Elect Magnet 14

8/2/2019 Elect Magnet 14

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10. 10. ((INDUCTANCE)INDUCTANCE) 10.1.10.1.

10.2.10.2. RLRL 10.3. 10.3.

10.4. 10.4.

10.5.10.5. LCLC

10.6.10.6. RLCRLC


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10.2.10.2.RLRL ((RLRL circuits)circuits)

t=0t=0

-- , ,

, , , , ,,


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44

IIII

dx=dx=--dIdI..

dI/dtdI/dt++,, EELL -- -- bb

L

dIL

dt E

dI IR L 0

dt E

x IR

E

L dx dx Rx 0 dt

R dt x L

0

x Rln t

x L


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55

t=0t=0 I=0I=0 xx00== EE /R/R

tt=L/R=L/R RLRL , ,

: : EE/R/R 6363%%--

0

x Rln t

x L

RtL

0

x x e

RtLI e

R R

E E

R t t

L I ( 1 e ) ( 1 e )R Rt

E E


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t=0t=0 I=0I=0 t=t=

EE/R/R

tt

EE/R/R

dI/dtdI/dt t=0t=0

t =t =


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dI/dtdI/dt t=0t=0

SS 11 EE/R/R

t=0t=0 22

RLRL IIII

t=0t=0 II00==EE/R/R tt=L/R=L/R

dI IR L 0

dt

t t

0 I e I eR

t t

E


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10.3 10.3 30 30

, , 6 6 ,, 12 12 . . t=0t=0 ..

a)a) . .

)) t=2t=2

3 L 30 10 5

R 6t

t 0 ,412 B I ( 1 e ) ( 1 e ) 0,659 AR 6

t E


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1010

UUmm

dUdUmm=LIdI=LIdI

LL II

QQ22/2C/2C

mdU dILIdt dt

mU I

2 2m m m

0 01 1U dU LIdI LI U LI 2 2


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1111

L=L=mm

00nn22VV

VV B=B=mm00 nInI II-- LL--

--

(1/2)(1/2)ee00EE22--

2 2 ~~

2m 1U LI

2

2

mm

0

U Bu

V 2m

2m 1U LI

2

22

20

0 0

1 B Bn V V2 n 2m

m m


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1212

10.4 10.4

. .

. . II (( )). . LL-- . .

The Coaxial CableThe Coaxial Cable. A long coaxial cable. A long coaxial cableconsists of two concentric cylindricalconsists of two concentric cylindricalconductors of radiiconductors of radii aa andand bb and lengthand lengthll, as in Fig. The inner conductor is, as in Fig. The inner conductor isassumed to be a thin cylindrical shell.assumed to be a thin cylindrical shell.Each conductor carries a currentEach conductor carries a current II(the(theouter one being a return path).outer one being a return path).Calculate the selfCalculate the self--inductanceinductance LL of thisof this

cable.cable. . .

ll,, aa bb


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:: 2 2

2 2

: :

: : r


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LL

ll,, bb--aa

drdr

ldrldr

BdA=BldrBdA=Bldr

bb

0 0 0m

aa

I Il Il dr b BdA ldr ln

2 r 2 r 2 a

m m m

m 0l b L ln I 2 a

m

0 bln2 a

m


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10.4. 10.4.

NN22 22 11 22

1212 22 11 MM2121

NN11 11 II11

2 2121

1

NM

I

.

2121 1

2

MI

N


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: :

II11 11 22

II22 11

-- EE== --LdI/dtLdI/dt

~~

EE11 == EE22 MM1212=M=M2121=M=M . .

21 12 21

d dIN M

dt dt

E2

212

dIM dt E

1

1dIM

dt

E22dIM

dt

E1

10 5 10 5


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10.5 10.5 A long solenoid of

length l has N1

turns,

carries a current I,

and has a cross-

sectional area A.

l , A

N1 I

.

A second coilcontaining N2turns is woundaround the centerof the first coil, as

in Fig.

Find the mutualinductance of thesystem.

N2

.

.


