p22 - 1 workshop: using visualization in teaching introductory e&m aapt national summer meeting,...
TRANSCRIPT
1P22-
Workshop: Using Visualization in Teaching Introductory E&M
AAPT National Summer Meeting, Edmonton, Alberta, Canada.
Organizers: John Belcher, Peter Dourmashkin, Carolann Koleci, Sahana Murthy
2P22-
Faraday’s Law Presentation Materials
P22- 3
MIT Class: Faraday’s Law
P22-
Faraday’s Law
Fourth (Final) Maxwell’s Equation
Underpinning of Much Technology
5P22-
Demonstration:Falling Magnet
P22-
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
Bd
dt
A changing magnetic flux induces an EMF
P22-
What is EMF?
d E s
Looks like potential. It’s a “driving force” for current
P22-
Faraday’s Law of Induction
Bdd
dt
E s
A changing magnetic flux induces
an EMF, a curling E field
P22-
Magnetic Flux Thru Wire Loop
c o sB B A BA B A
B
S
d Φ B A
Analogous to Electric Flux (Gauss’ Law)
(1) Uniform B
(2) Non-Uniform B
P22-
Minus Sign? Lenz’s Law
Induced EMF is in direction that opposes the change in flux that caused it
P22-19
Faraday’s Law of Induction
Bd
dt
Changing magnetic flux induces an EMF
Lenz: Induction opposes change
P22-
Ways to Induce EMF
• Quantities which can vary with time:
• Magnitude of B• Area A enclosed by the loop• Angle between B and loop normal
cosdBA
dt
P22-
Group Discussion:Magnet Falling Through a Ring
Falling magnet slows as it approaches a copper ring which has been immersed in liquid nitrogen.
P22-22
Magnet Falling Through a Ring
Falling magnet slows as it approaches a copper ring which has been immersed in liquid nitrogen.
P22-23
Example: Magnitude of B Magnet Falling Through a Ring
Falling magnet approaches a copper ringor Copper Ring approaches Magnet
P22-24
Moving Towards Dipole
As ring approaches, what happens to flux?Flux up increases
Move ring
down
P22-25
PRS Question:Faraday’s Law
P22-26
PRS: Faraday’s Law: Loop
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%
:00
1. Current clockwise; force up
2. Current counterclockwise; force up
3. Current clockwise; force down
4. Current counterclockwise; force down
P22-27
PRS Answer: Faraday’s Law: Loop
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).
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).
P22-28
PRS Question:Loop in Uniform Field
P22-29
PRS: Loop in Uniform Field
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:
v
Bout
0%
0%
0%
0%
0% 1. Current CW, Force Left, No Torque
2. Current CW, No Force, Torque Rotates CCW
3. Current CCW, Force Left, No Torque
4. Current CCW, No Force, Torque Rotates CCW
5. No current, force or torque
0
P22-30
PRS Answer: Loop in Uniform Field
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.
Answer: 5. No current, force or torque
v
Bout
P22-
Group Problem: Changing AreaConducting 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 BA B A
0B B A
P22-
The last of the Maxwell’s Equations (Kind of)
34P22-
Maxwell’s Equations
0
0
(Gauss's Law)
(Faraday's Law)
0 (Magnetic Gauss's Law)
(Ampere's Law)
in
S
B
C
S
enc
C
Qd
dd
dt
d
d I
Creating Electric Fields
E A
E s
Creating Magnetic FieldsB A
B s
P22-35
Experiment 5: Faraday’s Law of Induction
P22-36
Part 1: Current & Flux
Current?
Flux?
I > 0BLACK
RED
0
( ) ' 't
t R I t dt
P22-37
PRS Predictions:Flux & Current
P22-38
PRS: Flux Measurement
Moving from above to below and back, you will measure a flux of:
t
t
t
t
(A)
(C)
(
B)
(D)
0% 0% 0% 0%0%0%0%0%
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 B5. 5
6. 6
7. 7
8. 8
0
P22-39
PRS Answer: Flux Measurement
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.
t
(D)
P22-40
PRS: Current Measurement
Moving from above to below and back, you will measure a current of:
t
t
t
t
(A)
(C)
(
B)
(D)
NOTE: CCW is positive!
0% 0% 0% 0%0%0%0%0%
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 B5. 56. 67. 78. 8
0
P22-41
PRS Answer: Current Measurement
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.
t
(C)NOTE: CCW is positive!
P22-42
PRS: Flux Behavior
Moving from below to above, you would measure a flux best represented by which plot above (taking upward flux as positive)?
t
t
t
t
(1)
(3)
(
2)
(4)
NOTE: Magnet “Upside Down”
0% 0%0%0%
1. 1
2. 2
3. 3
4. 4
:0
P22-43
t
PRS Answer: Flux Behavior
Answer: 2.
