Today’s agendum: Induced emf

2

Today’s agendum:

Induced emf.You must understand how changing magnetic flux can induce an emf, and be able to determine the direction of the induced emf.

Faraday’s “Law.” You must be able to use Faraday’s “Law” to calculate the emf induced in a circuit.

Lenz’s “Law.”You must be able to use Lenz’s “Law” to determine the direction induced current, and therefore induced emf.

Generators (part 1).You must understand how generators work, and use Faraday’s “Law” to calculate numerical values of parameters associated with generators.

3

Induced emf and Faraday’s “Law”

Magnetic Induction

We have found that an electric current can give rise to a magnetic field…

I wonder if a magnetic field can somehow give rise to an electric current…

4

An electric current is induced if there is a closed circuit (e.g., loop of wire) in the changing magnetic flux.

It is observed experimentally that changes in magnetic flux induce an emf in a conductor.

A constant magnetic flux does not induce an emf—it takes a changing magnetic flux.

B

B

I

5

Passing the coil through the magnet would induce an emf in the coil.

They need to calibrate their meter!

6

Note that “change” may or may not not require observable (to you) motion.

A magnet may move through a loop of wire

N S

move magnet toward coilv region of

magnetic fieldchange area of loop

inside magnetic field

N S

rotate coil in magnetic field

this part of the loop is closest to your eyes

A magnet may move through a loop of wire, or a loop of wire may be moved through a magnetic field (as suggested in the previous slide). These involve observable motion.

I

7

In this case, nothing observable (to your eye) is moving, although, of course microscopically, electrons are in motion.

Induced emf is produced by a changing magnetic flux.

changing I

changing B

induced I

A changing current in a loop of wire gives rise to a changing magnetic field (predicted by Ampere’s “Law”) which can induce a current in another nearby loop of wire.

8

Today’s agendum:





9

We can quantify the induced emf described qualitatively in the last few slides by using magnetic flux.Experimentally, if the flux through N loops of wire changes by dB in a time dt, the induced emf is

Bd = - N .

dtε

Faraday’s “Law” of Induction is one of the fundamental laws of electricity and magnetism.

I wonder why the – sign…

Faraday’s “Law” of Magnetic Induction

Your text, pages 997-998, shows how to determine the direction of the induced emf. Argh! Lenz’s Law, coming soon, is much easier.

10

This is another version of Faraday’s “Law”:

Bd = - N

dtε

where the magnetic flux is

Faraday’s “Law” of Magnetic Induction

B B dA.

In a future lecture, we’ll work with

Web page with pictures of a whole bunch of applications:http://sol.sci.uop.edu/~jfalward/electromagneticinduction/electromagneticinduction.html

We’ll use this version in a later

lecture.dt

dsdE B

sdE

tp://sol.sci.uop.edu/~jfalward/electromagneticinduction/electromagneticinduction.htm

11

N SI

v

+ -

Example: move a magnet towards a coil of wire.

N=5 turns A=0.002 m2

dB = 0.4 T/s

dt

B dA

Bdd

= - N = - N dt dt

ε

d BA = - N (what assumption did I make here?)

dtε

dB = - N A

dtε

2 T = - 5 0.002 m 0.4 =-0.004 V

sε

12

Ways to induce an emf:

change B

change the area of the loop in the field

Homework hint: if B varies but loop B.

B Bd B dA B(t) dA

13

Ways to induce an emf (continued):

change the orientation of the loop in the field

=90 =45 =0

14

Example: a uniform (but time-varying) magnetic field passes through a circular coil whose normal is parallel to the magnetic field. The coil’s area is 10-2 m2 and it has a resistance of 1 m. B varies with time as shown in the graph. Plot the current in the coil as a function of time.

.01 T

=1 sFor 0 < t < 3:

Bd BAd dB

= - = - = - Adt dt dt

ε

A dB

= IR I = = - R R dtεε

.01dB B .01 A dB .01 = = I = - = - = - .0333 A

dt t 3 R dt .001 3

For 3 < t < 5: dB

= 0 I = 0dt

15

Example: a uniform (but time-varying) magnetic field passes through a circular coil whose normal is parallel to the magnetic field. The coil’s area is 10-2 m2 and it has a resistance of 1 m. B varies with time as shown in the graph. Plot the current in the coil as a function of time.

