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A straight wire segment
2aB  0i b
direction : 
 0i
L  d , B 
2d
-b
0i
a  R, B 
2 R
0i
R  0, B 
2a
The Magnetic Field of a Solenoid
 0iR 2
B
3
;
2( R 2  z 2 ) 2
 0 (nidz ) R 2
dB 
3
2
2 2
2[ R  ( z - d ) ]
B

 0 niR
2 L/2
2
 0 ni
2
 [R
- L/2
(
dz
2
 (z - d ) ]
2
L 2d
[R 2  (L 2  d )2 ]
If L>>R,
B  0 ni
1
3
2

2
L 2-d
[R2  (L 2 - d )2 ]
Bh  0 nhi
1
)
2
direction ?
Biot-Savart law


 00 qidvsrˆrˆ
dBB 
22
44 rr
Magnetic
Field
Symmetry
Moving
Electrical
charge
Current
 
 B  ds  0I
Ampere’s Law
A interesting question for interaction force between
electric / magnetic field and moving or resting charged particle
A charged particle at rest
Moving charged particle
Whether a charged particle
moves or not depends on
the chosen frame.
Electric field
Both electric and magnetic field
Whether can we conclude that
both electric and magnetic field
Depend on the chosen frame?
Einstein’s Postulates: The laws of physics are the same
in all inertial reference frames.
Are these two conclusions contradictory?
Is it true?
2017/5/7
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4
y
In the frame K
（at rest）
In the frame K Both electric and
magnetic force are zero.
’
In
the
frame
K
Magnetic force：
x
r
Ｉ
－ － － － － －
Fm = qvd B¢ =
vd
S Electric force：
＋ ＋ ＋ ＋ ＋ ＋
y’
K’
In the frame
（with moving
electrons）
B´
vd
Ｉ
vd
－
Fm
－ －
l+' = l+ / 1- (vd / c)2
L-' = L / 1- (vd / c)2
l-' = l- 1- (vd / c)2
ql+¢ vd2
FE = qE = q
=
2pe0 r
2pe0 rc 2
r
We have：
I ¢ = l+¢ vd , c2 =1/ (e0m0 )
S
＋ ＋＋ ＋ ＋ ＋
2017/5/7
L+' = L 1- (vd / c)2
l+¢ - l-¢
x’
－ － －
m0qvd I ¢
2p r
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Therefore
FE =
5
m0qvd I ¢
2p r
In different inertial reference frame, the results
come from the electric or magnetic field is same.
Transformation equations of E and B:
E x  E x
E y 
E z 
Bx  Bx
1
1- b
2
1
1- b
2
( E y - vBz )
( E z  vBy )
1
v
By 
( By  2 Ez )
2
c
1- b
1
v
Bz 
( Bz - 2 E y )
2
c
1- b
Magnetic
Field
Electrical
Current
Faraday’s Law of Induction
Chapter 34 Faraday’s Law of
Induction
Faraday’s Experiments
The experiment about the induction:
The common feature: change in flux
Faraday’s Law of Induction
Magnetic flux ΦB
 
dΦB  B  dA,
 
ΦB   B  dA
 
dΦB  B  dA  B cosdA
change in B
change in θ
change in A
EMF
dΦB
 
dt
EMF
Faraday’s law
Flux change
dΦB
 
dt
EMF
dΦB
 dt
induced
current
Lentz’ Law
Flux (cause)
increase
decrease
Flux (result)
induced
current
decrease
increase
Conservation of energy considerations:
i
N
S
i
Mechanic
energy
V
Electric
energy
Conservation of energy considerations:
Electric
energy
Electric
energy
Signs in Faraday’s law
sign of 
ddΦΦ
BB


 dd
tt
sign of dA
sign of flux ΦB
Motional EMF
dΦB
d
 
 ( BDx )  BDv
dt
dt
d  (v  B)  ds
   (v  B)  ds  BDv
F2  F3  0
From the signs of faraday' law , F1  0 W  0
dΦB the EMF is clockwise.
 dt the induced current is clockwise.
Generators and Motors
Generator is a device for
converting mechanical work (or other)
into electrical work in the load.
dΦB
d( BA cos t )
  BA sin t
dt
dt
alternating current (AC)
i

R

BA
R
sin t
chemical energy
electric energy
mechanical energy
atomic energy
electric energy
mechanical energy
Nuclear reactor at Tihange , Wind-powered electric generator
Belgium
at Sandia National laboratory.
•“Dynamic” Microphones
(E.g., some telephones)
–Sound
 oscillating pressure waves
 oscillating [diaphragm + coil]
 oscillating magnetic flux
 oscillating induced emf
 oscillating current in wire
after amplification  Sound
Induced Electric Field
In
In conductor
Space
Changing
magnetic
field
Induced electric fields.
E

dl


i

Ei
Electrical
Electrical
Field
Current
Example
r
r, R (resistance) are known,
Bz=-4.0-5.6t+2.2t2. Find induced
current in the loop at t=1s and t=2s.
 
