Chapter 28 Inductance and Electromagnetic Oscillations
Chapter 28
Inductance and
Electromagnetic Oscillations
1,Self and Mutual Induction
2,Energy Stored in a Magnetic Field
Chapter 28 Inductance and Electromagnetic Oscillations
New words and expressions
self induction 自感 应
inductance 感应系数
mutual induction 互感 应
Inductor 感应器,电感线圈
Chapter 28 Inductance and Electromagnetic Oscillations
28-2,Self-Induction
The process that an induced emf appear in any
coil in which the current is changing is called
self-induction (由于自己线路中的电流的变化而在自己的线路中产生感应电流的现象－－自感现象 ).
1,Self-Induction (自感 应 )(P645-647):
If we establish a current I in the
coil,the magnetic flux linkage
N? is proportional to I,
I
LIDefine proportionality constant is L,
Chapter 28 Inductance and Electromagnetic Oscillations
I
NL
Then the coefficient L is defined as the Self-Inductance.
A coil that has significant self-
inductance L is called an inductor,
It related to the geometry of the coils and
magnetic medium,自感系数与线圈的大小,形状,
磁介质,线圈密度有关,而与线圈中电流无关 。
1 henry (亨利 ) = 1 H = 1T.m2/AIts SI unit:
Chapter 28 Inductance and Electromagnetic Oscillations
The emf appears in self-induction is called self-
induced emf,It obeys Faraday’s law of induction.
The minus sign indicates that the self-induced
emf has the orientation such that it opposes the
change in I.
2,Self-Induction EMF
t
Ni
d
d
where LI
If L is constant
t
IL
i d
d
t
N
d
)(d
td
d
Chapter 28 Inductance and Electromagnetic Oscillations
3,Calculation of Self-Induction
① 假设线圈中的电流 I ；
② 求线圈中的磁通量 ；
③ 由定义求出自感系数 L。
IL
Chapter 28 Inductance and Electromagnetic Oscillations
Example,Inductance of a Solenoid,
nIlNIB 00
Magnetic field inside the
solenoid of length l is:
一长直螺线管，线圈密度为 n,长度为 l,横截面积为 S，
插有磁导率为? 的磁介质，求线圈的自感系数 L 。
l
S?
E
Solution,先设电流 I 根据安培环路定理求得 H B
Φ L,
Chapter 28 Inductance and Electromagnetic Oscillations
220 )( RlInn l B AN B
Magnetic flux linkage:
lSn
I
lnRI
I
L B
2
0
22
0N
l
S?
E
The inductance per unit length for a long solenoid
near its center,
Sn
l
L 2
0
Inductance L —— like capacitance —— depends
only on the geometry of the device.
Chapter 28 Inductance and Electromagnetic Oscillations
Example,Calculating self-inductance of co-axis
wire.
r
IB
2
0?
21 RrR
Solution:
有两个同轴圆筒形导体,其半径分别为 和,通过它们的电流均为,但电流的流向相反,设在两圆筒间充满磁导率为 的均匀磁介质,求 其自感 L.
1R 2R
I
L
1R
I
2R
lI
r
Chapter 28 Inductance and Electromagnetic Oscillations
如图在两圆筒间取一长为 的面,并将其分成许多小面元,
l P Q R S
SBΦ dd rBl d?
rlrIΦΦ R
R
dπ2d 2
1
1
2ln
π2 R
RIlΦ
1R
I
S
P
R
Q
2R
lI
r
rd
Chapter 28 Inductance and Electromagnetic Oscillations
1R
I
S
P
R
Q
2R
lI
r
rd
From the definition of L,
the self-inductance for a
length l is,
1
2ln
π2 R
Rl
I
ΦL
The inductance per unit
length is:
1
2ln
π2 R
R?
Chapter 28 Inductance and Electromagnetic Oscillations
1,Mutual-Induction (互感 应 )
A steady current in one coil will set up a magnetic
flux linking the other coil,If current changes with
time,an emf appears in the second coil,The
process is called mutual induction.
线圈 1 在线圈 2 中产生的磁链（磁通）：
21221 N?
12121 IM
SBN S d12 2 1I?
