第八章 磁能.ppt...
TRANSCRIPT
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University of Science and Technology of China
第八章磁能
静电能:
点电荷→连续分布→外场中→能量密度→非线性介质→静电能求力
磁能:
一个线圈→多线圈→外场中→磁能密度→非线性介质→磁能求力
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University of Science and Technology of China
K接通,L中的电流从0→I,达到稳定。--暂态过程
在此过程中:
说明:电源在dt时间内做功εidt,其中一部分转变为电阻R的焦耳热i2Rdt,另一部分反抗L的感应电动势做功,大小为Lidi,该能量转变为线圈的磁能:
LidiRdtiidtdtdiLiR
+=∴
+=
2ε
ε
m2
0 21
21
Φ=== ∫ IWLILidiW m
I
m 或:
一、一个载流线圈的磁能
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0
01
2
2
=+∴
=+ ∫
CI
dtIdL
IdtCdt
dIL电路方程:
解:
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LCt41π
=
LCLCTt2
3243
43
2π
π ===
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考虑纯电感情况(忽略电阻),各线圈电流由0 → Ii,过程中的某一时刻,第i个线圈中:
电源反抗εi在dt时间内做功:
∑≠
−−=ik
kki
iii dt
dIMdtdILε
∑ ∑∑
∑
= ≠==
≠
+=′
+=−=′
N
i
N
ikkkiki
N
iiii
ikkikiiiiiii
dIIMdIILAd
N
dIIMdIILdtIAd
1 )(11
:个线圈,电源做的总功
ε
二、N个载流线圈系统的磁能
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积分得:
此即系统的静磁能。
( )
( )∑∑<=
+=′
=+∴=N
kikikiiki
N
iii
kiikikikkikikiik
IIdMIdILAd
IIdMdIIMdIIMMM
)(,1
,Q
∑∑<=
+=′N
kikikiik
N
ii IIMILA
i)(,1
2
21
∑∑≠=
+=N
kikikiik
N
iim IIMILW
i)(,1
2
21
21
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• 令Mii = Li
• 令Φki = MkiIk=MikIk
第k个线圈在第i个线圈中产生的磁通。
∑=N
kikiikm IIMW
,21
的总磁通。:表示通过线圈则 iN
kkii ∑
=
Φ=Φ1
∑=
Φ=∴N
iiim IW
121
其它形式:
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对于两个线圈:
只考虑互感磁能:
物理上,这是将线圈1产生的磁场B1看成外磁场,因此W12可以看成线圈2在外磁场中的磁能。
2112222
211 2
121 IIMILILWm ++=
自感磁能 互感磁能
中产生的磁通。在是线圈 2112
212211212
Φ
Φ== IIIMW
三、载流线圈在外磁场中的磁能
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对于均匀外场(或外场的非均匀尺度远大于线圈尺度):
N个线圈在外磁场中:
∫∫∫∫⋅=∴
⋅=Φ=
Sm
S
SdBIW
SdBIIWrr
rr
外
21221212
:线圈的磁矩外 mBmBSIWmrrrrr
⋅=⋅=
:整个系统的磁矩。
均匀外场
t
t
N
kk
N
kSkm
m
BmBmSdBIWk
r
rrrrrr⋅=⋅ →⋅= ∑∑ ∫∫
== 11
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如图所示均匀磁场中有一磁矩为 的载流线圈,磁矩方向与磁场方向垂直。当该线圈偏转到磁矩与磁场平行的方向时:
1)该过程磁场做功,所以磁能减少;
2) ,偏转前Wm=0,偏转后Wm=mB,所以磁能增加。
思考题1
mr Br
mr
mr
BmWm
rr⋅=
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•答案2正确,因为答案1中没有考虑载流线圈中的电源为了保持电流不变而做的功。
•如果不是载流线圈,而是磁矩为m的基本粒子,那么就是答案1正确,此时磁能表达式为:W=-m⋅B,这就是顺磁效应微观机制中所用的公式。
探讨1mr B
r
mr
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•如果旋转的是超导线圈,情况会是怎样?
