13. 1 electromagnetic induction law
DESCRIPTION
Chapter 13 电磁感应 Electromagnetic Induction. 13. 1 Electromagnetic induction law. 13.2 Motional electromotive force. 13. 3 Induced electric field and electromotive force. 13. 4 Eddy current. 13. 5 Self-induction. 13. 6 Mutual induction. 13. 7 Energy of magnetic field. - PowerPoint PPT PresentationTRANSCRIPT
13.1 Electromagnetic induction law13.1 Electromagnetic induction law
Chapter 13 Chapter 13 电磁感应 电磁感应 Electromagnetic InductionElectromagnetic Induction
13.2 Motional electromotive force13.2 Motional electromotive force13.3 Induced electric field and 13.3 Induced electric field and electromotive forceelectromotive force13.4 Eddy current13.4 Eddy current
13.5 Self-induction13.5 Self-induction
13.6 Mutual induction13.6 Mutual induction13.7 Energy of magnetic field13.7 Energy of magnetic field13.8 Application13.8 Application
13.1 Electromagnetic induction law13.1 Electromagnetic induction law
1822 年实验物理学家法拉第提出了“磁能否产生电”的想法。历经多次失败,经过10 年探索,在人类历史上第一次发现了电磁感应现象,并总结得出了电磁感应定律。
迈克尔 · 法拉第,Michael Faraday ,( 1791 ~ 1867 )英国物理学家、化学家,著名的自学成才的科学家
1 Electromagnetic induction phenomenon 电磁感应现象
Electromagnetic induction Electromagnetic induction 电磁感应电磁感应 The change in magnetic flux produces thThe change in magnetic flux produces the induction current.e induction current.
InductionInduction electromotive force 感应电动势 The change in magnetic flux produces The change in magnetic flux produces the induction the induction electromotive force. For a l For a loop of wire, the induction current producoop of wire, the induction current produces.es.
2 Lenz’s law 楞次定律1833 年俄国物理学家楞次,提出了楞次定律 ( Lenz law ) 。The induction current always flows to create a magnetic field that opposite the change in flux through the circuit.
闭合回路中感应电流的方向,总是使得它所激发的磁场来阻止引起感应电流的磁通量的变化。
实质: Energy conservation
or : The polarity of the induction electromotive force is such as to oppose the change that caused it.
13.1 Electromagnetic induction law13.1 Electromagnetic induction law
13.1 Electromagnetic induction law13.1 Electromagnetic induction law3 The law of electromagnetic induction 电磁感应定律
Fig. 1 Fig 2 Fig 3 Fig 4
B 变 变 不变 不变
S 不变 不变 变 变
(II) 有 有 有 有
不管什么原因引起通过闭合回路的磁通量发生变化,都会在导体回路中产生电流。
A
SN
Fig1 A bar magnet inserts in or pulls out loop
A
Fig2 on and off
Fig3 : Rotating
A
Fig4 Moving
1845 年,纽曼( F.E.Neumann )给出了电磁感应定律的数学表达式。
iIt
0d
d )1( i iI
d(2)
d i iIt
d
di
Φ
t
Faraday’s law 法拉第电磁感应定律 The induction electromotive force (emf) is proportional to the change rate of magnetic flux through the loop
Basic idea 基本思想
d
d
t
Φi 国际单位制:
13.1 Electromagnetic induction law13.1 Electromagnetic induction law
EMF Point 感应电动势的指向
0,0,0)1(
idt
d 0,0d
d,0)2(
it
n
i
n
i
0,0d
d,0)3( it
ΦΦ 0,0
d
d,0)4( it
ΦΦ
n
i
n
i
1 标定回路绕行方向;2 根据回路的绕行方向,按右手螺旋定则定出回路所围面积的正法线方向;3 若磁感应强度的方向与的夹角小于,则穿过回路的磁通量为正;若与的夹角大于,则穿过回路的磁通量为负;4 有了磁通量的正负,根据磁通量变化率的正负确定感应电动势的正负。
13.1 Electromagnetic induction law13.1 Electromagnetic induction law
5 Correlated content 相关内容(1) Induction emf and induced current:
Related to resistance d
d
t
Φi
)-(1
d1
d 12
2
1
2
1
RRtIq
t
t ii
d
d1i
t
Φ
RRI i
(2) Induction charge and magnetic flux 感应电荷与磁通量 :
13.1 Electromagnetic induction law13.1 Electromagnetic induction law
(3) Multiturn winding ( loop) 多匝线圈
1 21 2
dd d d d( )
d d d d dN
i Nt t t t t
N 若 d d
d dN
t t
13.1 Electromagnetic induction law13.1 Electromagnetic induction law
(4) An induction emf can be created (The magnetic flux change can be created):
a) By changing the magnetic field;b) By changing S (the shape of the loop);c) By changing the orientation of the loop;d) By a combination of these.