.01 T

=1 s

For 5 < t < 11:

dB B -.01 = =

dt t 6.01A dB -.01

I = - = - R dt .001 6

= + .0167 A

I(t)

-.0333 A

16

Today’s agendum:





17

Lenz’s law—An induced emf always gives rise to a current whose magnetic field opposes the change in flux.*

Experimentally…

*Think of the current resulting from the induced emf as “trying” to maintain the status quo—to prevent change.

N SI

v

+ -

If Lenz’s law were not true—if there were a + sign in Faraday’s law—then a changing magnetic field would produce a current, which would further increase the magnetic field, further increasing the current, making the magnetic field still bigger…

18

…violating conservation of energy and ripping apart the very fabric of the universe…

19

In which direction is the current induced in the coil for each situation shown?

(counterclockwise) (no current)

Practice with Lenz’s Law.

20

(counterclockwise) (clockwise)

21

Rotating the coil about the vertical diameter by pulling the left side toward the reader and pushing the right side away from the reader in a magnetic field that points from right to left in the plane of the page.

(counterclockwise)

Now that you are experts on the application of Lenz’s “Law”, remember this:You can use Faraday’s “Law” to calculate the magnitude of the emf (or whatever the problem wants. Then use Lenz’s “Law” to figure out the direction of the induced current (or the direction of whatever the problem wants).

RememberBd

= - N ?dt

ε

22

Today’s agendum:




Generators.You must understand how generators work, and use Faraday’s “Law” to calculate numerical values of parameters associated with generators.

23

Motional emf: an overview

An emf is induced in a conductor moving in a magnetic field.Your text introduces four ways of producing motional emf. We will cover the first two in this lecture.

Bd= -

dt

side view

BA

1. Flux change through a conducting loop produces an emf: rotating loop.

= NBA sin tε

NBAI = sin t

R

P= INBA sin t

start with this

derive these

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2. Flux change through a conducting loop produces an emf: expanding loop.

Bd= -

dtv

ℓ

B

x=vdtdA

B

MF = I B

= B vεε B v

I = = R R

PP = F v =I Bv

start with these

derive these

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3. Conductor moving in a magnetic field experiences an emf: magnetic force on charged particles.

F= q E+v B

v

ℓ

B

B

–

+

= E (Mr. Ed)

Next time we will look at two more examples of motional emf…

start with these

derive this

ε = B v

You could also solve this using Faraday’s” Law” by constructing a “virtual” circuit using “virtual” conductors.

26

4. Flux change through a conducting loop produces an emf: moving loop.

Bd= -

dt

start with this

derive these

ε = B v

B v

I = P =I BvR

27

Generators and Motors: a basic introduction

Take a loop of wire in a magnetic field and rotate it with an angular speed .

S N

side view

BA

B =B A = BA cos

Choose 0=0. Then

0= t = t .

B = BA cos t

Bd = -

dtε

Generators are an application of motional emf.

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If there are N loops in the coil

Bd = - N

dtε

d BA cos t = - N

dtε

= NBA sin tε

side view

BA

|| is maximum when = t = 90° or 270°; i.e., when B is zero. The rate at which the magnetic flux is changing is then maximum. On the other hand, is zero when the magnetic flux is maximum.

The NBA equation!

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emf, current and power from a generator

= NBA sin tε

NBAI = = sin t

R Rε

P= I = INBA sin tε

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= N B A ω sin ωt ε

max = N B A ωε

maxN = B A ωε

-2 2 -1

170 VN =

0.15 T 2×10 m 2 ×60 s

N = 150 (turns)

Example: the armature of a 60 Hz ac generator rotates in a 0.15 T magnetic field. If the area of the coil is 2x10-2 m2, how many loops must the coil contain if the peak output is to be max = 170 V?

31

Recall that one of the ways to induce an emf is to change the area of the loop in the magnetic field. Let’s see how this works.

v

ℓ

B

vdtdA

A U-shaped conductor and a moveable conducting rod are placed in a magnetic field, as shown.The rod moves to the right with a constant speed v for a time dt.

The rod moves a distance v dt and the area of the loop inside the magnetic field increases by an amount

dA = ℓ v dt .

