chapter 30 inductance chương trình tiên tiến bách khoa.pdf

Chapter 30
Inductance
HUST – CTTT PH 1026
Energy through space for free??
• Left off with a puzzler!
Increasing
current in
time
Creates
increasing flux
INTO ring
Induce
counter-
clockwise
current and
B field OUT
of ring
• But… If wire loop has resistance R, current
around it generates energy! Power = i2/R!!
Increasing
current in
time
Induced current
i around loop of
resistance R
• Yet…. NO “potential difference”!
Increasing
current in
time
Induced current
i around loop of
resistance R
PH1026 Lecture's notes - by Prof.Dr. Vu Ngoc Tuoc

• Answer? Energy in B field!!
Increasing
current in
time
Induced current
i around loop of
resistance R
Goals for Chapter 30
• To learn how current in one coil can induce an
emf in another unconnected coil
• To relate the induced emf to the rate of change
of the current
• To calculate the energy in a magnetic field
Goals for Chapter 30
• Introduce circuit components called
INDUCTORS
• To analyze circuits containing resistors and
inductors
• To describe electrical oscillations in circuits and
why the oscillations decay
Introduction
• How does a coil induce a
current in a neighboring coil.
• A sensor triggers the traffic
light to change when a car
arrives at an intersection. How
does it do this?
• Why does a coil of metal
behave very differently from a
straight wire of the same metal?
• We’ll learn how circuits can be
coupled without being
connected together.

Mutual inductance
• Mutual inductance: A changing current in one coil
induces a current in a neighboring coil.
Increase current
in loop 1
Increase B flux
Induce current in
loop 2
Induce flux
opposing change
Mutual inductance
• EMF induced in single loop 2 = - d (ΦB2 )/dt
– Caused by change in flux through second loop of B field
– Created by the current in the single loop 1
• ΦB2 is proportional to i1
Increase current
in loop 1
Induce current in
loop 2
Induce flux
opposing change
Mutual inductance
• What affects mutual inductance?
– # turns of each coil, area, shape, orientation
(geometry!)
– Rate of change of current!
Increase current
in loop 1
Induce current in
loop 2
Induce flux
opposing change
Suppose that we have two
coils,
Coil 1 with N1 turns and
Coil 2 with N2 turns
dt
d
N
B2
2
2
Φ
−
=
ε
Coil 1 has a current i1 which
produces a magnetic flux, ΦΒ2 ,
going through one turn of Coil 2
If i1 changes, then the flux changes
and an emf is induced in Coil 2
which is given by
Mutual Inductance

Mutual Inductance
The flux through the second coil
is proportional to the current in
the first coil
1
21
2
2 i
M
N B =
Φ
where M21 is called the
mutual inductance
dt
di
M
dt
d
N B 1
21
2
2 =
Φ
or
dt
di
M 1
21
2 −
=
ε
Taking the time derivative of this we get
If we were to start with the second coil having a varying
current, we would end up with a similar equation with an
M12
We would find that M
M
M =
= 12
21
The two mutual inductances are the same because the mutual
inductance is a geometrical property of the arrangement of
the two coils
To measure the value of the mutual inductance you can use
either
dt
dI
M 1
2 −
=
ε or
dt
dI
M 2
1 −
=
ε
Mutual Inductance
Units of Inductance
2
Amp
J
1
Amp
sec
V
1
Henry
1 =
⋅
=
Mutual inductance examples
• Long solenoid with length l, area A, wound with N1 turns of wire
• N2 turns surround at its center. What is M?