1
I1 N1
S1 2
I2 N2
S2
28-1,Mutual Induction (P643-P645):
Chapter 28 Inductance and Electromagnetic Oscillations
Define the mutual inductance M21 of coil 2 with
respect to coil 1 as （线圈 1对 2的互感系数）：
1
21
21 IM
线圈 2 在线圈 1 中产生的磁链：
12112 N?
21212 IM
定义为线圈 2 对线圈 1 的互感系数
2
12
12 IM
MMM 1221两线圈的互感系数相等：
1 2
N1 N2
2I?
Chapter 28 Inductance and Electromagnetic Oscillations
The mutual inductance depends only on the
geometry of two devices,(If no magnetic
materials such as iron in their vicinity.)
互感系数与电流无关,只决定于两线圈本身性质 --几何尺寸,匝数,介质,相对位置 。
The SI unit for M (as for L ) is also the Henry.
In general,M should be measured in experiments.
Chapter 28 Inductance and Electromagnetic Oscillations
By similar way (interchanging the roles of coils 1
and 2)
According to Faraday’s law
td
d 21
21
线圈 1电流变化在线圈 2中产生的互感电动势
td
d 12
12
线圈 2电流变化在线圈 1中产生的互感电动势
t
IM
d
d 1
21
t
IM
d
d 2
12
2,Mutual emf
Chapter 28 Inductance and Electromagnetic Oscillations
l
S
n1 n2
例,长为 l,横截面积为 S 的长 直螺线管，插有磁导率为? 的磁介质，绕两个线圈，两线圈的线圈密度分别为 n1,n2,两线圈完全耦合，求两线圈的互感系数,
解,设线圈 1 中的电流为 I1
线圈 1 在线圈 2 中产生的磁链：
21221 mN
线圈 1 在线圈 2 中产生的互感系数：
1
21
21 IM
SBln 12? SInln 112
lSnn 21
Chapter 28 Inductance and Electromagnetic Oscillations
l
S
n1 n2I2
设线圈 2 中的电流为 I2
线圈 2 在线圈 1 中产生的磁链：
12112 mN
线圈 2 在线圈 1 中产生的互感系数：
2
12
12 IM
SBln 21?
SInln 221
lSnn 21
由此可看出，两线圈的互感系数相等。
1221 MM? M? lSnn 21
Chapter 28 Inductance and Electromagnetic Oscillations
例,在长直导线旁距 a 放置一长为 l,宽为 b 的矩形导线框，求两导体的互感系数。
解,设直导线中通有电流 I,
载流直导线在矩形线圈内产生的磁通量为：
SB d
ba
a
B ld x
ba
a
l d xxI2 0
a
baIl ln
2
0
a b
l
x dx
I
x
o
Chapter 28 Inductance and Electromagnetic Oscillations
互感系数：
1
21
21 IM
a
bal ln
2
0
I
N m
a
ba
I
lI ln
2
0
a
baIl ln
2
0
Chapter 28 Inductance and Electromagnetic Oscillations
Example,A coil of N2 turns wound
as shown around part of a toroid of
N1 turns,The toroid’s inner,outer
radius are a and b,its height is h.
Show M for the toroid-coil
combination.
Solution:
t
ctc
ct i
NM
ba ctct dAB?
a
bhNNM ln
2
210
ba tt h d rr iN 20 )l n (20 abhiN tt
)l n (20 abhNiiN tt
t
c
Chapter 28 Inductance and Electromagnetic Oscillations
28-3,Energy Stored in a Magnetic Field (P647-648)
Recall the energy stored in electric field:
C
QU
e 2
2
Energy of capacitor:
DEEu e 2121 2
Density of energy:
在右面的电路中，灯泡与电感线圈并联后，串在电源上，当电键 K 从闭合状态，变为打开状态时，灯泡并不是立即就熄灭，而是闪亮一下才熄灭。
L
K?
载流线圈具有能量。
Chapter 28 Inductance and Electromagnetic Oscillations
这是由于线圈中的磁场能量释放给灯泡 。 当电键 K 打开时,电路中电流迅速减小,在线圈中产生自感电动势,这个自感电动势比电源电动势要大,
所以灯泡比原来还亮一些,最后灯泡熄灭 。
线圈中的能量,是由于线圈在通电过程中,电流克服自感电动势作功,
使线圈具有能量 。
L
K?
