物理ii課件chapinduction前沿科技高端對(duì)話

前沿科技高端對(duì)話時(shí)間：2016年4月1日（週五）早上9:20-12:30地點(diǎn)：N101內(nèi)容：白春禮院士、葉培建教授、ErwinNeher教授及楊芙清院士進(jìn)行演講Mid-term

examination(期中考試)時(shí)間：2016年4月8日（週五）早上10:50-12:50地點(diǎn)：C301考試方式：閉卷內(nèi)容：本學(xué)期教授內(nèi)容，課件見公用郵箱（

;xpzhang)英文試卷/答題可帶計(jì)算器不可帶字典，電子詞典，手機(jī)，電腦Chapter

30Induction

and

Inductance實(shí)驗(yàn)一：插入或拔出條形磁鐵Someexperimentsaboutinduction實(shí)驗(yàn)二：插入或拔出載流線圈實(shí)驗(yàn)三：接通或斷開初級(jí)線圈的電流實(shí)驗(yàn)四：導(dǎo)線做切割磁感線運(yùn)動(dòng)When

the

magnetic

fluxpassingthrough

the

loop

is

changing,

an

emf

isinduced

in

the

loop.（emf：法語，電動(dòng)勢(shì)；英語：electromotive

force）當(dāng)穿過閉合導(dǎo)體回路的磁通量發(fā)生變化時(shí)，回路中就產(chǎn)生電流.Faraday’s

Law

of

Induction:

Magneticflux:

F

B

=

B

dSSSI：WeberLenz’s

Law

（楞次定律）：An

induced

current

has

a

direction

suchthat

the

magnetic

field

due

to

thecurrentopposes

the

change

in

the

magneticfluxthat

induces

the

current.閉合回路中感應(yīng)電流的方向，總是使得它所激發(fā)的磁場(chǎng)與引起感應(yīng)電流的磁通量的變化方向相反。The

direction

of

an

induced

emf is

that

of

theinduced

current.ΦB

Bεt(a)

Φ

>

0,

dΦ

BB

dtL>

0,

ε

<

0

Bε(b)

Φ

BL

ΦB>

0,

dΦ

B

<

0,

ε

>

0dttΦB

Bt(c)

Φ

<

0,

dΦ

BB

dtL<

0,

ε

>

0

B(d)

ΦBL

ΦB<

0,

dΦ

B

>

0,

ε

<

0dttThe

direction

of

emf

is

opposite

to

the

direction

ofchange

in

magnetic

flux.A

quantitative

treatmentAfter

accurate

experiments,

it

is

found

thatthe

magnitude

of

emf

is

proportional

to

thechange

rate

of

magnetic

flux.感應(yīng)電動(dòng)勢(shì)的大小與磁通量的變化率成正比。dte

=

-

dF

BExample

1:

A

rectangular

coil

of

N

turns

and

oflength

a

and

width

b

is

rotated

at

frequency

f

in

auniform

magnetic

field

B.

The

coil

is

connected

to

co-rotating

cylinders,

against

which

metal

brushes

slide

tomake

contact. Find

the

emf

induced

in

the

coil.

S

θ

Bw

ΦB

=

B S

=

BS

cos

θε

=

-

dΦB

=

BS

sin

θ

dθ

dt

dtε

=

BSω

sin

ωt

=

ε0

sin

ωtε0

=

BSω02effe=

e0=

0.707eEffective

emf:θ

=

ωt正弦交流電Solution:Example

2:

a

metal

rod

is

forced

to

movewith

constant

velocity

v

along

two

parallelmetal

rails,

the

magnetic

field

is

uniform.Find

the

induced

emf.Example

2:

a

metal

rod

is

forced

to

move

withconstant

velocity

v

along

two

parallel

metalrails,

the

magnetic

field

is

uniform.

Find

theinduced

emf.dtdtF

B

=

B

l

se

=

-

dF

B=

-Bl

ds

=

-Blv順時(shí)針Classwork

1:

A

square

loop

of

wire

has

sides

oflength

2.0

cm.

A

magnetic

field

is

directed

out

ofthe

page;

its

magnitude

is

given

by

B=4.0t2

y,where

B

is

in

teslas,

t

is

in

seconds,

and

y

is

inmeters.