• M = N2 ΦB2/i1
• We need ΦB2 from the first solenoid (B1 = μoni1)
• n = N1/l
• ΦB2 = B1A
• M = N2 μoi1 AN1/li1
• M = μoAN1 N2 /l
• All geometry!
• M = μoAN1 N2 /l
• If N1 = 1000 turns, N2 = 10 turns, A = 10 cm2, l = 0.50 m
– M = 25 x 10-6 Wb/A
– M = 25 μH
• Using same system (M = 25 μH)
• Suppose i2 varies with time as = (2.0 x 106 A/s)t
• At t = 3.0 μs, what is average flux through each turn of coil 1?
• What is induced EMF in solenoid?
Start Your Engines: The Ignition Coil
The gap between the spark plug in a
combustion engine needs an electric
field of ~107 V/m in order to ignite the
air-fuel mixture. For a typical spark
plug gap, one needs to generate a
potential difference > 104 V!
But, the typical EMF of a car battery is
12 V. So, how does a spark plug
work??
spark
12V
The “ignition coil” is a double layer solenoid:
• Primary: small number of turns -- 12 V
• Secondary: MANY turns -- spark plug
http://www.familycar.com/Classroom/ignition.htm
•Breaking the circuit changes
the current through “primary
coil”
• Result: LARGE change in flux
thru secondary -- large induced
EMF!

Start Your Engines: The Ignition Coil
Transformer: P=iV
• Suppose i2 varies with time as = (2.0 x 106 A/s)t
• At t = 3.0 μs, i2 = 6.0 Amps
• M = N1 ΦB1/i2 = 25 μH
• ΦB1= Mi2/N1 = 1.5x10-7 Wb
• Induced EMF in solenoid?
– EMF1 = -M(di2/dt)
– -50Volts
Calculating self-inductance and self-induced emf
• Toroidal solenoid with area A, average radius r, N turns.
• Assume B is uniform across cross section. What is L?
Self Inductance
Suppose that we have a coil having N turns carrying a
current I
That means that there is a magnetic flux through the
coil
This flux can also be written as being proportional to the
current
I
L
N B =
Φ
with L being the self inductance having the same units as the
mutual inductance

Self-inductance
Self-inductance: A varying
current in a circuit induces an
emf in that same circuit.
Always opposes the change!
Define L = N ΦB/i
Li = N ΦB
If i changes in time:
d(Li)/dt = NdΦB/dt = -EMF or dt
di
L
−
=
ε
If the current I is increasing, then
If the current I is decreasing, then
Calculating self-inductance and self-induced emf
• Toroidal solenoid with area A, average radius r, N turns.
• L = N ΦB/i
• ΦB = BA = (μoNi/2πr)A
• L = μoN2A/2πr (self inductance of toroidal solenoid)
• Why N2 ??
• If N =200 Turns, A = 5.0 cm2,
r = 0.10 m
L = 40 μH
Inductors as circuit elements!
• Inductors ALWAYS oppose change:
• In DC circuits:
– Inductors maintain steady current flow
even if supply varies
• In AC circuits:
– Inductors suppress (filter) frequencies
that are too fast.
Potential across an inductor
• The potential across an
inductor depends on the
rate of change of the
current through it.
• The self-induced emf
does not oppose current,
but opposes a change in
the current.
Vab = -Ldi/dt

If the current is i at a given instant and its rate of change is
given by di/dt then the power being supplied by the
external source is given by
Magnetic Field Energy
2
2 2
0
1 1
2 2 2
N A
U LI I
r
μ
π
= =
When a current is being established in a circuit, work has to be
done
dt
di
i
L
i
V
P L =
=
The energy supplied is given by Pdt
dU =
The total energy stored in the inductor is then
2
0
2
1
I
L
di
i
L
U
I
=
= ∫
2 2 2
0 2
0
1 1
2 2 (2 ) 2
U N I B
u U V
rA r
μ
π π μ
= = = = where
This energy that is stored in the magnetic field is available
to act as source of emf in case the current starts to decrease
We will just present the result for the energy density of the
magnetic field
0
2
2
1
μ
B
uB =
This can then be compared to the energy density of an
electric field
2
0
2
1
E
uE ε
=
Magnetic Field Energy
Magnetic field energy
• Inductors store energy in the magnetic field:
U = 1/2 LI2
• Units: L = Henrys (from L = N ΦB/i )
• N ΦB/i = B-field Flux/current through inductor that
creates that flux
Wb/Amp = Tesla-m2/Amp
• [U] = [Henrys] x [Amps]2
• [U] = [Tesla-m2/Amp] x [Amps]2 = Tm2Amp
• But F = qv x B gives us definition of Tesla
• [B] = Teslas= Force/Coulomb-m/s = Force/Amp-m
• Inductors store energy in the magnetic field:
U = 1/2 LI2
• [U] = [Tesla-m2/Amp] x [Amps]2 = Tm2Amp
• [U] = [Newtons/Amp-m] m2Amp
= Newton-meters = Joules = Energy!

