Chapter 26 Sources of Magnetic Field
Chapter 26
Sources of Magnetic Field
1,Biot-Savart’s Law
2,Magnetic Flux and Gauss’ Law
3,Ampere’s Law
4,Magnetic Materials
Chapter 26 Sources of Magnetic Field
New words and Expressions
Biot-Savart’s Law 毕奥 —萨伐尔定律
solenoid 螺线管
toroid 螺绕环
Ampere’s Loop Law 安培环路定理
paramagnetism 顺磁质
diamagnetism 抗磁质
ferromagnetism 铁磁质
magnetization 磁化强度
isotropic medium 各向同性介质
permeability of magnetic materials 磁导率
hysteresis 磁滞
magnetization Curve 磁化曲线
hysteresis Loop 磁滞回线
Chapter 26 Sources of Magnetic Field
26-6&26-5,Biot-Savart’s Law & Applications
(P613-616)
1,Biot-Savart’s Law毕奥 —萨伐尔定律,
(电流元在空间产生的磁场
The current element Idl
generates a magnetic
field dB given by,I
P *
lI?d
B?d
r?
lI?d
r?
B?d
2
0 s i nd
π4
d
r
lIB
3
0 d
π4
d
r
rlIB?
Chapter 26 Sources of Magnetic Field
真空磁导率 27
0 AN10π4
Magnetic permeability constant in a vacuum
The principle of superposition of magnetic field,
can be applied for a complete circuit:
3
0 d
π4
d
r
rlIBB?
任意载流导线在点 P 处的磁感强度磁感强度叠加原理
Chapter 26 Sources of Magnetic Field
1) A Long Straight Wire (载流长直导线的磁场 P614):
For the field near a
long straight wire
carrying a current I,
Determine the
magnetic field.
2,The Application of Biot-Savart Law:
yx
z
I
P
C
D
o 0r *
B?dr?z
zd
Solution:
方向均沿
x 轴的负方向
B?d
Chapter 26 Sources of Magnetic Field
yx
z
I
P
C
D
o 0r *
B?d
2
0 s i nd
π4
d
r
zIB
CD rzIBB 20 s i ndπ4d
s i n/,c o t 00 rrrz
20 s i n/dd rz?
1?
r?
2?
2
1
ds i n
π4 0
0?
r
IB
z
zd
The magnitude of the differential magnetic field
produced at P by the current-length element Idl
located a distance r0 from P is given by:
Chapter 26 Sources of Magnetic Field
）（ 21
0
0 c o sc o s
π4
r
I
的方向沿 x 轴的负方向,B?
2
1
ds i n
π4 0
0?
r
IB
1?
2?
P
C
D
yx
z
o
I
B?
+
Discussions:
(1) The magnetic field
from all current element
of straight wire are in
the same direction and
into the screen.
Chapter 26 Sources of Magnetic Field
π
0
2
1
0
0
π2 r
IB
）（ 21
0
0 co sco s
π4
r
IB
(2) For straight and infinite wire (无限长载流直导线 ),?1=0,?2=?,then
B?
I
B?
I
—— The magnitude of magnetic
field of infinite straight wire at a
point with distance R is in inverse
proportion to the distance.
Chapter 26 Sources of Magnetic Field
I
B
r
IB
π2
0
电流与磁感强度成 右螺旋关系半无限长 载流长直导线的磁场
r
I
B P
π4
0
无限长载流长直导线的磁场
r * P
I
o
π
2
π
2
1
I
BX
Chapter 26 Sources of Magnetic Field2) Current Loop
p615
I
x
真空中,半径为 R 的载流导线,通有电流 I,称 圆电流,求 其 轴线上一点 p 的磁感强度的方向和大小,
p
R
o *
Determine for point on the axis of a circular
loop of wire of radius R carrying a current I.
B?
Chapter 26 Sources of Magnetic Field
I
x
2
0 d
π4
d
r
lIB
r B
d
B?
B?
lI?d
p
R
o *
解 根据对称性分析
s ind BBB x
Chapter 26 Sources of Magnetic Field
x x
R
p*
2
0 dc o s
π4
d
r
lIB
x
l r lIB 20 dc o sπ4
222
c o s
xRr
r
R
R
l
r
IRB π2
03
0 d
π4
2
322
2
0
2 ）（ Rx
IR
B
2
0 d
π4
d
r
lIB
o
B?d
r
lI?d
Chapter 26 Sources of Magnetic Field
Discussions:
(1) At center O
(2) The direction of is along the
axis direction (right-hand rule) as
right figure and Fig.26-18 in P615.
