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Electromagnetism
Zhu Jiongming
Department of Physics
Shanghai Teachers University
Electromagnetism
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Electric Field
Conductors
Dielectrics
Direct-Current Circuits
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Magnetic Field
Electromagnetic Induction
Magnetic Materials
Alternating Current
Electromagnetic Waves
Chapter 6 Electromagnetic Induction
§1. Electromagnetic Induction
§2. Lenz’s Law
§3. Motional Electromotive Force (emf)
§4. Induced emf Induced Electric Field
§5. Self Induction
§6. Mutual Induction
§7. Eddy Current
§8. Transient State of RL Circuit
§9. Transient State of RC Circuit
§10. Magnetic Energy
§1. Electromagnetic Induction
1. Experiment A current can be induced in a loop as
the magnetic flux through the loop changes with time
2. Faraday’s Law of Induction

d
K
dt
A
Induced emf = K  time rate of change of flux
in SI unit（V、Wb、s） K = 1
d
|  |
dt
（ Direction --- Lenz’s Law）
§2. Lenz’s Law
1. Two statements
● Induced current always tends to oppose the flux
change
● Magnetic force on induced current always tends
to oppose the motion of conductor
2. Expression of Faraday’s Law
3. Examples
1. Two Statements (1)
Induced current always tends to oppose the flux
change —— flux of induced current
● in the opposite direction
for  > 0
● in the same direction
for  < 0

N
A
S
N
A
S
1. Two Statements (2)
Magnetic force on induced current always tends to
oppose the motion of conductor
I
∵ v towards right

∴ fAmpere left（ fAmpere=  Idl  B ） R
f安
v
∴ I counterclockwise
( or：  > 0 ∴ flux of I in opposite direction )
 Work and Energy：
Work by external force  I 2Rt ( heat energy )
（constant velocity，kinetic energy not changed）
If fAmpere towards right，energy conservation violated
2. Expression of Faraday’s Law

Positive directions for  and  ：

d
right hand rule


then
dt




0
d
0
dt
 0
0
d
0
dt
 0




0
d
0
dt
 0
0
d
0
dt
 0
Example （p.276 / 6-2-1）
A small N turns coil A of radius r is placed coaxially in
a long solenoid S of n turns per unit length. Current I
in S decreases from 1.5A to -1.5A at a steady rate in
0.05 s. Find emf in A.
Sol.：B = 0nI
  NBr 2  N 0 nr 2 I
d
2 dI
     0 nNr
dt
dt
dI  1.5  1.5


 60
dt
0.05
   600 nNr 2  0
Direction: same with I
and not changed
Exercises
p.276 / 6 - 2 -
2, 3
§3. Motional emf

d

dt
where :    B  dS
S
Three ways to change the flux  ：
 change the area ( move the loop )
—— motional emf
 change the field B
—— induced emf
 change both of them
§3. Motional emf
1. Motional emf and Lorentz Force
2. Calculation of Motional emf
3. Examples
4. Alternating-Current Generator
1. Motional emf and Lorentz Force
Magnetic force on electrons in conductor ab
a
I
d
f = - ev  B
（downward）
 I counterclockwise direction
（caused by emf on ab ）
f
emf = work on unit charge
c
b
by Lorentz force as moving it from b to a
a
a
a f

 dl   (v  B)  dl   vBdl vBl
b
b
b e
∵ vl is the rate at which the area changes
∴ vBl is the rate at which the flux changes ，
= d/dt
v
Motional emf
General formula for Motional emf ：