• The energy stored in an inductor is U = 1/2 LI2.
• The energy density in a magnetic field (Joule/m3) is
• u = B2/2μ0 (in vacuum)
• u = B2/2μ (in a magnetic material)
• Recall definition of μ0 (magnetic permeability)
• B = μ0 i/2πr (for the field of a long wire)
• μ0 = Tesla-m/Amp
• [u] = [B2/2μ0] = T2/(Tm/Amp) = T-Amp/meter
• The energy stored in an inductor is U = 1/2 LI2.
• The energy density in a magnetic field (Joule/m3) is
• u = B2/2μ0 (in vacuum)
• [u] = [B2/2μ0] = T2/(Tm/Amp) = T-Amp/meter
• [U] = Tm2Amp = Joules
• So… energy density [u] = Joules/m3
Energy Density in E and B Fields
uE =
ε0E2
2
uB =
B2
2μ0
The Energy Density of the Earth’s Magnetic
Field Protects us from the Solar Wind!

current through it.
the current.
current through it.
the current.
current through it.
the current.
current through it.
the current.

The R-L circuit
• An R-L circuit contains a
resistor and inductor and
possibly an emf source.
• Start with both switches open
• Close Switch S1:
• Current flows
• Inductor resists flow
• Actual current less than
maximum E/R
• E – i(t)R- L(di/dt) = 0
• di/dt = E /L – (R/L)i(t)
The R-L circuit
• Close Switch S1:
• E – i(t)R- L(di/dt) = 0
• di/dt = E /L – (R/L)i(t)
Boundary Conditions
• At t=0, di/dt = E /L
• i(∞) = E /R
Solve this 1st order diff eq:
• i(t) = E /R (1-e -(R/L)t)
Current growth in an R-L circuit
• i(t) = E /R (1-e -(R/L)t)
• The time constant for an R-L circuit is
τ = L/R.
• [τ ]= L/R = Henrys/Ohm
• = (Tesla-m2/Amp)/Ohm
• = (Newtons/Amp-m) (m2/Amp)/Ohm
• = (Newton-meter) / (Amp2-Ohm)
• = Joule/Watt
• = Joule/(Joule/sec)
• = seconds! ☺
Current growth in an R-L circuit
• i(t) = E /R (1-e -(R/L)t)
• The time constant for an R-L circuit is
τ = L/R.
• [τ ]= L/R = Henrys/Ohm
• EMF = -Ldi/dt
• [L] = Henrys = Volts /Amps/sec
• Volts/Amps = Ohms (From V = IR)
• Henrys = Ohm-seconds
• [τ ]= L/R = Henrys/Ohm = seconds! ☺
Vab = -Ldi/dt