B?
I
R
o
R
IB
2
0
0
x
0B
R
IB
2
00?x
Chapter 26 Sources of Magnetic Field
0// 21 BBrsid
Solution:
R
i
R
iB oo
84
)2/(
3
(direct into the screen)
R
iBBBB o
8321
(into the screen)
EXAMPLE 26-11 of P616:
One quarter of a circular loop of
wire carries a current I,and two
straight sections whose extensions
intersect the center C of the arc,
Find at point C?B?
1
3
2
For parts 1 and 2,
Chapter 26 Sources of Magnetic Field
3 A Stretched-out Solenoid
(载流直螺线管 轴线上 的磁场 )
II
如图所示，有一长为 l,半径为 R的载流密绕直螺线管，螺线管的总匝数为 N，通有电流 I,设把螺线管放在真空中，求管内轴线上一点处的磁感强度,
Chapter 26 Sources of Magnetic Field
The number of turns per unit length of the
solenoid(单位长匝数 ) n,As vertical cross section
(纵剖面 ) shown,(I,n,R,L,?1,?2)
+ + + +++ + + + + + +
p
R
+ +
*o
x
xd
x
Chapter 26 Sources of Magnetic Field
2/322
2
0
2 ）（ Rx
IRB
解 由圆形电流磁场公式
o p
1x
x
2x
+ + + +++ + + + + + ++ + +
2/322
2
0 d
2
d
xR
xInRB
c o tRx?
2222 c s cRxR
2
1
2/322
2
0 d
2
d
x
x xR
xRnIBB?
dc s cd 2Rx
2?
1?
Chapter 26 Sources of Magnetic Field
21 ds i n20 nI 2
1 dcs c
dcs c
2 33
23
0?
R
RnIB
120 c o sc o s
2
nIB
For a ―infinitely long‖ solenoid,The at the
axial line:
B?
0,21
nIB 0
Chapter 26 Sources of Magnetic Field
Summary of the Application of Biot-Savart Law:
A Long Straight Wire
）（ 21
0
0 co sco s
π4
r
IB
0
0
π2 r
IB
A Circular Arc of Wire
2
322
2
0
2 ）（ Rx
IR
B
A Stretched-out Solenoid
120 c o sc o s
2
nIB nIB 0
R
IB
2
0
Chapter 26 Sources of Magnetic Field
26-4&26-5,Ampere’s Loop Law p607-613
,安 培 分 子 环 流 假 说,,The source of
magnetism of all natural-matter come from
current（ 一切磁性的根源是电流 ）,
1,Ampere’s Loop Law 安培环路定理 (P607-611):
Any line integral of around a closed path was
proportional to the current encircled by the path,
B?
即在真空的稳恒磁场中，磁感应强度 沿任一闭合路径的积分的值，等于 乘以该闭合路径所包围的各电流的代数和,
B?
0?
Chapter 26 Sources of Magnetic Field
n
i
iIlB
1
0d?
电流 正负 的规定,与 成 右 螺旋时，
为 正 ； 反 之为 负,I
I LI
注意
Chapter 26 Sources of Magnetic Field
B Contributed by all current in space
Freely choose a closed loop
and assign a positive direction
L
Explanation:
en ci The net current encircled by the loop (与 L
套连的电流,回路所围面积截得 )accord with
right-hand rule (与 L绕行方向成右螺电流取正 )
ld Any differential elements on the L,
along the tangent direction to the loop
L
ld?1
I
2I
B?
Chapter 26 Sources of Magnetic Field
）（ 210 II
3I
2I
1I
L
1I
1I
)(d 21110 IIIIlB
L
问 1） 是否与回路 外电流有关？LB?
2） 若,是否回路 上各处？
是否回路 内无电流穿过？
0?B?L
0d lB
L
L
Chapter 26 Sources of Magnetic Field
It plays very important role in the theory of
magnetic field,One can also use Ampere’s law
to calculate with symmetrical distribution of
current.
B?
2,The Application of Ampere’s Law
Main Steps,P611
1) Symmetry analyses (由 I的分布分析分布的对称性 );
2) Choose a loop (使回路上各处 B相等，方向特殊，从而可从回路积分中提出 B)；
3) Put into the law and calculate B.
Chapter 26 Sources of Magnetic Field
Example,Find the inside and
outside a long straight wire of radius
R and uniformly distributed current I
over a cross section of the wire (无限长载流圆柱体的磁场 ).
