UCSD Physics 2B
Summer Session
Unit 5 Lecture Notes
Chapter 32
Inductance & Magnetic Energy
Introduction
We begin with some analyses and formulas, but we’ll concentrate in this unit
on analogies to physical systems. In this way we can try to understand the
behavior of circuits and their components from a non-mathematical point of
view and thereby gain further insight and intuition into a field in which we
seldom have any personal experience with our physical senses.
Please note that this unit brings together much of what we’ve learned up to this
point. As we proceed, then, it would be a very good idea to review the basic
concepts from previous units.
SKIP TO …
Section 2 Inductance
Consider the following diagram of a current-carrying wire with a single loop.
B
I
Let’s concentrate on the magnetic field produced just by the loop alone. Using
the right hand, we see that the magnetic field points out of the page as shown in
pink. If the current is constant in time, so is the field and, since the geometry is
fixed, the total flux threading the loop is also constant.
But what happens if the current changes? Now the space inside the loop sees
a change in the total flux threading it. It doesn’t know that the loop itself is
producing that change. All it knows is what Faraday’s Law, and especially
Lenz’s Law (the minus sign) wants to do: oppose the change in the flux.
1
In order to do so, Nature creates an Electromotive Force or Emf – a “circular”
electric field which wants to drive current in whatever direction is necessary to
provide a magnetic field opposing the change:
EMF = −
dΦ
dt
We’ve seen this before, but always in regards to an externally generated
magnetic field. This time, it’s the loop itself which is changing the flux.
The geometry of any device is fixed once we build it (the loop, in this
example) so the only thing that can change the magnetic flux is the current in
the wire. Because of this, it’s convenient to define the quantity (constant)
which doesn’t change with current, but depends only on the geometry:
L=
Φ
I
Self-Inductance
The unit of inductance is the Henry (H) and, from the formula, this is also
Weber
Tesla • meter 2
1 Henry = 1
=1
Ampere
Ampere
Example:
Solenoid
The flux through a solenoid is easy to calculate since the field inside is uniform
throughout. However, we must multiply the flux Φ for one loop by the
number of turns, since the field threads them all. Consider the diagram below
where we’ve “unwound” a solenoid and laid each winding side by side. Each
one has the same flux through it (field B x area A)
B
Φ
Φ
Φ
I
The magnetic field inside the solenoid is
B = μ0 nI and Φ = NBA so that Φ = μ0 nNIA
2
where n is the number of turns per unit length, N is the total number of turns, A
is the cross-sectional area and l is the length. Hence we have
L Solenoid = μ 0 n 2 Al
Aside: Note the product of the Area times the Length is just the Volume.
Section 3
Inductors in Circuits
Now that we’ve separated the current dependence by defining inductance, we
can now write
EMF = −
dΦ
d
dI
= − ( LI ) = − L
dt
dt
dt
This is the relation we want in order to treat the loop as a device, that is, as a
circuit component. In a circuit, such a device is called an inductor.
V = −L
dI
dt
Voltage drop across Inductor
This formula tells us that, in a circuit, an inductor L impedes or opposes
changes in current I like a resistor R impedes the current I itself.
See figures 32-8 and 32-9 on page 839
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Analyzing an Inductive Circuit
Let’s build a circuit as we did for a capacitor and see how it differs. By
applying Kirchhoff’s Loop rule, the sum of voltages around a closed loop must
be zero – only the battery voltage is positive, all others are negative:
⎛ t⎞
Let I ( t ) = exp ⎜ − ⎟ then
⎝ τ⎠
dI
L
1
1
⎛ t⎞
= − exp ⎜ − ⎟ = − I ( t ) ⇒ − IR + I = 0
dt
τ
τ
τ
⎝ τ⎠
L
− IR − L
R
τ=
dI
=0
dt
L
R
Time Constant for RL Circuit
The time-decay of the current in this circuit, assuming current was flowing at
time t = 0, is then given by
⎛ Rt ⎞
I ( t ) = I 0 exp ⎜ − ⎟
⎝ L ⎠
Current Decay for RL Circuit
Note that his formula has the same form as the “discharging” formula for the
Capacitor. Beware, however, since here we are describing the Current rather
than the Voltage!
L
R
S
V
⎡
⎛ Rt ⎞ ⎤
I ( t ) = I 0 ⎢1 − exp ⎜ − ⎟ ⎥
⎝ L ⎠⎦
⎣
Now in this circuit, we show the equivalent of
“charging” a capacitor. Some refer to this
process as “loading” the inductor with magnetic
field. Since the basic formulas for the capacitor
and inductor lead to the result above for
“discharging” or “unloading”, we should
immediately be able to write
“Loading” the inductor in an RL circuit
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Section 4
Magnetic Energy
Say we have an inductor with no current flowing through it. Then there is no
field inside either. Now suppose that we wish to turn on the current and
establish a field. What happens? The “back-emf” as it is commonly called
opposes the buildup of that field by opposing the current, so we have to do
work against this resisting force. Where does this work go? It must be stored in
the magnetic field itself, just like the work we did to establish an electric field
between the plates of a capacitor is stored in the electric field.
