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Chapter 10
Magnetic Field of a Steady Current
in Vacuum
§ 10-1 Magnetic Phenomena Ampere’s Hypothesis
§ 10-2 Magnetic Field Gauss’law in Magnetic Field
§10-3 Boit-Savart Law & Its Application
§ 10-4 Ampere’s Law & Its Application
§ 10-5 Motion of Charged Particles in Magnetic
§ 10-6 Magnetic Force on Current-carrying
Conductors
§ 10-7 The Hall Effect
§ 10-7 Magnetic Torque on a Current Loop
§ 10-1 Magnetic Phenomena
Ampere’s Hypothesis
1. Magnetic Phenomena
(1) the earliest magnetic phenomena that human
knew: the permanent magnet (Fe3O4) has N ,
S poles. Same poles repel each other and
different poles attract each other.
N pole
N pole
S pole
S pole
Magnetic monopole？
Never be seen!
(2) The magnetic field surrounding the earth
(3)The interaction between current and magnet
N
S
I
A
N
N
I
S
B
S

F
N
I
S
attraction
repellen
t
The motion of
electron in M-field
2. Ampere’s Hypothesis
Each molecule of the matter can be equated
with a closed current –called molecular current.
When the molecular currents arrange in same
direction, the matter appears magnetism in a
macroscopic size.
S
N
N
All Magnetic phenomena result from the
motion of the charge.
S
3. Magnetic field
magne
t
curren
t
Moving
charge
Mfield
Mfield
magnet
current
Moving
charge
§ 10-2 Magnetic Field
Gauss’law in Magnetic field
1. Magnetic field

Take a moving charge( v and q) as a test charge.
The characters of the force on the moving
charge by the magnetic field:
The direction of M-field at this point
each point in the M-field has a special direction.
when the q moves along this direction( or
opposite the direction), no force acts on it.
 The direction of the M-force acting on q always

perpendicular to v and M-field direction.
 M-force
depends
on
q,v
and
angle

between
 and M-field direction.
v
F  qv sin
Definition B 
F
--the magnitude
qv sin of M-field
F
or B  max tesla(T)
qv



B along the direction of Fmax  v
Superposition principle of M-field


B   Bi
i

2. Magnetic field line ( B line)

 tangential direction of B line--M-field direction.

 the magnitude of B
dN
B
dS


B -line is different with B line：

B -lines are always closed lines linked with
electric current. They have neither origin nor
termination.
3. Magnetic flux and Gauss’ Law in magnetics

M-flux：The number of B-lines through a
given surface.
 
d B  BdS  B  dS
 
 B   S B  dS


n
dS
unit：weber(Wb)T·m2
For any closed surface S ,
S
 
SB  dS  0 ---- Gauss’ Law in magnetics
----M-field is non-source field

B
§10-3
Calculation of the magnetic field set
up by a current
1. Boit-Savart Law

Idl--current element
The magnetic field

set up by Idl at
point P is

dB



P r
Idl
I
 
  0 Idl  r
dB 
--B-S Law
3
4 r
0  4  10 N A --permeability of
7
2
For any long current,


B   dB
l
 
 0 Idl  r

3

l
4
r

dB
 

P r
Idl
I
-- superposition principle of M-field
2. Application of B-S Law
[Example 1]Calculate the M-field of a straight
wire segment carrying a current I.
2
solution

Choose any Idl


Idl set up dB at P :
 
  0 Idl  r
dB 
3
4 r
I

l

r
a
1
direction：


all dB set up by all Idl have same direction
P
 0 I dl sin
 B   dB 
2

L
L
4
r
a
r
l  a  ctg
sin
a
dl  2 d
sin 
0 I 
sin

d

B 

4a 
2
1
0 I

(cos  1  cos  2 )
4a
2
I

l

r
a
1
P
discussion
 for infinite long current
1  0
0 I
B
(cos  1  cos  2 )
4a
2
2  
0 I
B 
2a
I

l
a
 semi-infinite current
0 I
B
4a
 on the prolong line of current
B0

r
1
P
[Example 2] Calculate the M-field on the axis of
a circle with radius R and carrying current I.

choose any Idl
Solution：

Idl
I
R
0

r
x

dB
P
 0 Idl
 dB  4 r 2
dB  

 dB  dB  dB//
dB//
x

 dB  0

IR
0
 B   dB//   dB sin 
dl
3

L
L
4r


 0 IR
2r
2
3
 0 IR

B
I
2
2( R  x )
2
----

r
2
3
R
0
x

dB dB

dB//

P
2
and I satisfy Right Hand Rule
x
discussion
at the center，x =0
B
 0 IR
2( R  x )
2
 B0 
0 I
2
2
2R
 the magnetic moment of the circular current