B?
R
I
The magnetic field must be the
same at all points that are the
same distance from the center of
the conductor.
I B?d
Id
.
B?
1) Cylindrical symmetry:
Solution:
Chapter 26 Sources of Magnetic Field
R
I
2） Choose a circular loop (选取回路 )
Rr?
IrB 0π2
r
IB
π2
0
I
R
rlBRr
l 2
2
0 π
πd0
I
R
rrB
2
2
0π2
2
0
π2 R
IrB
IlBl 0d
R
L
r
B?
的方向与 成右螺旋B? I
Chapter 26 Sources of Magnetic Field
,0 Rr
,Rr?
2
0
π2 R
IrB
r
IB
π2
0
R
I
R
I
π2
0?
B
Ro r
The field is zero at the center of the conductor
and increases linearly with r until r=R; beyond
r=R,B decreases as 1/r.
Chapter 26 Sources of Magnetic Field
0?B
例 无限长载流圆柱面的磁场
r
IB
π2
0IlB
l 0d
,Rr?
,0 Rr
0dl lB
RI
1L
r
2L
r
B
Ro r
R
I
π2
0?
解
Chapter 26 Sources of Magnetic Field
Read and discuss the example 26-5 and 26-6 on
page 610
Coaxial cable and a nice use for Ampere’s Law
Chapter 26 Sources of Magnetic Field
求长直密绕螺线管内磁场 P611
解 1 ) 对称性分析螺旋管内为均匀场,方向沿轴向,外 部磁感强度趋于零,即,0?B
Find inside a ―infinitely long‖ solenoid (The
number of turns per unit length of the solenoid n
and total number of coils N).
B?
Chapter 26 Sources of Magnetic Field
PMOPNOMNl lBlBlBlBlB ddddd
IMNnMNB 0 nIB 0
无限长载流螺线管内部磁场处处相等,外部磁场为零,
2 ) 选回路,L
+++ +++ ++++++
B?磁场 的方向与电流 成 右螺旋,
B?
I
L
M N
P O
Based on the feature of the distribution of,
we choose a rectangular loop MNOP,
B?
Chapter 26 Sources of Magnetic Field
R1R2
Example,Magnetic field of a
toroid (载流螺绕环内的磁场 ),
given i,N,R1,R2,
Solution:
From symmetry,the lines of
form concentric circles inside
the toroid,directed as figure.
B?
对称性分析；环内 线为同心圆，环外 为零,B?
B?
Chapter 26 Sources of Magnetic Field
Choose concentric (同心 ) circle loop L,(R1 < r < R2 ),
(Right-hand rule direction)
NIRBlBl 0π2d
r
NIB
π2
0
d
R
LNIB 0
当 时，螺绕环内可视为均匀场,dR2
RL π2?令
Chapter 26 Sources of Magnetic Field
Discussion:
2) In contrast to the situation for a solenoid,B is
not constant over cross section of a toroid,But if
its cross area is relatively small,then
It is as same as the expression of the one of long
solenoid.
(2) of toroid is limited in its interior.BB o u t,0
1) The magnetic field is not uniform within the
toroid,it is largest along the inner edge and
smallest at the outer edge.
nILNIB 00
Chapter 26 Sources of Magnetic Field
Two long parallel wires carrying currents exert
forces (平行载流导线 )on each other as the figure.
26-2 Force Between Two Parallel Wires (P605-606)
1I 2I
d
d
IB
π2
10
1
d
IB
π2
20
2
s i ndd 2212 lIBF?
1s i n,90
1B
2B
2dF
22dlI
11dlI
1dF
d
lIIlIBF
π2
ddd 2210
2212
Chapter 26 Sources of Magnetic Field
1I 2I
d
1B
2B
2dF
22dlI
11dlI
1dF
d
lIIlIBF
π2
ddd 1120
1121
d
II
l
F
l
F
π2d
d
d
d 210
1
1
2
2
Chapter 26 Sources of Magnetic Field
Think about how about the force between two
antiparallel wire 平行反向载流导线之间的力怎样
(方向,大小 )? —— repel each other!
Parallel currents attract,and antiparallel
currents repel.
d
II
l
F
l
F ba
b
ba
a
ab
2d
d
d
d 0
Acting force exerted on the AB wire in unit-length:
It is the same for both wires of AB and CD.
