Transcript W10D1_Presentation_answers_jwb

W10D1:
Inductance and Magnetic
Field Energy
Today’s Reading Assignment W10D1 Inductance &
Magnetic Energy Course Notes: Sections 11.1-3
1
Announcements
Math Review Week 10 Tuesday from 9-11 pm in 32-082
PS 7 due Week 10 Tuesday at 9 pm in boxes outside 32-082 or
26-152
Next Reading Assignment W10D2 DC Circuits & Kirchhoff’s
Loop Rules Course Notes: Sections 7.1-7.5
Exam 3 Thursday April 18
7:30 pm –9:30 pm
2
Outline
Faraday Law Problem Solving
Faraday Law Demonstrations
Mutual Inductance
Self Inductance
Energy in Inductors
Transformers
3
Faraday’s Law of Induction
If C is a stationary closed curve and S is a
surface spanning C then
d
E
×
d
s
=
òC
dt
B
×
d
A
òò
S
The changing magnetic flux through S
induces a non-electrostatic electric field
whose line integral around C is non-zero
4
Problem: Calculating Induced
Electric Field
d
E×ds = B×dA
ò
òò
dt open surface
closed path
Consider a uniform magnetic field
which points into the page and is
confined to a circular region with
radius R. Suppose the magnitude
increases with time, i.e. dB/dt > 0.
Find the magnitude and direction of
the induced electric field in the
regions (i) r < R, and (ii) r > R. (iii)
Plot the magnitude of the electric
field as a function r.
5
Faraday’s Law
Demonstrations
6
Demonstration: Electric Guitar H32
Pickups
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H%2032&show=0
7
Electric Guitar
8
Demonstration:
32-082 Aluminum Plate
between Pole Faces of a
Magnet H 14
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 14&show=0
26-152 Copper Pendulum
Between Poles of a Magnet
H13
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 13&show=0
9
Eddy Current Braking
What happened to kinetic energy of pendulum?
10
Eddy Current Braking
The magnet induces currents in the metal that
dissipate the energy through Joule heating:
1. Current is induced
counter-clockwise (out
from center)
2. Force is opposing motion
(creates slowing torque)
11
Eddy Current Braking
The magnet induces currents in the metal that
dissipate the energy through Joule heating:
1. Current is induced
clockwise (out from
center)
2. Force is opposing motion
(creates slowing torque)
3. EMF proportional to
angular frequency
12
Demonstration:
26-152 Levitating Magnet H28
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 28&show=0
32-082 Levitating Coil on an
Aluminum Plate H15
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 15&show=0
13
Mutual Inductance
14
Demonstration:
Two Small Coils and Radio H31
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 31&show=0
15
Mutual Inductance
Current I2 in coil 2, induces
magnetic flux F12 in coil 1.
“Mutual inductance” M12:
F12 º M12 I 2
M12 = M 21 = M
Change current in coil 2?
Induce EMF in coil 1:
dI 2
e12 º - M12
dt
16
Group Problem: Mutual Inductance
An infinite straight wire
carrying current I is placed to
the left of a rectangular loop
of wire with width w and
length l. What is the mutual
inductance of the system?
17
Self Inductance
18
Self Inductance
What if is the effect of putting
current into coil 1?
There is “self flux”:
F B º LI
Faraday’s Law

e
dI
= -L
dt
19
Calculating Self Inductance
L
F
total
B,self
I
V s
Unit: Henry 1 H = 1
A
1.
2.
3.
4.
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)
20
Worked Example: Solenoid
Calculate the selfinductance L of a
solenoid (n turns per
meter, length ,
radius R)
L
F
total
B,self
I
21
Solenoid Inductance
ò B×ds = B
= m0 I enc = m0 ( n ) I
B  0 nI
F B,turn = òò B × d A = BA = m0 nIp R
L
N F B,turn
I
2
 N 0 n R  0 n  R l
2
2
2
22
Concept Question: Solenoid
A very long solenoid consisting of N turns has
radius R and length d, (d>>R). Suppose the
number of turns is halved keeping all the other
parameters fixed. The self inductance
1. remains the same.
2. doubles.
3. is halved.
4. is four times as large.
5. is four times as small.
6. None of the above.
23
Concept Q. Ans.: Solenoid
Solution 5. The self-induction of the
solenoid is equal to the total flux
through the object which is the product
of the number of turns time the flux
through each turn. The flux through
each turn is proportional to the
magnitude of magnetic field which is
proportional to the number of turns per
unit length or hence proportional to the
number of turns. Hence the selfinduction of the solenoid is proportional
to the square of the number of turns.
If the number of turns is halved keeping
all the other parameters fixed then he
self inductance is four times as small.
24
Group Problem: Toroid
Calculate the selfinductance L of a toroid
with a square cross
section with inner radius a,
outer radius b = a+h,
(height h) and N square
windings .
L
F
total
B,self
I
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)
25
Energy in Inductors
26
Inductor Behavior
L
I