The field is downward so the flux is negative. It will increase then decrease as you move over the magnet.
(2)
P22-44
PRS: Current Behavior
Moving from above to below, you would measure a current best represented by which plot above (taking counterclockwise current as positive)?
t
t
t
t
(1)
(3)
(
2)
(4)
NOTE: Magnet “Upside Down”
0% 0%0%0%
1. 1
2. 2
3. 3
4. 4
0
P22-45
t
PRS Answer: Current Behavior
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)
(1)
P22-46
PRS Confirming Predictions?Flux & Current
P22-47
Part 2: Force Direction
Force when
Move Down?
Move Up?
Test with aluminum
sleeve
P22-48
PRS Question:Wrap-Up
Faraday’s Law
P22-49
PRS: Circuit
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. Clockwise2. Counterclockwise 3. Neither, the current is zero
0
P22-50
PRS Answer: Circuit
•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)
Answer: 1. Induced current is clockwise
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
P22-51
Faraday’s LawProblem Solving Session
P22-52
Technology
Many Applications of Faraday’s Law
P22-53
Metal Detector
P22-54
Induction Stovetops
P22-55
Ground Fault Interrupters (GFI)
P22-56
Electric Guitar
Pickups
P22-57
Electric Guitar
P22-58
Demonstration:Electric Guitar
P22-59
PRS Question:Generator
P22-60
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. Clockwise2. Counterclockwise 3. Neither, the current is zero4. I don’t know
:00
P22-61
PRS Answer: Generator
•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)
Answer: 1. Induced current is counterclockwise
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).
P22-62
Group Problem: GeneratorSquare loop (side L) spins with angular frequency 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?
P22-63
Demonstration:Levitating Magnet
P22-64
Brakes
P22-65
Magnet Falling Through a Ring
What happened to kinetic energy of magnet?
P22-66
Demonstration:Eddy Current Braking
P22-67
Eddy Current Braking
What happened to kinetic energy of disk?(link to movie)
P22-68
Eddy Current BrakingThe magnet induces currents in the metal that
dissipate the energy through Joule heating:
XXXX
1. Current is induced counter-clockwise (out from center)
2. Force is opposing motion (creates slowing torque)
P22-69
Eddy Current BrakingThe magnet induces currents in the metal that
dissipate the energy through Joule heating:
XXXX
1. Current is induced clockwise (out from center)
2. Force is opposing motion (creates slowing torque)
3. EMF proportional to 4. . 2
RF
P22-70
Faraday’s Law of Induction
Bd
dt
Changing magnetic flux induces an EMF
Lenz: Induction opposes change
P22-71
Today:Using Inductance
P22-72
First:Mutual Inductance
P22-73
Demonstration: Remote Speaker
P22-74
Mutual Inductance
12 12 2M I
212 12
dI
dtM
12 21M M M
Current I2 in coil 2, induces magnetic flux 12 in coil 1. “Mutual inductance” M12:
Change current in coil 2?Induce EMF in coil 1:
P22-75
TransformerStep-up transformer
;p p s s
d dN N
dt dt
s s
p p
N
N
Ns > Np: step-up transformerNs < Np: step-down transformer
Flux through each turn same:
P22-76
Demonstrations:
One Turn Secondary:Nail
Many Turn Secondary:Jacob’s Ladder
P22-77
Transmission of Electric Power
Power loss can be greatly reduced if transmitted at high voltage
P22-78
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 . Calculate the power loss if the power is sent at (a) 240 V, and (b) 24,000 V.
(a) 5
2
1.2 10500
2.4 10
P WI AV V
2 2(5.0 ) (0.40 ) 10LP I R A W
(b)
83% loss!!
0.0083% loss
2 2(500 ) (0.40 ) 100LP I R A kW
5
4
1.2 105.0
2.4 10
P WI AV V
P22-79
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?
