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Chapter 12
Electromagnetic
Induction
第十二章 电磁感应
§12-1 Nonelectrostatic Force, Source &
Electromotive Force
非静电力 电源 电动势
1. Nonelectrostatic Force
A
B
When S is turned off,
the charges on A and
B are constant,
the E-field between
A and B is constant.
SS
R
When S is turned on, there is a current in circle,
the charges on A and B decrease,  E-field 
 q=0, E=0, I=0
In order to sustain a constant
 current, some kind
of nonelectrostatic force ( F)n must be supplied to
move positive charges from B (low potential) to A
(high potential)

resisting Fe
A 
B
Fn

Fe
R

The device which supplies Fn is called source.
Nonelectrostatic Field


Fn
En 
q
---nonelectrostatic force on per unit
positive charge.
2. Electromotive Force 电动势
The work done by nonelectrostatic force for
moving per unit positive charge from B to A
within the source is defined as
 
A
B
 
En  dl ---emf
B
A
scalar
High potential Low potential
A B
+ _

R
R

If Fn exists in a closed loop, then


   En  dl
l
S
N
I
The emf in a complete circuit equals to the
line integral of the nonelectrostatic force
round the circuit .

B
3. Complete circuit Ohm law 全电路欧姆定律
A B
+ _

r
Source: emf =  ,
internal resistance = r
In a time interval dt,
emf moves dq from B to A
R
i
dq = idt
So the work done by emf is
dW=dq =  idt
Meanwhile, the resistances R and r consume
energy
Pdt = (i2R + i2r) dt
thermal energy
According to the principle of conservation of
energy, dW = Pdt
idt = (i2R + i2r) dt
i.e.

i
A B
+ _

a
i
Vab
Rr
Vab = iR
r
R

Vab =  - ir
b
---Complete circuit
Ohm law
§12-2
Faraday’s Law of Induction
电磁感应定律
1. Induction phenomena
S
magnet
moves
N
Ii
 Ii
B

Ii
wire moves
loop rotates
v

B

B
Conclusion：An induced current appears
 in a
closed conductor loop, only when the B-flux
that pass through the loop is changed.
2. Lenz’s Law
Determine the direction of an induced current.
The induced emf and current are in such a
direction as to oppose the change that
produces them.(感应电动势和电流的方向总是在
抵抗产生他们的变化的方向上)
S

B   B 


 Bi and B have opposite
direction
 I ’s direction is as Fig.
N
I
Bi
I

B
3. Faraday’s law of induction
For SI

d B

dt

d B

dt
The induced emf in a circuit is equal to the

rate at which the B-flux through that circuit
is changing with time.

1
d

B
Induced current I 

R
R dt
The induced charge passing through the loop
1
q   Idt  ( B1   B 2 )
R
Note ：
for a coil of N turns
d 1 d 2
  1   2    N  ( dt  dt  )
d N
N
  (   i )    ---total flux
 i
dt i 1
If  i   j   for any i, j
i 1
  N--number of flux linkages
d
 
dt
[Example] Alternating current (AC) generator
A coil of area S，N turns
rotating in a uniform M-field.
Angular velocity is ，and =0
when t=0. Find =?


B

  N  NBS cos 
 NBS cos  t
d
   NBS sin  t
dt
Let

 m  NBS
   m sin  t —alternating emf
[Example] A long, straight wire carries a
variable current I (t). Find the induced emf in
the rectangular loop.
dI
 0 、a 、b 、 l are known
dt
  ab 0 I
ldr
 B   B ds  a
2

r
dr
S
r
I
ds


B
l
 0 Il a  b

ln
2
a

a
b
 0 l a  b dI
d B
ln

2
a dt
dt
Direction：inverse-clockwise
§123 Motional electromotive force
动生电动势

d
d B


dt
dt
 
B

d
S

S

 B is a constant，S is a variable
----motional emf


S a constant，B is a variable
----induced emf
Motional emf：
a+
   ++  
En
 
 
v

 


B
F
The charge +q in the rod
When the rod is moved
with ,

v
+q suffers Lorentz force

 
F  qv  B

    
--nonelectrostatic force
  ---  
b
nonelectrostatic field is


F  
En   v  B
q
 
a
b
  
(v  B)  dl
Notes：

’s direction is along ba
d
l
v

 is the instantaneous velocity of dl .

B is the M-field in the spot that dl locates.
This formula can be used to calculate the
motional emf induced by any shape of
moving conductor in any M-field.
Question：
 
a
b
  
(v  B)  dl
is the work done by Lorentz force to
positive unit charge.



