Module 22: Inductance and Magnetic Field Energy
1 Module 22: Outline Self Inductance Energy in Inductors Circuits with Inductors: RL Circuit
2 Faraday’s Law of Induction dΦ = − B dt Changing magnetic flux induces an EMF
Lenz: Induction opposes change
3 Self Inductance
4 Self Inductance What if is the effect of putting current into coil 1? There is “self flux”:
ΦB ≡ LI
dI Faraday’s Law Æ ε = −L
dt 5 Calculating Self Inductance Φtotal Unit: Henry B,self V ⋅s L = 1 H = 1 I A 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)
6 Problem: Solenoid
Calculate tthehe self -inductance L of a solenoid ( n turns per meter, leng th l, radius R) Φtotal L = B, self REMEMBER I 1. Assume a currentcurrent I iiss fflowinglowing in your device 2. Calculate the B field due to that I 3. Calculate tthehe flux due to that B ffieldield 4. Calculate the self inductance (divide out I)
7 Solenoid Inductance
B G G Bs⋅dB==AAμμ I = nI v∫ 00enc ( ) w
l B = μ0nI G G Φ=⋅==BAdBAnIRμπ2 Bturn,0∫∫
NΦ L = B,turn = Nμ nπ R2 = μ n2π R2l I 0 0 8 Energy in Inductors
9 Inductor Behavior
L I dI ε = − L dt Inductor with cconstantonstant current does nothing
10 11 I t 0 d dI > L L
− ε , =
ε 0 < dt dI I 0 < Back EMF L ε , 0 > dt dI Energy To “Charge” Inductor 1. Start with “uncharged” inductor 2. Gradually increase current. Must work: dI dW = Pdt = εIdt = L Idt = LI dI dt 3. Integrate up to find total work done: I = = = 1 2 W ∫ dW ∫ LI dI 2 L I I =0
12 Energy Stored in Inductor U = 1 L I 2 L 2
But whe eere i ses en egystoedergy stored?
13 Example: Solenoid
Ideal solenoid, length l, radius R, n turns/length, current I: = μ = μ 2 π 2 B 0 nI L o n R l = 1 2 = 1 (μ 2 π 2 ) 2 U B 2 LI 2 o n R l I ⎛ B2 ⎞ U = ⎜ ⎟ π R2l B ⎝ μ ⎠ 2 o Energy Volume Density 14 Energy Density Energy is stored in the magnetic field! B2 u = : Magnetic Energy Density B 2μ o ε E 2 u = o : Electric Energy Density E 2 15 Problem: Coaxial Cable
Inner wire: r = a I I X Outer wwire:ire: r = b
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 toto dodo (2)(2) is to use ((1)1)
16 Technology
Many Applications of Faraday’s Law
17 Demos: Breaking Circuits
Big Inductor Marconi Coilil
The Question: What happens if big ΔI, small Δt
18 The Point: Big EMF dI ε = −L dt Big L Big dI Huge ε
Small dt 19 First: Mutual Inductance
20 Demonstration: Remote Speake r
21 Mutual Inductance
Current I2 in coil 2, induces magnetic flux Φ12 iil1in coil 1. “Mutual inductance” M12:
Φ12 ≡ M12 I2
M = M = M 12 21
Change current in coil 2? dI2 Induce EMF in coil 1: ε ≡ − M 12 12 dt 22 Transformer Step-up transformer Flux Φ through each tturnurn ssame:ame: dΦ dΦ ε = N ; ε = N p p dt s s dt ε N s = s ε p N p
Ns > Np:s: steptep -up transformer Ns < Np: step-down transformer 23 Demonstrations:
One Turn Secondary: Nail
Many Turn Secondary:Secondary: Jacob’s Ladder
24 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 , HHouse=Rightouse=Right
3. I don’t know 25 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)
26 Transmission of Electric Power
Power loss can be greatly reduced if transmitted aatt hhighigh vvoltageoltage
27 Example: Transmission lines An average of 120 kW of electric power is sent from a power plant . TThehe transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the power is sent at (a) 240 V , and (b) 2424, 000 V .
P 1. 2 ×105 W (a) I = = = 500A V 2.4 ×102V 83% loss!! = 2 = 2 Ω = PL I R (500A) (0.40 ) 100kW
P 1.2 × 105 W (b) I = = = 5.0 A V 2.4 × 104V 0.0083% loss = 2 = 2 Ω = PL I R (5.0 A) (0.40 ) 10W 28 Transmission lines
We just calculated that I2R is smaller for bbiggerigger voltages.
What about V2/R? Isn’t that bigger?
Why doesn’t that matter?
29 Brakes
30 Magnet Falling Through a Ring
Link to movie
What happ ened to kinetic energy of magnet? 31 Demonstration: Eddy Current Braking
32 Eddy Current Braking
What happened to kinetic energy of disk?
33 Eddy Current Braking
The magnet induces currents in the metal that dissipate the energy through Joule heating:
ω 1. Current is induced counter-clockwise (out from center) 2. Force is opp osing motion (creates slowing torque) XX XX
34 Eddy Current Braking
The magnet induces currents in the metal that dissipate the energy through Joule heating:
ω 1. Current is induced clockwise (out from center) 2. Force is opp osing motion (creates slowing torque) XX ω XX 3. EMF proportionalproportional to 4. . ε 2 F ∝ R
35 Demonstration: Levitating Magnet
Link to Movie
36 MIT OpenCourseWare http://ocw.mit.edu
8.02SC Physics II: Electricity and Magnetism Fall 2010
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