Chapter 20: Electromagnetic
Induction
•Motional EMF
•Electric Generators
•Faraday’s Law
•Lenz’s Law
•Transformers
•Eddy Currents
•Induced Electric Fields
•Mutual- and Self-Inductance
•LR Circuits
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§20.1 Motional EMF
Consider a conductor in a B-field moving to the right.
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FB = q(v × B )
An electron in the conductor
experiences a force downward.
eV
The electrons in the
bar will move toward
the bottom of the bar.
F
This creates an electric field in the bar and results in a
potential difference between the top and bottom of the
bar.
3
What if the bar were placed across conducting rails (in red)
so that there is a closed loop for the electrons to follow?
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In this circuit, the electrons flow clockwise;
the current is counterclockwise.
Öinduce emf because of motion
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The motional EMF is
ε = vBL
where L is the separation
between the rails.
The current in the rod is
ΔV ε vBL
I=
= =
R
R
R
where R is the resistance
in the “wires”.
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The rod has a current through it. What is the direction of
the magnetic force on the rod due to the external magnetic
field?
F = I (L × B )
The magnitude of the magnetic force on the rod is:
vBL
vB 2 L2
LB =
F = ILB sin 90° = ILB =
R
R
Using the right hand rule, the force on the bar is directed
to the left.
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To maintain a constant EMF, the rod must be towed to the
right with constant speed. An external agent must do work
on the bar. (Energy conservation)
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Faraday’s Law
At the moment one closes the
switch ammeter deflect and
returns immediately to zero.
Again deflection at the moment
opening the switch
There is no contact between
the coils. We called this
phenomenon a induce current.
If we induce a current we have
to have a potential difference:
induced emf (electromotive
force)
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Other way to show:
A coil experiences a current when magnetic passing through it
varies
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The magnetic flux is proportional to the number of B-field
lines that cross a given area.
Φ B = BA cos θ
Loop of wire
with area A
The unit of magnetic flux is
the weber: 1 Wb = 1 Tm2
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Faraday’s Law:
Φ final − Φ initial
ΔΦ
emf = − N
= −N
t final − tinitial
Δt
An induced EMF in a “coil” of N loops is due to a changing
magnetic flux.
Ways to induce an EMF:
1. Vary the magnetic field.
2. Vary the area of the coil.
3. Change the angle between B and A.
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12
Example: If the magnetic field in a region varies with time
according to the graph shown below, find the magnitude of
the induced EMF in a single loop of wire during the following
time intervals: (a) 0-2.0 ms, (b) 2.0-4.0 ms, and (c) 4.0-8.0
ms. The loop has area 0.500 m2 and the plane of the loop is
perpendicular to the B-field.
B (T)
0.50 T
t (ms)
2
4
8
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Example continued:
Using Faraday’s Law:
⎛ ΔB ⎞
⎛ ΔΦ B ⎞
ε = −⎜
⎟
⎟ = − A⎜
⎝ Δt ⎠
⎝ Δt ⎠
This is the slope of
the given B versus
time graph.
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Example continued:
(a) In the interval 0.0-2.0 ms,
⎛ ΔB ⎞
2 ⎛ 0.50T-0.00T ⎞
ε = − A⎜
⎟ = 130 V.
⎟ = 0.500 m ⎜
−3
⎝ 2.0 × 10 s ⎠
⎝ Δt ⎠
(
)
(b) In the interval 2.0-4.0 ms,
⎛ ΔB ⎞
2 ⎛ 0.50T-0.50T ⎞
ε = − A⎜
⎟ = 0 V.
⎟ = 0.500 m ⎜
−3
⎝ 2.0 × 10 s ⎠
⎝ Δt ⎠
(
)
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Example continued:
(c) In the interval 4.0-8.0 ms,
⎛ ΔB ⎞
2
ε = − A⎜
⎟ = 0.500 m
⎝ Δt ⎠
(
)
⎛ 0.00T-0.50T ⎞
⎜
⎟ = 63 V.
−3
⎝ 4.0 × 10 s ⎠
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Lenz’s Law
The direction of induced EMFs and currents always oppose
the change in flux that produced them.
That is, the induced I (and thus induced B) tries
to keep the total flux through the loop constant.
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Example: Towing the bar to the right produced an induced
current that was CCW. What is the direction of the induced
magnetic field?
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The induced B is out of the page to maintain the flux
originally through the loop before the bar started to move
to the right (the area of the loop is increasing).
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Example (text problem 20.12): A long straight wire carrying a
steady current is in the plane of a circular loop of wire. (a) If
the loop of wire is moved closer to the wire, what is the
direction of the induced current in the wire loop?
I
Wire loop
There is a magnetic field into the page at the location of the
loop. As the loop gets closer to the wire there is an increase
in flux. To negate this increase in flux, the induced B-field
must point out of the page. This requires a CCW current.
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Example continued:
(b) At one instant, the induced EMF in the loop is 3.5 mV.
What is the rate of change of the magnetic flux through
the loop in that instant?
ΔΦ B
ε =−
= 3.5 mV = 3.5 × 10 −3 Wb / s
Δt
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Electric Generators
A coil of wire is spun in a magnetic field. This produces
an EMF and also a current; both vary with time. (ACalternating current)
An energy source is needed to turn the wire coil. Examples
include burning coal or natural gas to produce steam; falling
water.
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The EMF produced by an AC generator is:
ε (t ) = ε 0 sin ωt
In the United States and Canada ε0 = 170 volts and f =
ω/2π = 60 Hz.
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§20.5 Transformers
Wrap an iron
core with wire.
Primary
coil
Secondary
coil
Apply a varying voltage to the primary coil. This causes a
changing magnetic flux in the secondary coil.
