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Moving
Electrical
Charge

 
FB  qv  B

 
FB  iL  B

 
dFB  idL  B
Magnetic
Field
Moving
Electrical
Charge
The net torque on
magnetic dipole moment
the loop is not zero.
  iAnˆ  B =  B
The Hall Effect
1 iB
V H 
nq 
1
Hall coefficient
nq
The magnetic force on moving charge
Moving Electrical Charge  B ?
Chapter 33 The Magnetic Field of
a Current
The Electric Field Due to a Charge

r
P

E
q
4 0 r
2
q
1
E  q, 2
r
rˆ
The Magnetic Field Due to a Motion Charge
q

r

P

v
  0 q1v  rˆ  0 qv  r
BB  q, 2 ,2 v, sin  3
4 r r
4 r
  01 qvqsin
rˆ  1 qr
B
(E

)
22
3
4
 0 rr
4 0 r
μ0 is the permeability constant.
 0  4 10 7 T  m/A
0
 10 7 T  m/A
4
The Magnetic Field of
a Current
Due
to a Motion Charge
A straight wire segment

r

P



  0 q1v  rˆ  0 qv  r
BB  q, 2 ,2 v, sin  3
4 r r
4 r
  0 dqv  rˆ  0 dqv  r
dB 

2
3
Biot-Savart law
4

r
4

r

q
v
   0 ids  rˆdq 0 ids  r
ddBqv dq 2  ds  ids3
4 drt dt 4 r
direction :  dB   0 idzr sin  00 idzd
iddz

3
44 ( zr2 3 d 2 ) 3 2
4
r
 0i
L  d , B 
L
2d

id
2d
i
Lz
2
0 0
B   dB 
44d0( L2( z 2  dd22))3122
4
Two Parallel Current
0i1
B1 
nˆ1
2 r1
0i2
B2 
nˆ2
2 r2
Superposition principle
Vector sum
  
B  B1  B2
The force exerted by one wire on another
 0i1
B1 
2d

 
F21  i2 L  B1
 0i1i2 L
F21  i2 LB1 
2d i1  i2
2dF21 The definition of
2
i 
the ampere
0 L
F
•
ib
F
ia
d
•
ib
ia
d
When F  2  10 7 N ,
d  1m, L  1m, i  1A
Parallel currents attract, and
antiparallel currents repel.
Example
0 di
0 i(dx / a)
dB y 

2 d  x 2 d  x
y
 0i a / 2 dx
By  dBy 
2a  a / 2 d  x

d
P
x
 0i d  a / 2

ln
2a d  a / 2
 0 di  0 i(dx / a)

dB 
d
2 d 2
Example
00ii((dx
dx // aa))
cos
dBx  dB cos 
2
22R
Rsec
sec
dR
cos
 R sec
0i0i (dx / a )
Bx  dBx    d 2

22aR sec 

2
x  R tan  dx  R sec d
 0i 1 a
 0i 
tan

Bx 
d 

a
2R
2a  
 0i
 0i
-1 a
a  R, tan
 a / 2 R, B 
R  0,   /2, B 
2R
2R
2a

The Magnetic Field of a Current
A circular current loop
Biot-Savart law
  0 ids  rˆ  0 ids  r
dB 

2
4 r
4 r 3
dBz  dB cos  ,
 dB　 0
 0iR0ids2RR
1
ds
B   dBz  
3

20
4
4r r( R 2  z 2 ) 2
 0iR 2
 0i

;
2( R  z )
2
2
3
when z  0, B 
2
 0 pm
B
2z 3
pe
pe  qd , E 
4 0 x 3
1
2R
 0iR 2
magnetic
dipole
If z  R
, B moment
2z3
 0i (R 2 )
B
2z 3
The Magnetic Field of a Solenoid
The Magnetic Field of a Solenoid
B
 0iR 2
2( R 2  z 2 )
dB 
3
;
2
 0 (nidz ) R 2
2[ R  ( z  d ) ]
2
B
 0 niR 2
2
L/2
 [R
3
2
dz
3
2

(
z

d
)
]
The field outside the ideal solenoid is zero.
 0 ni
L 2d
L 2d

(

)
1
1
2 [R 2  (L 2  d )2 ] 2 [R 2  (L 2  d )2 ] 2
2
- L/2
2
If L>>R,
2
B=μ0ni.
The Magnetic Field of a Solenoid
B
 0iR 2
2( R 2  z 2 )
dB 
3
;
2
 0 (nidz ) R 2
2[ R  ( z  d ) ]
2
B
 0 niR 2
2
L/2
 [R
3
2
dz
3
2

(
z

d
)
]
The field outside the ideal solenoid is zero.
2
- L/2
2
2
B
-L/2
L/2
z
parallel-plate capable
for electric field
solenoid for
magnetic field
E
B
E=s/0.
B  0nI
The Electric Field
Due to a Charge

E
q
rˆ
4 0 r
  q
 E  ds 
2
0
Gauss’ Law
 
 E  dl  0
s
The Magnetic Field Due
to a Moving Charge
  0 qv  rˆ
B
2
4 r 
  0 ids  rˆ
dB 
4
r
2
 
 B  ds  0
 
 B  dl  0i
Ampere’s Law
 
 B  dl  0i
Ampere’s Law
s is an Ampere loop,
i is net current passingthr ough
surface boundedby the loop.
 
 B  dl   B cosdl
L
L
0 I

rd
L 2r
 0 I 2

d

2 0
 0 I
 
 B  dl   B cosdl
L
L
0 I
 
rd
L 2r
 0 I 2

d

2 0
 0 I
dl cos  rd dl  cos   r d
   
B  dl  B  dl 
0 I 
' dl0 I' cos '

Bdl
cos


B
B  d(srd
 )
 0 i (  r ' d  )
2 r
2 r '
s
 0 Ampere’s Law

A straight wire segment
 0id L 2
dz
B   dB 
4  L 2 ( z 2  d 2 ) 3 2
 0i
L

4d ( L2  d 2 ) 12
4
 0i
when L  d , B 
2d
 
 B  ds   Bds  B(2d )  0i
 0i
B 
2d
Application of Ampere’s law
Long, straight wire (r<R)
 
r 2
 B  ds  B(2r )   0i R 2
 0ir
B
2R 2
 
r  R  B  ds  B(2r )   0i
B
 0i
B
2r
R
r
A solenoid
B2
 
 B  ds  0 nhi
loop2
B1
 
 B  ds  
ab

bc

cd

Bh   0 nhi
From loop2,
da
  Bds  Bh
ab
B   0 ni
hB1+ (-h)B2=0
B1= B2
A toroid
 
 B  ds  B(2r )  0iN
 0iN
B 
  0 ni
2r
Same as a solenoid
The field outside a solenoid
Bout 2 r  0i
Bout
0i

2 r
Bin  0 ni
Bout 0i 2 r
1


Bin
0 ni
2rn
Bout
 1, Bout  0
Bin
Exercises
P768-769 10, 13, 15
Problems
P772 8, 9
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?
2017/5/25
[email protected]
25
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
＋ ＋＋ ＋ ＋ ＋
2017/5/25
L+' = L 1- (vd / c)2
l+¢ - l-¢
x’
－ － －
m0qvd I ¢
2p r
[email protected]
Therefore
FE =
26
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 
2
1
1 
2
( E y  vBz )
( E z  vBy )
1
v
By 
( By  2 Ez )
2
c
1 
1
v
Bz 
( Bz  2 E y )
2
c
1 