(5) Motional electromotive force and induced electromotive force
由于导体移动或线圈转动产生的电动势称为动生电动势由于磁磁感应强度变化而产生的电动势称为感生电动势
d d cosS S
B S B S
13.2 Motional electromotive force13.2 Motional electromotive force
11 Motional electromotive force Motional electromotive force 动生电动势动生电动势
vBLi i
Pointing: a b
v
B
L
b
a2 Motional electromotive force and Motional electromotive force and Lorentz force
Befm
vMagnetic force:
Eefe
Electric force:
v
B
mf
a
b+
-
ef
By changing S and changing the orientation of the loop
em ffif
BeeE v BE v
BLVV ab vOpen circuit:
BLi v
运动导体相当于一个电源,洛伦兹力提供了非静电力
kEB
v 单位正电荷受到的洛伦兹力 - 非静电力
b
ai lB
dv
b
a
b
a ki lBlE
d)(d v
The Definition of MotionalMotional emf:
The pointing of emf: 电动势的指向
The linger line indicates the positive terminal, out of which current flow into an external circuit
13.2 Motional electromotive force13.2 Motional electromotive force
3 计算动生电动势的通用公式
说明: (1) 对于运动导线回路,电动
势也可以用法拉第电磁感应定律计算:
( 法拉第电磁感应定律 )
(2) 感应电动势做功,洛伦兹力做不做功?B
v
f V
'f
F
VF )'()'( vv
ff
vv '' ff
vvvv BeBe '' 0
e
'v
洛伦兹力做功为零
B
lv
l
d
Be
v
Be
'v
d
di
Φ
t
13.2 Motional electromotive force13.2 Motional electromotive force
例 1 长直导线通有电流 I ,一长为 l 的金属棒以速度平行于导线运动,如图所示。试计算移动金属棒上的动生电动势。 方法 1: 直接用动生电动势定义进行计算
方法 2 :利用法拉第电磁感应定律
13.2 Motional electromotive force13.2 Motional electromotive force
d ( ) d di B l B l
v v0 0d d d ln
2 2
a b a b a b
i ia a a
I I a bB l r
r a
vv v
0i 棒上动生电动势的指向与 的方向相同 dl
选定回路 ABCD:
0 0d d ln2 2
a b
S a
I Iy a by r
r a
B S
0 0d dln ln 0
d d 2 2i
Iy Ia b a b
t t a a
v
d
dy
t v
i 与 的方向相同dl
I b
Ex 1 A metal rod ab is pivoted at one end a and rotates with an angular speed in a uniform magnetic field B directed as shown. Find the electric potential difference Uab 。
Method 1: b
ai lB
dv
a
b
b
a
b
alBllB dd v
LlBl
0d 0
2
1 2 BL
B
v
l
d
lBLUUU baab 2
2
1 b 端电势高
Method 2 : Using Faladay’s law
22
1 LS 2
2
1 BLBSΦ
22i 2
1
d
d
2
1
d
d BL
tBL
t
Φ
a
bb'
td
d 随着时间变化减小
13.2 Motional electromotive force13.2 Motional electromotive force
a
b
c
dBad=bc=l1 ab=cd=l2
n
d)( a
bba lB
v
d)( c
ddc lB
v sind)sin( 20
2
lBlBl
vv 0 d)(
)(
)(
bc
dacdad lB
v
sindsin 20
2
lBlBl
vv
sin2 2lBdcbai v 2
1lv
sinsin 21 BSlBli 21 llS
The motional electromotive force is related to time.
0 t
Ex 2 Determine the motional electromotive force of a rectangular loop rotated in magnetic uniform field.
