Transcript Faraday`s Law Powerpoint

Workshop: Using Visualization
in Teaching Introductory E&M
AAPT National Summer Meeting, Edmonton, Alberta,
Canada.
Organizers: John Belcher, Peter Dourmashkin,
Carolann Koleci, Sahana Murthy
P22- 1
Faraday’s Law Presentation
Materials
P22- 2
MIT Class:
Faraday’s Law
P22- 3
Faraday’s Law
Fourth (Final) Maxwell’s Equation
Underpinning of Much Technology
P22-
Demonstration:
Falling Magnet
P22- 5
Magnet Falling Through a Ring
Falling magnet slows as it approaches a copper
ring which has been immersed in liquid
nitrogen.
P22-
Demonstration:
Jumping Rings
P22-
Jumping Ring
An aluminum ring jumps into the air when the
solenoid beneath it is energized
P22-
What is Going On?
It looks as though the conducting loops have
current in them (they behave like magnetic
dipoles) even though they aren’t hooked up
P22-
Faraday’s Law Applets
Discovery
P22-
Faraday’s Law Applets
Discovery Activity
P22-
Demonstration:
Induction
P22-
Electromagnetic Induction
P22-
Faraday’s Law of Induction

dB

dt
A changing magnetic flux
induces an EMF
P22-
What is EMF?
   E  ds
Looks like potential. It’s a
“driving force” for current
P22-
Faraday’s Law of Induction

dB
  E  ds  
dt
A changing magnetic flux induces
an EMF, a curling E field
P22-
Magnetic Flux Thru Wire Loop
Analogous to Electric Flux (Gauss’ Law)
(1) Uniform B
B  B A  BA cos  B  A
(2) Non-Uniform B
Φ B   B  dA
S
P22-
Minus Sign? Lenz’s Law
Induced EMF is in direction that opposes
the change in flux that caused it
P22-
Faraday’s Law of Induction

dB

dt
Changing magnetic flux induces an EMF
Lenz: Induction opposes change
P22- 19
Ways to Induce EMF

