*Superconductivity /Meissner effect/quantum levitation
•  Superconductor is COOL, usually really
really cool(ed by liquid nitrogen)!
•  Not only has ρ=0, but also has amazing
magentic property
•  It expels magnetic flux, causing levitation!
•  http://www.youtube.com/watch?v=CXppGO6okE
•  If an object is coated with a very thin layer
of superconductor material, the magentic
field can manage to penetrate through the
material defects, creating magnetic flux
tubs, and a 3D clamps on the object:
Quantum levitation
•  http://www.youtube.com/watch?
v=Ws6AAhTw7RA
If you can only learn one slide from this chapter

Changing E

Changing B

Another B arises (Ampere’s law)
 
∫ B • dl = µ (i + ε

Another E arises (Faraday’s

 law)
dΦ
E
•
d
l
=
0
−
∫
dt
0
C
dΦE
)
0
dt
B
€
€
€ FB=ILB
€
€
Fext>=ILB to keep
it from slowing down
(I=ε/R=vLB/R)
A sample problem-solving flow chart
Changing B flux
Induced EMF (E)
Generate Current
Magnetic force slowing the rod down
Come from work done by Fext
Electric energy
Cool video: https://www.youtube.com/watch?v=NqdOyxJZj0U&feature=g-subs-u
The actual CH29 Summary: Electromagnetic Induction
dφ B
ε=−
dt
Faraday’s law
Lenz’s law: induced current/emf
oppose or cancel the change!€
€
ε = vBL
Motional emf:
 
φB = B • A
= BAcos θ

 
ε = ∫ (v × B) • dl
Induced electric field
(non-conservative!)
€
Maxwell’s equation:
All three: B, A,
and θ can be changed
Isolated Conductor
Circuit loop


dΦB
(
E
)
•
d
l
=
−
∫
dt
€
Faraday’s law
Ampere’s law

 Qencl
∫ E • dA = ε
0


Gauss’s
∫ B • dA = 0


dΦB
E
•
d
l =0−
∫
dt
 
dΦ
B
∫ • dl = µ0 (iC + ε 0 dtE )
law
Displacement current
Chapter 30: Inductance
•  mutual inductance:
•  Time-varying current in one coil induces emf in an
unconnected coil
•  self-inductance:
•  Time-varying current in a circuit induces emf in iteself
•  magnetic-field energy:
•
Where is stored
•
Compare with electric-field energy in capacitor
•  Circuits with inductors:
•  R/L, L/C, R/L/C circuits (next Tuesday)
Opening question
•  Some traffic lights will sense
vehicles approaching. How?
•  Why is it that sometimes
bigger vehicles get through
those traffic lights faster?
Mutual inductance
Mutual inductance: the proportionality
A. between the induced emf in coil 2
due to changing current in coil 1
B. between the magnetic field flux
in coil 2 due to current in coil 1A
(Faraday’s law)
•  A coil in one device
generates a field that
creates a current in a
neighboring coil.
dΦB 2
dt
N 2ΦB 2 = M 21i1 ⇒
dΦB 2
di
N2
= M 21 1
dt
dt
ε 2 = −N 2
(M = M12 = M 21 =
ε 2 = −M
(Inductance definition)
(Take the derivative)
N1ΦB1 N 2ΦB 2
=
)
i2
i1
di1
dt
M: determined by geometry, material
€
Unit:
1H(henry) = 1Wb / A = 1V ⋅ s / A = 1Ω⋅ s = 1J / A 2
Application: Transformers, etc.
€
Mutual inductance—examples
A long solenoid magnet has a B field determined by the current:
B = µ0 ni
What is the mutual conductance of the two solenoids shown below?
N 2ΦB 2 N 2 (B1 A)
=
i1
i1
N
B1 = µ0 n1i1 = µ0 1 i1 ⇒
l
N
N 2 ( µ0 1 i1 A) µ AN N
l
M 21 =
= 0 1 2
i1
l
M 21 =
€
€
Can you see why M = M12 = M 21 ?
What if an iron core is added?
What changes in the above results?
€
i1
i1
€
Self-inductance
Definition:
L=
NΦB
⇒
i
Self-induced emf
ε = −N
dΦB
d(Li)
di
=−
= −L
dt
dt
dt
What’s the potential difference between point a and b?