v
)sin( 0 tBSi
13.2 Motional electromotive force13.2 Motional electromotive force
N turns:
sini NBS t
m NBS
sini m t
R
N
'o
o
ω i
Bne
tNBSN cosd
sindi NBS t
t
mmsin sini t I t
R
13.2 Motional electromotive force13.2 Motional electromotive force
(1) 感应电动势随时间变化的曲线是正弦曲线,称为简谐交变电动势,简称简谐交流电 (AC) 。
2
T
sin( )mi i t
:交变电流落后于交变电动势的相位
(2) Period:
(4) Principle of generator : 发电机就是利用电磁感应现象将机械能转化为电能的装置。 线圈中形成了感应电流时它在磁场中要受到安培力的作用,其方向是阻碍线圈运动的。为了继续发电,必须利用气轮机或水轮机来克服阻力矩作功。
Discussion:
(3) Alternating current :
13.2 Motional electromotive force13.2 Motional electromotive force
例 3 两个同心圆环,已知 r1<<r2, 大线圈中通有电流 I , 当小圆环绕直径以ω 转动时,求小圆环中的感应电动势 。
2r
1r
I
解 :
2
0
2rI
B
大圆环在圆心处产生的磁场
通过小线圈的磁通量
B S
cosπ2
cos 21
2
0 rr
IBS tr
r
I cosπ
22
12
0
2m 0 1
2
d πsin
d 2
I rt
t r
感应电动势
13.2 Motional electromotive force13.2 Motional electromotive force
13.3 Induced electric field and emf13.3 Induced electric field and emf
1 New concept
A
动生电动势是由洛伦兹力提供了非静电力。
在磁场变化而产生感生电动势的情况下,导体回路不动,其非静电力不可能是洛伦兹力。产生感生电动势的非静电力是什么?研究发现:不论回路的形状及导体的性质和温度如何,只要磁场变化导致穿过回路的磁通量发生了变化,就会有数值等于 |d/dt| 的感生电动势在回路中产生。这说明感生电动势的产生只是由变化的磁场本身引起的。在电磁感应现象分析的基础上,麦克斯韦指出:变化的磁场在其周围激发一种新的电场,产生感生电动势的非静电力就是由该电场提供的。
2 Induced electric field and emf
If a electric field is induced by changing magnetic field, it is called induced electric field (curl electric field) iE
有旋电场由变化的磁场产生,其电力线是闭合的:
0dlEi
产生感生电动势的非静电力就是有旋电场
St
B
tlEii
d
d
dd
感生电场沿任一闭合回路的线积分(即有旋电场的环流)等于穿过回路所围面积的磁通量的变化率的负值。
Induced emf 感生电动势
Induced electric field 感生电场
13.313.3 Induced electric field and emf Induced electric field and emf
静电场 感生电场场源 正负电荷 变化的磁场
场的性质有源场 有旋场
保守力场 非保守力场场(力)线 非闭合曲线 闭合曲线作用力做功 与路径无关 与路径无关
Properties of induced electric field (1) Induced by changing magnetic field (2) No sourced, non-conservative ;(3) Closed field lines ;(4) Work done related to path.
t
B
d
d
iE
iF qE
F qE
感生电场与静电场的比较
13.313.3 Induced electric field and emf Induced electric field and emf
感生电场方向的确定
例 1 均匀磁场垂直于半径为 0.05m 的线圈平面,线圈的电阻为 ,已知磁感应强度大小随时间增大的变化率为 , 1 )试求感生电动势的大小; 2 )计算在线圈上单位时间内释放的焦耳热。解:任一时刻通过线圈平面的磁通量
13.313.3 Induced electric field and emf Induced electric field and emf
6.036.0 10 T/s
BS
在线圈上产生的感生电动势的大小为: 3 2 5d d
6.0 10 0.05 4.7 10 Vd di
BS
t t
线圈上流动的感应电流为:5
64.7 107.85 10 A
6.0iI
R
在线圈上单位时间内释放的焦耳热为:5 6 104.7 10 7.85 10 3.69 10 J/siP I
Ex 2 A carrying current long solenoid of radius R , the magnetic field is uniformly distributed inside,Find the induced electric field .
0d
dC
t
B
Choose a circle path of radius r, as shown:
St
BlEi
d
d
dd rElE ii 2d
2
d
dd
d
dd
d
dr
t
BS
t
BS
t
B
rt
BEi d
d
2
1
r >R: rElE ii 2d 2
d
dd
d
dR
t
BS
t
B t
B
r
REi d
d
2
2
(r <R)
iE
rR
The negative sign means that the direction of field is as opposed to the direction of chosen.
负号表示电场方向与所选方向相反
13.313.3 Induced electric field and emf Induced electric field and emf
Ex 3 The magnetic field increase uniformly with time inside the straight solenoid of radius R , straight wire ab=R , as shown. Find the induced emf of induced emf of ab.
If dB/dt=K>0 St
BlEi
d
d
dd
t
BrrEi d
d2 2 )(,
2RrK
rEi
Rh2
3
ab iab iabab lElE cosddd
2
0 0
1 1 3d d 0
2 2 2 4
R Rr hK l Kh l KhR KR
r
Using Faladay’s law:
ab iOabO iab lElE
dd
2
4
3
2
1
d
dd
d
dKRKhRS
t
BS
t
B
B r R
O
h
a Ei
b
a b
13.313.3 Induced electric field and emf Induced electric field and emf
h
a b
RO
3 电子感应加速器
B EK
…….…….
…….…….
××××××××××
××××××××××
………………………
………………………
………………………
………………………
R
环形真空室
电子轨道O
B
F
v
由洛伦兹力和牛顿第二定律,有2
Re B mR
v
vR R
m pR
eB eB
v BR 为电子轨道所在处的磁感强度 .