From another formula F  qv  B ,
 
We get F  v .
Lorentz force does not do any work for q.
How can we explain their conflict？
q has two motions in the rod:

 
F  qv  B

 
F   qv   B
a
 
FR
 
F
  
v  vR
 
v

F

 
 
 b
The total work done by
resultant
force is


FR  v R
 
 
 
 ( F  F )  ( v  v )
   F  v  F  v  F   v  F   v
=0
Positive work
=0
=0
negative work
[Example]
 
l 、B 、v、 are known, find  =?
 
b 
solution：  ( v  B)  dl

a
 
v B   b 
  l 


  dl
a   
   
 
v 
 
 


  vB cos   dl
a
2

b
 vBl sin
Ub > Ua

[Example]：l、B 、
 are known, find
 
v  B
B
 
v
L
o l
A

dl
 
 =?
A
o
  
( v  B)  dl
L
   vBdl
0
1
2
 B  ldl    BL
0
2

“-”represents  ’s direction is opposite dl’s
direction, i.e. AO.
UO > UA
L
Another solution：using
 L
B 



d B
dt
Make a subsidiary line,
then the M-flux through
the fan area is
1
 B  B L  L
2
1 2
1 2 d
d B
 BL 
 BL
dt
2
dt
2
A metal disk with radius R is rotating around
its center in the uniform M-field.

B

o R
The motional emf from O
to any point P of its edge is
P
1
2
  B R
2
[Example]

I 、d、L、v are known, find =?

B
I
x
d

I
0
Solution B 
2 x

v
dx
a
b
L
 
vB
Ua> Ub
 
b
a
 
  
v  B dl
d L
d
 0I
vdx
2 x
 0 Iv d  L

ln
2
d
§12-4
Induced electric field
1. Maxwell ’s hypothesis
A changing M-field produces an

nonelectrostatic field -- induced electric field E n.

En
    
B
    
    

    
En

En

B
>0
t

En
2. Induced emf
 
L


En  dl


or   l En  dl
3 . The relation between induced emf and
changing M-field


   En  dl
L

d B
B 
  
 
 dS
   B  dS   S
dt
t
t



B 
En  dl   S dS
L
t

S —the area surrounded by L.
3 . The differences between electrostatic field
and induced electric field
 Electrostatic field is set up by static charges.

E-line originates on +q, terminates on –q.
It is never closed.

Induced electric field E n is produced by the
changing of M-field.

E n-line is closed, neither start, nor end.

 
--non-vortex field (无旋场)
E

d
l

0

L
E-potential can be introduced to describe
electrostatic field.

L


En  dl   --vortex field (有旋场)
Potential can not be introduced to describe
induced electric field.
[Example] A uniform
 M-field fills a cylindrical
B
volume of R. And
 0.
t

Find the distributions of E n



E n-lines are concentrical rings

 R 

L   
 B r
   .


En
P
r<R：Choose a closed circle L passing point P

L


En  dl   Endl  En  2 r
L

d B
B   B
2


r

 dS
S t
t
dt



B 
use
En  dl   S dS
L
t

B 2
 En  2 r    r
t
r B
En  
“ ”represents that the

2 t
direction of E n is determined
by Lenz’s Law
L

En


  

B
  

r>R：
R




r

L
P


En  dl  En  2 r

B
B 
2
S t  dS  t  R
En
r
R B
 En  
2r t
2
§12-5 Applications of induced electric field
1. Betatron 感应加速器
A device that can accelerate electrons to
high nenrgies by induced electric field.
S


N
B(r,t)
When an electron enters
the changing M-field,

 
Fm   ev  B









F
m

En


  
1 dB

, Fe  eEn 
En   r

2 dt



Fe  eE n
eE n
Tangent acceleration: a t 
m
t e
dB
t
e
r
dt 
rBt 
speed：v t   at dt  
0
0 2m
2m
dt
when B(t) , v(t)  .
( r keeps a constant. )
2.Vortex Current 涡电流
(1) formation
Alternating current I 


changing B  E  vortex ～I
n
current Iv

dB
dt

En
Iv
～I
(2) application
Using the hot effect of current, vortex
current can be used on electromagnetic stove
or induced stove to produce heat energy.
The energy loss of vortex current should be
decreased as less as possible on the
transformer systems.
I
~
§12-5 Self-induction & Mutual-induction
自感和互感
1. Self-induction
I