Vab = Va − Vb = ∫ a E c • dl
(conservative electric field)


 E c + E n = 0 (Assuming perfect Inductor, no resistance)

b
di
Changing-magneticE
•
d
l
=
ε
=
−L
∫a n
dt
Induced electric field En


b
b
di
∴ ∫ a E c • d l = − ∫ a E n • dl = L
dt is not conservative!
di
Vab = L We could still use kirchhoff’s loop rule for circuits including inductors!
dt
€
b
Q30.1
A small, circular ring of wire
(shown in blue) is inside a larger
loop of wire that carries a current I
as shown. The small ring and the
larger loop both lie in the same
plane. If I increases, the current
that flows in the small ring
Large loop
Small ring
I
I
A. is clockwise and caused by self-inductance.
B. is counterclockwise and caused by self-inductance.
C. is clockwise and caused by mutual inductance.
D. is counterclockwise and caused by mutual inductance.
Resistor/Inductor: Potential difference across both circuit elements
Resistor: Potential difference
is proportional to the current!
Inductor: Potential difference depends
on the rate of change of the current!
What do the signs mean in both
cases?
Imagine an induced battery that
opposes the current change)
Q30.2
A current i flows through an inductor L in
the direction from point b toward point a.
There is zero resistance in the wires of the
inductor. If the current is decreasing,
A. the potential is greater at point a than at point b.
B. the potential is less at point a than at point b.
C. the answer depends on the magnitude of di/dt
compared to the magnitude of i.
D. The answer depends on the value of the inductance L.
E. both C. and D. are correct.
example: Self-induced emf in a toroidal solenoid
µ0 Ni
A toroidal solenoid has magnetic field of: B = 2πr
What is the induced emf is the current increase from 0 to 6.0A in 3µs?
ε = −N
dΦB
d(Li)
di
=−
= −L
dt
dt
dt
€
L =?
L=
€
NΦB NBA
=
=
i
i
N
µ0 Ni
A
2
2πr = µ0 N A
i
2πr
Suppose N=200, A=5.0cm2,
r=0.10m, what is the value of L and ε?
Magnetic field energy in Inductor VS Electric field
energy in Capacitors
How is the energy stored in an inductor(such
as your car spark plug)?
Inductor:
P = Vab i = (L
Capacitor
di
)i
dt
dq
d(Cv)
)=v
dt
dt
dU = Pdt ⇒
P = vi = v(
dU = Pdt ⇒
U=
∫ dU =
∫
I
L
0
di
1
idt = LI 2
dt
2
(energy density in a toroidal solenoid)
€
1
u = U /volume = LI 2 /(2πrA)
2
2
µN A
L= 0
⇒
2πr
1 µ0 N 2 A 2
1 µ0 NI 2
u=
I /(2πrA) =
(
)
2 2πr
2 µ0 2πr
1 2
u=
B
2 µ0
U=
∫
V
0
v
d(Cv)
1
dt = CV 2
dt
2
(energy density in a parallel-plate capacitor)
€
True for em waves as well
€
∫ dU =
1
u = ε0 E 2
2
Q30.3
A steady current flows through an inductor. If the current is
doubled while the inductance remains constant, the amount of
energy stored in the inductor
A. increases by a factor of
.
B. increases by a factor of 2.
C. increases by a factor of 4.
D. increases by a factor that depends on
the geometry of the inductor.
E. none of the above
€
The R-L circuit
R-L circuit is similar to R-C circuit:
(loop rule)
di
=0⇒
dt
1
R
di = − dt
i − ε /R
L
i
t
1
R
∫ 0 i' −ε /R di' = ∫ 0 (− L )dt'
i − ε /R
R
ln(
)=− t
−ε /R
L
ε
ε
⇒ i = (1 − e −(R / L )t ) = (1 − e −t /τ ) = I(1 − e −t /τ )
R
R
ε − iR − L
€
(Or you could have guessed it)
Time constant τ=L/R
Comparing with a charging capacitor:
i=
ε −t / RC ε −t /τ
e
= e
= Ie −t /τ
R
R
We an also use this to derive the energy stored. Do you know how?
Current decay in an R-L circuit
If the previous circuit is
disconnected
From the battery, leaving only the
Resistor and conductor, how
would the current