ΔΦ B
ε 1 = − N1
Δt
ΔΦ B
ε 2 = −N2
Δt
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Since the flux through the coils is the same
ε 1 N1
=
ε 2 N2
The “turns ratio” gives
the ratio of the EMFs.
Depending on the turns ratio, a transformer can be used to
step-up or step-down a voltage.
The rate that power is supplied to both coils is the same
ε 1 I 2 N1
= =
ε 2 I1 N 2
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Example (text problem 20.25): A step-down transformer has
a turns ratio of 1/100. An AC voltage of amplitude 170 V is
applied to the primary. If the primary current is 1.0 mA, what
is the secondary current?
I 2 N1
=
I1 N 2
⎛ N1 ⎞
⎛ 100 ⎞
⎟⎟ I1 = ⎜
I 2 = ⎜⎜
⎟1.0 mA = 0.1 A
⎝ 1 ⎠
⎝ N2 ⎠
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Example (text problem 20.27): The primary coil of a
transformer has 250 turns and the secondary coil has 1000
turns. An AC voltage is sent through the primary. The EMF
of the primary is 16.0 V. What is the EMF in the secondary?
ε 1 N1
=
ε 2 N2
⎛ N2 ⎞
⎛ 1000 ⎞
ε 2 = ⎜⎜ ⎟⎟ε 1 = ⎜
⎟16.0 V = 64.0 V
⎝ 250 ⎠
⎝ N1 ⎠
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Eddy Currents
If a conductor is subjected to a changing magnetic flux, a
current will flow. (This includes sheets of metal, etc.)
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Consider a metal plate that swings through a magnetic field.
pivot
X
An external magnetic
field into the page
created by a magnet.
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As the plate swings through the region of magnetic field,
some regions of the plate are entering the B-field (increasing
flux), and other regions of the plate are leaving the B-field
(decreasing flux). There will be induced currents in the
conductor called eddy currents.
The eddy currents dissipate energy (according to I2R); this
results in the damping of the amplitude of the metal sheet.
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Induced Electric Fields
When a stationary conductor sits in a changing magnetic
field it is an induced electric field that causes the charges
in the conductor to move.
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Mutual- and Self-Inductance
A variable current
I1 flows in coil 1.
Coil 1
Coil 2
I1 then induces a
current in coil 2.
The flux (Φ21) through coil 2 due to coil 1 is
N 2 Φ 21 ∝ I1.
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Writing this as an equality,
N 2 Φ 21 = MI1
Where M is the mutual inductance. It depends only on
constants and geometrical factors. The unit of inductance is
Henry (1H = 1Vs/A).
ΔΦ 21
ΔI1
= −M
ε 2 = −N2
Δt
Δt
The induced EMF in the coils will be:
ΔΦ12
ΔI 2
ε 1 = − N1
= −M
Δt
Δt
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Self-inductance occurs when a current carrying coil
induces an EMF in itself.
The definition of self-inductance (L) is
NΦ = LI .
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Example (text problem 20.41): The current in a 0.080 Henry
solenoid increases from 20.0 mA to 160.0 mA in 7.0 s. Find
the average EMF in the solenoid during that time interval.
ΔΦ
ΔI
= −L
ε = −N
Δt
Δt
⎛ 160 mA − 20 mA ⎞
= −(0.080 H )⎜
⎟
7.0 s
⎝
⎠
= −1.6 × 10 −3 V
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An inductor stores energy in its magnetic field according to:
1 2
U = LI
2
The energy density in a magnetic field is:
uB =
1
2μ0
B2
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§20.9 LR Circuits
An inductor
and resistor
are connected
in series to a
battery.
As with an RC circuit, the current in the circuit varies
with time.
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ΔI
.
The voltage drop across an inductor is given by ε L = L
Δt
When an inductor is “charging” (the energy stored is
increasing) the current in the circuit is:
(
I (t ) = I f 1 − e − t /τ
)
Where τ = L/R is the time constant for the circuit and If =
εb/R maximum current in the circuit.
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Applying Kirchhoff’s loop rule to the circuit gives the EMF
in the inductor as:
ε L = ε b − IR = ε b e
− t /τ
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Plots of εL(t) and I(t) for this LR circuit:
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For a “discharging” inductor,
I (t ) = I 0 e
− t /τ
where I0 is the current in
the inductor when t=0.
The LR circuit time constant τ plays the same role as in
an RC circuit.
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Example: A coil has an inductance of 0.15 H and a
resistance of 33.0 Ω. The coil is connected to a 6.0 V ideal
battery. When the current reaches one-half the maximum
value:
(a) At what rate is the magnetic energy being stored in
the inductor?
⎛ I max ⎞⎛ Vmax ⎞
Power = P = IV = ⎜
⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠
Vmax= emf of the battery (εb)
= 6.0 Volts
I max =
εb
R
= 0.18 Amps
⎛ I max ⎞⎛ Vmax ⎞
P=⎜
⎟⎜
⎟ = 0.27 Watts
⎝ 2 ⎠⎝ 2 ⎠
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Example continued:
(b) At what rate is energy being dissipated?
Energy is dissipated in the resistor at a rate
2
⎛ I max ⎞
2
P= I R=⎜
⎟ R
⎝ 2 ⎠
2
⎛ 0.18 Amps ⎞
=⎜
⎟ (33.0Ω ) = 0.27 Watts.
2
⎝
⎠
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Example continued:
(c) What is the total power the battery supplies?
The battery must supply energy to the inductor and the
resistor. Part a and b calculate the rate at which energy
is delivered to the inductor and resistor respectively; the
battery must supply the sum of these: Pbattery = 0.54
Watts.
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Summary
•Motional EMF
•Faraday’s Law
•Lenz’s Law
•Transformers
•Eddy Currents
•Inductance and Inductors
•LR Circuits
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