d
   BA cos  
dt
• Quantities which can vary with time:
• Magnitude of B
• Area A enclosed by the loop
• Angle  between B and loop normal
P22-
Group Discussion:
Magnet Falling Through a Ring
Falling magnet slows as it approaches a copper
ring which has been immersed in liquid
P22-
Magnet Falling Through a Ring
Falling magnet slows as it approaches a copper
ring which has been immersed in liquid nitrogen.
P22- 22
Example: Magnitude of B
Magnet Falling Through a Ring
Falling magnet approaches a copper ring
or Copper Ring approaches Magnet
P22- 23
Moving Towards Dipole
Move
ring
down
As ring approaches, what happens to flux?
Flux up increases
P22- 24
PRS Question:
Faraday’s Law
P22- 25
PRS: Faraday’s Law: Loop
:00
A coil moves up
from underneath a
magnet with its
north pole pointing
upward. The
current in the coil
and the force on the
coil:
0%
0%
0%
0%
1.
2.
3.
4.
Current clockwise; force up
Current counterclockwise; force up
Current clockwise; force down
Current counterclockwise; force down
P22- 26
PRS Answer: Faraday’s Law: Loop
Answer: 3. Current is clockwise; force is down
The clockwise current
creates a self-field
downward, trying to offset
the increase of magnetic
flux through the coil as it
moves upward into stronger
fields (Lenz’s Law).
The I dl x B force on the coil is a force which is
trying to keep the flux through the coil from
increasing by slowing it down (Lenz’s Law again).
P22- 27
PRS Question:
Loop in Uniform Field
P22- 28
PRS:
Loop
in
Uniform
Field
0
Bout
v
A rectangular wire loop is pulled thru a uniform B field
penetrating its top half, as shown. The induced
current and the force and torque on the loop are:
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
Current CW, Force Left, No Torque
Current CW, No Force, Torque Rotates CCW
Current CCW, Force Left, No Torque
Current CCW, No Force, Torque Rotates CCW
No current, force or torque
P22- 29
PRS Answer: Loop in Uniform Field
Bout
v
Answer: 5. No current, force or torque
The motion does not change the magnetic flux, so
Faraday’s Law says there is no induced EMF, or
current, or force, or torque.
Of course, if we were pulling at all up or down there
would be a force to oppose that motion.
P22- 30
Group Problem: Changing Area
Conducting rod pulled along two conducting rails in a
uniform magnetic field B at constant velocity v
1. Direction of induced
current?
2. Direction of resultant
force?
3. Magnitude of EMF?
4. Magnitude of current?
5. Power externally
supplied to move at
constant v?
P22-
Changing Angle
B  B  A  BA
B  B  A  0
P22-
The last of the Maxwell’s
Equations (Kind of)
P22-
Maxwell’s Equations
Creating Electric Fields
Qin
(Gauss's Law)
 E  dA 
S
0
dB
C E  d s   dt
(Faraday's Law)
Creating Magnetic Fields
(Magnetic Gauss's Law)
 B  dA  0
S
 Bd s   I
0 enc
C
(Ampere's Law)
P22- 34
Experiment 5:
Faraday’s Law of Induction
P22- 35
Part 1: Current & Flux
BLACK
I>0
RED
Current?
Flux?
t
 (t )   R  I  t '  dt '
0
P22- 36
PRS Predictions:
Flux & Current
P22- 37
PRS: Flux Measurement
0
(A)
(B)
t
t
(D)
(C)
t
t
Moving from above to below and back, you will
measure a flux of:
0%
0%
0%
0%
8
0%
7
0%
6
0%
5
0%
4
5
6
7
8
B then B
D then D
B then D
D then B
3
5.
6.
7.
8.
5.
6.
7.
8.
2
A then A
C then C
A then C
C then A
1
1.
2.
3.
4.
P22- 38
PRS Answer: Flux Measurement
(D)
t
Answer: 6. D then D
The direction of motion doesn’t matter – the field and
hence flux is always upwards (positive) and it
increases then decreases when moving towards and
away from the magnet respectively.
P22- 39
PRS: Current Measurement
0
(A)
(B)
t
t
(D)
(C)
t
5.
6.
5
6
0%
6
0%
5
0%
4
0%
3
0%
2
1
0%
0%
0%
8
Moving from above to below and back, you will
measure a current of:
1. A then A 5. B then B
2. C then C 6. D then D
3. A then C 7. B then D
4. C then A 8. D then B
7
t
NOTE: CCW
is positive!
P22- 40
PRS Answer: Current Measurement
NOTE: CCW
is positive!
(C)
t
Answer: 2. C then C
The direction of motion doesn’t matter – the upward
flux increases then decreases so the induced current
will be clockwise to make a downward flux then
counterclockwise to make an upward one.
P22- 41
PRS: Flux Behavior
(1)
(2)
t
t
1.
2.
3.
4.
1
2
3
4
(4)
(3)
NOTE: Magnet
“Upside Down”
t
t
Moving from below to above, you would measure a
flux best represented by which plot above (taking
upward flux as positive)?
0%
0%
0%
0%
4
3
2
1
:0 P22- 42
PRS Answer: Flux Behavior
(2)
t
Answer: 2.
The field is downward so the flux is negative. It will
increase then decrease as you move over the
magnet.
P22- 43
PRS: Current Behavior
(1)
(2)
t
t
(4)
(3)
1.
2.
NOTE: Magnet
“Upside Down”
1
2
3.
3
4.
4
t
t
Moving from above to below, you would measure a
current best represented by which plot above (taking
counterclockwise current as positive)?
0%
4
0%
3
0%
2
1
0%
0
P22- 44
PRS Answer: Current Behavior
(1)
t
Answer: 1.
The field is downward so the current will first oppose it
(CCW to make an upward flux) then try to reinforce it
(CW to make a downward flux)
P22- 45
PRS Confirming Predictions?
Flux & Current
P22- 46
Part 2: Force Direction
Force when
Move Down?
Move Up?
Test with
aluminum
sleeve
P22- 47
PRS Question:
Wrap-Up
Faraday’s Law
P22- 48
PRS: Circuit
0
A circuit in the form of a
rectangular piece of wire is
pulled away from a long
wire carrying current I in
the direction shown in the
sketch. The induced
current in the rectangular
circuit is
0%
0%
0%
1.
2.
3.
Clockwise
Counterclockwise
Neither, the current is zero
P22- 49
PRS Answer: Circuit
Answer: 1. Induced current
is clockwise
•B due to I is into page; the flux
through the circuit due to that field
decreases as the circuit moves
away. So the induced current is
clockwise (to make a B into the
page)
Note: Iind dl x B force is left on the left segment and
right on the right, but the force on the left is bigger.
So the net force on the rectangular circuit is to the
left, again trying to keep the flux from decreasing by
slowing the circuit’s motion
P22- 50
Faraday’s Law
Problem Solving Session
P22- 51
Technology
Many Applications of
Faraday’s Law
P22- 52
Metal Detector
P22- 53
Induction Stovetops
P22- 54
Ground Fault Interrupters (GFI)
P22- 55
Electric Guitar
Pickups
P22- 56
Electric Guitar
P22- 57
Demonstration:
Electric Guitar
P22- 58
PRS Question:
Generator
P22- 59
PRS: Generator
A square coil rotates in a
magnetic field directed to
the right. At the time
shown, the current in the
square, when looking
down from the top of the
square loop, will be
0%
0%
0%
0%
1.
2.
3.
4.
Clockwise
Counterclockwise
Neither, the current is zero
I don’t know
:00P22- 60
PRS Answer: Generator
Answer: 1. Induced current
is counterclockwise
•Flux through loop decreases as
normal rotates away from B. To
try to keep flux from decreasing,
induced current will be CCW,
trying to keep the magnetic flux
from decreasing (Lenz’s Law)
Note: Iind dl x B force on the sides of the square loop
will be such as to produce a torque that tries to stop
it from rotating (Lenz’s Law).
P22- 61
Group Problem: Generator
Square loop (side L) spins with angular frequency
w in a field of strength B. It is hooked to a load R.
1) Write an expression for current I(t) assuming the
loop is vertical at time t = 0.
2) How much work from generator per revolution?
3) To make it twice as hard to turn, what do you
do to R?
P22- 62
Demonstration:
Levitating Magnet
P22- 63
Brakes
P22- 64
Magnet Falling Through a Ring
What happened to kinetic energy of
magnet?
P22- 65
Demonstration:
Eddy Current Braking
P22- 66
Eddy Current Braking
What happened to kinetic energy of disk?
(link to movie)
P22- 67
Eddy Current Braking
The magnet induces currents in the metal that
dissipate the energy through Joule heating:
w
XX
XX
1. Current is induced
counter-clockwise (out
from center)
2. Force is opposing motion
(creates slowing torque)
P22- 68
Eddy Current Braking
The magnet induces currents in the metal that
dissipate the energy through Joule heating:
w
XX
XX
1. Current is induced
clockwise (out from
center)
2. Force is opposing motion
(creates slowing torque)
3. EMF proportional to w
2
4. .