The R-L circuit
• E = i(t)R+ L(di/dt)
• Power in circuit = E I
• E i = i2R+ Li(di/dt)
• Some power radiated in resistor
• Some power stored in inductor
The R-L circuit example
• R = 175 Ω; i = 36 mA; current
limited to 4.9 mA in first 58 μs.
• What is required EMF
• What is required inductor
• What is the time constant?
The R-L circuit Example
• R = 175 Ω; i = 36 mA; current
limited to 4.9 mA in first 58 μs.
• What is required EMF
• What is required inductor
• What is the time constant?
• EMF = IR = (0.36 mA)x(175Ω )
= 6.3 V
• i(t) = E /R (1-e -(R/L)t)
• i(58μs) = 4.9 mA
• 4.9mA = 6.3V(1-e -(175/L)0.000058)
• L = 69 mH
• τ = L/R = 390 μs
Current decay in an R-L circuit
• Now close the second
switch!
• Current decrease is opposed
by inductor
• EMF is generated to keep
current flowing in the same
direction
• Current doesn’t drop to zero
immediately

• Now close the second
switch!
• –i(t)R - L(di/dt) = 0
• Note di/dt is NEGATIVE!
• i(t) = -L/R(di/dt)
• i(t) = i(0)e -(R/L)t
• i(0) = max current before
second switch is closed
• i(t) = i(0)e -(R/L)t
• Test yourself!
• Signs of Vab and Vbc when
S1 is closed?
• Vab >0; Vbc >0
• Vab >0, Vbc <0
• Vab <0, Vbc >0
• Vab <0, Vbc <0
• Test yourself!
S1 is closed?
• Vab >0; Vbc >0
• Vab >0, Vbc <0
• Vab <0, Vbc >0
• Vab <0, Vbc <0
• WHY?
• Current still flows
around the circuit
counterclockwise

• Test yourself!
S1 is closed?
• Vab >0; Vbc >0
• WHY?
around the circuit
counterclockwise
through resistor
• EMF generated in L is
from c to b
• So Vb> Vc!
• Test yourself!
S2 is closed, S1 open?
• Vab >0; Vbc >0
• Vab >0, Vbc <0
• Vab <0, Vbc >0
• Vab <0, Vbc <0
• Test yourself!
S2 is closed, S1 open?
• Vab >0; Vbc >0
• Vab >0, Vbc <0
• WHY?
counterclockwise
• di/dt <0; EMF
generated in L
is from b to c!
So Vb> Vc!
The L-C circuit
• An L-C circuit contains an inductor and a capacitor and is an
oscillating circuit.
• Initially capacitor fully charged; close switch
• Charge flows FROM capacitor, but inductor
resists that increased flow.
• Current builds in time.
• At maximum current, charge flow now
decreases through inductor
• Inductor now resists decreased flow, and
keeps pushing charge in the original direction
i

The L-C circuit
• Initially capacitor fully charged; close switch
• Charge flows FROM capacitor, but inductor
resists that increased flow.
• Current builds in time.
• Capacitor slowly discharges
• At maximum current, no charge is left on
capacitor; current now decreases through
inductor
• Inductor now resists decreased flow, and
keeps pushing charge in the original direction
i
The L-C circuit
The L-C circuit
• Now capacitor fully drained;
• Inductor keeps pushing charge in the
original direction
• Capacitor charge builds up on other side
to a maximum value
• While that side charges, “back EMF” from
capacitor tries to slow charge build-up
• Inductor keeps pushing to resist that change.
i

The L-C circuit
The L-C circuit
• Now capacitor charged on opposite side;
• Current reverses direction! System repeats
in the opposite direction
i
The L-C circuit
Electrical oscillations in an L-C circuit
• Analyze the current and
charge as a function of time.
• Do a Kirchoff Loop around
the circuit in the direction
shown.
• Remember i can be +/-
• Recall C = q/V
• For this loop:
-Ldi/dt – qC = 0

• -Ldi/dt – qC = 0
• i(t) = dq/dt
• Ld2q/dt2 + qC = 0
• Simple Harmonic Motion!
• Pendulums
• Springs
• Standard solution!
• q(t)= Qmax cos(ωt+φ)
where ω = 1/(LC)½
• q(t)= Qmax cos(ωt+φ)
• i(t) = - ω Qmax sin(ωt+φ)
(based on this ASSUMED
direction!!)
• ω = 1/(LC)½ =
angular frequency
The L-C circuit
Electrical and mechanical oscillations
• Table 30.1 summarizes the analogies between SHM and L-C
circuit oscillations.