Chapter 28 Inductance and Electromagnetic Oscillations
LR
K
1
IRdtdIL
Using Ohm’s on it:
To get magnetic energy,consider such a RL circuit:
tIt tRIILItI 0 200 ddd?
tt tRILItI 0 220 d21d?
Multiply each term by Idt and integrate,we have
2
2
1 LIU
B?
Energy stored by an inductor:
Chapter 28 Inductance and Electromagnetic Oscillations
221 LIU B
VnL 20
22021 VIn? VB
0
2
2
1
Take long solenoid as an example,
The energy stored per unit volume of the field,
density of energy (能量密度 ),is:
0
2
2
1
B
V
Uu B
B (magnetic energy density)
VBVuU
V
BB d2
1d 2
In general,one can use:
Chapter 28 Inductance and Electromagnetic Oscillations
例,计算半径为 R、长为 l,通有电流 I,磁导率为?
的均匀载流圆柱导体内磁场能量。
l
I
R
导体内沿磁力线作半径为 r 的环路，
解,由介质中安培环路定理确定导体内的磁感应强度 B,
r
IRrrH 2
2
2
IlH d
22 rRII IRr 2
2
22
IrH
HB 22 R
Ir
Chapter 28 Inductance and Electromagnetic Oscillations
l
R
r
将圆柱导体分割为无限多长为 l 厚度为 dr 的同轴圆柱面，
dr体积元处的磁场能量密度为：
2
2B
w m?
导体内的磁场能量为：
V mm dVwW 体体积元体积为,r l d rdV?2?体
22 R
IrB
,
2
2
Bw
m?,2 r l d rdV
体
Chapter 28 Inductance and Electromagnetic Oscillations
R
rrlB
0
2
d2
2
R
rrl
R
Ir
0
2
2 d222
1?
R drrR
lI
0
3
4
2
4?
4
4
2
16
R
R
lI
16
2lI
V mm VwW 体d
l
R
r
dr
Chapter 28 Inductance and Electromagnetic Oscillations
Homework:
P656,3,13,18
Chapter 28
Inductance and
Electromagnetic Oscillations
1,Self and Mutual Induction
2,Energy Stored in a Magnetic Field
Chapter 28 Inductance and Electromagnetic Oscillations
New words and expressions
self induction 自感 应
inductance 感应系数
mutual induction 互感 应
Inductor 感应器,电感线圈
Chapter 28 Inductance and Electromagnetic Oscillations
28-2,Self-Induction
The process that an induced emf appear in any
coil in which the current is changing is called
self-induction (由于自己线路中的电流的变化而在自己的线路中产生感应电流的现象－－自感现象 ).
1,Self-Induction (自感 应 )(P645-647):
If we establish a current I in the
coil,the magnetic flux linkage
N? is proportional to I,
I
LIDefine proportionality constant is L,
Chapter 28 Inductance and Electromagnetic Oscillations
I
NL
Then the coefficient L is defined as the Self-Inductance.
A coil that has significant self-
inductance L is called an inductor,
It related to the geometry of the coils and
magnetic medium,自感系数与线圈的大小,形状,
磁介质,线圈密度有关,而与线圈中电流无关 。
1 henry (亨利 ) = 1 H = 1T.m2/AIts SI unit:
Chapter 28 Inductance and Electromagnetic Oscillations
The emf appears in self-induction is called self-
induced emf,It obeys Faraday’s law of induction.
The minus sign indicates that the self-induced
emf has the orientation such that it opposes the
change in I.
2,Self-Induction EMF
t
Ni
d
d
where LI
If L is constant
t
IL
i d
d
t
N
d
)(d
td
d
Chapter 28 Inductance and Electromagnetic Oscillations
3,Calculation of Self-Induction
① 假设线圈中的电流 I ；
② 求线圈中的磁通量 ；
③ 由定义求出自感系数 L。
IL
Chapter 28 Inductance and Electromagnetic Oscillations
Example,Inductance of a Solenoid,
nIlNIB 00
Magnetic field inside the
solenoid of length l is:
一长直螺线管，线圈密度为 n,长度为 l,横截面积为 S，
插有磁导率为? 的磁介质，求线圈的自感系数 L 。
l
S?