13.313.3 Induced electric field and emf Induced electric field and emf
13.4 Eddy currents 13.4 Eddy currents 涡涡 (( 电电 ))流流1 Eddy currents 涡电流
大块导体处在变化磁场中,或者相对于磁场运动,在导体内部都会产生感应电流。这些感应电流的流线呈闭合的涡旋状,被称为涡电流或涡流。
The formation of eddy current
在圆柱形金属块上绕一组线圈,当线圈中通以交变电流时,金属块就处在交变磁场中。金属块可看成由一系列半径逐渐变化的圆柱状薄壳组成,每层薄壳自成一个回路。在交变磁场中,通过这些薄壳的磁通量都在不断地变化,则沿着一层层壳壁都会产生感应电流。磁场变化越快,感应电动势就越大,涡流就越强。
13.4 Eddy currents 13.4 Eddy currents 涡涡 (( 电电 ))流流 Heat effect: 由于大块金属的电阻很小,如果交变
电流频率很高,则涡电流可以是非常大的。强大的涡电流在金属内流动时,释放出大量的焦耳热。
2 Application 涡电流的有效利用 高频感应炉
热功率 2iP I
利用涡流释放的焦耳热,可用于冶炼金属—高频感应炉。这种冶炼方法的优点:温度高且易于控制;避免氧化和玷污(把坩埚放在真空中无接触地加热)。
13.4 Eddy currents 13.4 Eddy currents 涡涡 (( 电电 ))流流 Damped pendulum 阻尼摆
将一块铝片悬挂在电磁铁的一对磁极之间形成一个摆。由于穿过运动导体的磁通量发生变化,铝片内将产生感应电流。感应电流的效果总是反抗引起感应电流的原因,则铝片的摆动会受到阻力而迅速停止—电磁阻尼。利用该原理即可制作各种阻尼器。在一些电磁仪表中,常利用电磁阻尼使摆动的指针迅速停止在平衡位置上。电气火车中的电磁制动,也是利用了电磁阻尼原理。
13.4 Eddy currents 13.4 Eddy currents 涡涡 (( 电电 ))流流 减少涡流损耗 To reduce the eddy current loss
涡流的焦耳热有时也是有害的。在电机和变压器中,为了增大磁感应强度,都采用了铁芯。处在交变磁场中的电机或变压器中的铁芯内部,将产生很大的电流,白白地损耗了大量的能量(铁芯的涡流损耗),甚至发热量可达到烧毁这些设备。为了减少涡流及其损失,通常采用绝缘叠合起来的硅钢片代替整块铁芯,并且使硅钢平面与磁感应线平行。一方面由于硅钢片本身的电阻率较大,另一方面各片之间涂有绝缘漆或附有天然的绝缘氧化物,这样就可把涡电流限制在各薄片内,使涡流大大减小,从而也就减少了能量损耗。
13.4 Eddy currents 13.4 Eddy currents 涡涡 (( 电电 ))流流 趋肤效应 skin effect
I
i
引起趋肤效应的原因就是涡流。感应电流的效果总是反抗引起感应电流的原因的,所以涡流的方向在导体内部总是与原电流的变化趋势相反,即阻碍原来电流的变化;而在导体表面附近,却与原电流的变化趋势相同。于是,交变电流不易在导体内部流动,而易于在导体表面附近流动,这就形成了趋肤效应。 为改善涡流所造成的这种不利情形,通常采用两种方法:一种方法是采用相互绝缘的细导线束来代替总截面积与其相等的实心导线,这种方法实际上是抑制涡流;另一种方法是在导线表面镀银 , 这种方法实际上是降低导线表面的电阻率。
13.4 Eddy currents 13.4 Eddy currents 涡涡 (( 电电 ))流流 电磁炉工作原理简介
电磁炉主要由耐热陶瓷炉面、高频感应加热线圈(即励磁线圈)、高频电力转换装置、控制器及金属材料锅底炊具等部分组成。 使用时,加热线圈中通以交变电流,交变电流通过陶瓷板下方的线圈产生磁场,交变磁场的磁力线穿透灶台的陶瓷台板而作用于金属锅体,在烹饪金属锅体内因电磁感应就有强大的涡流产生,如图所示。 涡流克服锅体的内阻流动时完成电能向热能的转换,所产生的焦耳热就是烹调的热源,使锅底迅速发热,进而达到加热食品的目的。
1 Self induction phenomena Self induction phenomena 自感现象 If a current is suddenly turned on, a magnetic field proportional to the current is established inside the solenoid. The flux inside solenoid changes from zero to some value that depends on the current and on the solenoid. Because the flux changes, an emf is induced that opposes the change in flux. It is called self-induction phenomena.
当线圈中的电流发生变化时,它所激发的磁场穿过该线圈自身的磁通量也随之发生变化,从而在这个线圈中产生感应电动势。这种由于线圈中电流变化而在自身回路中所引起的感应现象称为自感现象 .
