Electromagnetic induction
So far we have found that
Electric charges create E-fields
Moving electric charges create B-fields
v kQ
E = 2 rˆ
r
v µ 0 qvv × rˆ
B=
4π r 2
But this is not the whole story!
Electromagnetic induction
Electric Fields can also be created by a changing Magnetic Field.
This is described by “Faraday’s Law”.
Magnetic Fields can also be created by a changing Electric Field.
This will help complete the set of equations called Maxwell’s
equations of electricity and magnetism.
Electricity and Magnetism can be unified in a single theory!
The electricity and magnetic forces are manifestations of a
single “electromagnetic interaction”
2
Electromagnetic induction
Magnetic Flux
Magnetic Flux
The flux through an element of area is:
To calculate the flux through the surface, we
need to integrate:
B
φ
For the simple case of a flat surface, we obtain:
Magnetic flux
Clicker Question
At what angle of orientation of the surface is the
flux through the surface half of the maximum
possible flux?
A) φ = 0 degrees
B) φ = 30 degrees
C) φ = 45 degrees
D) φ = 60 degrees
E) None of the above
B
φ
7
Faraday’s Law of Induction
A changing (time-dependent) magnetic field “induces” (generates) an EMF ε.
The magnitude of the EMF equals minus the rate of change of the flux.
dΦ B
ε =−
dt
Two new quantities introduced.
1. Magnetic Flux = ΦB
2. Electro Motive Force (EMF) = ε
8
Important observations:
• Recall: EMF is not a force, but a source of voltage
capable of generating power.
• The EMF is called “motional” EMF, to distinguish
from “chemical” EMF (batteries)
• Recall: The magnetic flux through a closed surface is
zero! (Gauss’ Law for B-fields)
• The surface for Faraday’s Law is an open surface!
Clicker Question
We have a square imaginary loop with side of length a.
There is a Magnetic Field that is uniform throughout the
region as shown. How does the Magnetic Flux magnitude
|ΦB| change if we halve the Magnetic Field strength and
double the sides of the square loop?
A) Flux goes to ½ original value
B) Flux goes to ¼ original value
X B
C) Flux goes to 2 times original value
D) Flux does not change
E) None of the above
10
Changing the magnetic flux
1. Change the strength of the Magnetic Field
2. Change the area of the loop
3. Change the orientation of the loop (A vector) and the B-field
11
Clicker Question
A loop of wire is moving rapidly through a uniform magnetic field as
shown. Is a non-zero EMF induced in the loop?
A) No, there in no EMF
B) Yes, there is an EMF
12
Clicker Question
A loop of wire is spinning rapidly about a stationary axis in uniform
magnetic field as shown.
True or False: There is a non-zero EMF induced in the loop.
A) False, zero EMF
B) True, there is an induced EMF
F = BA cos φ
and the angle φ is changing.
13
Direction of the EMF around a loop
Lenz’s Law
The induced EMF tends to induce a current in the
direction that opposes the changes in magnetic flux
ε = −
dΦ
dt
B
The (-) negative sign in Faraday’s Law is more of a reminder.
Lenz’s Law is just another way to express Faraday’s Law. Moreover,
it can be deduced from it.
15
Lenz’s Law
If we move the magnet closer to the loop, what is the direction of
the induced EMF and thus the direction of the current in the loop?
v v
Φ B = B ⋅ dA
1. The Magnetic
Flux is increasing.
surf
∫
2. Therefore
dΦ B
>0
dt
3. Induced current must create B-field
that fights the change!
16
Lenz’s Law
dΦ B
>0
dt
i
B
Induced current must create B-field that
fights the change!
An induced B-field down through the loop
will create a slight decrease in ΦB.
Thus, we have determined the direction of
the induced current and EMF.
17
Lenz’s Law
“The Magnetic Flux ΦB Up Through the Loop is Increasing.
To Fight this Change, an Induced B-field is made that
Decreases the Flux Up Through the Loop.
This is accomplished by the induced current.”
18
Lenz’s law (summary)
Clicker Question
A bar magnet is positioned below a loop of wire. The magnet is
pulled down, away from the loop. As viewed from above, is the
induced current in the loop clockwise, counterclockwise, or zero?
B
eyeball
i
A: clockwise
B: counter-clockwise
C: zero
Answer: Counterclockwise.
As the magnet is pulled away, the flux is decreasing. To
fight the decrease, the induced B-field should add to the
original B-field.
20
Clicker Question
A loop of wire is sitting in a uniform, constant magnet field as
shown. Suddenly, the loop is bent into a smaller area loop. During
the bending of the loop, the induced current in the loop is ...
A: zero
B: clockwise
C: counterclockwise
B(in)
B(in)
Answer: Clockwise. The flux into the page is decreasing as the
loop area decreases. To fight the decrease, we want the
induced B to add to the original B. By the right hand rule
(version II) , a clockwise induced current will make an induced B
into the page, adding to the original B.