F
R
P22- 69
Faraday’s Law of Induction

dB

dt
Changing magnetic flux induces an EMF
Lenz: Induction opposes change
P22- 70
Today:
Using Inductance
P22- 71
First:
Mutual Inductance
P22- 72
Demonstration:
Remote Speaker
P22- 73
Mutual Inductance
Current I2 in coil 2, induces
magnetic flux 12 in coil 1.
“Mutual inductance” M12:
12  M12 I 2
M12  M 21  M
Change current in coil 2?
Induce EMF in coil 1:
12  M 12
dI 2
dt
P22- 74
Transformer
Step-up transformer
Flux  through each turn same:
p
d
 Np
;
dt
s
p
s
d
 Ns
dt
Ns

Np
Ns > Np: step-up transformer
Ns < Np: step-down transformer
P22- 75
Demonstrations:
One Turn Secondary:
Nail
Many Turn Secondary:
Jacob’s Ladder
P22- 76
Transmission of Electric Power
Power loss can be greatly reduced if
transmitted at high voltage
P22- 77
Example: Transmission lines
An average of 120 kW of electric power is sent from
a power plant. The transmission lines have a total
resistance of 0.40 W. Calculate the power loss if the
power is sent at (a) 240 V, and (b) 24,000 V.
(a)
(b)
P 1.2 105W
I 
 500 A
2
V 2.4 10 V
PL  I 2 R  (500 A) 2 (0.40W)  100kW
P 1.2 105W
I 
 5.0 A
4
V 2.4 10 V
83% loss!!
0.0083% loss
PL  I 2 R  (5.0 A) 2 (0.40W)  10W
P22- 78
Group Discussion:
Transmission lines
We just calculated that I2R is smaller
for bigger voltages.
What about V2/R? Isn’t that bigger?
Why doesn’t that matter?
P22- 79
Self Inductance
P22- 80
Self Inductance
What if we forget about coil 2 and
ask about putting current into coil 1?
There is “self flux”:
  11  M11I1  LI
Faraday’s Law 

