Chapter 20: Electromagnetic Induction
PHY2054: Chapter 20
1
Topics
ÎElectromagnetic
Induction
‹ Magnetic
flux
‹ Induced emf
„ Faraday’s Law
„ Lenz’s Law
„ Motional emf
‹ Magnetic energy
‹ Inductance
‹ RL circuits
‹ Generators and transformers
PHY2054: Chapter 20
2
Reading Quiz 1
ÎMagnetic
‹ 1)
‹ 2)
‹ 3)
‹ 4)
‹ 5)
flux through a wire loop depends on:
thickness of the wire
resistivity of the wire
geometrical layout of the wire
material that the wire is made of
none of the above
Φ B = BA cosθ
Flux depends only on geometrical properties
PHY2054: Chapter 20
3
Reading Quiz 2
ÎAn
induced emf produced in a motionless circuit is due to
‹ 1)
‹ 2)
‹ 3)
‹ 4)
‹ 5)
a static (steady) magnetic field
a changing magnetic field
Faraday’s law
a strong magnetic field
the Earth’s magnetic field
a zero magnetic field
PHY2054: Chapter 20
4
Reading Quiz 3
ÎMotional
which is:
‹ 1)
‹ 2)
‹ 3)
‹ 4)
‹ 5)
emf relates to an induced emf in a conductor
long
sad
stationary
insulated
moving
Potential difference proportional to velocity
PHY2054: Chapter 20
5
Reading Quiz 4
Î
Faraday’s law says that
1. an emf is induced in a loop when it moves through an electric
field
2. the induced emf produces a current whose magnetic field
opposes the original change
3. the induced emf is proportional to the rate of change of magnetic
flux
Faraday’s law
PHY2054: Chapter 20
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Reading Quiz 5
ÎA
generator is a device that:
‹ a)
transforms mechanical into electrical energy
‹ b) transforms electrical into mechanical energy
‹ c) transforms low voltage to high voltage
PHY2054: Chapter 20
7
Magnetic Flux
Î Define
magnetic flux ΦB
Φ B = B ⋅ A = BA cosθ
‹
θ is angle between B and the
normal to the plane
‹
Flux units are T-m2 = “webers”
Î When
B field is not constant or
area is not flat
‹
Integrate over area
Φ B = ∫ B ⋅ dA
A
‹
Won’t cover this case here
PHY2054: Chapter 20
8
Φ B = B ⋅ A = BA cosθ
ΦB = 0
ΦB =
1
2
BA
PHY2054: Chapter 20
Φ B = BA
9
Experimental Observation of Induction
This effect can be quantified by Faraday’s Law
PHY2054: Chapter 20
10
Faraday’s Law of Induction
ΔΦ B
ε = −N
Δt
rate of change
of flux with time
induced
emf
number
of loops
¾ The
faster the change, the larger the induced emf
¾ Flux change caused by changes in either B, area, orientation
¾ Minus sign explained next slide
PHY2054: Chapter 20
11
Faraday’s Law of Induction
ΔΦ B
ε = −N
Δt
induced
emf
rate of change
of flux with time
number
of loops
¾
¾
Minus sign from Lenz’s Law:
Induced current produces a magnetic field which
opposes the original change in flux
PHY2054: Chapter 20
12
Comment on Lenz’s Law
ÎWhy
does the induced current oppose the change in flux?
ÎConsider
the alternative
‹ If
the induced current reinforced the change
‹ … then the change would get bigger
‹ … which would then induce a larger current
‹ … and then the change would get even bigger
‹ … and so on . . .
‹A
clear violation of conservation of energy!!
PHY2054: Chapter 20
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Direction of Induced Current
Bar magnet moves through coil
¾ Current induced in coil
Reverse pole
¾ Induced current changes sign
A
S
N
B
N
S
v
v
v
Coil moves past fixed bar magnet
¾
C
S
N
D
S
N
Current induced in coil as in (A)
Bar magnet stationary inside coil
¾ No current induced in coil
PHY2054: Chapter 20
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ConcepTest: Lenz’s Law
ÎIf
a North pole moves towards the loop from above the
page, in what direction is the induced current?
‹ (a)
clockwise
‹ (b) counter-clockwise
‹ (c) no induced current
Must counter flux change in
downward direction with upward B field
PHY2054: Chapter 20
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ConcepTest: Induced Currents
ÎA
wire loop is being pulled through a uniform magnetic
field. What is the direction of the induced current?