21
Clicker Question
Is there an induced current in this circuit? If so, what is its
direction?
A. Yes, clockwise
B. Yes, counterclockwise
C. No
* There is zero Magnetic Flux at all times through the loop.
22
Clicker Question
A current-carrying wire is pulled
away from a conducting loop in the
direction shown. As the wire is
moving, is there a clockwise current
around the loop, a counterclockwise
current or no current?
A. There is a clockwise current around the loop.
B. There is a counterclockwise current around the loop.
C. There is no current around the loop.
23
Area = (0.1m)2=0.01 m2
Example with numbers:
dB/dt = +0.1 Tesla/second
B (increasing)
| ε |=
cm
10
V
dΦ B d
dB
= ( BA) = A
dt
dt
dt
| ε |= (0.01)(0.1) = 0.001Volts
10 cm
Suppose there are 1000 loops to the wire coil.
| ε |=
dΦ B ( N )
dΦ B (1)
=N
= 1Volt
dt
dt
24
Generators
Generators
use Faraday’s Law to convert mechanical energy into
electrical energy.
This is the opposite of motors which convert electrical into
mechanical energy.
25
The alternator
• Wire loop in a constant external B-field.
• Turning a crank makes the loop rotate.
• This then induces a current in the wire loop.
B
B
X
X
i (induced)
26
The alternator
• Wire loop in a constant external B-field.
• Turning a crank makes the loop rotate.
• This then induces a current in the wire loop.
27
Clicker Question
Instead of turning a crank and inducing a current, what if you hook
up the wire loop to a battery and push current through it.
What happens to the loop?
B
X
B
X
i
A) Nothing
B) It feels a net force
C) It feels a net torque
D) The current stops flowing
Note that the wire loop is a rectangle that is slightly rotated
from the plane of the page.
28
Generator
Motor
Converts mechanical energy
into electrical energy.
Converts electrical energy into
mechanical energy.
Crank or otherwise
mechanically turns a wire loop
in an external B-field. This
induces an electric current in
the loop.
Battery creates a current in a
loop. This interacts with an
external B-field to create a
mechanical torque.
29
The “sliding bar generator”
Sliding metal bar on metal rails
L
X
B
v
X
x=vt
X
X
X
X
Bar is pulled by an
external agent to the
right at constant
velocity.
There is a uniform Bfield surrounding the
rails and loop
system.
30
The “sliding bar generator”
L
X
B
This creates a wire loop
whose area is growing with
time.
v
X
X
X
X
X
A = Lx = Lvt
x=vt
Thus, the Magnetic Flux is increasing with time!
d
d
d
Φ B = ( BA) = B ( Lvt ) = BLv
dt
dt
dt
31
Clicker Question
What is the direction of
the induced EMF and
thus current in the wire L
loop?
A) Clockwise
B) Counter Clockwise
Motional EMF induced
X
B
v
X
X
X
X
X
x=vt
ε = BLv
32
The “sliding bar generator”
There is another way to see why the current in the bar is up.
Charges in the bar are moving to the right
with the bar itself.
i
There is a B-field in this region.
v
Thus, there is an upward force on the charges.
v
v
v
F = qv × B = qvB (up)
X
B
Thus, there is an upward current in the bar!
33
The “sliding bar generator”
One more consequence….
FM
i FAgent
If there is an upward current in the wire and there
is an external B-field, the we get?
v
v v
FMag = IL × B = ILB (left)
v
X
B
But, there is also the external agent pulling the
rod to the right.
If the velocity is constant, what is true about the
forces? They must be equal and opposite!
34
Clicker Question
A conducting loop is halfway into a magnetic
field. Suppose the magnetic field begins to
increase rapidly in strength. What happens to
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 pulled to the left, into the magnetic field.
D. The loop is pushed to the right, out of the magnetic field.
E. The tension is the wires increases but the loop does not move.
Eddy currents
If a metal object (something in which electric
charges can move) and an external source of
Magnetic Field are in relative motion so that
there is changing Magnetic Flux through the
metal, then Faraday says there is an EMF and
thus an Eddy current.
If the metal is moving, then the direction of the
current is always such as to cause a force which
opposes the motion (slows the metal).
Eddy currents: Applications
Eddy Current Brakes on Trains
Since there is no direct contact, there is minimal friction.
Thus the brakes do not wear down!
Eddy currents: Applications
Metal detector
Portable metal detector
Eddy currents: Applications
Induction Cooktop
Works via changing B-field, inducing a current in the pot or wok (which must be
metal). This current then leads to power loss as heating of the pot. The cooktop
surface which is not metal and does not conduct, does not get hot.
Induced electric field
In the case where the loop or metal is stationary and the Magnetic
field is changing, then there is no Magnetic force on the charges in
the wire (since v=0)
v
v
v
FM = qv × B
Thus, the only way to explain the induced current in the wire is to
admit that there is the creation of a new Electric Field!