At

t=2.5s,

what

are

the

(a)

magnitude

and(b)

direction

of

the

emf

induced

in

the

loop?30lBF

=SF =

2

·

2.52

·

0.023

=

10-4

WebB

B

dS

=

4t

2

yldy

=

2t

2

ldte

=

-

dF

B=

-4tl

3e

=

-4

·

2.5·

0.023

=

-8·10-5

VSolution:Magnetic

flux:emf:順時(shí)針

f

=

iL

·

Bemf：ε

=

BLvi·

·

·

·

·

·

·

··

·

·

·

·

·

·

··

·B··

·

·

·

·

·

·

·

·

·

f

·

·

·v

-+-ε

=

BlvCDInduction

and

energy

transfersWhen

the

conducting

rodis

moving

in

the

magneticfield,

it

will

feel

theAmpere

’s

forceCurrent:i

=

e

=

BLvR

RB2

L2vRf

=

iBL

=B2

L2vRFex

=

f

=

iBL

=RB2

L2v2

e2P

=

Fex

v

==

RThe

external

force

equals

to

the

Lorentz’s

forcei.e.,

P

=

i

2

RThe

work

done

by

the

external

force

equals

to

thethermal

energy,

conversion

of

mechanical

energy

tothermal

energy.v

B

lLε

=

(

·

)

d

In

general, the

emf

for

a

moving

conductorrod

isExample

3:de

=

(v

·

B)

dl

=

vBdl

=

Bw

ldl

A

??v

=

w

·

l

L

B

O

2012LBw

Le

=de

=

Bw

ldl

=

2ABU

-U

=

1

Bw

L2

>

0Induced

electric

field(感生電場(chǎng))According

to

Faraday’s

law,

the induced

emfin

a

conducting

loop

depends

only

on

thechange

rate

of

the

magnetic

flux.The

induced

emf

is

due

to

the

inducedelectric

field. (感生電場(chǎng))A

new

description

of Faraday’s

Law:A

changing

magnetic

field

produces

anelectric

field.BdtdF=

-

Induced

emf：

e

=

E

i

dl

Total

electric

field

：

e

=

(

Es

+

Ei

)

dl

=

E

dldl

=

(

·

Es

)

dS

=

(

·(

V

))

dS

”

0

EsE

=

Es

+

EiFor

statisticfield

d

dt

S

E

dl

=

-

B

dS

LThenFor

a

given

loop,

S

?B?t

dS

E

dl

=

-

Ldt

d

dte

=

-

dF

B

=

-

S

B

dSOn

the

otherhand,iL

=

N

F

BInductor

and

InductanceInductor(電感)：任何線圈均可視為電感,通常有N匝。Inductance(電感系數(shù))：B

=

m0niInductance

of

a

long

solenoidL

=

N

F

B

=

NBSi

iL

=

Nm

nS

=

m

n2V0

0N

F

B

=

LiSelf-induced

emf:e

=

-

d

(

N

F

B

)

=

-L

didt

dtLC

oscillator

circuit (LC振蕩電路):L

di

+

q

=

0dt

Cdti

=

dqd

2q

qL

+ =

0dt

2

Cq

=

q0

cos(w

t

+j0

)1LCw

=LC

振蕩電路K12εEnergy

stored

in

a

magnetic

fielddte

=

L

di

+

iRdtei

=

Li

di

+

i

2

RAccording

to

the

conservation

of

energy,

theenergy

delivered

to

the

conductor

is

stored

inthe

magnetic

field.BdUBdtdt=

Li

di

dU=

Lidi2BU

=

1

Li

2thenBUuV=

B

Energy

density2B2VLi

2

m

n2Vi

2

m

n2i

2u

==

0

=

0

2VB

=

m0niUsingWe

obtain0B2uB

=

2mIt

is

the

generalexpression

ofmagnetic

fieldenergy.02e2u

=

e

EClasswork

1:

What

must

be

the

magnitude

of

a

uniformelectric

field

if

it

is

to

have

the

same

energy

density

as

thatpossessed

by

a

0.50

T

magnetic

field?Exercise:

A

long

solenoid

has

a

diameter

of

12.0

cm.When

a

current

i

exists

in

its

windings,

a

uniform

magneticfield

of

magnitude

B

=

30.0

mT

is

produced

in

its

interior.By

decreasing

i,

the

field

is

caused

to

decrease

at

the

rateof

6.50

mT/s.

Calculate

the

magnitude

and

direction

of

theinduced

electric

field

(a)

2.20

cm

and

(b)

8.20

cm

from

theaxis

of

the

solenoid.L

Sd

E

dl

=-

dt

B

dSClasswork

1:

What

must

be

the

magnitude

of

a

uniformelectric

field

if

it

is

to

have

the

same

energy

density

as

thatpossessed

by

a

0.50

T

magnetic

field?E20=

1

B2B

20u

=

1

e

E

2

=

umB0.50Te0

m0E

==(8.85·10-12

F

m)(4p

·10-7

H

m)=1.5·108

V m

.Exercise:L

Sd

E

dl

=

-

dt

B

dSFor

a

given

loop,L?t

S

?B

dS

E

dl

=-

L

LLAccording

to

symmetry

E

dl

=

Edl

=

E

dl

=

E(2p