎟
− IR − L = 0 ⇒
dt
R dt
R⎠
⎝
Solution to this equation when switch is closed at t = 0:
I (t ) =
ε
1− e )
(
R
− t /τ
L
τ = : LR time constant
R
P24- 15
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.
P24- 16
LR Circuit
c
Readings on Voltmeter
Inductor (a to b)
Resistor (c to a)
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.
P24- 17
General Comment: LR/RC
All Quantities Either:
Value(t ) = Value Final (1 − e − t /τ )
Value(t ) = Value0 e − t /τ
τ can be obtained from differential equation
(prefactor on d/dt) e.g. τ = L/R or τ = RC
P24- 18
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 circuit
3. Solve and plot I vs. t and voltmeters vs. t
P24- 19
PRS Questions:
LR Circuit & Problem…
P24- 20
Non-Conservative Fields
R=10Ω
I=1A
R=100Ω
G
G
dΦB
E
⋅
s
=
−
d
∫
dt
E is no longer a conservative field –
Potential now meaningless
P24- 21
This concept
(& next 3 slides)
are complicated.
Bare with me and try not to
get confused
P24- 22
Kirchhoff’s Modified 2nd Rule
G
G
dΦB
∑i ∆ Vi = − v∫ E ⋅ d s = + N d t
dΦB
⇒ ∑ ∆ Vi − N
=0
dt
i
If all inductance is ‘localized’ in inductors then
our problems go away – we just have:
dI
∑i ∆ Vi − L d t = 0
P24- 23
Ideal Inductor
• BUT, EMF generated
in an inductor is not a
voltage drop across
the inductor!
dI
ε = −L
dt
∆ Vi n d u c t o r
G
G
≡ −∫ E ⋅ d s = 0
Because resistance is 0, E must be 0!
P24- 24
Conclusion:
Be mindful of physics
Don’t think too hard doing it
P24- 25
Demos:
Breaking circuits with
inductors
P24- 26
Internal Combustion Engine
See figure 1:
http://auto.howstuffworks.com/engine3.htm
P24- 27
Ignition System
The Distributor:
http://auto.howstuffworks.com/ignition-system4.htm
(A) High Voltage Lead
(B) Cap/Rotor Contact
(C) Distributor Cap
(D) To Spark Plug
(A) Coil connection
(B) Breaker Points
(D) Cam Follower
(E) Distributor Cam
P24- 28
Modern Ignition
See figure:
http://auto.howstuffworks.com/ignition-system.htm
P24- 29
Energy in Inductor
P24- 30
Energy Stored in Inductor
dI
= + IR + L
dt
dI
2
Iε = I R + L I
dt
ε
d
Iε = I R +
dt
Battery
2
Supplies
Resistor
Dissipates
(
1
2
LI
2
)
Inductor
Stores
P24- 31
Energy Stored in Inductor
UL = L I
1
2
2
But where is energy stored?
P24- 32
Example: Solenoid
Ideal solenoid, length l, radius R, n turns/length, current I:
B = µ 0 nI
L = µo n π R l
2
U B = LI =
1
2
2
1
2
(µ n
o
2
π R l) I
2
2
2
⎛ B ⎞ 2
UB = ⎜
⎟π R l
⎝ 2µo ⎠
2
Energy
Density
Volume
P24- 33
Energy Density
Energy is stored in the magnetic field!
2
B
uB =
2µo
uE =
εo E
: Magnetic Energy Density
2
: Electric Energy Density
2
P24- 34
Group Problem: Coaxial Cable
I
X
I
Inner wire: r=a
Outer wire: r=b
1. How much energy is stored per unit length?
2. What is inductance per unit length?
HINTS: This does require an integral
The EASIEST way to do (2) is to use (1)
P24- 35
Back to Back EMF
P24- 36
PRS Question:
Stopping a Motor
P24- 37