Download • More on Self Inductance • Calculation of Self

LECTURE 18
Self Inductance
•  The magnetic field produced by the
current in the loop shown is
proportional to that current.
•  More on Self Inductance
•  Calculation of Self-Inductance for
Simple Cases
I
•  The flux is also proportional to the current.
•  RL Circuits
•  Energy in Magnetic Fields
•  Define the constant of proportionality between flux
and current to be the inductance, L
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RL Circuits
Self Inductance
•  At t=0, the switch is closed and
the current I starts to flow.
a
I
I
R
b
"
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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r
r << l
++++
d
d << A
- - - - -
3
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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.
4
1
RL Circuits
•  Find the current as a function of
time.
I=
ε
ε
1 − e− Rt / L = 1 − e−t /τ RL
R
R
(
)
(
)
a
I
RL Circuits ( on)
I
R
Current
b
I=
L
"
ε
1 − e− Rt / L
R
(
)
Max = /R
Voltage on L
VL = L
dI
= ε e− Rt / L
dt
Max = /R
37% Max at t=L/R
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2L/R
Q f( x ) 0.5
I
63% Max at t=L/R
•  What about potential differences?
L/R
/R 1
00
0
1
2
•  Why does RL increase for
larger L?
a
I
0.0183156
0
RL Circuits
I
R
After the switch has been in
position for a long time, redefined
to be t=0, it is moved to position b.
b
"
L
3
t
3
4
f( x V
) 0.5
L
0
1
2
0
RL Circuits
t
x
t/RC
1 1"
x
a
I
4
6
4
I
R
b
"
L
•  Why does RL decrease for
larger R?
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2
RL Circuits ( off)
Current
ε
I = e− Rt / L
R
Max = /R
37% Max at t=L/R
Voltage on L
VL = L
dI
= −ε e− Rt / L
dt
Inductor in Series
L/R
/R
1 1
i
i
2L/R
L1
L2
f( x ) 0.5
I
0.0183156
0
0
1
t
2
01 0
3
4
x
4
Q f( x ) V
0.5L
Max = -
37% Max at t=L/R
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0"
-
0
1
t
2
3
4
x
t/RC
Inductor in Parallel
i
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10
Energy in an Inductor
i2
•  How much energy is stored in
an inductor when a current is
flowing through it?
i1
L1
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L2
•  Start with loop rule:
ε = IR + L
dI
dt
a
I
I
R
b
L
"
•  Multiply this equation by I:
dI
dt
PL, the rate at which energy is being stored in the inductor:
ε I = I 2 R + LI
• 
PL =
dU
dI
= LI
dt
dt
•  Integrate this equation to find U, the energy stored in the inductor when
the current = I:
U
I
U = ∫ dU = ∫ LIdI
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0
0
U=
1 2
LI
2
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Where is the Energy Stored?
Mutual Inductance
A changing current in a
coil induces an emf in an
adjacent coil. The
coupling between the
coils is described by
mutual inductance M.
•  Claim: (without proof) energy is stored in the magnetic
field itself (just as in the capacitor / electric field case).
•  To calculate this energy density, consider the uniform field
generated by a long solenoid:
l
N
B = µ0
•  The inductance L is: L = µ 0
•  Energy U:
U=
l
I
r
N2 2
πr
l
N turns
⎞
1 2 1⎛ N
1
B
LI = ⎜ µ0
π r 2 ⎟ I 2 = π r 2l
2
2⎝
l
2
µ0
⎠
2
2
ε 2 = −M
•  The energy density is found by dividing U by the volume
containing the field:
2
u=
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U
1B
=
π r 2l 2 µ0
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Mutual Inductance Applications
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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Superconductors
Resistivity versus temperature
for an ordinary metal
Resistivity versus temperature
for ‘superconductor’
infinite
mobility!
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Magnetic Flux Through a Superconducting Ring
Superconductors
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 )
1.  Cool it down
2.  Move magnet
What will happen?
emf = −
dΦ mag
dt
Meissner effect (1939)
Current in the loop will produce its own B to compensate
for any changes in magnetic flux.
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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magnetic levitation of a type 1 superconductor
DEMO
Magnet: NdFeB
Magnetic levitation:
Repulsion between
the permanent
Magnetic field
producing the
applied field and the
magnetic field
produced by the
currents induced in
the superconductor
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Superconductor: YBa2Cu3O7
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