P22-80
Self Inductance
P22-81
Self Inductance
11 11 1 M I LI
dILdt
What if we forget about coil 2 and ask about putting current into coil 1?There is “self flux”:
Faraday’s Law
P22-82
Calculating Self Inductance
Total,selfLI
1. Assume a current I is flowing in your device2. Calculate the B field due to that I3. Calculate the flux due to that B field4. Calculate the self inductance (divide out I)
V s1 H = 1
A
Unit: Henry
P22-83
Group Problem: Solenoid
Calculate the self-inductance L of a solenoid (n turns per meter, length , radius R)
REMEMBER1. Assume a current I is flowing in your device2. Calculate the B field due to that I3. Calculate the flux due to that B field4. Calculate the self inductance (divide out I)
Self, totalL I
P22-84
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
P22-85
Review: Inductor Behavior
I
Inductor with constant current does nothing
dILdt
L
P22-86
I
Back EMF
0, 0L
dI
dt 0, 0L
dI
dt
I
dILdt
P22-87
Demos: Breaking Circuits
Big InductorMarconi Coil
The Question: What happens if big I, small t
P22-88
Internal Combustion Engine
P22-89
Ignition Overview
P22-90
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
P22-91
Energy in Inductors
P22-92
Inductor Behavior
I
Inductor with constant current does nothing
dILdt
L
P22-93
1. Start with “uncharged” inductor
2. Gradually increase current. Must work:
3. Integrate up to find total work done:
Energy To “Charge” Inductor
dIdW Pdt I dt L I dt LI dI
dt
212
0
I
I
W dW LI dI L I
P22-94
Energy Stored in Inductor
212LU L I
But where is energy stored?
P22-95
Example: Solenoid
2 2 2 21 12 2B oU LI n R l I
Ideal solenoid, length l, radius R, n turns/length, current I:
0B nI 2 2oL n R l
22
2Bo
BU R l
Energy
Density
Volume
P22-96
Energy Density
: Magnetic Energy Density
2
2o
E
Eu
: Electric Energy Density
Energy is stored in the magnetic field!
2
2Bo
Bu
P22-97
Group Problem: Coaxial Cable
1. How much energy is stored per unit length? 2. What is inductance per unit length?
HINTS: This does require an integralThe EASIEST way to do (2) is to use (1)
Inner wire: r = a
Outer wire: r = bXI I
P22-98
PRS Questions:Inductor in a Circuit
Stopping a Motor
P22-99
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. Increases2. Decreases3. Remains the Same4. I don’t know :20
P22-100
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).
P22-101
Think Harder about Faraday
P22-102
PRS Question:Faraday in Circuit
P22-103
PRS: Faraday CircuitA 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?).
R=100
R=10A
B
The potential difference VA-VB is:
0%
0%
0%
0%
0%
0%
0%
0%
0% 1. +10 V
2. -10 V
3. +100 V
4. -100 V
5. +110 V
6. -110 V
7. +90 V
8. -90 V
9. None of the above
0
P22-104
PRS Answer: Faraday CircuitAnswer: 9. None of the above
The question is meaningless. There is no such thing as potential difference when a changing magnetic flux is present.
R=100
R=10A
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.
P22-105
Non-Conservative Fields
R=100R=10
Bdd
d t
E s
E is no longer a conservative field – Potential now meaningless
I=1A
P22-106
Kirchhoff’s Modified 2nd Rule
ii
V d E s
0Bi
i
dV
d t
If all inductance is ‘localized’ in inductors then our problems go away – we just have:
0ii
d IV L
d t
Bd
d t
P22-107
Inductors in CircuitsInductor: Circuit element with self-inductance
Ideally it has zero resistance
Symbol:
P22-108
• BUT, EMF generated by an inductor is not a voltage drop across the inductor!
Ideal Inductor
d ILd t
i n d u c t o rV d E s
Because resistance is 0, E must be 0!
0
P22-109
Circuits:Applying Modified Kirchhoff’s(Really Just Faraday’s Law)
P22-110
LR Circuit
0ii
dIV L
dtIR
P22-111
LR Circuit
10
dI dIL Idt dt L R R
IR
P22-112
Need Some Math:Exponential Decay
P22-113
Exponential Decay1dAA
dt Consider function A where:
A decays exponentially:
0tA A e
P22-114
0 1 2 3 4 5 60.0A
f
0.5Af
1.0Af
A
Time t
Exponential Behavior
1f
dAA A
dt Slightly modify diff. eq.:
A “decays” to Af:
1 tfA A e
P22-115
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)
P22-116
LR Circuit
Solution to this equation when switch is closed at t = 0:
1dII
dt L R R
: time constantL
R
/( ) 1 tI t eR
(units: seconds)
P22-117
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.
P22-118
PRS Question:Voltage Across Inductor
P22-
PRS: Voltage Across Inductor
119
0%
0%
0%
0%
In the circuit at right the switch is closed at t = 0. A voltmeter hooked across the inductor will read:
/tLV e
/(1 )tLV e
0LV
1.
2.
3.
4. I don’t know 0
P22-120
PRS Answer: V Across Inductor
The inductor “works hard” at first, preventing current flow, then “relaxes” as the current becomes constant in time.
Answer: 1. tLV e
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.
P22-121
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.
Readings on VoltmeterInductor (a to b)Resistor (c to a)
c
P22-122
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+