Chapter 26 Sources of Magnetic Field
25-2,Magnetic Field lines and Flux (P581)
No origin,no termination and no
cross point(无头无尾不相交,闭合曲线 );
Wrap around with current与电流套连;? Right-hand rule with current
(与电流成右手螺旋关系 )
1,The features of magnetic field lines:
规定,曲线上每一点的 切线方向 就是该点的磁感强度 B 的方向,曲线的 疏密程度 表示该点的磁感强度
B 的大小,
Chapter 26 Sources of Magnetic Field
I
S
N
I
S N
I
Chapter 26 Sources of Magnetic Field
2,Magnetic flux & Gauss’ law:
B
dS
d B
Let –number of magnetic field line threading
area dS,B
Φd
B?S?
磁场中某点处垂直 矢量的单位面积上通过的磁感线数目等于该点 的数值,
B?
B?
Chapter 26 Sources of Magnetic Field
BSBSΦ?c o s
SeBSBΦ n
c o sdd SBΦ?
s d SBΦ
单位 2m1T1Wb1
SBΦ dd
B?
s
S?d
B
s?
B?
s
B?
ne
Chapter 26 Sources of Magnetic Field
(Gauss’ law of magnetic field)
The magnetic flux threading a closed surface is zero.
s d SBΦ
B?
S
0dd 111 SBΦ
0dd 222 SBΦ
0dc os SB
S
1dS
1? 1B?
2dS
2?
2B
磁场高斯定理物理意义,通过任意闭合曲面的磁通量必等于零
（ 故磁场是 无源的,）
Chapter 26 Sources of Magnetic Field
1d
2d
lI
xo
x
IB
π2
0 SB
//
xl
x
ISBΦ d
π2
dd 0
2
1
d
π2
d 0 ddS
x
xIlSBΦ
1
20 ln
π2 d
dIlΦ
例 如图载流长直导线的电流为,试求通过矩形面积的磁通量,
I
解 先求,对变磁场给出 后积分求,Φd Φ
B?
B?
Chapter 26 Sources of Magnetic Field
1,Classification of Magnetic Materials,P620
0B
B
r
When the interior of a solenoid is filled with some
material of relative permeability?r?相对磁导率 ),
26-10,Interaction between Magnetic Field and
Dielectrics Magnetic Materials:
'
0 BBB
介质磁化后的附加磁感强度真空中的磁感强度磁介质中的总磁感强度
Chapter 26 Sources of Magnetic Field
顺磁质
0BB
0BB
抗 磁质
0BB
铁 磁质 （铁、钴、镍等）
（铝、氧、锰等）
（铜、铋、氢等）
弱磁质
1r?
Ferromagnetism (铁磁质 )
Paramagnetism (顺磁质 )
1?r?
Diamagnetism (抗磁质 )
1?r?
Chapter 26 Sources of Magnetic Field
2,Magnetization(磁化 )of Magnetic Materials:
When the interior of a solenoid is fully filled with
some isotropic material?r,it will be magnetized (长直螺线管内部充满均匀的各向同性介质，将被均匀磁化 ),
磁介质是由大量分子或原子组成 分子内电子绕核旋转分子电流 i
i
分子磁矩?
Chapter 26 Sources of Magnetic Field
0?
无外场 Bo时，分子的磁矩排列杂乱无章，
有外场 Bo 时，分子磁矩沿外场转向，
分子磁矩的矢量和
0?
Bo
介质内分子磁矩的矢量和
Chapter 26 Sources of Magnetic Field
Cross area of
solenoid
Taking the paramagnetic material as an example,
Each molecular in it is
equivalent to a loop current,
I?
Chapter 26 Sources of Magnetic Field
'0 BBB
无外磁场顺磁质的磁化分子圆电流和磁矩
m?
I
0B
有外磁场
sI
Chapter 26 Sources of Magnetic Field
To express the extent to which a given medium is
magnetized,a vector quantity,magnetization
磁化强度,is introduced:
M?
The unit of is A/m.M?
V
M i
i
v?
0
lim
It can be proved,
L
ldMI
The net bound current enclosed in a closed loop
L equals to the circulation of magnetization along
the closed loop.
BM
r
r
0
1 and
意义 磁介质中单位体积内分子的合磁矩,
Chapter 26 Sources of Magnetic Field
3,Circulation Theorem with Magnetic Materials
L L ldMIldB
e n c0
0?
L IldMB e n c0
0
)(
MBH
0?
Let,
)( e n ce n c00 IIldBL
磁场强度
MBH
0?
where? is the permeability of magnetic materials.