ΦB   B  dA  B  dA  r 2 B
dΦB
2 d
  -r
B   r 2 (5.6 - 4.4t )
dt
dt
2
 r
I 
(5.6 - 4.4t )
R
R
I t 1  1.2
r
2
R
I t  2  -3.2
r
2
R
Example
E
E
rE
r, R (resistance) are known,
Bz=-4.0-5.6t+2.2t2. Find induced
current in the loop at t=1s and t=2s.
 
ΦB   B  dA  B  dA  r 2 B
E 
dΦB
2 d
0




r
B
E
d
l


Induced Electric Field
I
I ' 0
dt

R

r
R
I t 1  1.2
dt
2
Induced EMF
(5.6 - 4.4t )
r
2
R
I t  2  -3.2
r
2
R
Example
 ?
  
d  (v  B)  ds
d  Bvdr ,
R
R
0
0
   Bvdr  
1
2
Brdr  B R
2
suppose oab is a loop,
1 2
ΦB  BA  B( R  )
2
dΦB
1
2
  - BR
dt
2
What is the direction
of induced EMF?
ao
Example: As shown, find the emf ε when the rod (length L)
in location OM, and ON.
i
  
OM :    (v  B)  dl
M
ω
dl
 0i
 0i 2
l

l  dl 
L

r0
2

r
4

r
0
0
o
N

r
 
ON :    (v  B)  dl
 0i
Direction of EMF ?
  (r - r0 )
dr
2r
O  M,
ON
 0i
r0  L

( L - r0 ln
)
2
r0
Exercises
P796 23, 24, 27
Problems
P798 5, 9
Induced Electric Field
In
In conductor
Space
Changing
magnetic
field
 
dΦB
2REi   Ei  dl    dt
Induced Electric Field
Ei
Electrical
Electrical
Field
Current


F  qEs
F  qEi
Induced electric fields
In the closed loop, the work done on the charge by
 
the induced electric field is
W   qEi  ds

W

The induced emf in the loop is
    Ei  ds
q





q

d
Φ

B
d
-ds  0dA rˆ
Electrostatics
-  B EdsAEs 
  - B  Field
2

t 0 r 
dt
4
dt
Induced Electric Fields
 
B 
 Ei  ds  - t  dA  0
Electric fields
magnetic field
Calculation of Induced Electric Field
For the cylindrical field distribution (radius R), dB/dt>0,
 
   Ei  ds  Ei 2r
dB
dB
dB
 -
dt
 -
r dB
Ei  2 dt
Ei
dt
dA  -
dt
r
2
(r  R)
dB
dB
dB
r  R,    -
dA  -  R 2
dt
d
t
d
t
dB
 const.
R 2 dB
dt
Ei  (r  R)
2r dt
r
R
The betatron
 
dΦB
 Ei  ds    ΦB  Φo cos t
  Φo sin t
Betatron produces a pulsed
rather than a continuous beam.
dt
Induction furnace
 
dΦB
 Ei  ds    dt
Change current
EMF
w  i Rt
2
Electric energy
Heat
current
Eddy
currents
 
dΦB
 Ei  ds    dt
Electric energy
heat energy
Reduce heat
energy
Transformer
Materials are laminated.
Magnetic breaking
s
Magnetic breaking
N
Damping
Oscillation
STM
Magnetic
Damping
Induction and Relative Motion
In S frame
V  v  vd
dWN  N (vdt )  FB sin (vdt )
 (qVB)(vd /V )(vdt )  (qBv d )(vdt )
 (qBv )(vd dt )  qBvds
WN   qBv ds  qBvD
  WN q  BDv
Induction and Relative Motion
In S frame
V  v  vd
dWN  N (vdt )  FB sin (vdt )
 (qVB)(vd /V )(vdt )  (qBv d )(vdt )
 (qBv )(vd dt )  qBvds
WN   qBv ds  qBvD
  WN q  BDv
In S’ frame  '  --
   BDv
dBd
 ( Dx ' )
dt dt
N
´
Magnetic
Field
Induced EMF
Motional EMF
Induced E
Electrical
Current
dΦB
 d
t
  