‹ (a)
clockwise
‹ (b) counter-clockwise
‹ (c) no induced current
No change in flux, no induced current
x x x x x x x x x x x x x
x x x x x x x x x x x x x
x x x x x x x x x x x x x
x x x x x x x x x x x x x
x x x x x x x x x x x x x
x x x x x x x x x x x x x
x x x x x x x x x x x x x
PHY2054: Chapter 20
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ConcepTest: Induced Currents
x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x1x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x x x x x x x x x x
2
x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x x x x x x x x x x
x xCounterclockwise
x x x x x x x x x x x x x x x x xNone
x x x x x x
In the cases above, what is the
direction of the induced current?
PHY2054: Chapter 20
17
ConcepTest: Lenz’s Law
ÎIf
a coil is shrinking in a B field pointing into the page, in
what direction is the induced current?
‹ (a)
clockwise
‹ (b) counter-clockwise
‹ (c) no induced current
Downward flux is decreasing,
so need to create downward
B field
PHY2054: Chapter 20
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Induced currents
ÎA
circular loop in the plane of the paper lies in a 3.0 T
magnetic field pointing into the paper. The loop’s diameter
changes from 100 cm to 60 cm in 0.5 s
‹ What
is the magnitude of the average induced emf?
‹ What is the direction of the induced current?
‹ If the coil resistance is 0.05Ω , what is the average induced current?
ΔΦ B
= 3.0 ×
V =
Δt
‹ Direction
‹ Current
(
π 0.32 − 0.52
0.5
) = 3.016 Volts
= clockwise (Lenz’s law)
= 3.016 / 0.05 = 60.3 A
PHY2054: Chapter 20
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ConcepTest: Induced Currents
ÎA
wire loop is pulled away from a current-carrying wire.
What is the direction of the induced current in the loop?
‹ (a)
clockwise
‹ (b) counter-clockwise
‹ (c) no induced current
Downward flux through loop
decreases, so need to create
downward field
I
PHY2054: Chapter 20
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ConcepTest: Induced Currents
ÎA
wire loop is moved in the direction of the current. What
is the direction of the induced current in the loop?
‹ (a)
clockwise
‹ (b) counter-clockwise
‹ (c) no induced current
Flux does not change when
moved along wire
I
PHY2054: Chapter 20
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ConcepTest: Lenz’s Law
Î If
the B field pointing out of the
page suddenly drops to zero, in
what direction is the induced
current?
(a) clockwise
‹ (b) counter-clockwise
‹ (c) no induced current
‹
Upward flux through loop
decreases, so need to create
upward field
Î If
a coil is rotated as shown, in a
B field pointing to the left, in
what direction is the induced
current?
(a) clockwise
‹ (b) counter-clockwise
‹ (c) no induced current
‹
Flux into loop is increasing, so
need to create field out of loop
PHY2054: Chapter 20
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ConcepTest: Induced Currents
ÎWire
#1 (length L) forms a one-turn loop, and a bar
magnet is dropped through. Wire #2 (length 2L) forms a
two-turn loop, and the same magnet is dropped through.
Compare the magnitude of the induced currents in these
two cases.
‹ (a) I1
‹ (b) I2
= 2 I2
Voltage doubles, but R also
= 2 I1
doubles, leaving current the same
‹ (c) I1 = I2 ≠ 0
‹ (d) I1 = I2 = 0
‹ (e) Depends on the strength of the magnetic field
PHY2054: Chapter 20
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Motional EMF
ÎConsider
a conducting rod moving on metal rails in a
uniform magnetic field:
ε
d Φ B d ( BA) d ( BLx )
dx
=
=
=
= BL
dt
dt
dt
dt
Current will flow counter-clockwise in this “circuit”. Why?
x x x x x x x x x
x x x x x x x x x
x x x x x x x x x
v
L
ε
= BLv
x x x x x x x x x
x x x x x x x x x
x
PHY2054: Chapter 20
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Force and Motional EMF
Î
Pull conducting rod out of B field
Î
Current is clockwise. Why?
ε
i=
BLv
=
R
R
Î
Current within B field causes force
B 2 L2 v
F = iLB =
R
Force opposes pull (RHR)
‹ Also follows from Lenz’s law
‹
Î
We must pull with this force
to maintain constant velocity
PHY2054: Chapter 20
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Power and Motional EMF
B 2 L2 v
ÎForce required to pull loop: F = iLB =
R
B 2 L2 v 2
ÎPower required to pull loop: P = Fv =
R
ÎEnergy
dissipation through resistance
2
2 2 2
BLv
B
Lv
⎛
⎞
2
P=i R=⎜
⎟ R=
R
⎝ R ⎠
ÎSame
as pulling power! So power is dissipated as heat
‹ Kinetic
energy is constant, so energy has to go somewhere
‹ Rod heats up as you pull it
PHY2054: Chapter 20
26
Example
ÎPull
a 30cm x 30cm conducting loop of aluminum through
a 2T B field at 30cm/sec. Assume it is 1cm thick.