Chapter 26 Sources of Magnetic Field
then,
L IldH e n c0
The circulation of along any closed loop equals
to the algebraic sum of free current encircled by
the closed loop L —— Theorem of circulation,
H?
H?
磁介质 中的 安培环路 定理
IlHl d
各向同性磁介质 For isotropic medium,
HHB r0
Chapter 26 Sources of Magnetic Field
Explanation:
(1) If the magnetic material were not present,
then the above equation reduce to the Ampere-
loop law in vacuum.
0?M?
ED o(2) One may compare with,which is another supplementary physical quantity.H?
(3) If with symmetry in distribution of free current,
then
BHI0
Chapter 26 Sources of Magnetic Field
I
r?
r
例 有两个半径分别为 和 的“无限长”同轴圆筒形导体，在它们之间充以相对磁导率为 的磁介质,当两圆筒通有相反方向的电流 时，试 求
（ 1）磁介质中任意点 P 的磁感应强度的大小 ;（ 2）圆柱体外面一点 Q 的磁感强度,
r?
rR
I
解 对称性分析
Rdr IlH
l
d
IdH?π2
d
IH
π2
d
IHB
π2
r0
d
I
R
Chapter 26 Sources of Magnetic Field
0,0π2 HdH
0 HB?
0, Brd
同理可求
Rdr
d
IB
π2
r0
Rd? 0 II
lHld
I
r?
r
I
R
d
Chapter 26 Sources of Magnetic Field
R1R2
Solution:
Example,A toroid is fully filled with magnetic
medium,Find,interior the medium,BH,
L NIrHldH?2
nIrNIH2
As same as the one of long solenoid.
nIHB
Chapter 26 Sources of Magnetic Field
26-9,Magnetic Field in Magnetic Materials;
Hysteresis p618-620
当外磁场由 逐渐减小时，磁感强度 B并不沿起始曲线 0P 减小,而是沿 PQ比较缓慢的减小，
这种 B的变化落后于 H的变化的现象，叫做 磁滞现象,简称 磁滞,
mH?
mB
mH?
P
rB
cH
mH?
'P mB?
H
B
O
磁滞回线
Q
1,Magnetization Curve
(磁化曲线 ) (P618-620),
Chapter 26 Sources of Magnetic Field
2,Hysteresis Loop (磁滞回线 )(P619):
矫顽力
cH
由于磁滞,当磁场强度减小到零（即 ）
时,磁感强度,而是仍有一定的数值,
叫做剩余磁感强度 (剩磁 ).r
B
0?H
0?B
rB
当外磁场由 逐渐减小时，磁感强度 B并不沿起始曲线 0P 减小,而是沿 PQ比较缓慢的减小，
这种 B的变化落后于 H的变化的现象，叫做 磁滞现象,简称 磁滞,
mH?
mB
mH?
P
rB
cH
mH?
'P mB?
H
B
O
磁滞回线
Q
Chapter 26 Sources of Magnetic Field
Heating a magnetic too can cause a loss of
magnetism,Above a certain temperature known
as the Curie temperature (1043 K for iron),a
magnet cannot be made at all.
3,铁磁性材料
H
B
O
软 磁材料
H
B
O
硬 磁材料
H
B
O
矩 磁铁氧体材料实验表明，不同铁磁性物质的磁滞回线形状相差很大,
Chapter 26 Sources of Magnetic Field
Homework,p622—628:
3,14,15,16,27,28,31,32,33,50,61
Chapter 26
Sources of Magnetic Field
1,Biot-Savart’s Law
2,Magnetic Flux and Gauss’ Law
3,Ampere’s Law
4,Magnetic Materials
Chapter 26 Sources of Magnetic Field
New words and Expressions
Biot-Savart’s Law 毕奥 —萨伐尔定律
solenoid 螺线管
toroid 螺绕环
Ampere’s Loop Law 安培环路定理
paramagnetism 顺磁质
diamagnetism 抗磁质
ferromagnetism 铁磁质
magnetization 磁化强度
isotropic medium 各向同性介质
permeability of magnetic materials 磁导率
hysteresis 磁滞
magnetization Curve 磁化曲线
hysteresis Loop 磁滞回线
Chapter 26 Sources of Magnetic Field
26-6&26-5,Biot-Savart’s Law & Applications
(P613-616)
1,Biot-Savart’s Law毕奥 —萨伐尔定律,
(电流元在空间产生的磁场
The current element Idl
generates a magnetic
field dB given by,I
P *
lI?d
B?d
r?
lI?d
r?
