13.1 Electromagnetic 13.1 Electromagnetic induction law induction law Chapter 13 Chapter 13 电电电电 电电电电 Electromagnetic Induction Electromagnetic Induction Motional electromotive force Motional electromotive force Induced electric field and Induced electric field and tromotive force tromotive force 3.4 Eddy current 3.4 Eddy current 13.5 Self-induction 13.5 Self-induction .6 Mutual induction .6 Mutual induction 7 Energy of magnetic field 7 Energy of magnetic field 13.8 Application 13.8 Application
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 Presentation
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13.1 Electromagnetic induction law13.1 Electromagnetic induction law
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
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
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
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
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
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.
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
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
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
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.
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。
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: