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### Transcript of CHAPTER 21 ELECTROMAGNETIC INDUCTION - 21.pdf  1 CHAPTER 21 ELECTROMAGNETIC INDUCTION BASIC...

• 1

CHAPTER 21

ELECTROMAGNETIC

INDUCTION

BASIC CONCEPTS

Lenzs Law

Motional emf

• 2

Electric Charge Produces an

Electric Field

Moving Electric Charge Produces a

Magnetic Field

Now Changing Magnetic Field Produces an

mf

• 3

a potential difference (voltage)

What is changing and causing

the emf is the magnetic flux.

• 4

We defined Electric Flux

=

Magnetic Flux is similar

=

Flux through an area

• 5

We will define a vector to

represent area. The direction

of the vector will be

perpendicular to the surface.

The length of the vector will be

the area of the surface. The

vector will be .

The flux will be

• 6

First, flat area perpendicular to

= 0 =

• 7

For a surface parallel to

= 90 = 0

And for in between

=

=

• 8

If this flux changes an emf will

be induced in the area.

Consider a copper coil with one

loop

B A

r

• 9

=

For N loops

=

B A

r

N loops

• 10

The emf induced in a circuit is

directly proportional to the

time rate of change of the

magnetic flux through the

circuit.

Or

=

• 11

If changes with time an emf, , will be produced.

Coil in magnetic field.

• 12

Flux through coil

=

Flux can change:

1. Magnitude of changes.

2. Area of loop changes.

• 13

3. The angle changes.

4. Any combination of the

three.

• 14

Example:

Consider a coil with 200 turns

on a rectangular frame, 20 cm

by 30 cm. The resistance of the

coil is 2 .

• 15

If increases uniformly from zero to 0.5"#/% in 0.8 s what is the current in the coil?

= &0.2%(&0.3%( = 0.060%

At = 0 = 0

At = 0.8 = =&0.5"#/%(0.060% = 0.03"#

• 16

|| = & = 0( & = 0.8(0.8

|| = 200&0.03"#(0.8

|| = 7.5/

The current will be

0 = 1 =7.5/2 = 3.75

• 17

Lenzs Law

A changing magnetic flux will

cause an emf and if the emf is

in a conductor there will be a

current.

What will be the direction of

the current?

question.

• 18

The polarity of the induced emf

is such that it tends to produce

a current that will create a

magnetic flux to oppose the

change in magnetic flux

through the loop.

• 19

a. Magnet moves toward loop.

Magnet field is down and

increasing. Induced emf and

thus current will be ccw to

oppose the increase.

• 20

b. Magnet moves away from

loop. Magnetic field is up and

but decreasing.

Induced current gives magnetic

field to oppose the decrease.

• 21

Motional emf

x

• 22

In time t the rod will move x.

The area covered in t will be

= 34

4 = 5

= 35

• 23

=

= = 35

= 35 = 35

A conducting rod moving in a

magnetic field will have an emf

Induced across it given by this

equation.

• 24

Example: My airplane

Luscombe 8E

Built 1947

Wingspan 22 feet = 6.7m

Speed 100 mph = 44.7 m/s

• 25

If I am flying it at a point on

earth where the vertical

magnetic field is 5410789 what is the voltage drop across the

wings?

= :5

= &5410789(&6.7%(&44.7%/(

= 1.54107/

• 26

Consider a loop around a long solenoid.

• 27

The current in the solenoid is increasing

so the flux through the loop is

increasing.

An emf is induced in the loop by

.

• 28

INDUCTANCE

BASIC CONCEPTS

Mutual Inductance

Self Inductance

Magnetic Field Energy

Circuits with Inductors

• 29

Mutual Inductance

I1 #1

= >

If I1 changes with time t

0>

?@ABCCCD > E=

• 30

Put second coil at P

I1 #1 #2

F

?@ABCCCDGHG:#2

• 31

?@ABCCCD 0

If I1 changes with time

For example 0> = 0JGHK

Then > = JGHK

And

= LM

• 32

Or = JKK

And

0 = 1 =JK

1 K

I1 #1 #2

I2

• 33

N = >0> =PJ >: 0>

0>

N = PJ >: =PJ>

:

Only depends on geometry, number of

turns, area and length.

• 34

Self Inductance

If the current is changing then there will be

a changing magnetic flux through the coil.

• 35

A changing magnetic flux will induce an

emf.

= @L@

Therefore we define in analogy to the

mutual inductance

3 = G

This is Self-Inductance

Solve for

• 36

= 3G

= QLRQ = 3QQ

Self-Induced emf

= 3 G

• 37

What is the inductance, 3, of the black coil?

3 = S0

TUU@ = PJ >: 0>

= PJ >: 0>

3 = >PJ>: 0>

0> = PJ

>:

Area

A

• 38

Magnetic Field Energy

Consider the coil

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box.]

Current starts at zero and increases to 0.

• 39

=VW = 01 = /

1 = 0/ Or

= = 0 And

= 3 0

|=| = 0 X3 0Y = 030

Power is work per unit time.

= = "

• 40

" = = = 03 0 = 300 How much work to bring current through

inductor from zero to 0?

To calculate the work integrate from zero

current to current I

We wont do it but you may see how on the

next page.

• 41

" = Z 30[0 = 3 Z 0[0

\

J

\]\

\]J

" = 12 30

"^ _UA UBCCCD `W[ HWab

• 42

The energy stored in the magnetic field is

c = 12 30

Apply to solenoid

• 43

/:d%W = :

= PJ >: 0 So

0 = :PJ>

3 = PJ >

:

Put into equation for c

c = 12 30 = 12 PJ

>

: X

:PJ> Y

• 44

c = 12

:

PJ

c = 12PJ &:(

And

/:d%W = : So

HWab/:d%W =

12

PJ &:(&:(

d = 12PJ

Energy stored in magnetic field of coil.

• 45

R-L Circuit

Consider this circuit

• 46

First consider that switch > is closed and calculate how the current will increase with

time.

Second, then while there is a current in the

circuit consider opening > and closing ` and calculate how the current will decay.

Close >

• 47

At time = 0 close switch. Kirchkoffs loop starting at the switch and

going ccw.

• 48

G1 3 G = 0

G =

G13

G =

3

13 G

At t = 0 i=0

Then

X0Y]J = 3

After long time

0 = 0

• 49

And

0 = 1

• 50

• 51

The L-C Circuit

The capacitor has been charged to an initial

charge e and then placed across the inductor.

• 52

The L-C Circuit

Remember energy in

Capacitor cf = >ghi

And

Inductor c = > 3G

Then study

• 53

• 54

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