B?d
2
0 s i nd
π4
d
r
lIB
3
0 d
π4
d
r
rlIB?
Chapter 26 Sources of Magnetic Field
真空磁导率 27
0 AN10π4
Magnetic permeability constant in a vacuum
The principle of superposition of magnetic field,
can be applied for a complete circuit:
3
0 d
π4
d
r
rlIBB?
任意载流导线在点 P 处的磁感强度磁感强度叠加原理
Chapter 26 Sources of Magnetic Field
1) A Long Straight Wire (载流长直导线的磁场 P614):
For the field near a
long straight wire
carrying a current I,
Determine the
magnetic field.
2,The Application of Biot-Savart Law:
yx
z
I
P
C
D
o 0r *
B?dr?z
zd
Solution:
方向均沿
x 轴的负方向
B?d
Chapter 26 Sources of Magnetic Field
yx
z
I
P
C
D
o 0r *
B?d
2
0 s i nd
π4
d
r
zIB
CD rzIBB 20 s i ndπ4d
s i n/,c o t 00 rrrz
20 s i n/dd rz?
1?
r?
2?
2
1
ds i n
π4 0
0?
r
IB
z
zd
The magnitude of the differential magnetic field
produced at P by the current-length element Idl
located a distance r0 from P is given by:
Chapter 26 Sources of Magnetic Field
）（ 21
0
0 c o sc o s
π4
r
I
的方向沿 x 轴的负方向,B?
2
1
ds i n
π4 0
0?
r
IB
1?
2?
P
C
D
yx
z
o
I
B?
+
Discussions:
(1) The magnetic field
from all current element
of straight wire are in
the same direction and
into the screen.
Chapter 26 Sources of Magnetic Field
π
0
2
1
0
0
π2 r
IB
）（ 21
0
0 co sco s
π4
r
IB
(2) For straight and infinite wire (无限长载流直导线 ),?1=0,?2=?,then
B?
I
B?
I
—— The magnitude of magnetic
field of infinite straight wire at a
point with distance R is in inverse
proportion to the distance.
Chapter 26 Sources of Magnetic Field
I
B
r
IB
π2
0
电流与磁感强度成 右螺旋关系半无限长 载流长直导线的磁场
r
I
B P
π4
0
无限长载流长直导线的磁场
r * P
I
o
π
2
π
2
1
I
BX
Chapter 26 Sources of Magnetic Field2) Current Loop
p615
I
x
真空中,半径为 R 的载流导线,通有电流 I,称 圆电流,求 其 轴线上一点 p 的磁感强度的方向和大小,
p
R
o *
Determine for point on the axis of a circular
loop of wire of radius R carrying a current I.
B?
Chapter 26 Sources of Magnetic Field
I
x
2
0 d
π4
d
r
lIB
r B
d
B?
B?
lI?d
p
R
o *
解 根据对称性分析
s ind BBB x
Chapter 26 Sources of Magnetic Field
x x
R
p*
2
0 dc o s
π4
d
r
lIB
x
l r lIB 20 dc o sπ4
222
c o s
xRr
r
R
R
l
r
IRB π2
03
0 d
π4
2
322
2
0
2 ）（ Rx
IR
B
2
0 d
π4
d
r
lIB
o
B?d
r
lI?d
Chapter 26 Sources of Magnetic Field
Discussions:
(1) At center O
(2) The direction of is along the
axis direction (right-hand rule) as
right figure and Fig.26-18 in P615.
B?
I
R
o
R
IB
2
0
0
x
0B
R
IB
2
00?x
Chapter 26 Sources of Magnetic Field
0// 21 BBrsid
Solution:
R
i
R
iB oo
84
)2/(
3
(direct into the screen)
R
iBBBB o
8321
(into the screen)
EXAMPLE 26-11 of P616:
One quarter of a circular loop of
wire carries a current I,and two
straight sections whose extensions
intersect the center C of the arc,
Find at point C?B?
1
3
2
For parts 1 and 2,
Chapter 26 Sources of Magnetic Field
3 A Stretched-out Solenoid
(载流直螺线管 轴线上 的磁场 )
II
如图所示，有一长为 l,半径为 R的载流密绕直螺线管，螺线管的总匝数为 N，通有电流 I,设把螺线管放在真空中，求管内轴线上一点处的磁感强度,
Chapter 26 Sources of Magnetic Field
The number of turns per unit length of the
solenoid(单位长匝数 ) n,As vertical cross section
(纵剖面 ) shown,(I,n,R,L,?1,?2)
+ + + +++ + + + + + +
p
R
+ +
*o
x
xd
x
Chapter 26 Sources of Magnetic Field
2/322
2
0
2 ）（ Rx
IRB
解 由圆形电流磁场公式
o p
1x
x
2x
+ + + +++ + + + + + ++ + +
2/322
2
0 d
2
d
xR
xInRB
c o tRx?
