Physics 122B
Electricity and Magnetism
Lecture 22 (Knight: 33.5 to 33.7)
Faraday’s Law
Martin Savage
Lecture 21 Announcements
 Lecture HW has been posted and is due on
Wednesday at 10 PM.
 Next Friday we will have Exam 3 in this room.
It will consist of multiple-choice questions on
the Laboratory (25 pts) and Lecture (35 pts).
Bring a Scatron sheet, a double-sided page of
notes, and a calculator with good batteries.
7/19/2016
Physics 122B - Lecture 22
2
Question
What is the ranking of the forces in the figure?
(a) F1=F2=F3=F4;
(d) F1=F4<F2=F3;
7/19/2016
(b) F1<F2=F3>F4;
(e) F1<F2<F3=F4;
Physics 122B - Lecture 22
(c) F1=F3<F2=F4;
3
Magnetic Flux
The number of arrows
passing through the loop
depends on two factors:
(1) The density of arrows,
which is proportional to B
(2) The effective area
Aeff = A cos q of the loop
We use these ideas to
define the magnetic flux:
Flux :  m  Aeff B  AB cos q
Flux units : 1 weber = 1 Wb = 1 Tm 2
7/19/2016
Physics 122B - Lecture 22
4
Area Vector
Define the area vector A of a loop such that it has the loop area
as its magnitude and is perpendicular to the plane of the loop. If a
current is present, the area vector points in the direction given by
the thumb of the right hand when the fingers curl in the direction of
current flow. If the area is part of a closed surface, the area vector
points outside the enclosed volume. With this definition:
m  Aeff B  AB cosq  A  B
7/19/2016
Physics 122B - Lecture 22
5
Example: A Circular Loop
Rotating in a Magnetic Field
The figure shows a 10 cm
diameter loop rotating in a
uniform 0.050 T magnetic field.
What is the magnitude of
the flux through the loop when
the angle is q=00, 300, 600, and
900?
A   R 2   (0.005 m)2  7.85 103 m2
 3.93 104 Wb for q  0

4
3.40 10 Wb for q  30
 m  AB cos q  
4
1.96

10
Wb for q  60


0 Wb for q  90
7/19/2016
Physics 122B - Lecture 22
6
Magnetic Flux
in a Nonuniform Field
So far, we have assumed that
the loop is in a uniform field. What
if that is not the case?
The solution is to break up the
area into infinitesimal pieces, each
so small that the field within it is
essentially constant. Then:
d m  B  dA
m 

B  dA
area of loop
7/19/2016
Physics 122B - Lecture 22
7
Example: Magnetic Flux from
a Long Straight Wire
The near edge of a 1.0 cm x 4.0 cm
rectangular loop is 1.0 cm from a long
straight wire that carries a current of
1.0 A, as shown in the figure.
What is the magnetic flux through
the loop?
dA  b dx
B
0 2 I
4 x
d  m  B  dA 
0
dx
2 Ib
4
x
ca
0