B
I
Induced emf
M-field
B
varies
B
I Varies

I  B    N B
I
  LI
L --Self inductance of the circle. It is determined
by the size, the shape, the number of turns,
and the magnetic properties of the material
around the circuit. It has nothing with I.
The SI unit of L is Henry（H）亨利
Self-induced emf ：when I is changing,
dI
 L  L
dt
Self inductance：

L
I
or
L
L
dI
dt
[Example]：A long and thin straight solenoid
with l S  r N. Find its L=?
rS
l
N
Solution
Assume the solenoid carries
current I，
then
B   r  0 nI

 r 0 N
l
I
  NBS
 0  r N S  0  r lSN 2

L


2
I
l
l
2
L  n V
2
V --the volume of the solenoid.
[Example]：Find L=? for the unit length of
coaxial cable. (a,b,0 are known)
a
b

Then the distributions of B is


B

I
Solution assume it carries I ,
I
B
0
（r<a）
0I
2 r
（a<r<b）
0
（r>b）
0 I
 B   d B  
 h  dr
a 2 r
b
a
b

B
h1
I
0 I b

ln
2
a
The self inductance for unit
length cable is
I
r dr
 B 0 b
L

ln
I
2 a
2. Mutual induction
I1
I2
1
B1
N112  12
2

B2
N 2 21  21
Mutual induced emf :
when the currents I1, I2 change,
d12
1  
dt
d21
2  
dt
Circle 1：
12  I 2
12  M12 I 2
M12 :the mutual inductance of loop 1 relative to 2.
It is determined by the sizes, the shapes, the
number of turns, and the magnetic properties of
the material around the loops, the relative
position of the two loops. It has nothing with I2.
The SI unit of M is Henry（H）

dI 2
 1   M12
dt
Circle 2：
21  I1
21  M 21I1
M 21 :the mutual inductance of loop 2 relative to 1.
dI1
  2   M 21
dt
It was proved that
M12  M 21  M
dI1
2  M
dt
dI 2
 1   M
dt
Calculate M:
or
12 21
M

I2
I1
M 
1
dI 2
dt

2
dI1
dt
[Example]：A long and thin straight solenoid
with l S  r and two coils N1 N 2 . Find their M=?
Solution
N2
S
r
N1
l
Assume the coil 1carries
current I1，
then the M-field
produced by I1 is
B1   r  0 nI1 
 r  0 N1
l
I1
The number of magnetic flux linkages through
the coil 2 is
 0  r N1 N 2 S
21  N 2 B1 S2 
I1
l
21
0  r N1 N 2 S
 M

I1
l
If the coils 2 carries current I2，M=？
§12-7 Energy of the M-field
R

L
。。
K
Turn on the switch K , the current in the circle
0 i(t)，
At any time,
   L  iR
di
  L  iR
dt
di
dt

  iR
L
R

t

di
dt

1  e L 


i
t



0   iR 0 L

R 


When t，i t   I  (constant)
R
i t 
t
During the current changes from 0 to ，Mfield changes from 0 to .
The work done by the battery：
di


W    dq    idt   iR  L  idt

dt 
t
I
  i Rdt   Lidi
2
0
0
Energy of M-field
Joule energy dissipated on R
i.e. the part of work is transferred in the
energy of M-field.
Wm  
I
0
1 2
Lidi  LI
2
Energy of M-field is
1 2
Wm  LI
2
The energy density of M-field
（the density of magnetic energy）
is
dWm
wm 
dV
For long, thin straight solenoid,
L  n V ,
2
B  nI
2
1 2 1 B V
 Wm  LI
2 
2
The density of magnetic energy is
2
dWm 1 B
1
1
2
  H  BH

wm 
2
2
2 
dV
The energy of M-field is dWm  wmdV
Wm   dWm 

V
wm dV
the volume filled by M-field
[Example]：Find Wm=? for the unit length of
coaxial cable. (I,a,b,0 are known)

The distributions of B is
a
b
0
B
0I
2 r
0
h1
I
I
（r<a）
（a<r<b）
（r>b）
choose a section of the
cable with h=1,
on the region of r<a and r>b ,
we have W m
0
1B

I
0
a<r<b： w m 
 2 2
2  0 8 r
2
0 I
dWm  wm dV 2 2 1 2rdr
8 r
2
b  I2
0 I
b
0
Wm   dWm  

ln
2

rdr
2
2
a
8 r
4
a
2
2
1 2
Or usingW m LI :
2

b
0
We have known that L 
ln
2 a
2
0 I
b
 Wm 
ln
4
a