  (v  B)  dl
motional
suitable also for unclosed conductor，but no I，
only emf


v
o

B in the formula not change with time
2. Calculation of Motional emf
Two ways to calculate Motional emf ：
(1)  motional   (v  B )  dl
where v、B might not be uniform，so could not be
brought out in front of the integral sign
d
where :    B  dS
(2)   
S
dt
If not closed，add an imaginary
v
curve to make it closed
or get d/dt by the area swept
per unit time
Example （p.229/ [Ex.1]） (1)
A straight wire of length L is rotating at angular
velocity  in a uniform magnetic field B. Find emf ab
and potential difference Uab .
b
Sol.：(1) take dl at a distance l from a
dl

v  B same direction with dl , v =  l
l
a
b
L
1
v
2
 ab  a (v  B)  dl  0  Bl dl  BL
2
(  0,
direction
：apotential，negative
 b)
a： low
charges
ab
b： high potential，positive charges
U ba   ab
1
 U ab   ab   BL2
2
Example （p.229/ [Ex.1]）(2)
(2)
assume rotating d in d t
the area swept is
1
dS   L  Ld
2
a
1 2
d  BdS  BL d
2
d
1 2 d 1 2

 BL
 BL 
dt
2
dt
2
b
L
d
direction：imaginary loop counterclockwise，
ab
Ld
4. Alternating-Current Generator

Loop plane is perpendicular to
magnetic field at t = 0
（ S is in the same direction
with B ， = 0 at t = 0 ）
   B  dS  BS cos t
S
d
  BS  sin t
dt
   BS sin t
B
N
S
Exercises
p.277 / 6- 3 -
1, 2
§4. Induced emf
Induced Field
1. Induced Electric Field
2. Properties of Induced Electric Field
3. Induced Electric Field Due to the Changing
Magnetic Field in a Solenoid
4. Examples
1. Induced Electric Field
Loop does not move，magnetic field is changing
d
B changing  flux =B·dS changing    
0
dt
 induced current  force F on q  F/ q = E
Coulomb's law  Coulomb's electric field E C
 changing magnetic field  induced electric field E I
Total electric field E = E C + E I
or
d
  EC  dl
 
dt

2. Properties of Induced Electric Field
Potential field
 E  dl  0
Source field
 E  dS  q / 
d
d
 E  dl     dt   dt  B  dS
B
  
 dS  0
Eddy field
t
Sourceless field
 E  dS  0
EC :
L
S
EI :
L
C
C
in
0
I
S
S
S
I
Total electric field
 E  dS  q
S
in
/ 0
E = EC + EI
B
LE  dl  S t  dS
3. Induced Electric Field in a Solenoid
symmetry（infinite long），E
perpendicular to the axis
—— no axial component
 symmetry（circle）， E I
—— no radial component

 E
S
感
I
is in the plane
S
 dS  0
Conclusion： E I has only tangent component ，
E I lines are concentric circles
Example 1 （p.236/ [Ex.1]）
In a long solenoid of radius R, dB/dt is known, find EI .
Sol.： EI  dl  EI dl  EI dl  EI  2r


L

L

L
B
B
 dS  
dS
S t
S t
dB

 r 2 (r  R)
dB
dt

d
S
dB
dt S

 R 2 (r  R)
dt
r dB
 Inside EI  
2 dt
R 2 dB
outside EI  
2r dt
R
r
Example 2（p.237/ [Ex.2]）
In a long solenoid of radius R, dB/dt, h, L are known.
Find  MN .   N E  dl
I

M
MN
Sol.：
o
L r dB
r
h
 
cosdl ( r cos  h )
0 2 dt
L
M
N
1 dB
h dB L

dl   hL

2
dt
2 dt 0
1
Another way： OMN = BS  hLB
2
emfs are zero on OM and ON
d
1 dB
  MN   OMN     hL
dt
2
dt
Exercises
p.279 / 6- 4 -
2
§5. Self Induction
1. Self Induction
2. Self Inductance
3. Examples
1. Self Induction
IB 
if
d
0
dt
then

B
I
d

0
dt
An induced emf appears in any coil in which the
current is changing
—— Self-induced emf
2. Self Inductance
tightly wound coil：every turn as a loop，the same
flux S passes through all the turns
 N - turn coil：flux linkage S = NS
S  S  B  I
Definition： S = L I
L：Self inductance
d S
dS
dI
S  N