假设初态时超导线圈电流为I0,并假设LI0>BS。那么在旋转过程中为保持超导线圈磁通量不变,电流会减小。但是由于LI0>BS,电流不会反向。因此旋转过程中仍然是磁场对线圈做功。
由于维持外磁场B的电源会对体系做功,因此从这个例子还无法看出线圈在外场中的磁能表达式W=m⋅B对超导线圈是否仍然成立。
探讨20mr B
r
1mr
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讨论:超导线圈的磁能
1 1L I 2 2L IM
载流线圈
超导线圈
( )1 1 21
1 1 1 1 1 1 2
2 2 1
2 22 2 2
1 1 1 1 1 1 1 2 22 2
2 2 21 1 2 2 2 2
1+
=- =-
=- = +2 + =0 0
1 1 1= - = = - = -2 2 2
1 1 1= + - =2 2 2
m
m
t dt
d L I MIddt dt
dA I dt L I dI MI dIL I MI
M MdA L I dI W A L I L I L IL L
W L I L I L I
ε
ε
→
Φ
∴ ⇒
时间内,
载流线圈 中的感应电动势:
电源反抗感应电动势做功:
对于超导线圈 : (超导线圈初始磁通为 )
或: 2 21 1 2 2 1 2
1+ +2
L I L I MI I
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一些结论
•磁能表达式对超导线圈仍然成立,但是由于超导线
圈自感磁能会发生变化,因此外场中的磁能公式W=m⋅B并没有什么意义。
•超导线圈放入磁场中,总磁能会减少,减少量等于
超导线圈的“自感磁能”。
2 21 1 2 2 1 2
1 1= + +2 2mW L I L I MI I
2 21 1 2 2
1 1= -2 2mW L I L I
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超导线圈进入磁场,若要保持源磁场不变,电源做功A:
•电源做功是磁能变化的两倍。因此在应用虚功原理时,虽然超导线圈的电流会发生变化,但只要外磁场不变,仍然适用“I不变”的条件。•将超导线圈放入磁场,虽然磁能减小,但由于电源“吸收”了两倍的能量,因此仍然需要外界提供能量。
( ) ( )
( )
1 1 2 21
21 1 1 2 1 2 2 2
+=- =-
=- = = =-
d L I MI d MIdt dt
dA I dt I d MI A MI I L I
ε
ε ∴
电源做功问题
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如图所示,x>0区域
有一均匀磁场,一个自
感L的矩形线圈(电阻
忽略)以初速度v0进入
磁场。当线圈中电流为
I时,线圈速度v是多少?
练习
x
y
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螺线管:
H=nIB=µrµ0nIL=Φ/I= µrµ0nNS= µrµ0n2V
Sl
rµn
HBBHV
Ww
BHVVInLIW
mm
rm
rr⋅===
===∴
21
21
21
21
21 22
02
能量密度:
µµ
四、磁场的能量和能量密度
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可推广:
( )
( )
( ) ( ) ( ) ( )[ ]( ) ( )[ ]∫∫∫
∫∫∫
∫∫∫
∫∫∫
×⋅∇+×∇⋅=
×∇⋅−×∇⋅=×⋅∇×∇⋅=
×∇=⋅=
→⋅=⋅=Φ=
dVAHAH
baabbadVHA
HjdVjA
dVjlIdldAISdBIIW
rrrr
rrrrrrrr
rrrr
rrrrrr
21
21
21
21
21
21
BHwm
rr⋅=∴
21∫∫∫ ⋅ dVBH
rr
21 ( )
01~
~1~1~ 22
→⋅×∴
⋅×=×⋅∇ ∫∫∫∫∫∞→
rSdAH
rdSrHrA
SdAHdVAHr
rrr
rrrrr
,,
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(假设导线的µr=1)
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计算自感系数的方法
• 通过磁能求自感
• 通过自感定义求自感
• 通过感应电动势求自感
22 2
21
IWLLIW m
m =⇒=
ILLI m
mΦ
Φ =⇒=
dtdIL
dtdIL LL εε −=⇒−=
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介质存在时,电源对线圈做功。
t→t+dt B→B+dB电源做功:
( )
HVdBNSdBNHl
NHlINIldHINSdB
NSdBdIddtdIdtAd
==
=∴=⋅=
=ΦΦ=
Φ
−=−=′
∫
,rr
εε
五、非线性介质及磁滞损耗
nS
lrµ
ε
![Page 30: 第八章 磁能.ppt [兼容模式]staff.ustc.edu.cn/~xuck/%bd%b2%d2%e5%20%b5%e7%b4%c5%d1%a7%d3%eb%b… · University of Science and Technology of China K接通,L中的电流从0fiI,](https://reader030.vdocuments.pub/reader030/viewer/2022033120/5e0edecbf24bc168dc10cb8c/html5/thumbnails/30.jpg)
