Chapter 20.1 Induced EMF and magnetic flux
Electric current gives rise to magnetic fields
B
I
Can a magnetic field give rise to a current?
The answer is yes as
discovered by Michael
Faraday.
Michael Faraday
1791-1867
Chapter 20.1 Induced EMF and magnetic flux
×
Moving magnetic fields induce
currents
B
×
Change the strength of B:
Induced current
B
×
B
Induced EMF!
Moving current loop induces currents
v
Chapter 20.1 Induced EMF and magnetic flux
SI units: weber (Wb)
Fig. 20.2
20.2&20.4 Faraday’s law of induction
Faraday’s law:
The induced emf, ɛ, in a closed wire is
equal to the time rate of change of the
magnetic flux, ΦB.
An induced emf, ɛ, always gives rise to a
current whose magnetic field opposes the
original change in magnetic flux.
Lenz’s law
Faraday’s Law and Lenz’s Law both
state that a loop of wire will want its
magnetic flux to remain constant.
Michael Faraday
1791-1867
2. A circular loop with a radius of 0.20 m is placed in a
uniform magnetic field B=0.85 T. The normal to the loop
makes an angle of 30o with the direction of B. The field
increases to 0.95 T what is the change in the magnetic flux
through the loop?
Change in flux?
∆Φ B = B ' A 'cosθ '− BA cosθ
30o
B=0.85T
A ' = A = π R2
θ ' = θ = 30o
∆Φ B = A cos θ (B '− B ) = π R 2 cos30(B '− B )
B’=0.95T
∆Φ B = π (0.2)2 cos30(0.95 − 0.85)
∆Φ B = 1.1x10 −2Wb
8. A circular coil with a radius of 20 cm is in a field of 0.2 T
with the plane of the coil perpendicular to the field. If the
coil is pulled out of the field in 0.30 s find the average
emf during this interval
∆Φ B
BA cos θ − 0
=N
ε =N
∆t
∆t
N= 1
B
cosθ= 1
πR2
A=
B=0
Bπ R 2 0.2π (0.2)2
ε=
=
0.3
∆t
ε = 8.4 x10−2V
20.3 Motional emf
A voltage is produced by a conductor moving in
a magnetic field
Charges in the conductor
experience a force upward
x
x
F
x
x
x
x
x
x
x
x
x
x
x
x
∆V
x
v
L
x
x
x
F = qvB
The work done in moving
a charge from bottom to
top
W = FL = qvBL
x
B into the page
x
The potential difference is
∆V =
W
= vBL
q
20.3 Motional emf
A voltage is produced by a conductor moving in
a magnetic field
Charges in the conductor
experience a force upward
x
x
x
F
x
x
x
x
x
∆V
x
x
v
x
x
x
x
L
x
x
x
F = qvB
x
V
o
lt
a
g
e
velocity
W = FL = qvBL
x
B into the page
x
The potential difference is
∆V =
W
= vBL
q
20.3 Motional emf
The potential difference can drive a current through a circuit
The emf arises from changing flux due to changing area
according to Faraday’s Law
∆x
∆AB ∆Φ B
∆V = vLB =
=
=ε
LB =
∆t
∆t
∆t
wire
x
R
∆V
x
I
F
x
x
x
x
x
x
x
x
v
ε
L
x
x
x
x
x
x
x
x
x
x
Changing Magnetic Flux
B into the page
BLv
=
I=
R
R
18. R= 6.0 Ω and L=1.2 m and B=2.5 T. a) What
speed should the bar be moving to generate a current
of 0.50A in the resistor? b) How much power is
dissipated in R? c) Where does the power come from?
ε
wire
x
R
∆V
x
x
a)
I
F
x
x
x
x
x
x
v
x
x
x
x
x
x
L
x
BLv
I= =
R
R
IR 0.5(6.0)
=
v=
BL 2.5(1.2)
v = 1.0m / s
x
x
x
B into the page
x
b)
P = I 2R = (0.5)2 (6.0)
P = 1.5W
c) Work is done by the
force moving the bar
20.4 Lenz’s law revisited
Lenz’s Law
The polarity of the induced emf is such that it
induces a current whose magnetic field opposes the
change in magnetic flux through the loop. i.e. the
current flows to maintain the original flux through the
loop.
B increasing in loop
Bin
B
S
N
V
ε
I
Bin acts to oppose the
-
change in flux
+
Current direction that
produces Bin is as
shown (right hand rule)
Emf has the polarity shown. ε drives current in
external circuit.