The L-R-C series circuit
• An L-R-C circuit exhibits
damped harmonic motion
if the resistance is not too
large.
2
2
If we add a resistor in an circuit (see figure) we must
modify the energy equation, because now energy is
being dissipated on the resistor: .
E B
RL
dU
i R
dt
q
U U U
= −
= + =
Damped Oscillations in an Circuit
RCL
2
2
2 2
Li dU q dq di
Li i R
C dt C dt dt
+ → = + = −
( )
2
2
2
/2
2
2 2
1
0. This is the same equation as that
of the damped harmonics osc 0, which has the
illator:
The angular f
solution
re
( ) c que
:
os .
bt m
m
d x dx
m b kx
dt dt
x
dq di d q d q dq
i L R q
dt dt dt dt dt
t x e t
C
ω φ
−
= → = → + + =
+ + =
′
= +
( )
2
/2
2
2
2
ncy
For the damped circuit the solution is:
The angular freque
1
( ) cos . .
4
ncy
.
4
Rt L R
q
k b
RC
t Qe t
m
L
L
C L
m
ω φ ω
ω
−
′ ′ −
−
+
=
= =
′
(31-6)
/ 2
Rt L
Qe−
/ 2
Rt L
Qe−
( )
q t
Q
Q
−
( )
q t
( )
/2
( ) cos
Rt L
q t Qe t
ω φ
−
′
= +
2
2
1
4
R
LC L
ω′ = −
/2
2
2
The equations above describe a harmonic oscillator with an exponentially decaying
amplitude . The angular frequency of the damped oscillator
1
is always smaller than the angular
4
Rt L
Qe
R
LC L
ω
−
′ = −
2
2
1
frequency of the
1
undamped oscillator. If the term we can use the approximation .
4
LC
R
L LC
ω
ω ω
=
′ ≈

τRC = RC τRL = L / R τRCL = 2L / R
AC Driven Circuits:
1) A Resistor:
0
=
− R
v
emf
vR = emf = Em sin(ωd t)
iR =
vR
R
=
Em
R
sin(ωd t)
Resistors behave in AC very much as in DC,
current and voltage are proportional
(as functions of time in the case of AC),
that is, they are “in phase”.
For time dependent periodic situations it is useful to
represent magnitudes using Steinmetz “phasors”. These
are vectors that rotate at a frequency ωd , their
magnitude is equal to the amplitude of the quantity in
question and their projection on the vertical axis
represents the instantaneous value of the quantity.
Charles
Steinmetz

AC Driven Circuits:
2) Capacitors:
vC = emf = Em sin(ωd t)
qC = Cemf = CEm sin(ωd t)
iC =
dqC
dt
= ωdCEm cos(ωd t)
iC = ωdCEm sin(ωd t + 900
)
iC =
Em
X
sin(ωd t + 900
)
reactance

1
where
C
X
d
ω
=
im =
Em
X looks like i =
V
R
Capacitors “oppose a resistance” to AC
(reactance) of frequency-dependent magnitude 1/ωd C
(this idea is true only for maximum amplitudes,
the instantaneous story is more complex).
AC Driven Circuits:
3) Inductors:
vL = emf = Em sin(ωd t)
L
dt
v
i
dt
i
d
L
v L
L
L
L
∫
=
⇒
=
iL = −
Em
Lωd
cos(ωd t) =
Em
Lωd
sin(ωd t − 900
)
iL =
Em
X
sin(ωd t − 900
)
im =
Em
X d
L
X ω
=
where
Inductors “oppose a resistance” to AC
(reactance) of frequency-dependent magnitude ωd L
(this idea is true only for maximum amplitudes,
the instantaneous story is more complex).

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