How can we calculate the work since there is no Force x Distance? Recall that
work = power x time. Electric power P = VI and the work the battery does
against the inductor’s opposition is positive, so let’s integrate
dI ⎞
1
⎛
Work = ∫ Pdt = ∫ IVdt = ∫ I ⎜ + L ⎟ dt = L ∫ IdI = LI 2 z
dt ⎠
2
⎝
UL =
1 2
LI
2
Energy stored in an Inductor.
Compare this with the Potential Energy “stored” in a Capacitor
UC =
1
CV 2
2
Energy stored in a Capacitor.
We can now use this to calculate the energy density in the magnetic field:
UL
B2
1 2 LI 2 μ0 n 2 I 2 Al μ02 n 2 I 2
=
=
=
=
uL =
2 Al
2 μ0
2 μ0
Volume
Al
B2
uB =
2 μ0
Magnetic Energy Density
Compare this with the energy density “stored” in the electric field
1
uE = ε 0 E 2
2
Electrostatic Energy Density
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Chapter 33
Section 1
AC Circuits
Alternating Current
The circuits we studied in Unit 3 all had Direct Current (DC) voltage sources
driving the current. That is, the polarity of the batteries remained fixed for all
time (ideal batteries, of course… real ones fade away.) So, as long as the
battery was connected, the current in the circuits always moved in the same
direction. An alternating voltage source changes polarity at regular intervals
and generally does so smoothly (sine function) and drives the current in circuit
backwards and forwards. Since the direction alternates, we call it Alternating
Current (AC).
Recall from Unit Exam 4 the exercise where a square loop immersed in an
external magnetic field rotates at constant angular velocity ω . Such a device is
called a generator and, since the flux is constantly increasing and then
decreasing, the output Emf voltage alternates at frequency ω as well.
We will consider a common form of AC, one in which the voltage varies in
time sinusoidally
V ( t ) = VPeak sin (ωt )
Sections 2 & 3
AC Voltage Source
Circuit Elements in AC Circuits - REACTANCE
We will consider three simple AC circuits and see if we can compare them.
1. RESISTOR
This circuit consists of an AC voltage source
R connected to a single resistor. Here the current
VAC
and the voltage are in phase since
V (t ) 1
= VP sin ωt
V ( t ) = VP sin ωt ⇒ I ( t ) =
R
R
In other words, we still have Ohm’s Law – the linear relation VP = I P R for the
resistor even though the current is changing in time.
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2. CAPACITOR
VAC
Here we have the same source connected to a
C single capacitor. The current leads by 90˚ (out
of phase) since
Q = CV
I ( t ) = Q ( t ) = CV ( t ) = ω CVP cos ωt
Note, however, that the peak voltage is related to the peak current by
1
VP =
I =X I
(ω C ) P C P so we have a kind of Ohm’s Law for a capacitor
1
takes the place of the AC
ωC
resistance “R” and is called the capacitive reactance.
driven by an AC source, where X C =
3. INDUCTOR
VAC
Now the source drives a single inductor. The
L current lags by 90˚ (again out of phase) since
VP
cos ωt
ωL
V ( t ) = − L dI dt ⇒ I ( t ) = − ∫ VP sin ωt dt = −
Again, we can write VP = I P X L but here, X L = Lω takes the place of the AC
resistance and is called the inductive reactance.
IMPORTANT CONCEPT
Note that both Capacitive Reactance and Inductive Reactance depend on the
driving frequency. Hence, in the FILTER circuit below, an inductor would
restrict (filter) high frequencies, while a capacitor would restrict low ones.
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R, L, C or ?
VAC
Section 3
RLOAD
LC Circuits
Consider the following circuit consisting of a single capacitor and inductor.
L
C
Suppose we charge the capacitor to a voltage V first
and then connect it to the inductor. What happens?
Qualitatively, the separated charge on the capacitor
“sees” a return path through the inductor and current
begins to flow. However, Lenz’s Law in the coil
creates a “back emf” which fights the current so it
takes time for the capacitor to discharge.
Meanwhile, back in the coil, a magnetic field is building up due to the current.
Once the capacitor has discharged, it pushes no more current. But the current
continues to flow since, once again, Lenz goes to work – this time trying to
maintain the decaying field. Finally, the magnetic field has spent itself, but not
before pushing excess charge onto the capacitor opposite to the initial charge.
Here the process begins again, this time in reverse – with the capacitor fully
charged and the inductor idle.
We can look at the process in another dynamic way. Recall that there is energy
stored in the electric field of a charged capacitor. When it discharges, where
does that energy go? It cannot be dissipated since there is no resistor in the
circuit. It must become the energy stored in the inductor’s magnetic field.
Viewed in this way, the energy oscillates or “sloshes” back-and-forth between
the two circuit elements. A circuit will not oscillate without both of these.
We can find the oscillation frequency ω by applying Kirchhoff’s Law.