‹ Circumference
= 120cm = 1.2m, cross sectional area = 10-4 m2
‹ R = ρL/A = 2.75 x 10-8 * 1.2 / 10-4 = 3.3 x 10-4 Ω
ÎEMF
ε = BLv = 2 × 0.3 × 0.3 = 0.18V
ÎCurrent
i = ε / R = 0.18 / 3.3 × 10−4 = 545A
ÎForce
F = iLB = 545 × 0.3 × 2 = 327 N
ÎPower
P = i R = 98 W
2
74 lbs!
About 0.330 C per sec
(from specific heat, density)
PHY2054: Chapter 20
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Electric Generators
ÎRotate
a loop of wire in a uniform magnetic field:
θ ⇒ changing flux ⇒ induced emf
= B A cos θ = B A cos(ωt)
‹ changing
‹ ΦB
Rotation:
PHY2054: Chapter 20
θ = ωt
28
Electric Generators
ÎFlux
is changing in a sinusoidal manner
‹ Leads
ε
to an alternating emf (AC generator)
dΦB
d cos(ωt )
=N
= NBA
= NBAω sin(ωt )
dt
dt
¾ This is how electricity is generated
¾ Water or steam (mechanical
power) turns the blades of a
turbine which rotates a loop
¾ Mechanical power converted to
electrical power
PHY2054: Chapter 20
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ConcepTest: Generators
ÎA
generator has a coil of wire rotating in a magnetic field.
If the B field stays constant and the area of the coil
remains constant, but the rotation rate increases, how is
the maximum output voltage of the generator affected?
ε
‹ (a)
Increases
‹ (b) Decreases
‹ (c) Stays the same
‹ (d) Varies sinusoidally
PHY2054: Chapter 20
= NBAω sin(ω t )
30
Induction in Stationary Circuit
ÎSwitch
closed (or opened)
‹ Current
ÎSteady
‹ No
induced in coil B (direction shown is for closing)
state current in coil A
current induced in coil B
A
B
PHY2054: Chapter 20
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Self - Inductance
ÎConsider
a single isolated coil:
‹ Current
(red) starts to flow clockwise due to the battery
‹ But the buildup of current leads to changing flux in loop
‹ Induced emf (green) opposes the change
This is a self-induced emf (also called “back” emf)
ε
dΦ
di
= −N
= −L
dt
dt
12V
L is the self-inductance
units = “Henry (H)”
PHY2054: Chapter 20
induced
emf
32
Inductance of Solenoid
ÎTotal
flux (length l)
B = μ 0in
N Φ B = ( n l )( B A ) = μ 0 n 2 A li
ε
dΦB
di
di
2
= −N
= − μ0n Al = − L
dt
dt
dt
L = μ0n Al
2
To make large inductance:
¾ Lots of windings
¾ Big area
¾ Long
PHY2054: Chapter 20
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LR Circuits
ÎInductance
ÎStart
and resistor in series with battery of EMF V
with no initial current in circuit
‹ Close
switch at t = 0
‹ Current is initially 0 (initial increase
causes voltage drop across inductor)
ÎFind
i(t)
ΔVR = Ri
‹ Inductor: ΔVL = L di/dt
‹ Use back emf of inductor
‹ Resistor:
V
R
L
V − ΔVL V − Ldi / dt
i (t ) =
=
R
R
PHY2054: Chapter 20
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Analysis of LR Circuit
ÎGeneral
solution
(
i = (V / R ) 1 − e
−tR / L
)
Rise from 0 with
time constant τ = L / R
Final current (maximum)
PHY2054: Chapter 20
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L-R Circuits (2)
ÎSwitch
off battery: Find i(t) if current starts at i0
i = i0e −tR / L
Exponential fall to 0 with
time constant τ = L / R
Initial current (maximum)
PHY2054: Chapter 20
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Exponential Behavior
Îτ
= L/R is the “characteristic time” of any RL circuit
‹ Only
Ît
t / τ is meaningful, just like for RC circuit
=τ
‹ Current
falls to 1/e = 37% of maximum value
‹ Current rises to 63% of maximum value
Ît
=2τ
‹ Current
falls to 1/e2 = 13.5% of maximum value
‹ Current rises to 86.5% of maximum value
Ît
=3τ
‹ Current
falls to 1/e3 = 5% of maximum value
‹ Current rises to 95% of maximum value
Ît
=5τ
‹ Current
falls to 1/e5 = 0.7% of maximum value
‹ Current rises to 99.3% of maximum value
PHY2054: Chapter 20
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ConcepTest: Generators and Motors
ÎA
current begins to flow in a wire loop placed in a
magnetic field as shown. What does the loop do?
