1 w10d1: inductance and magnetic field energy today’s reading assignment w10d1 inductance &...
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W10D1:Inductance and Magnetic
Field Energy
Today’s Reading Assignment W10D1 Inductance & Magnetic Energy Course Notes: Sections 11.1-3
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
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Outline
Faraday Law Problem SolvingFaraday Law DemonstrationsMutual InductanceSelf Inductance
Energy in Inductors
Transformers
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Faraday’s Law of InductionIf C is a stationary closed curve and S is a surface spanning C then
The changing magnetic flux through S induces a non-electrostatic electric field whose line integral around C is non-zero
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Problem: Calculating Induced Electric Field
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.
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Demonstration: Electric Guitar H32
Pickups
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H%2032&show=0
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Demonstration:32-082 Aluminum Plate
between Pole Faces of a Magnet H 14
26-152 Copper Pendulum Between Poles of a Magnet
H13
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 14&show=0
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 13&show=0
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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)
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Eddy Current BrakingThe 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
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Demonstration:
26-152 Levitating Magnet H28
32-082 Levitating Coil on an Aluminum Plate H15
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 28&show=0
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 15&show=0
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Demonstration:Two Small Coils and Radio H31
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 31&show=0
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Mutual Inductance
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:
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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?
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Self Inductance
What if is the effect of putting current into coil 1?There is “self flux”:
Faraday’s Law
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Calculating Self Inductance
L
B,selftotal
I
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)
1 H = 1
V sA
Unit: Henry
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Worked Example: Solenoid
Calculate the self-inductance L of a solenoid (n turns per meter, length , radius R)
L
B,selftotal
I
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Concept Question: SolenoidA 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.
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Concept Q. Ans.: SolenoidSolution 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 self-induction 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.
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Group Problem: ToroidCalculate the self-inductance L of a toroid with a square cross section with inner radius a, outer radius b = a+h, (height h) and N square windings .
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)
L
B,selftotal
I
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Demos: 26-152 Back “emf” in Large
Inductor H17 32-082 Marconi Coil H12
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 17&show=0
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 12&show=0
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1. Start with “uncharged” inductor
2. Gradually increase current. Must do work:
3. Integrate up to find total work done:
Energy To “Charge” Inductor
W dW LI dI
I 0
I
12
L I 2
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Example: Solenoid
Ideal solenoid, length l, radius R, n turns/length, current I:
B 0nI
L on2 R2l
U
B
B2
2o
R2l
Energy Density Volume
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Energy Density
Magnetic Energy Density
u
E
oE 2
2 Electric Energy Density
Energy is stored in the magnetic field
u
B
B2
2o
Energy is stored in the electric field
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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.
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Solution: Energy Stored in Toroid
The magnetic field in the torus is given by
The stored energy is then
The self-inductance is
B
0NI
2r
Umag
1
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B2 dVvol
all space
1
20
B2h2r dra
b
h
0
0NI
2r
2
r dra
b
h
0N 2 I 2
4dr
ra
b
h
0N 2 I 2
4ln
b
a
L 2Umag
I 2h
0N 2
2ln
b
a
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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
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TransformerStep-up transformer
Ns > Np: step-up transformerNs < Np: step-down transformer
Flux through each turn same:
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Demonstrations:
26-152 One Turn Secondary: Nail H10
26-152 Many Turn Secondary: Jacob’s Ladder H11
32-082 Variable Turns Around a Primary Coil H9
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 10&show=0
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 11&show=0
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 9&show=0
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1. House=Left, Line=Right2. Line=Left, House=Right3. I don’t know
Concept Question: Residential Transformer
If the transformer in the can looks like the picture, how is it connected?
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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)
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Electrical Power
Power is change in energy per unit time
So power to move current through circuit elements:
P
d
dtU
d
dtqV dq
dtV
VIP
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Power - ResistorMoving across a resistor in the direction of current decreases your potential. Resistors always dissipate power
R
VRIVIP
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dissipated
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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) I
P
V
1.2 105W
2.4 102V500A
PLI 2 R (5.0A)2(0.40) 10W
(b)
83% loss!!
0.0083% loss
PLI 2 R (500A)2(0.40) 100kW
I
P
V
1.2 105W
2.4 104V5.0A
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