40
Motional EMF revisited
Technically, the EMF around a closed Loop is defined as:
v v
ε = ∫ E ⋅ dl
Loop
But, recall that Voltage is defined as:
v v
∆V = − ∫ E ⋅ dl
Was not this integral around a closed loop always zero?
The motional EMF is not always zero around a closed loop !
41
Faraday’s Law reformulated
ε = −
∫
Loop
v
v
E ⋅ dl = −
dΦ
dt
d
dt
B
∫
v
v
B ⋅ dA
surface
How do the loop and the surface relate?
42
Motional EMF revisited
∫
v
v
E ⋅ dl = −
Loop
d
dt
∫
v
v
B ⋅ dA
surf
Bexternal
v
E (2π R ) = −
R
[
d v
B ext ( π R 2 )
dt
43
Motional EMF revisited
Wire Loop
B
V
If B = constant,
Then Magnetic Flux = constant
dΦB/dt = 0
EMF = 0 and Voltmeter reads 0.
If B =increasing,
Then Magnetic Flux = changing
|dΦB/dt| > 0
EMF > 0 and Voltmeter reads > 0.
What if the wire loop was not there?
44
]
Induced electric field
A changing B-field creates a new kind of E-field.
B
If B is decreasing
E-field is CCW
If B is increasing
E-field is Clockwise
E
B
E
45
Induced electric field
The (non-Coulomb) E-field is very different from the Coulomb Efield created by single electric charges.
E
q
v v
∆V = − ∫ E ⋅ dl = 0
B
E
v v
ε = ∫ E ⋅ dl ≠ 0
Loop
46
Time dependent B-fields
So far we have seen that a time varying flux can be generated by moving a magnet, or by
mechanically changing the surface area…
But what about changing the magnitude of the field?
From Ampere’s Law:
Replacing in Faraday’s Law we obtain:
Videos
http://www.youtube.com/watch?v=jBrneA-wQGA
Tesla
Ampere’s Law revisited
Maxwell realized that Ampere’s Law must be incomplete
because he discovered a situation where it gave the wrong
answer!
v v
∫ B ⋅ dl = µ0ienc + ?
49
Ampere’s Law revisited
Consider a charging capacitor. The charge builds up in the plates, and there
is a time-dependent current flowing through the wires I(t)
E
B
Current I(t)
+
+
+
+
+
-
B
Current I(t)
1. As the currents are flowing, one has Magnetic Fields around the wires.
2. Charge is also building up on the plates, creating an increasing Electric Field
between them.
50
Clicker Question
Current I
-
+
+
+
+
+
B=?
Current I
v
µ I dL × rˆ
Btot = ∫ dB = 0 ∫ 2
4π
r
v
v
Consider the Biot-Savart Law for Magnetic Fields.
Is there a Magnetic Field at the point labeled between the plates?
A) Yes, there is a B-Field
B) No, there is zero B-field
51
Clicker Question
Current I
+
+
+
+
+
-
Current I
B=?
Now consider an Amperian Loop as drawn. According to
Ampere’s Law is there a Magnetic Field at the point labeled
between the plates?
Something must be wrong with
A)Yes, there is a B-Field
Ampere’s Law!
B)No, there is zero B-field
52
Ampere’s Law revisited
Current I
+
+
+
+
+
-
Current I
Both surfaces have the same boundary, if Ampere’s law applies, then the
result should be the same, independently of the surface we choose.
Ampere’s Law revisited
Both surfaces have the same boundary, if Ampere’s law applies, then the
result should be the same, independently of the surface we choose.
Ampere’s law: solving the paradox
The changing Electric Field inside the capacitor must act like a current enclosed
for Ampere’s Law.
v
v
⎛
∫ B ⋅ dl = µ0 ⎜ I enc + ε 0
⎝
Where
v
dΦ E
dt
⎞
⎟
⎠
v
Φ E = ∫ E ⋅ dA
Maxwell defined a fictitious “displacement current”:
ID = ε0
dΦ E
dt
Thus, a changing Electric Field creates a Magnetic Field !
The displacement current
Q = CV =
ε0 A
d
ID = ε0
( Ed ) = ε 0 EA = ε 0 Φ E
dΦ E dQ
=
= IC
dt
dt
Between the plates:
Outside the capacitor:
dΦ E
; IC = 0
dt
I D = 0; I C ≠ 0
ID = ε0
The B-field in between the plates is exactly the same as the field created by
the current outside.
The fictitious “displacement current” gives an idea of continuity of the current,
even though there is no charge flowing in the space separating the plates
Maxwell’s equations
r
r
∫ E ⋅ dA =
Surf
r
Qencl
ε0
r
∫ B ⋅ dA = 0
Surf
r
r
⎛
Gauss’ Law for E-fields
Gauss’ Law for B-fields
∫ B ⋅ dl = µ0 ⎜ I thru + ε 0
loop
r
⎝
r
∫ E ⋅ dl = −
loop
dΦ B
dt
dΦ E
dt
Faraday’s Law
⎞
⎟
⎠
Ampere’s Law