B

Another Kind of Generator: A Slidewire Generator

x

32

Bd = - N

dtε

d BA = 1

dtε

B dA =

dtε

dx

= B dt

ε

= B v . ε

B and v are vector magnitudes, so they are always +. Wire length is always +. You use Lenz’s law to get the direction of the current.

v

ℓ

B

vdtdA

B

x

BAAdBB The loop is perpendicular to the magnetic field, so

the magnetic flux through the loop is . The emf induced in the conductor can be calculated using Faraday’s law:

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The induced emf causes current to flow in the loop.

v

ℓ

B

vdtdA

Magnetic flux inside the loop increases (more area).

System “wants” to make the flux stay the same, so the current gives rise to a field inside the loop into the plane of the paper (to counteract the “extra” flux).Clockwise current!

I

Direction of current?

x

34

As the bar moves through the magnetic field, it “feels” a force

ℓ

B

I

A constant pulling force, equal in magnitude and opposite in direction, must be applied to keep the bar moving with a constant velocity.

MF = I B

P MF = F = I B

v

FM FP

x

35

If the loop has resistance R, the current is

v

ℓ

B

(as expected).I

Power and current.

ε B vI = = .

R R

And the power is

PP = F v =I Bv

2P = I IR = I R

Mechanical energy (from the pulling force) has been converted into electrical energy, and the electrical energy is then dissipated by the resistance of the wire.

x

36

Today’s agendum:

Motional emf.You must be able to apply Faraday’s and Lenz’s “Laws” to calculate motional emf, as well as current and power in circuits powered by motional emf.

Motors and Generators (part 2).In the previous lecture, we used Faraday’s “Law” to calculate numerical values of parameters associated with generators. You must also understand conceptually how motors and generators work.

Back emf.You must be able to use Lenz’s “Law” to explain back emf.

37

Motional emf

An emf is induced in a conductor moving in a magnetic field.In the last lecture you learned about two examples of motional emf.

Bd= -

dt

side view

BA

1. Flux change through a conducting loop produces an emf: rotating loop.

= NBA sin tε

NBAI = sin t

R

P= INBA sin t

start with this

derive these

38

2. Flux change through a conducting loop produces an emf: expanding loop.

Bd= -

dtv

ℓ

B

x=vdtdA

B

MF = I B

= B vεε B v

I = = R R

PP = F v =I Bv

start with these

derive these

39

3. Conductor moving in a magnetic field experiences an emf: magnetic force on charged particles.

F= q E+v B

v

ℓ

B

B

–

+

= E (Mr. Ed)

Today we’ll look at two more examples of motional emf.

start with these

derive this

ε = B v

You could also solve this using Faraday’s “Law” by constructing a “virtual” circuit using “virtual” conductors.

40

4. Flux change through a conducting loop produces an emf: moving loop.

Bd= -

dt

start with this

derive these

ε = B v

B v

I = P =I BvR

41

Let’s work out these two examples now.

Example 3 of motional emf: moving conductor in B field.

Example 4 of motional emf: flux change through conducting loop. (Entire loop is moving.)

Remember, it’s the flux change that produces the emf. Flux has no direction associated with it. However, the presence of flux is due to the presence of a magnetic field, which does have a direction, and allows us to use Lenz’s “Law” to determine the “direction” of current and emf.

42

Motional emf is the emf induced in a conductor moving in a magnetic field.

Example 3 of motional emf: moving conductor in B field.

v

ℓ

B

B

If a conductor (purple bar) moves with speed v in a magnetic field, the electrons in the bar experience a force

MF = qv B= - ev B

The force on the electrons is “up,” so the “top” end of the bar acquires a net – charge and the “bottom” end of the bar acquires a net + charge.

“up”

–

+

43

The separated charges in the bar produce an electric field pointing “up” the bar. The emf across the length of the bar is

v

ℓ

B

B

The electric field exerts a “downward” force on the electrons:

EF = qE= - eE

An equilibrium condition is reached, where the magnetic and electric forces are equal in magnitude and opposite in direction.

“up”

–

+

ε

evB= eE= e

ε = E

ε = B v

44

Example 4 of motional emf: flux change through conducting loop. (Entire loop is moving.)

I’ll include some numbers with this example.

45

B = 0.6 T

5 c

mA square coil of side 5 cm contains 100 loops and is positioned perpendicular to a uniform 0.6 T magnetic field. It is quickly and uniformly pulled from the field (moving to B) to a region where the field drops abruptly to zero. It takes 0.10 s to remove the coil, whose resistance is 100 .

46

B = Bf - Bi = 0 - BA = - (0.6 T) (0.05 m)2 = - 1.5x10-3 Wb.

(a) Find the change in flux through the coil.

BAAdB :Initial Bi

0 :Final Bf

47

The current must flow clockwise to induce an “inward” magnetic field (which replaces the “removed” magnetic field).

initialfinal

(b) Find the current and emf induced.

Current will begin to flow when the coil starts to exit the magnetic field. Because of the resistance of the coil, the current will eventually stop flowing after the coil has left the magnetic field.

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The induced emf is

Bd d(BA) dA = - N = - N = - N B

dt dt dtε

v

A

d xdA dx = = = v

dt dt dt

x

49

m = 100 0.6 T 0.05 m 0.5

sε

= 1.5 Vε

“uniformly” pulled

= - N B vε

x 5 cm m

v = = = 0.5 t 0.1 s s

The induced current is

I = = = 15 mA .R 100 Ωε 1.5 V

50

Current flows “only*” during the time flux changes.

W = P t = I2R t = (1.5x10-2 A)2 (100 ) (0.1 s) = 2.25x10-3 J

*Remember: if there were no resistance in the loop, the current would flow indefinitely. However, the resistance quickly halts the flow of current once the magnetic flux stops changing.

The loop has to be “pulled” out of the magnetic field, so there is a pulling force, which does work.

The “pulling” force is opposed by a magnetic force on the current flowing in the wire. If the loop is pulled “uniformly” out of the magnetic field (no acceleration) the pulling and magnetic forces are equal in magnitude and opposite in direction.

(c) How much energy is dissipated in the coil?

(d) Discuss the forces involved in this example.

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The flux change occurs only when the coil is in the process of leaving the region of magnetic field.

No flux change.No emf.No current.No work (why?).

52

Flux changes. emf induced. Current flows. Work done.

D

Fapplied

53

No flux change. No emf. No current. (No work.)

54

Remember, a force is needed only when the coil is partly in the field region.

(e) Calculate the force necessary to pull the coil from the field.

I

55

magF = N IL B

where L is a vector in the direction of I having a magnitude equal to the length of the wire inside the field region.

I

L1

L2 L3

F1

F2

F3

Sorry about the “busy” slide!The forces should be shown acting at the centers of the coil sides in the field.There must be a pulling force to the right to overcome

the net magnetic force to the left.

Fpull

L4=0

Multiply by N because there are N loops in the coil.

56

-2 -2 -2mag pullF = NILB= 100 1.5 10 5 10 0.6 =4.5 10 N=F

Magnitudes only (direction shown in diagram):

I

L1

L2 L3

F1

F2

F3

Fpull

This calculation assumes the coil is pulled out “uniformly;” i.e., no acceleration, so Fpull = Fmag.

57

ˆ ˆ

pull pullW =F D = NILB i L i

Work done by pulling force:

IFpull

D

-2 -2 -2 -3pullW = 100 1.5 10 5 10 0.6 5 10 =2.25 10 J

x

The work done by the pulling force is equal to the electrical energy provided to (and dissipated in) the coil.

58

If you build a generator and it doesn’t seem to be working…

Handy Hint: Debugging Your Homemade Generator

…or if you want to test a wall socket to see if it is “live…”

…simply purchase a Vilcus Plug Dactyloadapter from ThinkGeek ( http://www.thinkgeek.com/stuff/41/lebedev.shtml )

and follow the instructions.

European model shown. US model also available.

http://www.thinkgeek.com/stuff/41/lebedev.shtml






59

Today’s agendum:

Motional emf.You must be able to apply Faraday’s and Lenz’s “Laws” to calculation motional emf, as well as current and power in circuits powered by motional emf.



60

electric motors and web applets

Generator: source of mechanical energy rotates a current loop in a magnetic field, and mechanical energy is converted into electrical energy.

Electric motor: a generator “in reverse.” Current in loop in magnetic field gives rise to torque on loop.

Other useful animations here.

A dc motor animation is here.

Details about ac and dc motors at hyperphysics.