Adding the voltage drops through the two devices around the loop, we get
−
Q
dI
dQ
+ 1 Q = 0
−L
= 0 but I =
so that LQ
C
dt
dt
C
8
We turn the differential equation into an algebraic one by guessing the
iω t
oscillating solution Q ( t ) = e which, upon substitution, gives
ω=
Section 5
1
LC
Power in AC Circuits
Suppose we measure the voltage on our home power outlet as a function of
time. We’ll see the following sine wave which oscillates 60 times per second
or 60 Hertz (Hz).
V
t
Now we ask, just in the simple case of a resistor, what will be the average
power dissipated over time? Surely it cannot be the simple product of the peak
voltage and the peak current, but rather the integral (area under the curve)
divided by time. Here is the product of the sinusoidal Voltage and Current
P ( t ) = V ( t ) I ( t ) = (VP sin ωt )( I P sin ωt ) = VP I P sin 2 ω t
P
t
If we “bulldoze” the peaks and dump the contents into the valleys, we get a flat
surface representing the average or “root mean square” power. We define
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VRMS =
1
VPEAK
2
and
I RMS =
1
I PEAK
2
The power is then very easy to calculate – it’s just the product of the RMS
voltage and current
PAVE = (VRMS )( I RMS )
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Section 6
Mutual Inductance, Transformers & Power Transmission
Suppose you lived in New York a hundred years ago.
Q: Why are New Yorkers always so depressed?
A: Because the light at the end of their tunnel is New Jersey.
Back then, electric power was delivered to local residents by Direct Current,
generated at a number of local Edison power generating stations. Why local?
Up to now we have considered circuits where the wires connecting the
elements were ideal conductors, that is, true equipotentials. What happens
when we replace these ideal wires with real ones?
Let’s say you have twelve 100-Watt light bulbs turned on in your house. The
power you are using is 1200 Watts and, if your house is wired for 120 Volts,
the current your house draws through the power line must be
I=
P 1200 W
=
= 10 A
120 V
V
But the power line wire has some resistance because it is not “ideal”. If you
live next to the power station, the resistance might be R = 1Ω so the voltage
drop across the line will be V = IR = (10 A )(1Ω ) = 10 V . Hence, if the power
station generates 120 Volts, your house will see only 110 Volts.
But what if you live further away where the total line resistance RTOT = 10Ω ?
Then the voltage loss must be 100 Volts, and you only get 20 Volts. In fact,
since it is the fixed resistance of the bulb and the assumed line voltage that
determine its power, your 100-Watt light bulbs will glow very dimly indeed!
To make matters worse, some of the generated power is lost in transmission,
PLOST = I 2 RLINE = (10 A ) (10Ω ) = 1000 W
2
In other words, half the generated power is lost along the way before it even
gets to your house.
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Transformers
The solution to this dilemma was provided by George Westinghouse who
began supplying power via Alternating Current. The big advantage comes from
the fact that, for a given power – say 1200 Watts – you have the freedom to
choose the voltage and current, provided that their product is fixed.
George used an invention of Nikolai Tesla who made use of the principle of
mutual induction to change the transmission voltage at will. Recall that a coil
of wire (solenoid) has a self-inductance due to the changing magnetic flux
inside it but, if that flux also threads another coil, then the two coils become
magnetically “coupled”. Suppose we arrange to conserve the flux by guiding it
through an iron core as shown below:
Recall that the flux threading a
solenoid is the product of the
field, area and number of turns
Φ B = NBA
?
Hence the ratio of the fluxes
through the two coils equals
the ratio of the turns.
An alternating current through the top coil – called the primary – produces an
alternating magnetic flux through the bottom coil – called the secondary – so,
by Faraday’s Law, there should be an alternating electromotive force (voltage)
across its windings. Ah, but Faraday’s Law says
emf = −
d ΦB
dt
so the output voltage must depend on the frequency of the input. So it does.
But, when the frequency is low enough (60 Hertz will do) we get a particularly
easy-to-use formula where N is the total number of turns
VSecondary
VPrimary
=
N Secondary
N Primary
Transformer Equation
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EXAMPLE
Consider the following situation. You own a power plant and must deliver One
Megawatt of power to a small town far away. You must choose your
transmission voltage carefully in order to minimize power being wasted along
the way. This loss is due to heating of the wires as current passes through them,
and is commonly know as “I-squared-R loss” from the formula P = I 2 R
We can model this system as a circuit where each “leg” of the transmission line
has resistance R representing a single strand of wire suspended between two
adjacent towers. Suppose R = 10−3 Ω (this is very small)
R
1MW
Sears
1
2
R
R
N
· · ·
R
R
R
RLOAD
6
To deliver a total of One Megawatt P = 10 Watt , we can choose between
or
I = 106 Ampere
I = 1 Ampere
V = 1 Volt
V = 106 Volt
at
at
The power lost just through one leg of the line then is either
(
) (10 Ω ) = 10 Watt
R = (1A ) (10 Ω ) = 10 Watt
P = I 2 R = 106 A
or
P = I2
2
2
−3
−3
9
−3
The first choice clearly loses all the power before getting to the first tower
while, in the second case, a thousand towers away only One Watt is lost. This
is why transmission lines are maintained on high towers at very high voltages.
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