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10 電磁 III
How is an aurora so thin yet
so tall and wide?
Sections
1. 磁場
2. 電流與磁場
10-1 磁場
The electric field
and the magnetic
field
Electromagnets
and permanent
magnets
10-1.1 The definition of B

 
FB  qv  B

 F
E  (N/C) (V  m)
q0
The tracks in a bubble
chamber
The SI unit for B
1 tesla = 1T =1 N/A‧m=104 gauss
108 T
1.5 T
10-2 T
10-4 T
10-10 T
10-14 T
Magnetic Field Lines

Magnetic vs. electric dipoles
A horseshoe and a Cshaped magnets
例 1 A 5.3 MeV proton
v
B = 1.2 mT
2 K / m  3.2  10 m/s
7
FB  qvB sin   6.1  10
15
a  FB / m  3.7  10 m/s
12
N
10-1.2 Crossed Fields: Discovery
of the Electron
A
cathode ray tube
 Thomson’s procedure:
 設定E = 0, B = 0, 並記錄光點位置
 開啟電場
 開啟磁場,並調至與電場相等
Calculation
2
qEL
y
, qE  qvB
2
2mv
2 2
m B L
v  E/B, 
q
2 yE
the charge-to-mass ratio of the electron
1.75881961011 C·kg-1
F QE
ay 

m
m
1
2
y  a y t , L  vxt
2
QEL2
y
2mvx2
10-1.3 Crossed Fields: The
Hall Effect
 By
the conduction electrons in copper:
V  Ed
eE  evd B
J
i
vd 

ne neA
Bi
A
n
(l  )
Vle
d
例 2 A cube generator
d = 1.5 cm, v = 4.0m/s, B = 0.05T
eE  evB
V  Ed
V  dvB
V  3.0mV
10-1.4 A Circulating Charged
Particle
F  ma  mv / r
2
qvB  mv / r
2
r  mv / qB
T  2r / v
 2m / qB
頻率與軌跡
The frequency and angular frequency
1
qB
qB
f  
  2f 
T
2m
m
The magnetic bottle machine
Helical Paths
V∥ and V⊥
v11  v cos  v  v sin 
The pitch (螺距) of the helical path
2m
p  v11T  v cos 
qB
極光橢圓圈
例 3 The Mass Spectrometer
(質譜儀)
1
2
mv  qV
2
v  2qV / m
mv 1

r
qB B
x  2r
2mV
q
質譜儀
x = 1.6254m, V = 1000.0V, B = 80.000mT
2
x  2r 
B
2
2mV
q
2
B qx
m
 203.93u
8V
Isotope Separation
Centrifuge and diffusion chamber
10-1.5 Cyclotrons and
Synchrotrons
(迴旋加速器與同步加速器)
Fermilab: 6.3km ring
Synchrotrons
 The
f  f osc
resonance condition:
 When proton energy > 50Mev:
 Out of resonance (relativistic effect)
 A huge magnet (4×106 m2) is needed for
high energy (500Gev) protons
 The proton sychrotron at Fermilab can
produces 1Tev proton
CERN LHC
The LHC is 27km long and sits 100m below the surface.
10-1.6 Magnetic Force on a
Current-Carrying Wire
Magnetic Force
q  it  iL / vd
FB  qvd B sin   iLB

 
FB  iL  B
For a wire segment:



dFB  idL  B
例 4 A length of wire with a
semicircular arc
Calculation
F1  F3  iLB
dF  iBdL  iB ( Rd  )
F2 


0

dF sin   iBR  sin d
 iBR cos 
0

0
 2iBR
F  F1  F2  F3  2iB ( L  R )
線圈
10-1.7 Torque on A Current
Loop
 F2
and F4 cancel
 F1 and F3 form a force couple
F2  ibB sin( 90   )  ibB cos 

b
b
   (iaB sin  )  (iaB sin  )
2
2
 iabB sin 
  N   ( NiA) B sin 
例 5 A galvanometer for analog
meters
  NiAB sin   
NiAB sin 