E
Solution,先设电流 I 根据安培环路定理求得 H B
Φ L,
Chapter 28 Inductance and Electromagnetic Oscillations
220 )( RlInn l B AN B
Magnetic flux linkage:
lSn
I
lnRI
I
L B
2
0
22
0N
l
S?
E
The inductance per unit length for a long solenoid
near its center,
Sn
l
L 2
0
Inductance L —— like capacitance —— depends
only on the geometry of the device.
Chapter 28 Inductance and Electromagnetic Oscillations
Example,Calculating self-inductance of co-axis
wire.
r
IB
2
0?
21 RrR
Solution:
有两个同轴圆筒形导体,其半径分别为 和,通过它们的电流均为,但电流的流向相反,设在两圆筒间充满磁导率为 的均匀磁介质,求 其自感 L.
1R 2R
I
L
1R
I
2R
lI
r
Chapter 28 Inductance and Electromagnetic Oscillations
如图在两圆筒间取一长为 的面,并将其分成许多小面元,
l P Q R S
SBΦ dd rBl d?
rlrIΦΦ R
R
dπ2d 2
1
1
2ln
π2 R
RIlΦ
1R
I
S
P
R
Q
2R
lI
r
rd
Chapter 28 Inductance and Electromagnetic Oscillations
1R
I
S
P
R
Q
2R
lI
r
rd
From the definition of L,
the self-inductance for a
length l is,
1
2ln
π2 R
Rl
I
ΦL
The inductance per unit
length is:
1
2ln
π2 R
R?
Chapter 28 Inductance and Electromagnetic Oscillations
1,Mutual-Induction (互感 应 )
A steady current in one coil will set up a magnetic
flux linking the other coil,If current changes with
time,an emf appears in the second coil,The
process is called mutual induction.
线圈 1 在线圈 2 中产生的磁链（磁通）：
21221 N?
12121 IM
SBN S d12 2 1I?
1
I1 N1
S1 2
I2 N2
S2
28-1,Mutual Induction (P643-P645):
Chapter 28 Inductance and Electromagnetic Oscillations
Define the mutual inductance M21 of coil 2 with
respect to coil 1 as （线圈 1对 2的互感系数）：
1
21
21 IM
线圈 2 在线圈 1 中产生的磁链：
12112 N?
21212 IM
定义为线圈 2 对线圈 1 的互感系数
2
12
12 IM
MMM 1221两线圈的互感系数相等：
1 2
N1 N2
2I?
Chapter 28 Inductance and Electromagnetic Oscillations
The mutual inductance depends only on the
geometry of two devices,(If no magnetic
materials such as iron in their vicinity.)
互感系数与电流无关,只决定于两线圈本身性质 --几何尺寸,匝数,介质,相对位置 。
The SI unit for M (as for L ) is also the Henry.
In general,M should be measured in experiments.
Chapter 28 Inductance and Electromagnetic Oscillations
By similar way (interchanging the roles of coils 1
and 2)
According to Faraday’s law
td
d 21
21
线圈 1电流变化在线圈 2中产生的互感电动势
td
d 12
12
线圈 2电流变化在线圈 1中产生的互感电动势
t
IM
d
d 1
21
t
IM
d
d 2
12
2,Mutual emf
Chapter 28 Inductance and Electromagnetic Oscillations
l
S
n1 n2
例,长为 l,横截面积为 S 的长 直螺线管，插有磁导率为? 的磁介质，绕两个线圈，两线圈的线圈密度分别为 n1,n2,两线圈完全耦合，求两线圈的互感系数,
解,设线圈 1 中的电流为 I1
线圈 1 在线圈 2 中产生的磁链：
21221 mN
线圈 1 在线圈 2 中产生的互感系数：
1
21
21 IM
SBln 12? SInln 112
lSnn 21
Chapter 28 Inductance and Electromagnetic Oscillations
l
S
n1 n2I2
设线圈 2 中的电流为 I2
线圈 2 在线圈 1 中产生的磁链：
12112 mN
线圈 2 在线圈 1 中产生的互感系数：
2
12
12 IM
SBln 21?
