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 24

7:30 pm –9:30 pm

Conflict Exam 3 Friday April 25 8-10 am, 10-12 noon If you have a regularly scheduled MIT activity that conflicts with Exam 3, contact [email protected] and explain conflict and request time to take conflict exam. 2

Outline Faraday Law Demonstrations Mutual Inductance Self Inductance Energy in Inductors

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Faraday’s Law of Induction If C is a stationary closed curve and S is a surface spanning C then

  d E ⋅ d s = − C∫ dt

  B ⋅ d A ∫∫ S

The changing magnetic flux through S induces a non-electrostatic electric field whose line integral around C is non-zero 4

Faraday’s Law Demonstrations

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Demonstration: Electric Guitar H32 Pickups

http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H%2032&show=0 6

Electric Guitar

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Demonstration: 26-152 Aluminum Plate between Pole Faces of a Magnet H 14 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 14&show=0

32-082 Copper Pendulum Between Poles of a Magnet H13 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 13&show=0

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Eddy Current Braking

What happened to kinetic energy of pendulum? 9

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 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



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Demonstration: 32-082 Levitating Magnet H28 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 28&show=0

26-152 Levitating Coil on an Aluminum Plate H15 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 15&show=0 12

Mutual Inductance

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Demonstration: Two Small Coils and Radio H31

http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 31&show=0 14

Mutual Inductance Current I2 in coil 2, induces magnetic flux F12 in coil 1. “Mutual inductance” M12:

Φ12 ≡ M12 I 2 M12 = M 21 = M Change current in coil 2, induces emf in coil 1.

ε12

dI 2 ≡ − M12 dt 15

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

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Self Inductance What if is the effect of putting current into coil 1? There is “self flux”:

Φ B ≡ LI Faraday’s Law

→

ε

dI = −L dt 18

Calculating Self Inductance

L= Unit: Henry 1.  2.  3.  4. 

Φ self I 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) 19

Worked Example: Solenoid Calculate the selfinductance L of a solenoid (n turns per meter, length , radius R)

L=

Φ self I 20

Solenoid Inductance

  ∫ B ⋅ d s = B = µ0 I enc = µ0 ( n) I

B = µ0 nI Φturn

L=

Φ self I

  2 = ∫∫ B ⋅ d A = BA = µ0 nI π R

N Φturn 2 2 2 = = N µ0 nπ R = µ0 n π R l I 21

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. 22

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=

Φ 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) 23

Energy in Inductors

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Inductor Behavior

L I

ε

dI = −L dt

Inductor with constant current does nothing 25

Back emf

ε

dI = −L dt

ε

dI = −L dt

I

dI >0⇒ dt

εL < 0

I

dI < 0 ⇒ εL > 0 dt

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Demos: 32-082 Back “emf” in Large http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 17&show=0

Inductor H17 26-152 Marconi Coil H12 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 12&show=0 27

Marconi Coil: On the Titanic Another ship Same era Titanic

Marconi Telegraph

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Marconi Coil: Titanic Replica

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The Point: Big emf

ε Big L Big dI

Small dt

dI = −L dt Huge emf 30

Energy To “Charge” Inductor 1. Start with “uncharged” inductor 2.  Gradually increase current. Must do work:

dI dW = (−ε )dQ = L I dt = LI dI dt 3. Integrate up to find total work done:

W = ∫ dW =

I=I f

∫

LI dI = L I f 1 2

2

I =0 31

Energy Stored in Inductor

UL = L I 1 2

2

But where is energy stored?

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Example: Solenoid Ideal solenoid, uniform magnetic field, length l, radius R, turns per unit length n, current I:

B = µ0 nI

L = µo n π R l 2

2

U B = LI = ( µo n π R l)I 1 2

2

1 2

2

2

2

⎛ B2 ⎞ 2 UB = ⎜ π R l ⎟ ⎝ 2µo ⎠ Energy Density

Volume 33

Energy Density Energy is stored in the magnetic field 2

B uB = 2µo

Magnetic Energy Density

Energy is stored in the electric field

εo E uE = 2

2 Electric Energy Density 34

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 µ0 NI B= 2π r The stored energy is then U mag =

1

∫ 2 µ0 all space

B 2 dVvol =

2

1 2 µ0

hµ0 N 2 I 2 hπ ⎛ µ0 NI ⎞ = r dr = ∫ ⎜ ⎟ µ0 a ⎝ 2π r ⎠ 4π b

b

2 B ∫ h2π r dr a

2 2 dr hµ0 N I b ∫ r = 4π ln a a b

The self-inductance is L=

2U mag I2

µ0 N 2 b =h ln 2π a 36

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) 37