13.5 13.5 自感自感 (( 应应 )) 和自感系数 和自感系数 Self-inductioSelf-induction n
13.5 13.5 自感自感 (( 应应 )) 和自感系数 和自感系数 Self-inductioSelf-induction n
2 自感自感 ((系数系数 )) 和自感电动势 Self-inductance and emf Self-inductance and emf IB
t
d
di IΦ LIΦ
I
ΦL 自感 (系数)在数值上等于线圈中电流
强度为一单位时通过线圈的磁通量
)d
d
d
dI(
d
d
t
LI
tL
t
Φi 当线圈形状、大小和周围介
质的磁导率都保持不变时:
t
IL
d
d
t
IL
d
d/ 自感系数在数值上等于线圈中电流变化为一
个单位时此线圈中产生的感应电动势的大小The unit of Self-inductanceSelf-inductance 自感的单位: Henri H亨利 (SI)
The self-induction emf is proportional to the rate of the current in the circuit itself
13.5 13.5 自感自感 (( 应应 )) 和自感系数 和自感系数 Self-inductioSelf-induction n Ex Determine the Self-inductance of Self-inductance of a solenoid.n , N , 0 , l , S
nIB 0 nISBS 0
The flux through N turns (flux linkage 磁链 )
NnISNBSN 0 VnlSnI
L 20
20
The Self-inductance depends on the geometry of the Self-inductance depends on the geometry of the solenoid and not on the current, so it is constant for a given solenoid.
自感系数L与磁导率、单位长度上的线圈数及 n 线圈体积V 有关,而与电流强度I无关 ,反映线圈本身特性。计算自感时可直接假设线圈中通有电流I
13.5 13.5 自感自感 (( 应应 )) 和自感系数 和自感系数 Self-inductioSelf-induction n 自感现象应用实例 自学
P.84
1 Mutual inductance phenomenon Mutual inductance phenomenon 互感现象A changing magnetic flux in one circuit causes an induction emf in a second circuit.
2 Mutual inductance Mutual inductance 互感系数 互感系数 Let I1 denote the current in the first circuit and let 21 denote the flux in the second circuit due to to I1 若线圈的形状、大小和相对位置均保持不变:
12121 IMΨ 1
2121 I
ΨM 21212 IMΨ
2
1212 I
ΨM
It is proved by theory and experiment: MMM 1221
2
12
1
21
I
Ψ
I
Ψ
I
ΨM
两个线圈的互感系数M,在数值上等于一个线圈中通有单位电流时通过另一线圈所包围的面积的磁通量。
13.6 Mutual inductance13.6 Mutual inductance
33 Mutual induction electromotive forceMutual induction electromotive force 互感电动势互感电动势
t
IM
t
Ψi d
d
d
d若 M
不变 :)
d
d
d
d(
d
d
t
MI
t
IM
t
Ψi
Mutual inductance Mutual inductance M M depends on how the flux of the depends on how the flux of the first circuit is intercepted by the second. The further the first circuit is intercepted by the second. The further the circuits are apart, the less flux passes from one through circuits are apart, the less flux passes from one through the other. the other.
Mutual inductance Mutual inductance M M is the emf in circuit 2 is the emf in circuit 2 caused by the changing current caused by the changing current II11 in curren in curren
t 1, divided by the rate of change of t 1, divided by the rate of change of II11. .
It has the same unit as self-inductance.It has the same unit as self-inductance.tI
M i
dd
13.6 Mutual inductance13.6 Mutual inductance
实际上,当线圈中的电流发生变化时,除了在邻近线圈中产生互感电动势外,还会在自身线圈中产生自感电动势,因此,计算感应电动势时,在线圈中产生的总的电动势是自感电动势和互感电动势之和,即:
1 21 1
d d
d d
I IL M
t t
2 12 2
d d
d d
I IL M
t t
互感现象被广泛地应用于无线电技术及电磁测量中。通过互感线圈能够使能量或信号从一个线圈传递到另一个线圈。各种电源变压器、中轴变压器、输入或输出变压器以及电压和电流互感器等,都是利用互感现象制成的。
13.6 Mutual inductance13.6 Mutual inductance
Ex 1 A straight wire is coplanar with a rectangle loop, as shown in figure. Find the mutual inductanceinductance.
)()(
ddSS
SBSBΦ
a
bILrL
r
Ib
aln
2d
200
a
b
LI
ΦM
a
bLln
20
M 由线圈的几何形状、大小、匝数、相对位置、磁导率等决定,与电流无关。The mutual inductance depends on the geometry and inductance depends on the geometry and relative position of the both two loopsrelative position of the both two loops and not on the current.