2222 c s cRxR
2
1
2/322
2
0 d
2
d
x
x xR
xRnIBB?
dc s cd 2Rx
2?
1?
Chapter 26 Sources of Magnetic Field
21 ds i n20 nI 2
1 dcs c
dcs c
2 33
23
0?
R
RnIB
120 c o sc o s
2
nIB
For a ―infinitely long‖ solenoid,The at the
axial line:
B?
0,21
nIB 0
Chapter 26 Sources of Magnetic Field
Summary of the Application of Biot-Savart Law:
A Long Straight Wire
）（ 21
0
0 co sco s
π4
r
IB
0
0
π2 r
IB
A Circular Arc of Wire
2
322
2
0
2 ）（ Rx
IR
B
A Stretched-out Solenoid
120 c o sc o s
2
nIB nIB 0
R
IB
2
0
Chapter 26 Sources of Magnetic Field
26-4&26-5,Ampere’s Loop Law p607-613
,安 培 分 子 环 流 假 说,,The source of
magnetism of all natural-matter come from
current（ 一切磁性的根源是电流 ）,
1,Ampere’s Loop Law 安培环路定理 (P607-611):
Any line integral of around a closed path was
proportional to the current encircled by the path,
B?
即在真空的稳恒磁场中，磁感应强度 沿任一闭合路径的积分的值，等于 乘以该闭合路径所包围的各电流的代数和,
B?
0?
Chapter 26 Sources of Magnetic Field
n
i
iIlB
1
0d?
电流 正负 的规定,与 成 右 螺旋时，
为 正 ； 反 之为 负,I
I LI
注意
Chapter 26 Sources of Magnetic Field
B Contributed by all current in space
Freely choose a closed loop
and assign a positive direction
L
Explanation:
en ci The net current encircled by the loop (与 L
套连的电流,回路所围面积截得 )accord with
right-hand rule (与 L绕行方向成右螺电流取正 )
ld Any differential elements on the L,
along the tangent direction to the loop
L
ld?1
I
2I
B?
Chapter 26 Sources of Magnetic Field
）（ 210 II
3I
2I
1I
L
1I
1I
)(d 21110 IIIIlB
L
问 1） 是否与回路 外电流有关？LB?
2） 若,是否回路 上各处？
是否回路 内无电流穿过？
0?B?L
0d lB
L
L
Chapter 26 Sources of Magnetic Field
It plays very important role in the theory of
magnetic field,One can also use Ampere’s law
to calculate with symmetrical distribution of
current.
B?
2,The Application of Ampere’s Law
Main Steps,P611
1) Symmetry analyses (由 I的分布分析分布的对称性 );
2) Choose a loop (使回路上各处 B相等，方向特殊，从而可从回路积分中提出 B)；
3) Put into the law and calculate B.
Chapter 26 Sources of Magnetic Field
Example,Find the inside and
outside a long straight wire of radius
R and uniformly distributed current I
over a cross section of the wire (无限长载流圆柱体的磁场 ).
B?
R
I
The magnetic field must be the
same at all points that are the
same distance from the center of
the conductor.
I B?d
Id
.
B?
1) Cylindrical symmetry:
Solution:
Chapter 26 Sources of Magnetic Field
R
I
2） Choose a circular loop (选取回路 )
Rr?
IrB 0π2
r
IB
π2
0
I
R
rlBRr
l 2
2
0 π
πd0
I
R
rrB
2
2
0π2
2
0
π2 R
IrB
IlBl 0d
R
L
r
B?
的方向与 成右螺旋B? I
Chapter 26 Sources of Magnetic Field
,0 Rr
,Rr?
2
0
π2 R
IrB
r
IB
π2
0
R
I
R
I
π2
0?
B
Ro r
The field is zero at the center of the conductor
and increases linearly with r until r=R; beyond
r=R,B decreases as 1/r.
Chapter 26 Sources of Magnetic Field
0?B
例 无限长载流圆柱面的磁场
r
IB
π2
0IlB
l 0d
,Rr?
,0 Rr
0dl lB
RI
1L
r
2L
r
B
Ro r
R
I
π2
0?