Behave as a function of time?
di
=0⇒
dt
1
R
di = − dt
i
L
i 1
t
R
∫ε / R i' di' = ∫ 0 (− L )dt'
i
R
ln(
)=− t
ε /R
L
ε ε
⇒ i = = e −t /τ
R R
−iR − L
Not 0!
€
Q30.4
An inductance L and a resistance
R are connected to a source of
emf as shown. When switch S1 is
closed, a current begins to flow.
The final value of the current is
A. directly proportional to RL.
B. directly proportional to R/L.
C. directly proportional to L/R
D. directly proportional to 1/(RL).
E. independent of L.
Q30.5
An inductance L and a resistance
R are connected to a source of
emf as shown. When switch S1 is
closed, a current begins to flow.
The time required for the current
to reach one-half its final value is
A. directly proportional to RL.
B. directly proportional to R/L.
C. directly proportional to L/R.
D. directly proportional to 1/(RL).
E. independent of L.
Q30.6
An inductance L and a resistance
R are connected to a source of
emf as shown. Initially switch S1
is closed, switch S2 is open, and
current flows through L and R.
When S2 is closed, the rate at
which this current decreases
A. remains constant.
B. increases with time.
C. decreases with time.
D. not enough information given to decide
The L-C circuit
The L-C circuit: Mathematical results
Kirchhoff’s rule:
di q
Recall:
− =0
dt C
dq
i≡
⇒
dt
d 2q q
−L 2 − = 0
dt
C
d 2q
1
1 2
€
=
−
q
=
−(
)
2
dt
LC
LC
q = ?sin(
€
d sin ωt
= ω cos ωt
dt
d 2 sin ωt
= −ω 2 sin ωt
2
dt
−L
1
t)
LC
q = ?cos(
OR
1
t)
LC
If the circuit starts with q=Q, fully charged then
Total energy? € 1 q 1
(Qcos ωt)
€
E = Li +
= (L(−ωQsin ωt) +
)
2
2C
2
C
Conserved!
t) = Qcosωt
2
2
tot
1
q = Qcos(
LC
dq
i=
= −ωQsin ωt
dt
Lω 2 = L(
E tot
€
€
2
2
1 2 1
) ≡ ⇒
LC
C
Q2
Q2
2
2
=
(sin ωt + cos ωt) =
2C
2C
L-C system VS Mass-Spring system
Q30.7
An inductor (inductance L) and
a capacitor (capacitance C) are
connected as shown.
If the values of both L and C
are doubled, what happens to
the time required for the
capacitor charge to oscillate
through a complete cycle?
A. It becomes 4 times longer.
B. It becomes twice as long.
C. It is unchanged.
D. It becomes 1/2 as long.
E. It becomes 1/4 as long.
Q30.8
An inductor (inductance L) and
a capacitor (capacitance C) are
connected as shown. The value
of the capacitor charge q
oscillates between positive and
negative values. At any instant,
the potential difference between
the capacitor plates is
A. proportional to q.
B. proportional to dq/dt.
C. proportional to d2q/dt2.
D. both A and C.
E. all of A, B, and C.
The L-R-C circuit
The amplitude of the charge oscillation
Is proportional to the total E/M energy.
If there is resistance, then the part of the E/M
Energy will be dissipated, leading to smaller
Amplitude!
di q
−iR − L
dt
−
C
R2 <
4L
C
€
=0
dq
⇒
dt
d 2q R dq 1
+
+
q=0
dt 2 L dt LC
i≡
1
R2
q = Ae
cos(
−
t + φ)
LC 4L2
q = Ae −(R / 2L )t cos(ω ' t + φ )
R2 >
4L
C
R 2 >>
4L
C
−(R / 2L )t
ω'=
€
€
1
R2
−
LC 4L2
€
CH30 Summary: Inductance
Mutual Inductance:
(example of two solenoid)
ε 2 = −N 2
dΦB 2
dt
M = M12 = M 21 =
N1ΦB1 N 2ΦB 2
=
i2
i1
NΦB
⇒
i
dΦB
d(Li)
di
ε = −N
=−
= −L
dt
dt
dt
L=
Self Inductance
€
1
U = LI 2
2
1 2
u=
B
2 µ0
Magnetic field energy
€
Stored in inductor (VS electric field)
R-L circuits: exponential
L-C circuits:
€
Oscillating, energy conservation
ε
ε
(1 − e −(R / L )t ) = (1 − e −t /τ )
R
R
€
τ = L /R
i=
q = Qcos(
1
t) = Qcosωt
LC
dq
i=
= −ωQsin ωt
dt
L-R-C circuits: Oscillating/Or decaying
€
(damped oscillations)
€
ω=
1
R2
− €2
LC 4L
ω=
1
LC