dI
 L
dt
P22- 81
Calculating Self Inductance
L
1.
2.
3.
4.
Total,self
I
Unit: Henry
V s
1H=1
A
Assume a current I is flowing in your device
Calculate the B field due to that I
Calculate the flux due to that B field
Calculate the self inductance (divide out I)
P22- 82
Group Problem: Solenoid
Calculate the self-inductance L of a
solenoid (n turns per meter, length ,
radius R)
REMEMBER
1. Assume a current I is flowing in your device
2. Calculate the B field due to that I
3. Calculate the flux due to that B field
4. Calculate the self inductance (divide out I)
L  Self, total I
P22- 83
Group Problem: Torus
Calculate the inductance of the above
torus (square cross-section of length a,
radius R, N total turns)
1) For assumed current I, what is B(r)?
2) Calculate flux, divide out I
P22- 84
Review: Inductor Behavior
L
I

dI
 L
dt
Inductor with constant current does nothing
P22- 85
Back EMF 
dI
 L
dt
I
dI
 0,
dt
L  0
I
dI
 0,
dt
L  0
P22- 86
Demos: Breaking Circuits
Big Inductor
Marconi Coil
The Question:
What happens if big DI, small Dt
P22- 87
Internal Combustion Engine
P22- 88
Ignition Overview
P22- 89
The Workhorse: The Coil
Primary Coil:
~200 turns heavy Cu
DC (12 V) in to GND
Secondary Coil:
~20,000 turns fine Cu
Usually no voltage…
When primary breaks
up to ~45,000 V
P22- 90
Energy in Inductors
P22- 91
Inductor Behavior
L
I

dI
 L
dt
Inductor with constant current does nothing
P22- 92
Energy To “Charge” Inductor
1. Start with “uncharged” inductor
2. Gradually increase current. Must work:
dI
dW  Pdt   I dt  L I dt  LI dI
dt
3. Integrate up to find total work done:
W   dW 
I

LI dI  L I
1
2
2
I 0
P22- 93
Energy Stored in Inductor
UL  L I
1
2
2
But where is energy stored?
P22- 94
Example: Solenoid
Ideal solenoid, length l, radius R, n turns/length, current I:
B  0 nI
L  o n  R l
2
U B  LI 
1
2
2
1
2
 n  R l I
2
2
2
2
o
 B2  2
UB  
 R l
 2o 
Energy
Density
Volume
P22- 95
Energy Density
Energy is stored in the magnetic field!
2
B
uB 
2o
uE 
o E
: Magnetic Energy Density
2
: Electric Energy Density
2
P22- 96
Group Problem: Coaxial Cable
Inner wire: r = a
Outer wire: r = b
I
X
I
1. How much energy is stored per unit length?
2. What is inductance per unit length?
HINTS: This does require an integral
The EASIEST way to do (2) is to use (1)
P22- 97
PRS Questions:
Inductor in a Circuit
Stopping a Motor
P22- 98
PRS: Stopping a Motor
Consider a motor (a loop of wire rotating in a B
field) which is driven at a constant rate by a
battery through a resistor.
Now grab the motor and prevent it from
rotating. What happens to the current in the
circuit?
0%
0%
0%
0%
1.
2.
3.
4.
Increases
Decreases
Remains the Same
I don’t know
:20P22- 99
PRS Answer: Stopping a Motor
Answer: 1. Increases
When the motor is rotating in a magnetic field
an EMF is generated which opposes the
motion, that is, it reduces the current. When
the motor is stopped that back EMF disappears
and the full voltage of the battery is now
dropped across the resistor – the current
increases. For some motors this increase is
very significant, and a stalled motor can lead to
huge currents that burn out the windings (e.g.
your blender).
P22-100
Think Harder about Faraday
P22-101
PRS Question:
Faraday in Circuit
P22-102
PRS: Faraday Circuit
0
A magnetic field B penetrates this
circuit outwards, and is increasing
at a rate such that a current of 1 A
is induced in the circuit (which
direction?).
The potential difference VA-VB is:
0%
0%
0%
0%
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
6.
7.
8.
9.
+10 V
-10 V
+100 V
-100 V
+110 V
-110 V
+90 V
-90 V
None of the above
A
R=10W
R=100W
B
P22-103
PRS Answer: Faraday Circuit
Answer: 9. None of the above
A
R=10W
The question is meaningless.
There is no such thing as
potential difference when a
changing magnetic flux is present.
R=100W
B
By Faraday’s law, a non-conservative E is
induced (that is, its integral around a closed
loop is non-zero). Non-conservative fields
can’t have potentials associated with them.
P22-104
Non-Conservative Fields
R=10W
I=1A
R=100W
dB
E

