第8章恒定電流與穩(wěn)恒磁場(chǎng)小結(jié)

穩(wěn)恒磁場(chǎng)總結(jié)畢奧—薩伐爾定律安培環(huán)路定律磁

通

量安培力與磁力矩一、畢奧-薩伐爾定律用于計(jì)算任意載流體激發(fā)的磁場(chǎng)強(qiáng)度Id

lradBr

3dB

=4pm

Id

l

·

rr

24p

m

Idl

sin

a大小dB

=元電流：r

3L

L

4pm0

·lId

r整個(gè)導(dǎo)線：B

=

dB

=重要結(jié)論2paB

=

m0

I“無限長(zhǎng)”載流直導(dǎo)線:有限長(zhǎng)載流直導(dǎo)線:I+

b

21-

b2

14paB

=

mI

[sin

b

-

sin(b

)]IR圓載流導(dǎo)線中心的場(chǎng)：LB

=

mI

4pR

2圓弧長(zhǎng)度ABCDO12-

sin

b

)4pRB

=

mI

(sin bAB=

mI

[sin

0

-

sin(

-

p

)]=

mI BBC=

mI

p

=

mI2R

2p

4RBCD4pR

2

4pR=

BAB例1.計(jì)算O點(diǎn)磁感應(yīng)強(qiáng)度B

=

BAB

+

BBC

+

BCD例2.內(nèi)外半徑分別為R1、R2，面電荷密度為s的均

勻帶電非導(dǎo)體平面圓環(huán)，繞軸線以勻角速度w轉(zhuǎn)動(dòng)，求圓環(huán)中心的磁感應(yīng)強(qiáng)度。解：利用圓電流在中心出產(chǎn)生的磁場(chǎng)結(jié)論2RB

=

mI2rdB

=

mdIT

12RodB

=B

=R2

m

swdr2= ms

w(

R2

-

R1

)2p

/

wdI

=

dq

=

s

2prdr

=

swrdr21dB

=

mswdrIda例3.計(jì)算平面電流附近P點(diǎn)的磁感應(yīng)強(qiáng)度。xxdxI

dxdB

=

mdI

=

m2px

2px

dxmI

a

+d

dxB

=

dB

=

2pd

a=

mI

ln

a

+

d2pd

a解：利用長(zhǎng)直電流的磁場(chǎng)結(jié)論2pxdB

=

mdIddI

=

I

dx二、安培環(huán)路定律用于計(jì)算載流體激發(fā)對(duì)稱場(chǎng)的磁場(chǎng)強(qiáng)度NL

H

d

l

=

I

ii

=1

ILILH由環(huán)流內(nèi)外傳導(dǎo)電流共同決定圖示環(huán)流：L

H

dl

=

?

=

0IL例1r

<

RIpR

2Ir2pR

2H

=圓柱通有總電流為I，求各點(diǎn)磁場(chǎng)R2

p

r

＝

I解：由

L

H d

l

=

Ir

>

R

HIH

=2

p

rm

I

0

2

p

rB

=2pR

2pr

2mIrB

=H

2pr

=Ir2

p

R

2H

=r

<

Rj例2“無限大”導(dǎo)體薄平板，其上的面電流密度（即通過與電流方向垂直的單位長(zhǎng)度的電流）為j，求該平面電流的磁場(chǎng)分布。abcd解：作長(zhǎng)方形環(huán)路a-b-c-d，垂直穿過平板dacdH

dl+

H dl

+

H

dl

=

ab

H dl

+

bc

H

dl

=

2Hab=

jabH

=

j

B

=

m

j2

2dSF

m

=

S

B

dS

三、磁通量的計(jì)算IB由安培環(huán)路定律求得例1.ds

=

ldxB

=

m

I

2

p

xF =

mldxB

dsa

+

b

m

Ia=2

p

x=

m

Il

ln

a

+

b求直導(dǎo)線旁矩形面積的磁通量2

p

a注意：矩形面積若是變?yōu)槿鐖D形式，則F =

mIl

ln

bm

2

p

aIxbb

aba?·例2F =

B

dsmds

=

ldxB

mI

=

2

px=

mIl

ln

b2p

aIba求同軸無限長(zhǎng)載流圓筒的磁通量aa

+bmI

ldx=2pxxx注意：如果面積改成如圖形式，則磁通量是多少？不變。例3ds

=

ldx2

pR

2B

=

m

Ix

x

<

R

:F

=

0mIl2pR2Bds

=

ldx

=R

mIx

x<RIxRRlB

=

m

I

2

pxx

>

R

:F

m

=F

x<R

+F

x>RF

=4pldx

=

ln

2Bds

=2R

mI

mIl2px

2pRx>R求載流圓柱體的磁通量。四、安培定律及其應(yīng)用[一]安培定律B

dF

=

dF

cosgF

=dF

=

dF

cosaF

=zzF

Fy

=

dFy

=

dF

cosbxxdF

=

Idl B

sinqId

lqdFxyzLF

=

Idl

·

B

[二]磁場(chǎng)對(duì)載流線圈的作用（1）磁矩：描述線圈本身特性的物理量ISnPm

=

IS

n電流環(huán)繞面積與I成右旋關(guān)系（2）磁力矩：線圈在磁場(chǎng)中受力ISnBqM

=

Pm

·

B大小:

M

=

Pm

B

sinq例1計(jì)算[1]AB所受的磁場(chǎng)力；[2]AB所受的磁場(chǎng)力對(duì)O點(diǎn)的力矩。1I2IaboxaF

=

b

IdlB

d

F

=

Id

l

·

B?aF

=

b

I2

p

xm

I

12mI

1

I

2ln

b

a2

pdx

=dx例2計(jì)算[1]AB所受的磁場(chǎng)力；[2]AB所受的磁場(chǎng)力對(duì)O點(diǎn)的力矩。2

d

M

=

r

·

d

FdF

=

I

d