dx 0
ca
ca
m 
2 Ib 

2 Ib ln x c  0 2 Ib ln

4
x 4
4
c
c
 m  5.55 109 Wb
7/19/2016
Physics 122B - Lecture 22
8
Lenz’s Law (1)
Heinrich Friedrich Emil Lenz
(1804-1865)
In 1834, Heinrich Lenz announced a rule for determining the direction
of an induced current, which has come to be known as Lenz’s Law.
Here is the statement of Lenz’s Law:
There is an induced current in a closed conducting loop if and only if
the magnetic flux through the loop is changing. The direction of the
induced current is such that the induced magnetic field opposes the
change in the flux.
7/19/2016
Physics 122B - Lecture 22
9
Lenz’s Law (2)
If the field of the bar magnet is already in
the loop and the bar magnet is removed, the
induced current is in the direction that tries
to keep the field constant.
Superconducting
loop
If the loop is a superconductor, a persistent
standing current is induced in the loop, and the
field remains constant.
7/19/2016
Physics 122B - Lecture 22
10
Six Induced Current Scenarios
7/19/2016
Physics 122B - Lecture 22
11
Example: Lenz’s Law 1
-
+
- +
The switch in the circuit shown has been closed for a long time.
What happens to the lower loop when the switch is opened?
7/19/2016
Physics 122B - Lecture 22
12
Example: Lenz’s Law 2
+
-
The figure shows two solenoids facing each other.
When the switch for coil 1 is closed, does the current in coil 2
flow from right to left or from left to right?
7/19/2016
Physics 122B - Lecture 22
13
Example: A Rotating Loop
A loop of wire is initially in
the xy plane in a uniform
magnetic field in the x
direction. It is suddenly
rotated 900 about the y axis,
until it is in the yz plane.
In what direction will be the
induced current in the loop?
Initially there is no flux through the coil. After rotation the coil will
be threaded by magnetic flux in the x direction. The induced current in
the coil will oppose this change by producing flux in the –x direction. Let
your thumb point on the –x direction, and your fingers will curl clockwise.
Therefore, the induced current will be clockwise, as shown in the figure.
7/19/2016
Physics 122B - Lecture 22
14
Faraday’s Law
Consider the loop shown:
 m   B  dA  BA  Bl x
d m d
dx
 Bl x  Bl
dt
dt
dt
E  Blv  Bl
dx
dt
d m
Therefore, E 
dt
This is Faraday’s Law. It can be stated as follows:
An emf E is induced in a conducting loop if the magnetic flux m
through the loop changes with time, so that E = |dm/dt| for the
loop. The emf will be in the direction that will drive the induced
current to oppose the flux change, as given by Lenz’s Law.
7/19/2016
Physics 122B - Lecture 22
15
Example: Electromagnetic
Induction in a Circular Loop
The magnetic field shown in the figure
decreases from 1.0 T to 0.4 T in 1.2 s. A 6.0
cm diameter loop with a resistance of 0.010 W
is perpendicular to the field.
What is the size and direction of the
current induced in the loop?
I
d m
d ( r 2 B)
dB
E

  r2
dt
dt
dt
dB B 0.6 T


 0.50 T/s
dt
t
1.2 s
E   r2
dB
  (0.03 m) 2 (0.50 T/s)  1.4110-3 V
dt
E (1.4110-3 V)
I 
 0.141 A
R
(0.010 )
7/19/2016
Physics 122B - Lecture 22
The current direction is
such as to reinforce the
diminishing B field.
Therefore, the current I
will be clockwise.
16
Example: Electromagnetic
Induction in a Solenoid
A 3.0 cm diameter loop with a resistance
of 0.010 W is placed in the center of a
solenoid. The solenoid is 4.0 cm in diameter,
20 cm long, and is wound with 1000 turns of
square insulated wire. The current through
the solenoid wire as a function of time is
shown in (b).
Find the induced current in the loop.
B
dI sol
 10 A/s until t=1.0 s and =0 after that.
dt
0 NI sol
l
 m  BA 
0 ANI sol
E  1.97 105 V until t=1.0 s and =0 after that.
l
d  m 0 AN dI sol I loop  E / R  1.97 mA until t=1.0 s and =0 after that.

dt
l
dt
dI
 1.97 106 sol
dt
E
7/19/2016
Physics 122B - Lecture 22
17
What does Faraday’s
Law Tell Us?
Faraday’s Law tells us that all induced currents are the associated
with a changing magnetic flux. There are two fundamentally
different ways to change the magnetic flux through a loop:
(1) The loop can move, change size, or rotate, creating motional emf;
(2) The magnetic field can change in magnitude or direction.
We can write:
E
d m
d ( B  A)
dA