d
s



dt
E is no longer a conservative field –
Potential now meaningless
P22-105
Kirchhoff’s Modified 2nd Rule
dB
i D Vi    E  d s   d t
dB
  D Vi 
0
dt
i
If all inductance is ‘localized’ in inductors then
our problems go away – we just have:
dI
i D Vi  L d t  0
P22-106
Inductors in Circuits
Inductor: Circuit element with self-inductance
Ideally it has zero resistance
Symbol:
P22-107
Ideal Inductor
• BUT, EMF generated
by an inductor is not a
voltage drop across
the inductor!
dI
  L
dt
D Vi n d u c t o r    E  d s  0
Because resistance is 0, E must be 0!
P22-108
Circuits:
Applying Modified Kirchhoff’s
(Really Just Faraday’s Law)
P22-109
LR Circuit
dI
i Vi    IR  L dt  0
P22-110
LR Circuit
dI
dI
1  
  IR  L  0   
I  
dt
dt
L R
R
P22-111
Need Some Math:
Exponential Decay
P22-112
Exponential Decay
dA
1
Consider function A where:
 A
dt

A decays exponentially:
A
1.0A0
A  A0 e
0.5A0
t 
A0/e =
0.368 A0
0.0A0
0
1
2
3
Time t
4
5
6
P22-113
Exponential Behavior
dA
1
Slightly modify diff. eq.:
   A  Af
dt

A “decays” to A :

f
A
1.0Af

A  Af 1  e
0.5Af
0.0Af
0
1
2
3
Time t
4
t 
5

6
P22-114
This is one of two differential
equations we expect you to
know how to solve (know the
answer to).
The other is simple harmonic
motion (more on that next week)
P22-115
LR Circuit
dI
1 



I  
dt
L R
R
Solution to this equation when switch is closed at t = 0:
I (t ) 

1 e 

R
t /
L
  : time constant
R
(units: seconds)
P22-116
LR Circuit
t=0+: Current is trying to change. Inductor works as
hard as it needs to to stop it
t=∞: Current is steady. Inductor does nothing.
P22-117
PRS Question:
Voltage Across Inductor
P22-118
PRS: Voltage Across Inductor
In the circuit at right the
switch is closed at t = 0. A
voltmeter hooked across
the inductor will read:
0%
0%
0%
0%
 t /
1. VL   e
t /
2. VL   (1  e )
3. VL  0
4. I don’t know
0
P22-119
PRS Answer: V Across Inductor
Answer: 1. VL   e
t 
The inductor “works hard” at
first, preventing current flow,
then “relaxes” as the current
becomes constant in time.
Although “voltage differences” between two
points isn’t completely meaningful now, we
certainly can hook a voltmeter across an
inductor and measure the EMF it generates.
P22-120
LR Circuit
c
Readings on Voltmeter
Inductor (a to b)
Resistor (c to a)
t=0+: Current is trying to change. Inductor works as
hard as it needs to to stop it
t=∞: Current is steady. Inductor does nothing.
P22-121
Group Problem: Circuits
For the above circuit sketch the currents
through the two bottom branches as a
function of time (switch closes at t = 0, opens
at t = T). State values at t = 0+, T-, T+
P22-122