2
pm  ISn  IR n

B 

 0 pm
2 ( R  x )
2
2
3
2
3
2
[Example 3] Calculate the M-field on the axis of
a solenoid with radius R. The number of turns
per length of solenoid is n, its carrying current I.
Solution
l dl
Take dl along axis and
its distance to P is l.
1 r
A2
A1
2
The number of
R P
turns on length dl ,
dN  ndl
the current on dl,
dI  IdN
 dB 
 0 InR dl
2
2( R  l )
2
2
3
direction：
2
Rd
dl   2
sin 
l  R ctg 
l
2
R
R l  2
sin 
2
2
 B   dB

0
2
1
A1
P
R
L
2
0
1
2
nI  sin d 
dl
r
2
A2
nI (cos  2  cos  1 )
0
Discussion
B  nI (cos  2  cos  1 )
2
 Solenoid “infinite long”：
1  
2  0
 B   0 nI
l
1
A1
R
dl
r
2
P

B ’s direction: satisfy right-hand rule with I.
-- the M-field on the axis of a solenoid with
infinite length.
A2
[Example 4]A long straight plate of width L
carrying I uniformly. P and plate current are at
same plane. Find B=? of P.
Solution I  dI
Straight line current
I
dI  dx
a
 0dI
 dB 
2x
 0 I dx

2a x
L
P
I
dI
d
x
dx

All dB have same direction.
 B   dB
 0 I L d dx


2a L x
0 I L  d

ln
2a
L
L
P d
x
I
dI
dx
Direction :
[Example 5]A half ring with radius R, uniform

charge Q and angular speed . Find B  ? at O.
Solution
Take an any dl ,

It charge,
dQ  dl
Q

Rd
R
Q
 d

r
dl
R
x
O
When dQ is rotating, it equates with
2dQ Q
dI 
 2 d
T

dB 
R
 0 r dI
2
2( r  x )
2
r  x  R
2
r
dl
dI set up M-field at O,
2
2
2

x
O
3
2
 0 Q
r

d
3
2
2 ( r 2  x 2 ) 2
2
r  R sin

All dB have same direction
 B   dB
r
dl
L

2
0

 0 Q
r
d
3
2
2 ( r 2  x 2 ) 2
2
R
x
O

 0 Q 2 2
 0 Q

sin

d


2

2 R 0
8R
Direction : 
3. M-field set up by moving charge

Take Idl，it set up
 0
dB 
4
0

4
IS
 
Idl  r

3
r
 
( qnSv )dl  r
3
r






  
v
n
dl
The number of moving charges in dl,
dN  nSdl
 
Idl , v same
direction


  0 q  dN  v  r
 dB 

3
4
r


B set up by each moving charge ( q , v ):



 dB  0 qv  r
B

 3
dN 4
r
§10-4 Ampere’s Law
1. Ampere’s Law
 
Question： B  dl  ?

B
L
Special example, infinite
straight line current I
0 I
B
2 r
I
 Choose a circle L is just
along B-line.
 
 B  dl  Bdl  B dl
L

L


B
I
L
 B  2 r   0 I
L
L
--does not depend on r
choose is any closed line L surrounding I and
in the plane perpendicular to I
 
B

d
l


I
0

L
Any L surrounding I
 

 
 B  dl   B  (dl//  dl )
L
L
  Bdl
L
 0 I

dl//

dl

dl 

dl
 L does not surround I
 
0 I
B2  dl2  B2dl2 cos 2  B2r2d 
d
2
 
B1  dl1  B1dl1 cos1   B1r1d

B1
0 I

d
2
 
  B  dl
I
L

L1
 
 
B  dl1   B  dl2  0
L
2

dl1
r1
d
L1
a

B2 dl2
r2
L2
b L
 
Conclusion B  dl   0  I --Ampere’s Law
L
Notes：

Amperian loop
 I is the algebraic sum of all currents closed
by L.
 I >0 when it satisfy right-hand rule with L.
otherwise, I <0.
 
 I has not contribution to B  dl if it is outside L

l
Set up by all I (inside or outside L)

 B is non-conservative field.
2. Application of Ampere’s Law
[Example1]A long straight
wire with R,uniform I. Find
B=? inside and outside it..
Solution
Analyze the symmetry of B
--Axis symmetry
I
R

B
r
L
Take L to be a circle with r, same direction with B,
r>R：
L

 

B
dr

B
2

r
LB  dl   Bdl
L
B  2 r  0 I
R
0 I
B 
2 r
L
r<R:the current closed by L,
I
2
 I  R 2  r
 B  2r   0  I

 0 Ir
R
2
2
 0 Ir
B
2
2R
B
0
R
r
[Example 2] Find the M-field of a infinite
solenoid. The number of turns per length of it is
n, its carrying current I.