SInln 221
lSnn 21
由此可看出，两线圈的互感系数相等。
1221 MM? M? lSnn 21
Chapter 28 Inductance and Electromagnetic Oscillations
例,在长直导线旁距 a 放置一长为 l,宽为 b 的矩形导线框，求两导体的互感系数。
解,设直导线中通有电流 I,
载流直导线在矩形线圈内产生的磁通量为：
SB d
ba
a
B ld x
ba
a
l d xxI2 0
a
baIl ln
2
0
a b
l
x dx
I
x
o
Chapter 28 Inductance and Electromagnetic Oscillations
互感系数：
1
21
21 IM
a
bal ln
2
0
I
N m
a
ba
I
lI ln
2
0
a
baIl ln
2
0
Chapter 28 Inductance and Electromagnetic Oscillations
Example,A coil of N2 turns wound
as shown around part of a toroid of
N1 turns,The toroid’s inner,outer
radius are a and b,its height is h.
Show M for the toroid-coil
combination.
Solution:
t
ctc
ct i
NM
ba ctct dAB?
a
bhNNM ln
2
210
ba tt h d rr iN 20 )l n (20 abhiN tt
)l n (20 abhNiiN tt
t
c
Chapter 28 Inductance and Electromagnetic Oscillations
28-3,Energy Stored in a Magnetic Field (P647-648)
Recall the energy stored in electric field:
C
QU
e 2
2
Energy of capacitor:
DEEu e 2121 2
Density of energy:
在右面的电路中，灯泡与电感线圈并联后，串在电源上，当电键 K 从闭合状态，变为打开状态时，灯泡并不是立即就熄灭，而是闪亮一下才熄灭。
L
K?
载流线圈具有能量。
Chapter 28 Inductance and Electromagnetic Oscillations
这是由于线圈中的磁场能量释放给灯泡 。 当电键 K 打开时,电路中电流迅速减小,在线圈中产生自感电动势,这个自感电动势比电源电动势要大,
所以灯泡比原来还亮一些,最后灯泡熄灭 。
线圈中的能量,是由于线圈在通电过程中,电流克服自感电动势作功,
使线圈具有能量 。
L
K?
Chapter 28 Inductance and Electromagnetic Oscillations
LR
K
1
IRdtdIL
Using Ohm’s on it:
To get magnetic energy,consider such a RL circuit:
tIt tRIILItI 0 200 ddd?
tt tRILItI 0 220 d21d?
Multiply each term by Idt and integrate,we have
2
2
1 LIU
B?
Energy stored by an inductor:
Chapter 28 Inductance and Electromagnetic Oscillations
221 LIU B
VnL 20
22021 VIn? VB
0
2
2
1
Take long solenoid as an example,
The energy stored per unit volume of the field,
density of energy (能量密度 ),is:
0
2
2
1
B
V
Uu B
B (magnetic energy density)
VBVuU
V
BB d2
1d 2
In general,one can use:
Chapter 28 Inductance and Electromagnetic Oscillations
例,计算半径为 R、长为 l,通有电流 I,磁导率为?
的均匀载流圆柱导体内磁场能量。
l
I
R
导体内沿磁力线作半径为 r 的环路，
解,由介质中安培环路定理确定导体内的磁感应强度 B,
r
IRrrH 2
2
2
IlH d
22 rRII IRr 2
2
22
IrH
HB 22 R
Ir
Chapter 28 Inductance and Electromagnetic Oscillations
l
R
r
将圆柱导体分割为无限多长为 l 厚度为 dr 的同轴圆柱面，
dr体积元处的磁场能量密度为：
2
2B
w m?
导体内的磁场能量为：
V mm dVwW 体体积元体积为,r l d rdV?2?体
22 R
IrB
,
2
2
Bw
m?,2 r l d rdV
体
Chapter 28 Inductance and Electromagnetic Oscillations
R
rrlB
0
2
d2
2
R
rrl
R
Ir
0
2
2 d222
1?
R drrR
lI
0
3
4
2
4?
4
4
2
16
R
R
lI
16
2lI
V mm VwW 体d
l
R
r
dr
Chapter 28 Inductance and Electromagnetic Oscillations
Homework:
P656,3,13,18