 B

dt
dt
dt
motional
emf
A
dB
dt
new
physics
The second term says that an emf can be created simply by changing
a magnetic field, even if nothing is moving.
7/19/2016
Physics 122B - Lecture 22
18
An Unanswered Question
A very long solenoid with no
field outside passes through a
conducting loop. The current in
the solenoid is increased so that
the B field inside the solenoid
increases. (B outside = 0).
There is no B-field at the loop wire. Is a current induced in the
loop?
YES! Since the flux through the loop changes, an emf is induced
in the loop, even though the field that produces the flux does not
touch the loop.
How can this happen? Faraday would say that when the number of
lines of force in the solenoid increases, they must “come in” from
infinity and must cut through the loop on their way in.
7/19/2016
Physics 122B - Lecture 22
19
Question
A conducting loop is half way into a magnetic field. Suppose that the
field begins to increase rapidly in strength.
Which statement describes the behavior of the loop?
(a) The loop is pushed upward, toward the top of the page;
(b) The loop is pushed downward, toward the bottom of the page;
(c) The loop is pushed to the left, into the magnetic field;
(d) The loop is pushed to the right, out of the magnetic field;
(e) The tension in the wire increases, but the wire does not move.
7/19/2016
Physics 122B - Lecture 22
20
Induced Fields and
Electromagnetic Waves
There is still a puzzle piece missing.
Faraday’s Law allows us to calculate an induced
current, but what causes the current? What
force pushes the electrons around in the wire? If the wire is stationary,
there can be no motional vxB magnetic force. Therefore, there must be an
induced electric field. Thus, there are two ways to create an electric field:
(1) A Coulomb electric field that is created by positive or negative charges;
(2) A non-Coulomb electric field that is created by a changing magnetic field.
7/19/2016
Physics 122B - Lecture 22
21
Maxwell’s Theory
Maxwell produced a mathematical
formulation of Faraday’s lines of force
picture.
He reasoned from this that if a
changing magnetic field produces an
electric field, then a changing electric
field should be equivalent to a current in
producing a magnetic field.
Otherwise, there is a paradox. An
Amperian loop near a charging capacitor
will predict a different magnetic field,
depending on whether the surface
enclosed by the loop passes through
the current (a) or through the
capacitor gap (b). If the changing
electric field is effectively a current
(called the “displacement current”)
there is no paradox.
7/19/2016
Physics 122B - Lecture 22
James Clerk Maxwell
(1831-1879)
22
Electromagnetic Waves
Maxwell’s formulation of electricity and
magnetism has an interesting consequence.
The equations can be manipulated to give a
wave equations for E and B of the form:
d 2E
d 2E
  0 0 2
2
dx
dt
This can be recognized as describing an
electromagnetic wave traveling through space
with a velocity of:
vEM wave 
(4  9.0 109 Nm 2 /C 2 )
8


3.0

10
m/s
7
2
(4 10 N/A )
 0 0
1
This is quite a remarkable result. Somehow, equations for charges and
currents making stationary electric and magnetic fields are telling us about
electromagnetic waves traveling through space at the speed of light!
7/19/2016
Physics 122B - Lecture 22
23
Generators
The figure shows a coil with
N turns rotating in a magnetic
field, with the coil connected to
an external circuit by slip rings
that transmit current
independent of rotation. The
flux through the coil is:
 m  A  B  AB cos q
 AB cos t
Ecoil  N
d m
d
 ABN  cos t    ABN sin t
dt
dt
Therefore, the device produces emf and current that will vary
sinusoidally, alternately positive and negative. This is called an alternating
current generator, producing what we call AC voltage.
7/19/2016
Physics 122B - Lecture 22
24
Example: An AC Generator
A coil with area 2.0 m2 rotates in a 0.10 T magnetic field at a
frequency of 60 Hz. How many turns are needed to generate an AC emf
with a peak voltage of 160 V?
Ecoil   ABN sin t
Emax   ABN
  2 f
Emax
(160 V)
N

 21 turns
2
2 f AB 2 (60 Hz)(2.0 m )(0.10 T)
7/19/2016
Physics 122B - Lecture 22
25
Transformers
When a coil wound around an
iron core is driven by an AC voltage
V1cos wt, it produces an oscillating
magnetic field that will induce an
emf V2cos wt in a secondary coil
wound on the same core. This is
called a transformer.
The input emf V1 induces a
current I1 in the primary coil that
is proportional to 1/N1. The flux in
the iron is proportional to this, and
it induces an emf V2 in the secondary
coil that is proportional to N2. Therefore, V2 = V1(N2/N1). From
conservation of energy, assuming no losses in the core, V1I1 = V2I2.
Therefore, the currents in the primary and secondary are related by the
relation I1 = I2(N2/N1).
A transformer with N2>>N1 is called a step-up transformer, which
boosts the secondary voltage. A transformer with N2<<N1 is called a stepdown transformer, and it drops the secondary voltage.
7/19/2016
Physics 122B - Lecture 22
26
The Tesla Coil
A special case of a step-up
transformer is the Tesla coil. It
uses no magnetic material, but
has a very high N2/N1 ratio and
uses high-frequency electrical
current to induce very high
voltages and very high
frequencies in the secondary.
There is a phenomenon called
“the skin effect” that causes
high frequency AC currents to
reside mainly on the outer
surfaces of conductors. Because
of the skin effect, one does not
feel (much) the electrical
discharges from a Tesla coil.
7/19/2016
Physics 122B - Lecture 22
27
Metal Detectors
Metal detectors like those used at
airports can detect any metal objects,
not just magnetic materials like iron.
They operate by induced currents.
A transmitter coil sends high
frequency alternating currents that
will induce current flow in conductors
in its field. Because of Lenz’s Law,
the induced current opposes the field
from the transmitter, so that net
field is reduced. A receiver coil
detects the reduction in the magnetic
fields from the transmitter and
registers the presence of metal.
7/19/2016
Physics 122B - Lecture 22
28
(Self-) Inductance
We define the inductance L of a coil of wire producing flux m as:
m
L
I
The unit of inductance is the henry: 1 henry = 1 H = 1 T m2/A = 1 Wb/A
The circuit diagram symbol used to represent inductance is:
Example: The inductance of a long solenoid with N turns of
cross sectional area A and length l is:
0 NI
 per turn  BA
B
 m  N  per turn  NBA 
7/19/2016
0 N A
l
2
l
I
Physics 122B - Lecture 22
Lsolenoid
 m 0 N 2 A


I
l
29
Example: Length of an Inductor
An inductor is made by tightly winding 0.30 mm diameter wire around
a 4.0 mm diameter cylinder.
What length cylinder has an inductance of 10 mH?
L
0 N 2 A
l

0 (l / d )2 ( r 2 )
l

0 r 2l
d
2
 1.0 105 H
d 2L
(3.0 104 m) 2 (1.0 105 H)
l

 0.057 m  5.7 cm
2
7
3
2
0 r
(4 10 Tm/A) (2.0 10 m)
7/19/2016
Physics 122B - Lecture 22
30
Potential Across an Inductor
Ecoil  N
d  per turn
dt
Ecoil  L
7/19/2016
d m

dt
dI
dt
Physics 122B - Lecture 22
31
Potential Across an Inductor (2)
Ecoil  L
7/19/2016
dI
dt
VL   L
Physics 122B - Lecture 22
dI
dt
32
Lecture 21 Announcements
 Lecture HW has been posted and is due on
Wednesday at 10 PM.
 Next Friday we will have Exam 3 in this room.
It will consist of multiple-choice questions on
the Laboratory (25 pts) and Lecture (35 pts).
Bring a Scatron sheet, a double-sided page of
notes, and a calculator with good batteries.
7/19/2016
Physics 122B - Lecture 22
33