M 与电流强度I无关,计算时可直接假设一个线圈中通有电流I,计算通过另一个线圈的磁通量,得到 M
13.6 Mutual inductance13.6 Mutual inductance
Ex 2 A coil consisting of N2=20 turns of wire encircles a long solenoid that has N1=1000 turns and cross-sectional area S =10cm2 ,and length l= 1.0m , (1) Find the mutual inductance of the two coils. (2) If dI1/dt= 10A/s, find the induced emf in the coil N2. The field of coil 1 is given by 11= I
l
NnIBB 1
00
SIl
NBS 1
1== 0
The flux linkage through coil 2 is
SIl
NNΦN 1
2102 Ψ
)H(1025 6210
1
Sl
NN
I
ΨM
The induced emf in the coil 2 is
t
IM
d
d 12 (V)10250101025 66
parallel to axes
The flux through coil 2 is
13.6 Mutual inductance13.6 Mutual inductance
例例 3 3 如图所示,一无限长直导线与一长为 、如图所示,一无限长直导线与一长为 、宽为 的矩形线框共面放置,且矩形线框的一宽为 的矩形线框共面放置,且矩形线框的一端距离直导线为 ,磁导率为 。求:端距离直导线为 ,磁导率为 。求:(( 11 )两线圈的互感系数;()两线圈的互感系数;( 22 )当直导线)当直导线中通有电流 时,在矩形线圈中产生中通有电流 时,在矩形线圈中产生的互感电动势。的互感电动势。
ab
d 0
0sinI I t
解(解( 11 ))
(( 22 ))
0
2π
IB
x
d dΦ B S
0 0d ln( )2 π 2 π
d b
d
I Ia b dΦ a x
x d
0 ln( )2 π
aΦ b dM
I d
00
dln( ) cos
d 2 πi
aI b dM I t
t d
讨论:讨论: P89P89
13.6 Mutual inductance13.6 Mutual inductance
13.7 The energy of magnetic field13.7 The energy of magnetic field
1 自感磁能当电流开始流过电路时,穿过此回路的磁通量发生变化,由于自感的作用,电路中出现反抗电流增加的自感电动势: 设电源电动势为 ,内阻为零,在接通电源之后的任何瞬时,电路中的总电动势为:
Questions : (1) Energy ?energy density ?(2) How to calculate the energy ?
d
di
iL
t
d
di
iL
t
d
d
iL Ri
t
d
d
iRi L
t
外电源不仅要供给电路中产生焦耳热所需要的能量,而且还要反抗自感电动势做功。
欧姆定律
2 2
0 0
1d d
2
t ti t Ri t LI
2d d di t Ri t Li i
回路电阻所放出的焦耳热
电源做功
电源反抗自感电动势做的功
电源电动势反抗自感电动势作功的结果建立了磁场The work done is equal to the energy stored in the inductor ( 感应线圈)
21
2LW LI
d
d
iRi L
t
13.7 The energy of magnetic field13.7 The energy of magnetic field
自感磁能 Energy:
电源反抗互感电动势所做的功:
13.7 The energy of magnetic field13.7 The energy of magnetic field
2 互感磁能
2 11 2 12 1 21 2 12 1 21 2
0 0 0 0
d dd d d d
d d
t t t t
M
i iA A A i t i t M i t M i t
t t
1 2
2 11 2 1 2 1 2 1 2
0 0 0
d d dd d d
d d d
I It ti iM i i t M i i t M i i MI I
t t t
与自感一样,电源反抗互感电动势所做的功,也以磁能的形式储存了起来,称为互感磁能
1 2MW MI I
两个相邻的载流线圈中所储存的总磁能 2 2
1 2 1 1 2 2 1 2
1 1
2 2m L L MW W W W L I L I MI I
需要注意的是,自感磁能总是正的,但互感磁能可能是负的。例如,当线圈 1 中的电流 1 所产生的通过线圈 2
的磁通量与线圈 2 中的电流 2 在自身回路中所产生的磁通量同号时,电流 1
与电流 2 同号,互感磁能为正;反过来,当电流 1 与电流 2异号时,互感磁能为负。
13.7 The energy of magnetic field13.7 The energy of magnetic field
13.7 The energy of magnetic field13.7 The energy of magnetic field
3 磁场的能量螺绕环中流有电流 : I 0B nI
20 0
NBSL N nS n V
I I
2 2 2 22 2 2 0
00 0
1 1 1 1
2 2 2 2m L
n I BW W LI n VI V V
磁能密度 Energy density:
2
0
1
2m
m
W Bw
V
Question (1)0
2
2
1
B
wm 对非均匀磁场适用否?
(2) 对非均匀磁场,如何计算磁能?
Note:
0
2
2
1
B
wm Although the formula for energy density was derived for a solenoid, it is true for any magnetic field.