Solution：analyze the distribution of B
Choose a rectangular loop abcda
 
LB  dl  Bbc  Bda
0
a
b
d
c
 B  B   0 nI
--uniform field
exterior： B  0
R
[Example 3] A  straight cylinder conductor with R. A
hole with radius a is far b from the central axis of
cylinder. The conductor has current I，Find B=? at
point P.
Solution
Assume a current I
in conductor 
Current density：
I
j
2
2
 (R  a )
Compensatory
R
a
O
b
method ：


Imagine there are j and  j in the hole.
P

The B set by the
conductor with a
hole = the B1 set up
by one without hole
+ the B2 set up by
the hole’s negative
current

B1
R
a
O
b
0
2
 B1 
  (a  b) j
2 ( a  b)
0 (a  b)

j
2
Direction :see Fig.
P

B1
For hole’s -j :
0
2
B2 
 a j
2a
 0a

j
2
R
O
P
a

b B2
Direction: see Fig.
 0bI


a  b

a
0
0
 BP  B1  B2 
j
j
2
2
2

(
R

a
)
2
2
Direction: 
[Example 4] A  conductor flat carries current
The current density is j per unit length along the
direction of perpendicular to j. Find the
distribution of B outside the flat.

j

B dl1
dl2

dB1
P

B

 dB
dB2
Solution
Take a rectangle path abcda
 
   
  B  dl   B  dl   B  dl  2 Bl   0 jl
bc
L
B 
0 j
2
At the two side of the
flat, M-field has same
magnitude and
opposite direction.
da

B
a
b
d
l
c 
B
§10-5 Motion of Charged Particles in M-field
1. Lorentz force
--Magnetic force acting on
the moving charge.

 
Fm  qv  B

v

Fm

 
v// B

v
Notes
 
 Fm v

 Fm does not do work to q.

--Change v ’s direction,

don’t change v’s magnitude.

there are E-field +M-field in the space，
a moving charge q sustains:
  
  
F  Fe  Fm  q( E  v  B)
2. Moving charge in uniform M-field

Let q goes into M-field

with initial velocity v
 
 v // B：
q

--straight line motion with uniform velocity.

B

  v 
 
 v  B:
--Circle motion with
uniform speed
 in the
plane of B
 F  qvB  m v R
  O  
 R  
   
2
mv
R 
qB
2R 2 m
period T 

qB
v
Application: mass spectrometer （质谱仪）
A charged particle
from S is speeded
U
up by U
1
2
qU  mv (1)
2
S2
 Enter M-field
2R
B
mv
R
(2)
qB
Combine (1) and (2)
q
2U
 2 2
m B R
Application: cyclotron （回旋加速器）
2 m
T
qB
do not depend
on R
E: speed up q
B: change the
velocity
direction of q


 v and B with any angle

v//  v cos  ---- // B uniform, straight line

v  v sin ---- B uniform, circle

v

v

v

 v //

v

v//

B
Moving
path---helix
mv 
Revolving radius R 
qB
 helical distance
B
h
2 mv //
h  v//T 
qB
Application: magnetic focusing （磁聚焦）
B
A·
· A
h
The particles have same v//
same h
They focus on point A again
3. Moving charge in non-uniform M-field
mv

As R 
qB
2 mv //
h
qB
R, h are different when B is not constant.
plasma
Magnetic restraint
---Magnetic bottle
M-field of the earth
Van Allen radiation belts
beautiful aurora
§10-6
Magnetic force on a currentcarrying conductor

B
1. Ampere’s Law

Idl is in M-field
The force acting on
each electron is



IS   
  
 n
v

 
Fm   ev  B

The numbers of electron in Idl is
dN  nSdl
The resultant force acting on the dN electrons is
 
 
 
dF  Fm dN  ( ev  B)nSdl  enSvdl  B


 vdl  vdl
I
 

 dF  Idl  B --Ampere’s Law of M-force

for any shape of current-carrying wire,
 


F   dF   Idl  B
L
L
2. The application of Ampere’s Law
[Example 1] A straight wire with length
 L
carrying I is in a uniform B . Find F =?
Take

Idl
force dF  BIdl sin
direction：


I

B
Idl
L
 The M-force acting on L is
F   dF   BI sin dl  BIL sin 
L
l
0
[example 2] A curved wire segment with I is in
the plane which
  B . Suppose AB=L is
known.Find F =?
Set up a coordinate system,
 


take any Idl
dF  Idl  B
dFx  dF cos 
 dF sin
 IBdl sin
  IBdy

 I
 
y
A


dFy dF
  dF
x
Idl 
 
B x
Similar, dFy  dF sin  IBdx
 Fx   dFx   IB
l

yB
yA
dy  0
Fy   dFy  IB  dx  IBL
l
xA


  
Vector express: F  IBLj
xB
Same as the straight
wire from A to B.
 I
 
A

 
 