True Fact you didn’t know: all electrical motors operate on smoke. Every motor has the correct amount of smoke sealed inside it at the factory. If this smoke ever gets out, the motor is no longer functional. I can even provide the source of this true information. http://www.xs4all.nl/~jcdverha/scijokes/2_16.html#subindex

http://home.a-city.de/walter.fendt/phe/phe.htm

http://www.walter-fendt.de/ph14e/electricmotor.htm

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/mothow.html#c1

61

Let’s begin by looking at a simple animation of a generator. http://www.wvic.com/how-gen-works.htm

Here’s a “freeze-frame.”

Normally, many coils of wire are wrapped around an armature. The picture shows only one.

Brushes pressed against a slip ring make continual contact.The shaft on which the armature is mounted is turned by some mechanical means.

detailed look at a generator (if time permits)

brush

slip ring

Disclaimer: there are some oversimplifications in this analysis which an expert would consider “errors.” Anyone who is an expert at generators is invited to help me correct these slides!

http://www.wvic.com/how-gen-works.htm






62

Let’s look at the current direction in this particular freeze-frame.

B is down.

Put your fingers along the direction of movement. Stick out your thumb.

Bend your fingers 90°. Rotate your hand until the fingers point in the direction of B. Your thumb points in the direction of conventional current.

B is down. Coil rotates counter-clockwise.

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One more thing…

This wire…

…connects to this ring…

…so the current flows this way.

Another way to generate electricity with hamsters: give them little magnetic collars, and run them through a maze of coiled wires.http://www.xs4all.nl/~jcdverha/scijokes/2_16.html#subindex

64

Later in the cycle, the current still flows clockwise in the loop…

…but now this* wire…

…connects to this ring…

…so the current flows this way.

Alternating current! ac!

Without commutator—“dc.” *Same wire as before, in different position.

65

Today’s agendum:

Motional emf.You must be able to apply Faraday’s and Lenz’s “Laws” to calculation motional emf, as well as current and power in circuits powered by motional emf.



66

A changing magnetic field in wire produces a current. A constant magnetic field does not.

We saw how changing the magnetic field experienced by a coil of wire produces ac current.

But the electrical current produces a magnetic field, which by Lenz’s law, opposes the change in flux which produced the current in the first place.

http://campus.murraystate.edu/tsm/tsm118/Ch7/Ch7_4/Ch7_4.htm

back emf (also known as “counter emf”) (if time permits)









67

The effect is “like” that of friction.

The counter emf is “like” friction that opposes the original change of current.

Motors have many coils of wire, and thus generate a large counter emf when they are running.

Good—keeps the motor from “running away.” Bad—”robs” you of energy.

68

If your house lights dim when an appliance starts up, that’s because the appliance is drawing lots of current and not producing a counter emf.

Motors have design speeds their engineers expect them to run at. If the motor runs at a lower speed, there is less-than-expected counter emf, and the motor can draw more-than-expected current.

When the appliance reaches operating speed, the counter emf reduces the current flow and the lights “undim.”

If a motor is jammed or overloaded and slows or stops, it can draw enough current to melt the windings and burn out. Or even burn up.

69

Induced emf on an airplane wing.

Two brief examples (if time permits):

Blood flow measurement.

70

v

Example: An airplane travels 1000 km/h in a region where the earth’s field is 5x10-5 T and is vertical. What is the potential difference induced between the wing tips that are 70 m apart?

71

The electrons “pile up” on the left hand wing of the plane, leaving an excess of + charge on the right hand wing.

v

-

+

The equation for at the bottom of slide 10 gives the potential difference. (You’d have to derive this on a test.)

= B vε

-5 = 5×10 T 70 m 280 m/sε

= 1 Vε

No danger to passengers! (But I would want my airplane designers to be aware of this.)

72

= B ℓ v

v = / (B ℓ)

If B is applied to the blood vessel, then B is to v. The ions flow along the blood vessel, but the emf is induced across the blood vessel, so ℓ is the diameter of the blood vessel.

v = (0.1x10-3 V) / ( (0.08 T)(0.2x10-3 m) )

v = 0.63 m/s

Example: Blood contains charged ions, so blood flow can be measured by applying a magnetic field and measuring the induced emf. If a blood vessel is 2 mm in diameter and a 0.08 T magnetic field causes an induced emf of 0.1 mv, what is the flow velocity of the blood?

73

Today’s agendum:

Induced Electric Fields.You must understand how a changing magnetic flux induces an electric field, and be able to calculate induced electric fields.