d  (v  B)  ds
  
   (v  B)  ds

 
B 
 Ei  ds  - t  dA
Magnetic
Field
Electrical
Field
Example B  6t - 5 Ei  ?
when r < R,
2
d
(2BrdB
B)
2 rEi   ii  -- r
dtdtdt
r d2BdB
2rE
Ei i--r
Ei  -3r
2 dt dt
When r > R,
d B
2 dB
i   -R
dt
dt
2 dB
2rEi   i  -R
dt
R 2 dB
3R 2
Ei  2r dt
r
Example
 0i dr
d B
 Dv
dt
2r dt
r  i
 0i
r
0
D ln
B  
Ddr 
r0 2r
2
r0
  
d  (v  B)  ds  vBds
 ?
i
D
r
v
0
 
D
0
r
direction of 
 0i
vBds 
Dv
2r
r0 r0 DD
d B  -- 00ii dL
v ln(ln(
))
 ?  22 dt
r0 r0
dt
r0  D  i
 0i
r0  D
0
L ln(
)
B  
Ldr 
r0
2
r0
2r
Example
r dr
D
i
L
r0
v
direction of 
  
d  (v  B)  ds  vBds
 
r0  D
r0
vBds  
r0  D
r0
 0i
v
dr
2r
 0i
r0  D

v ln(
)
2
r0
Example
y
Ei
As shown, R, L, h, dB/dt are known,
find emf in the rod.

b
B
r
 o θh L
R
a
 
B
2 dB
2rEi  Ei dl  -  ds  -r
t
dt
r dB
x
Ei  2 dt
 
rh dB h
d i  Ei  dy  Ei dy cos  dydy
dB
0
dt
Direction of emf a→b
i
2 dt r
L h dB
b
hL dB
2
dy   d i -- L
a
2 2 dt
2 dt

Example

y
c
As shown, R, L, h, dB/dt are known,
find emf in the rod.
b
hLhLdBdB
 i  d i  b
a
2 2dtdt
Loop “a-c-b-a”
B
 o h L
R
a
dB
0
dt
1 2 1
Φ

BA

B
(
R   Lh)
B
acb
x
2
2
dΦB
R 2 hL dB
  acb   ab
 '  -(
 )
dt
2
2 dt
 acb
Direction of emf a→b
2
R
 dB
dB


2
 (-R
)

2 dt
dt 2
2
 ab   '- acb
hL dB
2 dt
Example
As shown, R, L, h, dB/dt are known,
find emf in the rod.
b
hLhLdBdB
y
 i  d i  b
a
2 2dtdt
Loop “a-b-d-a”
B
1
1
 o h L d x
ΦB  BAabd  B( R 2 - Lh)

c
R
a
dB
0
dt
2
2
dΦB
R 2 hL dB
 '  -(
- )   bda   ab
dt
2
2 dt
2

R
 dB
dB

2
 bda  
 (-R
)
2 dt
dt 2
2
Direction of emf a→b
 ab   '- acb
hL dB

2 dt
Example
Ei
c
As shown, R, L, h, dB/dt are known,
find emf in the rod.
b
hLhLdBdB
 i  d i  b
a
2 2dtdt
Loop “a-o-b-a”

y
B
 o h L
x
R
a
dB
0
dt
 
Ei  dl  0
Direction of emf a→b
1
ΦB  BAaob  B( Lh)
2
dΦB
hL dB hL

i   -( )
3
dt
2 dt
 oa   ob  0
2
B  2- t
3
emf b→a
dB
2
dt
3
Example
v  v0 
b
dΦB
hL dB hL
 i1  
(b  a)
dt
2 dt
3
 
b 
i 2   (v  B)  ds  -v0 BL  -2v0 L (b  a)
a
B
 o
h L
R
a
Loop “a-o-b-a”
x
2
B  2- t
3
t  0, B  2
hL
 i   i1   i 2   2v0 L (b  a)
3
hL
hL
d( B)
d( )
dΦB
hL dB
2
2
i  Bdt
dt
dt
2 dt
dh BL hL
BL hL
hL
  v0

 v0 L 
dt 2
3
2
3
3
(b  a)
Example
v  v0 
b
 oa  0,  ob  0
B
 o
h L
R
a
x
Exercises
P797 29, 31,
Problems
P799 10, 11, 14