(2) Uniform field: VB
VwW mm0
2
2
1
(3) Non- Uniform field:
磁场中 dV 小区域内的磁能: VB
VwW mm d2
1dd
0
2
总磁能 )(
0
2
)()(d
2
1dd
VV mV mm VB
VwWW
mVWV
BLI )(
0
22 d
2
1
2
1
the L can be gotUsing
(4)0
2
2
1
B
wm 某点的磁能密度只与该点的磁感应强度和介质的性质有关,也说明了磁能是定域在磁场中的这个客观事实。
(1)
202
1LIWm
202
1LIWm
不能直接用密度乘以体积计算能量
13.7 The energy of magnetic field13.7 The energy of magnetic field
13.7 The energy of magnetic field13.7 The energy of magnetic field
例例 1 1 一长为 的直螺线管,绕有 匝线圈,截面积为一长为 的直螺线管,绕有 匝线圈,截面积为( ),载有电流 ,将此螺线管当作一理想螺线( ),载有电流 ,将此螺线管当作一理想螺线管,求它的的磁能。管,求它的的磁能。解:解:长直螺线管内部的磁感应强度
磁能密度
自感磁能
l N SS l I
0 0
NB nI I
l
2 2 2 2 220 0
2 20 0
1 1
2 2 2m
N I N IBw
l l
2 2 2 20 0
22 2m m
N I N I SW w V lS
l l
I o
I
Ex A coaxial cable consists of two metal shells of inner radius R1 and outer radius R2. Current I flows from inner shell to outer shell. Find the magnetic energy stored in coaxial cable of length l.Using Ampere’s circuital theorem to calculate the B :
i
iLIlB 0)(
d
r
drIrB 02
r
IB
2
0
Energy density 0
22
220
0
2
82
r
IBwm
半径为 r 与 r+dr ,长为 l 的圆柱壳体的体元: rdrldV 2
半径为 r 与 r+dr ,长为 l 的圆柱壳体内储存的磁能:
VwW mm dd rrlI
rlrI
d4rd28
20
22
20
1
22
02
0 ln4
d4
d 2
1 RRlI
rrlI
VwWR
RV mm
The energy stored in coaxial cable of length l:
To calculate the self-inductance
mVWV
BLI )(
0
22 d
2
1
2
1
1
202
ln2
2RRl
I
WL
Addition 1
20 ln2 RRl
L
1
22
02 ln42
1RRlILIWm
2
21 LIWm
IL
利用 mVWV
BLI )(
0
22 d
2
1
2
1
在已知自感时可以计算能量,在已知能量时也可以计算自感
Ex 1 A rectangular coil with edges L and b is placed in a field of wire. The rectangular coil moves with velocity v. Find the emf at the position as shown in Fig .
a b
B C
I
A D
r
v L
Clockwise
Method 1: Using the definition of emf
Li lB
d)(v
)(
)(
)(
)(d)(d)(
D
C
B
AlBlB
vv
LBLB vv 21 Lbaa
Iv)
11(
20
Method 2 : Using Faladay’s law: 选顺时针回路
SBSBΦΦ mm ddd
ba
ar
r
ILd
20
aba
ILln)ln(
20
vv
baa
IL
aba
IL
t
a
at
a
ba
ILi
11
2
11
2d
d1
d
d1
2000
电动势的方向与所选定的方向一致—顺时针方向。0i
Ex 2 一无限长直载流导线,通以电流 I,导线旁有一长为 L的金属棒与之共面,金属棒绕其一端 O以角速度 顺时针转动, O点与导线的垂直距离为 a,求当金属棒转至与长直导线垂直的位置时(如图所示),棒内感应电动势的大小和方向。
dl
r
The B at position dl: rIB
2
0
I
a
O
棒上所有线元在图示位置时速度均向下,但大小不同, B
v 沿 OA
Av
Speed of dl v=l
Emf : lBi
d)(d v
llBlB ddd i v
l
arl rl dd
L
OAi ldlBlB0
d)( �
v
L
ldlr
I0
0
2
La
arar
r
Id)(
2 0
)ln(2
d 2
00
a
LaaL
Ir
r
arI La
a
Direction: o A.
Ex 3 A metal rod of length L0 is rotated around axis OZ with an angular speed in a uniform field B directed as shown. Angle between rod and axis is . Find the electric potential difference in rod OA.
dl
l
Method 1: Using definitionlBi
d)(d v
B
v sind)90cos(d lBlB vv
r
v sinlv
0
0
2 dsindl
ii lBl 0sin2
1 220 Bl
Method 2 : Using Faladay’s law: ⊙ l0sin
a '
O a
B
Vertical view
2)(21 OaBSBΦm 22
0 sin21 Bl
Z
B
O
AChoosing dl , the emf:
sin2
1
d
dsin
2
1
d
d 20
20 Bl
tBl
t
Φmi
Choosing path Oaa'O
(1) 在磁场中运动(平动或转动)的线圈或闭合导体回路产生的电动势,通常可直接利用法拉弟电磁感应定律 计算,较为方便;(2) 一段在磁场中运动的导体产生的动生电动势,更多地利用动生电动势定义公式来计算。具体计算步骤为:
动生电动势的计算方法:
a 由题意作出示意图,在运动导体上取一线元 dl ;b 在图中标出导体线元的运动方向 v 、线元所在处的磁感应强度 B 的方向,并标出 v与 B 的夹角以及( vB)与 dl之间的夹角。c 写出该线元所产生的动生电动势的表达式;d 确定积分上、下限,统一积分变量对整个导体积分;e 根据积分的结果,确定电动势的方向。
Ex 4 A B is distributed uniformly Inside a cylinder of radius R. A metal rod of length AB=l is shown in Fig.. If find the induced emf in the rod.
