1 W10D1: Inductance and Magnetic Field Energy Today’s Reading Assignment W10D1 Inductance & Magnetic Energy Course Notes: Sections 11.1-3.

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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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Faraday’s LawDemonstrations

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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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Electric Guitar

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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


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Eddy Current Braking

What happened to kinetic energy of pendulum?

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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



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Mutual Inductance

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Demonstration:Two Small Coils and Radio H31


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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

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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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Solenoid Inductance

B 0nI

L

NB,turn

IN

0n R2

0n2 R2l

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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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Energy in Inductors

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Inductor Behavior

I

Inductor with constant current does nothing L

dI

dt

L

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I

Back EMF

dI

dt 0

L0

dI

dt 0

L 0

I

L

dI

dt L

dI

dt

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Demos: 26-152 Back “emf” in Large

Inductor H17 32-082 Marconi Coil H12



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Marconi Coil: On the Titanic

Marconi

Telegraph

Titanic

Another ship

Same era

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Marconi Coil: Titanic Replica

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The Point: Big EMF

L

dI

dt

Big L

Big dI

Small dt

Huge EMF

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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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Energy Stored in Inductor

U

L 1

2L I 2

But where is energy stored?

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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




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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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Transmission of Electric Power

Power loss can be greatly reduced if transmitted at high voltage

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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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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?

1 W10D1: Inductance and Magnetic Field Energy Today’s Reading Assignment W10D1 Inductance & Magnetic Energy Course Notes: Sections 11.1-3.

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