B
Conclusion

 in a uniform B , the M-force acting on any
shape wire = the M-force acting on the
equivalent straight wire .

 for a closed wire, F=0 in uniform B

[Example 3] I1  I2 . AB=L . Find F =? acting on
AB.
L  dl
the force acting on dl is

dF
I1
A
d
x
I2
L
dF  IBdl
 0 I1
B
2x
F   dF   I 2 Bdl
B
L
 0 I1 I 2

2

dL
d
dx  0 I 1 I 2 d  L

ln
x
2
d
3. The interaction between two parallel currents

1 set up B1at 2 ,
d
 0 I1
B1 
2d

dF21
The M-force
acting

on I 2 dl ,
 

dF21  I 2dl  B1
I1
I2

I 2 dl

B1

I1 I 2
0
Magnitude dF21  I 2 B1dl 
dl
2d
direction：21

I
I
0
1
2
Similarly F12 
dl
2

d
,

dF21 , dF12 have opposite
d
 
B2 dF21 
I 2 dl


B1
dF12
direction.
The force for per unit
length wire,
dF  0 I1 I 2

dl
2d
I1
I2
§10-7 The Hall effect
1. The Hall effect
--there is an electric
 potential difference on the
direction of  B when a current-carrying plate
is put in M-field.
Experiment result,

B
d
b
1
I
VH   H
IB
d
V H：Hall coefficient.
2
Depends on the material.
2. Theoretical explanation

 Let the velocity of free electrons is v ,
number density is n

 
Fm   ev  B
In equilibrium state,

 
ev  B  eEH  0

 
 EH  v  B


Fe  eEH

B
1
    
Fm I

E Hv 

F
e  
   
2
V
Hall E-P-difference,
VH  U1  U 2  
2
1
 2   

EH  dl  1 ( v  B)  dl
2
   vBdl  vBh

B 1
1
 I  nevbh
1 IB
VH  
ne b
1
And  H  
ne

EH
I
V
2

For moving positive charges,
VH  U1  U 2
  
  ( v  B)  dl
2
1

B 1
    


FL 

I
v
EH

Fe
  
2
2
  vBdl  vBh
1
1 IB
VH 
nq b
1
H 
nq
V
Notes：
n has large magnitude in conductors
(~1029/m3). The Hall effect is not obvious.
The Hall effect is obvious in semiconductor
 n type semiconductor：electron conduction.
 p type semiconductor：hole conduction.
Positive charge
to measure H(or VH) can judge the moving
charges and find n.
§10-8 Magnetic torque on a current loop
1. The torque acting on a loop by M-field

The normal direction of loop n ：
Satisfy right hand rule with I

Fad
a
l1
l2
I
b

d B

 I
n
c

Fbc
Fad  Il 2 B sin(    )
 Il 2 B sin
Fbc  Il 2 B sin
Same magnitude, opposite
direction, locate on a line.


 Fad  Fbc  0

Fcd

B
d (c)

Fab  Fcd  Il1 B
Do not locate a line.

 Set up a torque
l2
l2
M  Fab sin  Fcd sin
2
2
 Il1l2 B sin
 ISB sin
direction：
n
a(b)
Fab
For the loop with N turns, M  NISB sin
Vector express:
Define

 
M  NISn  B


pm  NISn
--M-moment of a current
loop
 

 M  pm  B

Fcd

B
d (c)


a (b)

Fab
--can be used for any shape plane loop in
uniform M-field
n
Discussion

   : M  M max pm B
2
 =0： M=0 --stable equilibrium position.
 = : M =0 -- unstable equilibrium position
When suffers disruption, it turns  =0
 The resultant force acting on loop=0 in uniform
M-field. But the torque 0
--only rotation, not translation
 In non-uniform M-field, M0, F 0.
--rotation and translation
2. Potential energy of current loop


A current loop has pm  ISn
It suffers：M  pm B sin

M ’s direction:


 n and B


n

B
I
M makes  decreasing
Increase  1 to  2，external force does work:
2
2
1
1
W   Md    pm B sind
 pm Bcos1  cos  2 
= The increment
of
potential
energy
of
the

loop in B
 Wm 2  Wm1


pm is put in B, the
A loop with M-moment
potential energy of the system ( loop + M-field) is
 
Wm   pm B cos   pm  B