Eddy Currents.You must understand how induced electric fields give rise to circulating currents called “eddy currents.”

Displacement Current and Maxwell’s Equations.Displacement currents explain how current can flow “through” a capacitor, and how a time-varying electric field can induce a magnetic field.

Leftovers.Any assorted topics I need to “cover” before finishing magnetism.

74

Time-Varying Magnetic Fieldsand Induced Electric Fields

This suggests that a changing magnetic flux produces an electric field. This is true not just in conductors, but any-where in space where there is a changing magnetic field.

A Changing Magnetic Flux Produces an Electric Field?

Bd = - N

dtε

b aV=V - V - Ed=

V= - Edε=

Bd - N = - Ed

dt

75

The previous slide uses an equation derived specifically for the electric field between two large parallel plates. Let’s see what a more general analysis gives us.Consider the conducting loop of radius r around (but not in) a region where the magnetic field is into the page and increasing.

But the changing magnetic flux induces an emf around the loop.

r

Bd = -

dtε

The charged particles in the conductor are not in a magnetic field, so they experience no magnetic force.

This could be a wire loop around the outside of a

solenoid.

76

r

The induced emf causes a counter-clockwise current (charges move).But the magnetic field did not accelerate the charged particles (they aren’t in it). Therefore, there must be a tangential electric field around the loop.

The work done moving a charged particle once around the loop is.

r

W=q =qV ε

The sign is positive because the particle’s kinetic energy increases.

I

E

EE

E

77

r

We can look at work from a different point of view.

The electric field exerts a force qE on the charged particle. The instantaneous displacement is always parallel to this force.

Thus, the work done by the electric field in moving a charged particle once around the loop is.

r

The sign is positive because the particle’s displacement and the force are always parallel.

I

E

EE

Eds

dsqEqEr2W

78

r

Summarizing…

r

I

E

EE

Eds

Combining and generalizing…

dsqEqEr2W

qWdt

d B

sdEqdsqEq

dt

dsdE B

dt

dsdE B

79

Generalizing still further…

r

I

E

EE

Eds

The loop of wire was just a convenient way for us to visualize the effect of the changing magnetic field.

The electric field exists whether or not the loop is present.

Was there anything in this discussion that bothered you?

A changing magnetic flux gives rise to an electric field.

dt

dsdE B

80

r

I

E

EE

Eds

This should bother you: where are the + and – charges in this picture?

Answer: there are no + and – charges. Instead, there are electric field lines that form continuous, closed loops.

2

qE= k , away from +

r

+-

Huh?

81

But wait…there’s more!

A potential energy can be defined only for a conservative force.*

*The work done by the force is independent of path.

A potential energy is a single-valued function.

If this electric field E is due to a conservative force, then the potential energy of a charged particle must be unchanged when it goes once around the loop.

E

82

But the work done is

If we tried to define a potential energy, it would not be single-valued:

Work depends on the path!

U is not single-valued! We can’t define a U for this E!

(*%&^#!)

EI and F

r

r2qEsdEqW

F I ifeven 0dt

dsdEqUU B

IF

sdEqUU IF

83

Induced Electric Fields: a summary of the key ideas

A changing magnetic flux induces an electric field, as given by Faraday’s Law:

This is a different kind of electric field than the one you are familiar with; it is not the electrostatic field caused by the presence of charged particles.

Unlike the electrostatic electric field, this “new” electric field is nonconservative.

“conservative,” or “Coulomb”

“nonconservative”

dt

dsdE B

NCC EEE

84

Stated slightly differently: we have “discovered” two different ways to generate an electric field.

Coulomb Electric Field

“Faraday” Electric Field

Both “kinds” of electric fields exert forces on charged particles. The Coulomb force is conservative, the “Faraday” force is not.

Both “kinds” of electric fields are part of Maxwell’s Equations.

dt

dsdE B

fromaway , r

qkE

2

85

Direction of Induced Electric Fields

The direction of E is in the direction a positively charged particle would be accelerated by the changing flux.

Use Lenz’s Law to determine the direction the changing flux would cause a current to flow. That is the direction of E.

dt

dsdE B

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Example

A long thin solenoid has 900 turns per meter and a radius of 2.50 cm. The current is increasing at a steady rate of 60 A/s. What is the magnitude of the induced electric field near the center of the solenoid 0.5 cm from the axis of the solenoid?