University of Science and Technology of China
单位体积做功:
电源做功等于宏观磁能的改变加上磁化功。
( )
MdHHd
MHBBdHHdBVAdad
rr
rrrrr
⋅+
=
+=⋅==′
=′
020
0
2
µµ
µ
宏观磁能密度磁化功
![Page 31: 第八章 磁能.ppt [兼容模式]staff.ustc.edu.cn/~xuck/%bd%b2%d2%e5%20%b5%e7%b4%c5%d1%a7%d3%eb%b… · University of Science and Technology of China K接通,L中的电流从0fiI,](https://reader030.vdocuments.pub/reader030/viewer/2022033120/5e0edecbf24bc168dc10cb8c/html5/thumbnails/31.jpg)
University of Science and Technology of China
• 线性无损介质
( )
( )
( )m
m
wdHBd
MHHHd
MHdHdad
MHdMdH
MHdHdMMdH
=
⋅=
⋅+⋅=
⋅+
=′∴
⋅=⋅
⋅=⋅=⋅∴
rr
rrrr
rr
rrrr
rrrrrr
21
21
21
21
21
21
0
02
0
00
µ
µµ
µµ
χ
磁化能密度
磁化功:
常数,
电源做功等于磁能改变。(无损耗)
![Page 32: 第八章 磁能.ppt [兼容模式]staff.ustc.edu.cn/~xuck/%bd%b2%d2%e5%20%b5%e7%b4%c5%d1%a7%d3%eb%b… · University of Science and Technology of China K接通,L中的电流从0fiI,](https://reader030.vdocuments.pub/reader030/viewer/2022033120/5e0edecbf24bc168dc10cb8c/html5/thumbnails/32.jpg)
University of Science and Technology of China
• 非线性介质磁滞回线,循环一周做功:
磁滞损耗
磁滞回线面积=
⋅=′=′ ∫∫
0 MdHAdArr
µ
H
M
![Page 33: 第八章 磁能.ppt [兼容模式]staff.ustc.edu.cn/~xuck/%bd%b2%d2%e5%20%b5%e7%b4%c5%d1%a7%d3%eb%b… · University of Science and Technology of China K接通,L中的电流从0fiI,](https://reader030.vdocuments.pub/reader030/viewer/2022033120/5e0edecbf24bc168dc10cb8c/html5/thumbnails/33.jpg)
University of Science and Technology of China
例:小线圈在外场中受力
( ) ( )
∂∂
−=
∇−=
∂∂
=
∇=
Φθ
Φ
θ θθm
m
I
m
Im
WL
WF
WL
WFrr
m
m
)(或
)(或
( ) ( )[ ]( ) ( ) ( )
BmemBeWL
BmBmBm
BmWF
mBBmW
m
mm
m
rrrrr
rrrrrr
rrr
rr
×=−=
∂∂
=
∇⋅=×∇×+∇⋅=
⋅∇=∇=
=⋅=
θθθ
θ θθ
θ
sin
cos
六、利用磁能求磁力
![Page 34: 第八章 磁能.ppt [兼容模式]staff.ustc.edu.cn/~xuck/%bd%b2%d2%e5%20%b5%e7%b4%c5%d1%a7%d3%eb%b… · University of Science and Technology of China K接通,L中的电流从0fiI,](https://reader030.vdocuments.pub/reader030/viewer/2022033120/5e0edecbf24bc168dc10cb8c/html5/thumbnails/34.jpg)
University of Science and Technology of China
Ø载流线圈的磁能
其它形式:
Ø在外场中的磁能
∑∑≠=
+=N
kikiik
N
iim IIMILW
i 21
21
1
2
自感磁能 互感磁能
∑∑=
Φ==N
iiim
N
kikiikm IWIIMW
1, 21
21
外均匀外场
外外 BmSdBIIWm
rrrr⋅ →⋅=Φ= ∫∫
小结
![Page 35: 第八章 磁能.ppt [兼容模式]staff.ustc.edu.cn/~xuck/%bd%b2%d2%e5%20%b5%e7%b4%c5%d1%a7%d3%eb%b… · University of Science and Technology of China K接通,L中的电流从0fiI,](https://reader030.vdocuments.pub/reader030/viewer/2022033120/5e0edecbf24bc168dc10cb8c/html5/thumbnails/35.jpg)
University of Science and Technology of China
Ø能量密度
Ø磁滞损耗
Ø磁能求力
BHwm
rr⋅=
21
( ) ( )
∂∂
−=
∇−=
∂∂
=
∇=
Φθ
Φ
θ θθm
m
I
m
Im
WL
WF
WL
WFrr
m
m
)(或
)(或
MdHHdadrr
⋅+
=′ 0
20
2µ
µ
宏观磁能密度磁化功