20.4 Lenz’s law revisited
Now reverse the motion of the magnet
The current reverses direction
B decreasing in loop
Bin
S
N
V
B
ε
I
Bin acts to oppose the
+
change in flux
-
Current direction that
produces Bin is as
shown (right hand rule)
Emf has the polarity shown.
20.4 Lenz’s law revisited
Lenz’s Law and Reaction Forces
Fmc
Fcm
Bin
B
S
N
V
N
A force is exerted by the
magnet on loop to produce the
current
S
I
A force must be exerted by
the current on the magnet to
oppose the change
The current flowing in the direction shown
induces a magnetic dipole in the current loop
that creates a force in the opposite direction
20.5 Generators
Flux through a rotating loop in a B field
Normal to the plane
θ
B
ε
The flux through the loop
Φ B = BA cos θ
θ = ωt
ω
= angular velocity (radians/s)
B
20.5 Generators
∆Φ B
Relation between ΦB and
∆t
BA
Φ B = BA cos ωt
ΦB
t
-BA
BAω
∆Φ B
∆t
-BAω
∆Φ B
= −BAω sin ωt
∆t
t
proportional to ω
20.5 Generators
The emf generated by a loop of N turns rotating at constant
angular velocity ω is
∆Φ B
ε = −N
∆t
ε = NBAω sin ω t
NBAω
ε
0
t
-NBAω
20.5 Generators
35. In a model ac generator, a 500 turn rectangular coil
8.0 cmx 20 cm rotates at 120 rev/min in a uniform magnetic
field of 0.60 T. a) What is the maximum emf induced in the
coil?
ε = NBAω sin ω t
The maximum value of ε
ε max = NBAω
ε max
(120 x 2π )
= (500)(0.6)(0.08 x 0.2)
= 60V
60
20.5 Generators
Alternating Current (AC) generator
Rotational
Work
20.5 Generators
Direct Current (DC) generator
20.5 Generators
A generator is motor acting in reverse
I
ε drives rotation
DC motor
20.6 Self-Inductance
• a property of a circuit carrying a current
• a voltage is induced that opposes the change in current
• used to make devices called inductors
Self- inductance of a circuit
a reverse emf is produce
by the changing current
∆Φ B
ε =−
∆t
20.6 Self-Inductance
Self-inductance of a coil
+
ε
-
B increases,
B
∆B
∆t
changes magnetic flux in
the coil,
I
Current
increasing
∆Φ B A∆B
=
∆t
∆t
Produces emf in coil
∆Φ B
A∆B
ε = −N
= −N
∆t
∆t
The direction of the induced emf opposes the change in
current.
20.6 Self-Inductance
A changing current in a coil induces an emf that
opposes the change
+ -
-
ε
+
I
I
I increasing
induced emf
opposes I
I decreasing
induced emf
supports I
ε
20.6 Self-Inductance
Inductance L is a measure of the
The self-induced emf is
self-induced emf
∆Φ B
ε = −N
∆t
ε
but
I
Current
increasing
∆Φ B ∆I
∝
∆t
∆t
proportionality constant is L
∆I
ε = −L
∆t
L is a property of the coil, Units of L , Henry (H)
Vs
A
20.6 Self-Inductance
Inductance of a solenoid with N turns and length ℓ,
wound around an air core (assume the length is much
larger than the diameter).
∆B
∆t
l
A
∆I
∆t
N
Φ B = BA = µo IA
l
∆Φ B
N ∆I
= µo
A
∆t
l ∆t
∆Φ B
N 2 ∆I
∆I
= − µo
A
= −L
ε = −N
∆t
l
∆t
∆t
N2
L = µo
A
l
inductance proportional to N squared x area/length
20.6 Self-Inductance
An air wound solenoid of 100 turns has a length of 10 cm
and a diameter of 1 cm. Find the inductance of the coil.
l= 10 cm
d=1 cm
I
2
2
2
N
N
d
L = µo
A = µo
π
l
l
4
2
2
−7
4π 10 (100) π (0.01)
−5
L=
= 1.0 x10 H
0.1(4)
20.7 RL circuits
The inductor prevents the
rapid buildup of current
∆I
ε = −L
∆t
But at long time does not
reduce the current, ∆I
∆t
=0
at t=∞
−
t
τ
I = Io (1 − e )
L
τ=
Applications of Inductors: R
Reduce rapid changes of
current in circuits
Produce high voltages in
automobile ignition.
20.8 Energy stored in a magnetic field
Energy is stored in a magnetic field of an inductor.
Bo
B increasing
B=0
I
ε
I=Io
Work is done against ε to produce the B field.
This produces a change in the PE of the inductor
1 2
PEL = LI
2
This stored PE can be used to do work