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Some Revolutionary Applications of Faraday’s Law

Electric Guitar Pickup Coils

Magnetic Tape Readers

Phonograph Cartridges

Ground Fault Interruptors

Alternators

Generators

Transformers

Electric Motors

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The Big Picture

These equations can also be written in differential form:

0

E

B 0

B×E=-

t

You have now learned Gauss’s Law for both electricity and magnetism… and Faraday’s Law of Induction:You have now learned Gauss’s Law for both electricity and magnetism…

dt

dsdE B

o

enclosedqAdE

0AdB

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Today’s agendum:




Leftovers.Any assorted topics I need to “cover” before Spring Break.

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Eddy Currents

You have seen how a changing magnetic field can induce a “swirling” current in a conductor (the beginning of this lecture).

These currents are called “eddy currents.”

If a conductor and a magnetic field are in relative motion, the magnetic force on charged particles in the conductor causes circulating currents.

Eddy currents give rise to magnetic fields that oppose any external change in the magnetic field.

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Eddy Currents

Eddy currents are useful in generators, microphones, and roller coaster brakes (among other things).

However, the I2R heating from eddy currents causes energy loss, so if you don’t want energy loss, you probably think eddy currents are “bad.”

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An ac current in a coil in the stove top produces a changing magnetic field at the bottom of a metal pan.

The changing magnetic field gives rise to a current in the bottom of the pan.

Because the pan has resistance, the current heats the pan. If the coil in the stove has low resistance it doesn’t get hot but the pan does.

An insulator won’t heat up on an induction stove.

Remember the controversy about cancer from power lines a few years back? Careful studies showed no harmful effect. Nevertheless, some believe induction stoves are hazardous.

Conceptual Example: Induction Stove

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Today’s agendum:





94

Displacement Current

+ --q+q

ICIC

dsApply Ampere’s “Law” to a charging capacitor.

CoIsdB

95

+ --q+q

ICIC

ds

Apply Ampere’s “Law” to the charging capacitor.

The shape of the surface used for Ampere’s “Law” shouldn’t matter, as long as the “path” is the same.

Imagine a soup bowl surface, with the + plate resting near the bottom of the bowl.

The integral is zero because no current passes through the “bowl.”

0sdB

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Hold it right there pal! You can’t have it both ways. Which is it that you want?

CoIsdB

0sdB

Evidently, our understanding is incomplete.

97

+ --q+q

ICIC

ds

As the capacitor charges, the electric field between the plates changes.

E

εA

q=C V= Edd

ε εE= EA=

As the electric field changes, the electric flux changes.

ε ε E E

dq d d= =

dt dt dt

This term has units of current.

Note: on this slide is 0, not an emf.

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+ --q+q

ICIC

ds

We define the displacement current to be

E

ε .D E

dI =

dt

The changing electric flux through the “bowl” surface is equivalent to the current IC through the flat surface.

The generalized (“always correct”) form of Ampere’s “Law” is then

Magnetic fields are produced by both conduction currents and time varying electric fields.

dt

dI II sdB E

oencloenclDCo

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The “stuff” inside the gray boxes is a new starting equation and serves as your official starting equation for the displacement current ID.

In a vacuum, replace with 0.

dt

dI II sdB E

oencloenclDCo

100

The Big Picture

These equations can also be written in differential form:

0

E

02

1 dEB= +μ J

c dt

B 0

dB×E=-

dt

You have now learned Gauss’s “Law” for both electricity and magnetism, Faraday’s “Law” of Induction, and the generalized form of Ampere’s “Law”:

dt

dI sdB E

oenclo

0AdB

o

enclosedqAdE

dt

dsdE B

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Today’s agendum:





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leftover: torque on generator coils

In lecture 15, I derived an expression for the torque on a current-carrying coil in a magnetic field.

FL

FR

B

IL

IR

W2

W

sin 2

net R L= =WILB sin =IBA sin

A

The source of external mechanical energy provides this torque.

Recall that for linear motion,

P = F V

For rotational motion,

P =

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The torque and angular velocity vectors are parallel, so

FL

FR

B

IL

IR

A

P =

P = IBA sin t

For a coil of N loops:

P = INBA sin t

If there are no frictional torques or other dissipative forces present, the generator’s power output is equal to the external power input.

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Today’s agendum: Induced emf