 L
dt
dt
dt

1 Weber
Unit： 1 Henry 
1 Ampere
Example （p.245/ [Ex.]）
A long solenoid of volume ，n turns per unit length.
Find it’s L .
Sol.：B = 0nI
S = N S
= nl S
= 0n2IS l
= 0n2 I
= LI
 L = 0n2
Exercises
p.280 / 6- 5 - 1, 2, 3
§6. Mutual Induction
1. Mutual Induction
2. Mutual Inductance
3. Two Coils in Series
● connect in the same direction
● connect in the opposite direction
1. Mutual Induction
I1  11 ，12
I1 11
12
12 ： flux through coil 2
by the field of I1
d11
Self induced emf
dt
2
d12
1
Mutual induced emf
dt
 in the same way
21 ： flux through coil 1 by the field of I2
total flux through coil 1
1= 11  21 + for I1 , I2 in same direction
- for opposite direction

B
2. Mutual Inductance
flux linkage：11 = N111 , 21 = N121 , 1 = N11
 1= 11  21
（ 1= 11  21 ）
11 = L1 I1 ， 21 = M21 I2
M21： flux through coil 1 by the field of I2  I2
M12： flux through coil 2 by the field of I1  I1
 can be prove： M12 = M21 = M
Mutual Inductance

d1
dI1
dI 2
1  
 ( L1
M
) M ：interaction
dt
dt
dt
of two coils
d2
dI 2
dI1
and  2  
 ( L2
M
)
dt
dt
dt
3. Two Coils in Series
Connect in the same direction
 Connect in the opposite direction

Connect in the Same Direction
1= 11 + 21
2= 22 + 12
dI
dI
I
1  ( L1  M )
dt
dt
dI
 ( L1  M )
dt
dI
and  2  ( L2  M )
dt
dI
  1   2  ( L1  L2  2M )
dt
 L  L1  L2  2M
I
Connect in the Opposite Direction
1= 11 - 21
2= 22 - 12
dI
1  ( L1  M )
dt
dI
 2  ( L2  M )
dt
I
dI
  1   2  ( L1  L2  2M )
dt
 L  L1  L2  2M
I
Exercises
p.280 / 6- 6 -
1, 2
§7. Eddy Current
 Applications
Harmfulness
 Electromagnetic Damping
 Skin Effect

§8. Transient State of RL Circuit
1. Connect RL Circuit with an emf
2. Remove the emf from RL Circuit
R
L
Steady state：I = 0 （K open）
and I =/R（K close）
 Transient State ：i ：0  I

K
capitals： I、U steady quantities
little：
i、u varying quantities
Ohm’s Law, Kirchhoff’s Rules are workable ,
as long as the quasi-steady condition satisfied

1. Connect RL Circuit with an emf （1）
Equation：（ Kirchhoff’s loop rule）
L
R
d
i
   S  iR
S  L
S
dt
di
i
   L  iR
dt
—— deferential equation of i
Initial condition ：i  0 at t  0

K
di
General solution ： L  iR  
dt
R
 t
di
R

  dt
 i   Ae L
i  / R
L
R
R
ln( i   / R)   t  K（constant ）
L
Connect RL Circuit with an emf （2）

Spetial solution ：i   A  0 at t  0
R

A
R

R
 t

i  (1  e L )
R
 I (1  e
(t 0)
R
 t
L

)
R
L
i
自
K
i (t)

I
R
0
t
Connect RL Circuit with an emf （3）
i(t )  I (1  e
R
 t
L
i (t)
)
I

Discussion：
R
0.63I
(1) if t   ，then i  I ，
actually need only little time
0 1  2
t
(2) Inductive time constant 
R
R the time constant  is，
 t
The
less
e L  0 rate governed by
L rapidly the current
the more
L
goes
to it’s equilibrium
value.
Definition ：  （ time
constant
）
R
when t   , i  I (1  e1 )  0.63I
2. Remove the emf from RL Circuit（1）
L
R
The switch closes at t = 0，consider
the uper part of the circuit ( loop )
自
i
di
 S  iR
S  L
dt
di
  L  iR
dt