dI
 L
dt
Inductor with constant current does nothing
27
Back EMF

dI
 L
dt

dI
 L
dt
I
dI
0
dt
L  0
I
dI
 0  L  0
dt
28
Demos:
26-152 Back “emf” in Large
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 17&show=0
Inductor H17
32-082 Marconi Coil H12
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 12&show=0
29
Marconi Coil: On the Titanic
Another ship
Same era
Titanic
Marconi
Telegraph
30
Marconi Coil: Titanic Replica
31
The Point: Big EMF

Big L
Big dI
Small dt
dI
 L
dt
Huge EMF
32
Energy To “Charge” Inductor
1. Start with “uncharged” inductor
2. Gradually increase current. Must do work:
dI
dW = Pdt = e 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
33
Energy Stored in Inductor
UL  L I
1
2
2
But where is energy stored?
34
Example: Solenoid
Ideal solenoid, length l, radius R,
n turns/length, current I:
B  0 nI
L  o n  R l
2
2
U B = LI = ( mo n p R l)I
1
2
2
1
2
2
2
2
 B2 
2
UB  
R l

 2 o 
Energy Density
Volume
35
Energy Density
Energy is stored in the magnetic field
2
B
uB 
2o
Magnetic Energy Density
Energy is stored in the electric field
o E
uE 
2
2
Electric Energy Density
36
Worked Example: Energy Stored in
Toroid
Consider a toroid with a square cross section with
inner radius a, outer radius b = a+h, (height h) and
N square windings with current I. Calculate the
energy stored in the magnetic field of the torus.
37
Solution: Energy Stored in Toroid
The magnetic field in the torus is given by
0 NI
B
2 r
The stored energy is then
U mag 
1
2 0

B 2 dVvol 
all space
2
1
2 0
h 0 N 2 I 2
h  0 NI 

r dr 



0 a  2 r 
4
b
b
2
B
 h2 r dr
a
2 2
dr h0 N I
b

ln
r
4
a
a
b
The self-inductance is
L
2U mag
I2
0 N 2 b
h
ln
2
a
38
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)
39
Transformer
Step-up transformer
Flux F through each turn same:
ep
dF
= Np
;
dt
dF
e s = Ns
dt
es Ns
=
ep Np
Ns > Np: step-up transformer
Ns < Np: step-down transformer
40
Demonstrations:
26-152 One Turn Secondary: Nail H10
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 10&show=0
26-152 Many Turn Secondary:
Jacob’s Ladder H11
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 11&show=0
32-082 Variable Turns Around a
Primary Coil H9
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 9&show=0
41
Concept Question: Residential
Transformer
If the
transformer in
the can looks
like the picture,
how is it
connected?
1. House=Left, Line=Right
2. Line=Left, House=Right
3. I don’t know
42
Answer: Residential Transformer
Answer: 1. House on left, line on right
The house needs
a lower voltage,
so we step down
to the house
(fewer turns on
house side)
43
Transmission of Electric Power
Power loss can be greatly reduced if
transmitted at high voltage
44
Electrical Power
Power is change in energy per unit time
So power to move current through circuit elements:


d
d
dq
P U 
qV 
V
dt
dt
dt
P  I V
45
Power - Resistor
Moving across a resistor in the direction of current
decreases your potential. Resistors always
dissipate power
V
Pdissipated  I V  I R 
R
2
2
46
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)
P 1.2  105W
I 
 500A
2
V 2.4  10 V
83% loss!!
PL  I 2 R  (500A)2 (0.40W)  100kW
5
P
1.2

10
W
(b) I 

 5.0 A
4
V 2.4  10 V
PL  I 2 R  (5.0A)2 (0.40W)  10W
0.0083% loss
47
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?
48