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Solution:Solution: If the solenoidIf the solenoid

carries a currentcarries a current II11, the, themagnetic field at itsmagnetic field at its

center is given bycenter is given by

Since the fluxSince the flux 2121 througtthrougtcoilcoil 22 due to coildue to coil 11 isis BABA,,the mutual inductance isthe mutual inductance is

For example, ifFor example, ifNN11=500=500turns,turns, A=3A=31010--33mm22, l=0.5, l=0.5 mmandandNN22=8=8 turns, we getturns, we get

:: II11

11 22 2121 BABA

:: NN11=500=500,,A=3A=31010--3322,,l=0.5l=0.5,,NN22=8=8

0 1 1N IBl

m

2 21 2 1 20

1 1

N N BA N N AM

I I l

m

7 3 26( 4 10 Wb / A m )( 500 )( 8 )( 3 10 m ) M 30 10 H 30 H

0.5 m

m

10 510 5 LCLC


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10.5.10.5.LCLC

, ,

QQmm t=0t=0

. . t=0t=0 , ,

2

mQ

2C


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2020

, , 2 2--

. .

. .

. .

maxmax, ,

LCLC


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2121

LCLC tt

QQ,, II

2 2 UU

dU/dt=0dU/dt=0

Q, IQ, I

Q, IQ, I I=dQ/dtI=dQ/dt

22

C L

Q 1U U U LI

2C 2

22dU d Q 1 Q dQ dI LI LI 0

dt dt 2C 2 C dt dt


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2222

Q=QQ=Qmmcos(cos(wwt+t+dd)) QQmm-- LCLC

22dU d Q 1 Q dQ dI LI LI 0

dt dt 2C 2 C dt dt

2

2

dI d Q

dt dt

2 2

2 2

d Q Q d Q 1 L 0 Q

dt C dt LC

1

LCw


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QQ

Q=QQ=Qmmcos(cos(wwt+t+dd)) --

dd

t=0t=0 I=0I=0.. 0=0= --wwQQmmsinsindd.. dd=0=0

t=0t=0 Q=QQ=Qmm QQ,, II--

Q=QQ=Qmmcoscoswwtt I=I= --wwQQmmsinsinwwt=t= -- IImmsinsinwwtt

QQmm --QQmm,, IImm --IImm

, , 9090

I=0I=0,,Q=0Q=0

mdQ I Q sin( t )dt

w w d

LCLC


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LCLC

UUCC+U+ULL

Q,IQ,I-- --

, ,

22

C L

Q 1U U U LI

2C 2

2 22

2 2 2m mQ LIQ 1U LI cos t sin t 2C 2 2C 2

w w

2

mQ / 2C

21m2

LI


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, ,

coscos22wwt+sint+sin22wwt=1t=1 UUCC+U+ULL

UU

2

2mmQ 1LI

2C 2

2 22 2m m

C L

Q QU U U (cos t sin t )

2C 2C w w


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29.

6

An LC circuit has an

inductance of 2.81

mH and a capacitance

of 9 pF (Fig.). Thecapacitor is initially

charged with a 12-V

battery when the switch

S is open. The battery is

then removed from the

circuit, and the switch isclosed so that the

capacitor is shorted

across the inductor. (a)

Find the frequency of

oscillation.

LC 2.81 mH

, 9 pF

(). S

12V

.

S

.

(a)

.

10 610 6 RLCRLC


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10.6.10.6.RLCRLC

LCLC

QQmm

22

C L

Q 1U U U LI

2C 2

2dU / dt I R

2dU dI Q dQ LI I Rdt dt C dt

2

2

dQ dI d QI ,

dt dt dt

II--

2

2

d Q dQ Q L R 0

dt dt C


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R=0R=0 LCLC

, ,

RR

wwdd

I=dQ/dtI=dQ/dt

Rt/2L

m dQ Q e cos t w

12

2

d1 R

LC 2Lw

R 4L / C

1 / LC


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RR

--

RR

R=RR=RCC

R>RR>RCC

QQ II--

C R 4L / C

Elect Magnet 14

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