R’
di
R
  dt
at t  0 i  A  I 0
i
L

R

ln i   t  K（constant ）
R  R'
L
 i  Ae
R
 t
L
 i(t )  I 0e
R
 t
L
Remove the emf from RL Circuit（2）
i (t )  I 0 e
R
 t
L
i (t)
I0
Discussion：
(1) if t   ，then i  0 ，
0.37I0
actually need only little time
0 1 2
t
(2) time constant ：
L
The less the time constant  is，

the more rapidly the current
R
goes to it’s equilibrium value 0 .
i(t )  I e t /
0
when t   , i  I 0e  0.37 I 0
1
Exercises
p.280 / 6- 8 -
1, 5
§9. Transient State of RC Circuit
1. Connect RC Circuit with an emf
2. Remove the emf from RC Circuit
3. Summary
1. Connect RC Circuit with an emf（1）
Charging a capacitor：uc：0  
Equation（Kirchhoff’s loop rule）
R
C
uR
uC
uR  uC  
i
dq
q  CuC
uR  iR i 
dt
duC

K
 RC
 uC  
dt
duC
dt

uC  
RC
uC  0 at t  0
1
ln( uC   )  
tK
 A 0
RC
 t / RC
 t / RC

u
(
t
)


(
1

e
)
C
 uC    Ae
Connect RC Circuit with an emf（2）
uC (t )   (1  e
Discussion：
 t / RC
)
uC (t)

(1) if t   ，then uC   ，
t
actually need only little time 0
(2) Capacitive time constant： = RC
The less the  , the more rapidly it goes to equilibrium .
 R smaller，current larger
Need less time
 C smaller，need less charge
2. Remove the emf from RC Circuit
The switch 2  1 at t = 0
R
（discharging a capacitor）
u R  uC  0
i
duC
RC
 uC  0
dt
duC
dt


uC
RC
uC (t)
 t / RC
 uC  Ae

at t  0 , uC    A
 uC (t )  e t / RC
 e
t /
0
C
1
2
K
t
3. Summary
Steps to treat transient process ：
differential equation  general solution
（in exponential fashion）
initial condition  constant
 Features ：
 C  voltage can not change suddenly
 L  current can not change suddenly
 energy（L、C can store energy，need time）
 C： W  1 Cu 2
L： WM  1 Li 2
E
2
2
（ R does not store energy，I, U change suddenly ）

Exercises
p.281 / 6- 9 -
1, 2
§10. Magnetic Energy
1. Magnetic Energy of a Self-Inductor
2. Magnetic Energy of Mutual-Inductors
1. Magnetic Energy of a Self-Inductor

Steady：I constant， = IR
Power of emf
 I = I2R
L
Thermal energy
transient ：
di
  L  iR
dt
 idt：  idt = i 2R dt + Lidi

R

K
work of emf = thermal energy + energy to build up field in L
（this is magnetic energy）
i ：0  I
Wm  
I
0
1 2
1
Lidi  LI ( compare ：WE  CU 2 )
2
2
（ I lager ，B stronger， Wm lager ）
2. Magnetic Energy of Mutual-Inductors
di1
di2
1  L1  M  i1R1 ①
dt
dt
di2
di1
 2  L2  M  i2 R2 ②
dt
dt
①×i1dt + ②× i2dt：
1i1dt   2i2dt 
L1
M
K1
R1
L2
K2
R2
1
2
2
2
 i1 R1dt  i2 R2dt  L1i1di1  L2i2di2  M (i1di2  i2di1 )
work of emfs = thermal energy in Rs+ magnetic energy Wm in Ls
I1I 2
Wm   L1i1di1   L2i2di2   M d(i1i2 )
0
0
0
1
1
2
2
 L1 I1  L2 I 2  MI1I 2
2
2
I1
I2
Exercises
p.282 / 6- 11 -
2