 [( 250)(100  10-6 A )( 2.52  10-4 m 2 )
 (0.23T )(sin 90 )] / 28
 5.2  108 M  m / degree
10-1.8 The Magnetic Dipole
 The
magnetic dipole moments
  NiA    B sin 

 
  
     B (cf :   p  E )
 The magnetic potential energy
 
 
U ( )   p  E  U ( )     B
磁能
U  (  B )  (  B )
 2 B
10-2 Magnetic Fields due
to Currents
Conventional rocket
EM Rail Gun
10-2.1 Calculating the Magnetic
Field due to a current
dE 

dE 
1
4 e
0
1
4 e
0
dq
2
r
dq 
3 r
r
 0 ids sin 
dB 
4
r2
 0 id s  r
dB 
4
r3
The law of Biot and Savart
Magnetic Field Due to a Current in a
Long Straight Wire
 0  4 10 7 T  m/A permeabili ty
0 ids sin 
dB 
2
4
r
B


dB

2
dB



 0i

2
0


0
sin ds
2
r
Integration
r
s  R , sin  =
2
R
2
s R
2
2
Rds
 0i 
B
2 3/ 2
2

0
(s  R )
2
0i
s
 0i

] 
[ 2
=
2 1/ 2 0
2R
2R ( s  R )
Magnetic Field Due to a Current in a
Circular Arc of Wire
0 ids sin 90 
0 ids
dB 
=
2
4
R
4 R 2
 
0 iRd 
B   dB  
0 4
R2
 0i 
0i
=
d 

4R 0
4R
例 6 What B does the current
produce?
 
ds  r  0
B1  B2  0
 0i ( / 2)  0i
B3 
=
4R
8R
10-2.2 Two Parallel Currents
0ia
Ba 
2d

 
Fba  ib L  Ba
0 Liaib
Fba  ib LBa sin 90 
2d

例 7 The Field Between Two wires
B ( x)  Ba ( x)  Bb ( x)
=
 0i
2 ( d  x)

 0i
2 ( d  x)

 0id
2 ( d 2  x 2 )
10-2.3 Ampere’s Law
Comparing
Gauss’ lawand Ampere’s law

Ampere’s law
 B  ds 0ienc
The Magnetic Field Outside a Long
Straight Wire with Current
 
 B  ds  B cos ds
 B  ds  B(2r )
B(2r )  0i
The Magnetic Field Inside a Long
Straight Wire with Current
 
 B  ds  B (2r )
ienc
r 2
i
R 2
r 2
B ( 2r )   0i
R 2
例 7 A hollow conducting cylinder
r
ienc   JdA   cr ( 2rdr )
2
a

c( r  a )
4
4
2
B( 2r )  
c( r  a )
4
2
4
10-2.4 Solenoids and Toroids
Magnetic
Field of a Solenoid (螺線管)
Magnetic Field of a Toroid (螺線環)
Magnetic Field of a Solenoid
 
b 
 c 
 B  ds  a B  ds  b B  ds
d 
 a 
  B  ds   B  ds
c
Bh  0inh
d
B  0in
Magnetic Field of a Toroid
0iN
B(2r )   0iN B 
2r
磁圍阻核融合反應器
Tokamak Fusion Test Reactor
10-2.5 A Current Carrying Coil as
a Magnetic Dipole
A
current loop and a bar magnet
Magnetic Field of a Coil
B
 dB cos 
 0ids R

4r 2 r
 0iR

2R
2
2 3/ 2
4 ( R  z )
 0iR 2

2( R 2  z 2 ) 3 / 2
 0i

(at center of coil)
2R
Coulomb’s Law
•Using Gauss’s law to
take advantage of
special symmetry
situations
•Gaussian surfaces
•高斯面上各點電場與
面內總電荷相關
Gauss’ Law
•Flux  enclosed
charge
e 0
 
 e 0  E  dA
 qenc
Gauss’Law and Coulomb’Law
•From G.L. to C.L.


e 0  E  dA  e 0  EdA  q
e 0 E  dA  q
e 0 E ( 4r )  q
2
1
q
E
2
4e0 r
敬請期待
電磁 IV