‹ (a)
moves to the right
‹ (b) moves up
‹ (c) rotates around horizontal axis
‹ (d) rotates around vertical axis
‹ (e) moves out of the page
B
This is how a motor works !!
PHY2054: Chapter 20
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Electric Motors
¾ Current is supplied from an
external source of emf (battery
or power supply)
¾ Forces act to rotate the wire
loop
¾ A motor is essentially a
generator operated in reverse!
PHY2054: Chapter 20
39
Motor
Î
Forces act to rotate the loop
towards the vertical.
Î
When loop is vertical, current
switches sign and the forces
reverse, in order to keep the
loop in rotation.
Î
This is why alternating current is
necessary for a motor to
operate.
PHY2054: Chapter 20
40
Motors
Generators
Electrical ⇒ mechanical energy
Mechanical ⇒ electrical energy
PHY2054: Chapter 20
41
Energy of an Inductor
Î Energy
is stored in inductor when current flows through it
1 2
U L = Li
2
‹ Energy
Î Similar
UC
is stored in the B field
to energy in charged capacitor
q2 1
=
= 2 CV 2
2C
‹ Energy
is stored in the E field (between plates)
PHY2054: Chapter 20
42
Energy of Solenoid
ÎRecall
B field of solenoid (current i, n turns per meter)
B = μ 0 ni
ÎInductance
NΦB
L=
i
Φ B = BA = μ 0 niA
of solenoid
nl ) ( μ 0 niA )
(
=
= μ 0 n 2 Al
i
ÎEnergy
of solenoid can be written in terms of B field and
volume (V = lA)
2
2
⎛
⎞
B
B
2
2
U L = 12 Li = 12 μ 0 n lA ⎜
Al
⎟ =
2μ0
⎝ μ0n ⎠
(
)
Energy density
of B field
PHY2054: Chapter 20
Volume
43
Energy of Solenoid (2)
ÎTypical
MRI magnet
‹B
= 0.5 – 1.5 T
‹ R = 0.25 m, l = 2.5 m
V = Al = π R 2l ≅ 0.5m3
ÎAssume
UB
B = 1 T typically
B2
12
=
V =
( 0.5 ) ≈ 200 kJ
−
7
2μ0
2 × 4π × 10
‹ This
is a lot of energy
‹ Needs to be held in by strong materials
PHY2054: Chapter 20
44
Gigajoule Magnet at CERN
ÎCMS
experiment at CERN
‹ p-p
collisions at world’s highest energy in 2007
‹ Hope to discover new particles, find the origin of mass and new
fundamental forces
Compact Muon Solenoid
PHY2054: Chapter 20
45
Human
PHY2054: Chapter 20
46
CMS Experiment Magnet
ÎLarge
central solenoid magnet to study particle production
‹B
= 4T, R = 3.15 m, l = 12.5 m
‹ UB = 2.6 x 109 J = 2.6 gigajoules!!
UB
(
)
B2
42
2
=
π
×
Al =
3.15
(12.5 )
−
7
2μ0
2 × 4π × 10
http://www.spacedaily.com/news/energy-tech-04b.html
PHY2054: Chapter 20
47
CMS Articles and Pictures
ÎHome
page
‹ http://cmsinfo.cern.ch/Welcome.html/
ÎPictures
of detector
‹ http://cmsinfo.cern.ch/Welcome.html/CMSdetectorInfo/CMSdetect
orInfo.html
‹ http://cmsinfo.cern.ch/Welcome.html/CMSmedia/CMSmedia.html
ÎInteresting
article on solenoid, with pictures
‹ http://cmsinfo.cern.ch/Welcome.html/CMSdocuments/MagnetBroc
hure/MagnetBrochure.pdf
ÎOther
documents & pictures about CMS
‹ http://cmsinfo.cern.ch/Welcome.html/CMSdocuments/CMSdocume
nts.html
PHY2054: Chapter 20
48