LECTURE 18
ANNOUNCEMENT
*Exam 2: Monday November 5, 2012, 8 PM - 10 PM
*Location: Elliot Hall of Music
*Covers all readings, lectures, homework from Chapters 25
through 28.
*The exam will be multiple choice (15-18 questions).
•  More on Self Inductance
•  Calculation of Self-Inductance for
Simple Cases
Be sure to bring your student ID card and
your own two-page (two-side) crib sheet (one
from exam 1 and a new one).
•  RL Circuits
•  Energy in Magnetic Fields
NOTE THAT FEW EQUATIONS WILL BE GIVEN – YOU ARE REMINDED
THAT IT IS YOUR RESPONSIBILITY TO CREATE WHATEVER TWO-SIDED
CRIB SHEET YOU WANT TO BRING TO THIS EXAM.
The equation sheet that will be given with the exam is
posted on the course homepage. Click on the link on the
left labeled “Equation sheet”
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1
Self Inductance
•  The magnetic field produced by the
current in the loop shown is
proportional to that current.
Self Inductance
I
•  The flux is also proportional to the current.
•  Define the constant of proportionality between flux
and current to be the inductance, L
•  Archetypal inductor is a long solenoid, just as a pair of
parallel plates is the archetypal capacitor
A
l
N turns
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3
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r
r << l
++++
d
d << A
- - - - -
4
1
Calculation of Inductance (from last class)
Example 1
l
r
Φ M = NBA = N µo nIA
N turns
L=
ΦM
= N µo nA = µo n 2lA
I
n=
N
l
•  At t=0, the switch is closed and
the current I starts to flow.
I
(a) L2 < L1
2N turns
R
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•  Find the current as a function of
time.
ε
ε
I = (1 − e− Rt / L ) = (1 − e−t /τ RL )
R
R
L
(c) L2 > L1
(b) L2 = L1
RL Circuits
I
b
"
r
N turns
What is the relation between L1 and L2?
5
a
2l
r
r
– Inductor 2 has length 2l, 2N total
turns and has inductance L2.
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RL Circuits
l
•  Consider the two inductors shown:
– Inductor 1 has length l, N total
turns and has inductance L1.
*Long solenoid with N turns, radius r, length L:
6
a
I
I
R
b
"
L
•  What about potential differences?
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Initially, an inductor acts to oppose changes in current through
it. A long time later, it acts like an ordinary connecting wire.
7
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8
2
RL Circuits ( on)
Current
ε
I = (1 − e− Rt / L )
R
Max = /R
RL Circuits
/R 1
VL = L
dI
= ε e− Rt / L
dt
Max = /R
37% Max at t=L/R
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00
I
a
•  Why does RL increase for
larger L?
2L/R
I
R
b
L
"
Q f( x ) 0.5
I
63% Max at t=L/R
Voltage on L
L/R
0
1
2
t
3
t
3
•  Why does RL decrease for
larger R?
4
x
t/RC
1 1"
f( x V
) 0.5
L
0.0183156
0
0
1
2
0
RL Circuits
After the switch has been in
position for a long time, redefined
to be t=0, it is moved to position b.
x
a
I
4
9
10
RL Circuits ( off)
I
R
Current
b
"
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4
ε − Rt / L
e
R
Max = /R
I=
L
37% Max at t=L/R
Voltage on L
VL = L
dI
= −ε e− Rt / L
dt
L/R
/R
1 1
2L/R
f( x ) 0.5
I
0.0183156
0
0
1
01 0
2
t
3
x
4
4
Q f( x ) V
0.5L
Max = -
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11
37% Max at t=L/R
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0"
-
0
1
2
x
t/RC
t
3
4
12
3
Inductor in Series
i
Inductor in Parallel
i
L1
L2
L1
13
•  How much energy is stored in
an inductor when a current is
flowing through it?
•  Start with loop rule:
dI
ε = IR + L
dt
a
I
R
B = µ0
•  Energy U:
dU
dI
= LI
dt
dt
•  Integrate this equation to find U, the energy stored in the inductor when
the current = I:
U
I
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0
0
U=
1 2
LI
2
l
I
r
N2 2
•  The inductance L is: L = µ 0
πr
l
dI
dt
PL, the rate at which energy is being stored in the inductor:
U = ∫ dU = ∫ LIdI
14
•  To calculate this energy density, consider the uniform field
generated by a long solenoid:
l
N
L
ε I = I 2 R + LI
PL =
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•  Claim: (without proof) energy is stored in the magnetic
field itself (just as in the capacitor / electric field case).
•  Multiply this equation by I:
•
L2
Where is the Energy Stored?
I
b
"
i2
i1
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Energy in an Inductor
i
U=
N turns
1 2 1 ⎛ N 2 2 ⎞ 2 1 2 B2
LI = ⎜ µ0
πr ⎟ I = πr l
2
2⎝
l
2
µ0
⎠
•  The energy density is found by dividing U by the volume
containing the field:
2
u=
15
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U
1B
=
2
π r l 2 µ0
16
4
Mutual Inductance
Mutual Inductance Applications
A changing current in a
coil induces an emf in an
adjacent coil. The
coupling between the
coils is described by
mutual inductance M.
ε 2 = −M
dI1
dI
and ε1 = − M 2
dt
dt
M depends only on geometry of the coils (size,
shape, number of turns, orientation, separation
between the coils).
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17
Superconductors
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18
Magnetic Flux Through a Superconducting Ring
1.  Cool it down
2.  Move magnet
What will happen?
Resistivity versus temperature
for an ordinary metal
emf = −
dΦ mag
dt
Current in the loop will produce its own B to compensate
for any changes in magnetic flux.
Resistivity versus temperature
for ‘superconductor’
infinite
mobility!
5
magnetic levitation of a type 1 superconductor
Superconductors
DEMO
Resistivity ρ = 0 for temperature T < Tc .
A type 1 superconductor is a perfect
diamagnetic material with χ m = −1.
 
B = Bapp (1+ χ m )
Magnetic levitation:
Repulsion between
the permanent
Magnetic field
producing the
applied field and the
magnetic field
produced by the
currents induced in
the superconductor
Meissner effect (1939)
B field becomes =0 because
superconducting currents on
the surface of the
superconductor create a
magnetic field that cancel the
applied one for T<Tc
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Magnet: NdFeB
21
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Superconductor: YBa2Cu3O7
22
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