,0d
d
t
B
A
O
B
B
0d
d
t
B
•
l
Using Faladay’s law:
lEt iAB
d
d
d )(
ddAB
ii lElE
t
BlR
lS
t
B
t
BS
t d
d
42d
d
d
)d(
d
d 22
t
BlR
lAB d
d
42
22 Direction: A→B
Calculate it using )(
dAB
i lE
Choosing a closed path OABO :
by yourself.
Ex 5 直线 OA 、 OB 的夹角为 60 ,如图所示,以 O 为圆心的范围内磁感应强度为 B(into page) ,由 t=0开始,以不变的速率增加,半圆形导线环在导轨 AOB 上以速度v沿半径方向向圆心匀速运动, t 时刻半环形导线环心已与 O 点重合 , 其半径为 r ,求此时闭合导线回路中的感应电动势。
A
O
B
v
BUsing Faladay’s law:
)(d
d
d
dBS
tt
Φmi )
d
d
d
d(
t
SB
t
BS
The area of closed path 2
6
1rS
vv rrt
rr
t
S 3
12
6
1
d
d2
6
1
d
d
)3
1
d
d
6
1()
d
d
d
d( 2 vrB
t
Br
t
SB
t
BSi
vt
r
d
d
注意:1 感生电动势的计算方法通常有两种:
SL Ki t
SBlE dd
dd
B
A KAB lE d
(1) 当涉及回路和线圈时,可以直接利用电磁感应定律求解;(2) 在一些具有某种对称性的简单情形下,可以先求出由变化的磁场产生的感生电场,再利用式 可计算出。
2 由以上几例可以看出,感应电动势的计算最基本的方法是法拉弟电磁感应定律,在具体应用时要先选定回路,计算出磁通量,再对上式其对时间的变化率。此时应注意积分和求导的意义:积分是对整个闭合回路内各微元面积的磁通进行求和,求导将整个面积中的磁通是对时间求导。
由于变化的磁场在闭合回路中产生的感生电动势,对于变化磁场中的一段导体,则有
3 感应电动势的方向判定可用三种方法:(1) 用法拉弟电磁感应定律中的负号;(2) 用楞次定律。由于感生电流的磁场总是阻碍原磁通的变化,因此可先确定感生电流磁场的方向,尔后再确定感生电动势、感生电流的方向;(3)对动生电动势还可用右手定则判定。此时应注意:电动势的方向在电源内部是由低电势指向高电势。
Ex 6 如图所示,一长金属线半径为 a ,置于一半径为b 的长金属圆柱面内部,二者中心线重合,且一端相连,求该导体系统单位长度内的自感系数( a < r < b) 。
abI
IMethod 1 Using I
ΦL
The B of wire of radius a : r
IB
2
0
The magnetic flux through the area of shown in Fig:
rdr
rr
ISBΦm d
2dd 0
The total flux :
a
bIr
r
IΦ
b
am ln2
d12
00
a
b
I
ΦL m ln
20
Method 2Using 2
2
1LIWm
a
bIrr
r
IVwW
b
amm ln4
d28
d2
022
20
a
b
I
WL m ln
2
2 02
Ex 7 A rectangular coil with edges 3a and b is placed coplanar with a straight wire, as shown, (1) Find the mutual inductimutual induction of wire and coil. (on of wire and coil. (2) When the current of rectangular coil is I=I0sint, find the induced emf in straight wire.
a 2a
b
(1) The B of straight wire with I' ,
r
IB
2
'0
The flux: 21d ΦΦs SBΦ
2ln2
'd
2
'd
2
' 0
0
02
0
0
bI
rr
bIr
r
bI aa
Mutual inductance: 2ln2
0
b
IM m
(2) Mutual induced emf: 00
dln 2 cos
d 2i
bIM I t
t
Ex 8 Two paralleled straight wires of radius a are separated by a distance d , each carrying a current I flowing in the opposite direction. (1) Find the self-inductance for unit length of two wire.Outside the two wire the total magnetic flux is zero. Inside :The B at distance x:
)(2200
xd
I
x
IB
x
1 I 2
A dx B
l
I l
O D C x
d
The magnetic flux through the section ABCD as shown in Fig is:
xlxd
I
x
ISB
ad
aSm d)(22
d 00
a
adIl ln0
a
ad
Ila
adIl
IlL m
lnln
So 0
0
相距为 r 时 l长的导线受到另一导线的作用力
r
IlBlIF
2
0
rFAA
dd
取 l= 1m ,二者距离增大的过程中,磁力作正功,且
单位长度上磁能的改变ad
adIr
r
Iad
ad
2ln
2d
2
20
22
0
ad
adIAWm
2ln
2
20
这些能量从何而来?当导线间距离增大时,导线中会出现与原来电流方向相反的感应电动势,如维持导线中的电流不变,处接电源必须多作功,意即电源要多输出电能,其中一部分转化为磁场的能量,另一部分通过磁力作功而变为其它形式的能量。
(2) If the distance is 2d, find the work done on unit length wire by magnetic field and the increment of magnetic energy.
导线间距离增大时 , 自感增大 , 磁场能量增加 , 磁力作正功