解
Chapter 26 Sources of Magnetic Field
Read and discuss the example 26-5 and 26-6 on
page 610
Coaxial cable and a nice use for Ampere’s Law
Chapter 26 Sources of Magnetic Field
求长直密绕螺线管内磁场 P611
解 1 ) 对称性分析螺旋管内为均匀场,方向沿轴向,外 部磁感强度趋于零,即,0?B
Find inside a ―infinitely long‖ solenoid (The
number of turns per unit length of the solenoid n
and total number of coils N).
B?
Chapter 26 Sources of Magnetic Field
PMOPNOMNl lBlBlBlBlB ddddd
IMNnMNB 0 nIB 0
无限长载流螺线管内部磁场处处相等,外部磁场为零,
2 ) 选回路,L
+++ +++ ++++++
B?磁场 的方向与电流 成 右螺旋,
B?
I
L
M N
P O
Based on the feature of the distribution of,
we choose a rectangular loop MNOP,
B?
Chapter 26 Sources of Magnetic Field
R1R2
Example,Magnetic field of a
toroid (载流螺绕环内的磁场 ),
given i,N,R1,R2,
Solution:
From symmetry,the lines of
form concentric circles inside
the toroid,directed as figure.
B?
对称性分析；环内 线为同心圆，环外 为零,B?
B?
Chapter 26 Sources of Magnetic Field
Choose concentric (同心 ) circle loop L,(R1 < r < R2 ),
(Right-hand rule direction)
NIRBlBl 0π2d
r
NIB
π2
0
d
R
LNIB 0
当 时，螺绕环内可视为均匀场,dR2
RL π2?令
Chapter 26 Sources of Magnetic Field
Discussion:
2) In contrast to the situation for a solenoid,B is
not constant over cross section of a toroid,But if
its cross area is relatively small,then
It is as same as the expression of the one of long
solenoid.
(2) of toroid is limited in its interior.BB o u t,0
1) The magnetic field is not uniform within the
toroid,it is largest along the inner edge and
smallest at the outer edge.
nILNIB 00
Chapter 26 Sources of Magnetic Field
Two long parallel wires carrying currents exert
forces (平行载流导线 )on each other as the figure.
26-2 Force Between Two Parallel Wires (P605-606)
1I 2I
d
d
IB
π2
10
1
d
IB
π2
20
2
s i ndd 2212 lIBF?
1s i n,90
1B
2B
2dF
22dlI
11dlI
1dF
d
lIIlIBF
π2
ddd 2210
2212
Chapter 26 Sources of Magnetic Field
1I 2I
d
1B
2B
2dF
22dlI
11dlI
1dF
d
lIIlIBF
π2
ddd 1120
1121
d
II
l
F
l
F
π2d
d
d
d 210
1
1
2
2
Chapter 26 Sources of Magnetic Field
Think about how about the force between two
antiparallel wire 平行反向载流导线之间的力怎样
(方向,大小 )? —— repel each other!
Parallel currents attract,and antiparallel
currents repel.
d
II
l
F
l
F ba
b
ba
a
ab
2d
d
d
d 0
Acting force exerted on the AB wire in unit-length:
It is the same for both wires of AB and CD.
Chapter 26 Sources of Magnetic Field
25-2,Magnetic Field lines and Flux (P581)
No origin,no termination and no
cross point(无头无尾不相交,闭合曲线 );
Wrap around with current与电流套连;? Right-hand rule with current
(与电流成右手螺旋关系 )
1,The features of magnetic field lines:
规定,曲线上每一点的 切线方向 就是该点的磁感强度 B 的方向,曲线的 疏密程度 表示该点的磁感强度
B 的大小,
Chapter 26 Sources of Magnetic Field
I
S
N
I
S N
I
Chapter 26 Sources of Magnetic Field
2,Magnetic flux & Gauss’ law:
B
dS
d B
Let –number of magnetic field line threading
area dS,B
Φd
B?S?
磁场中某点处垂直 矢量的单位面积上通过的磁感线数目等于该点 的数值,
B?
B?
Chapter 26 Sources of Magnetic Field
BSBSΦ?c o s
SeBSBΦ n
c o sdd SBΦ?
s d SBΦ
单位 2m1T1Wb1
SBΦ dd
B?
s
S?d
B
s?
B?
s
B?
ne
Chapter 26 Sources of Magnetic Field
(Gauss’ law of magnetic field)
The magnetic flux threading a closed surface is zero.
s d SBΦ
B?
S
0dd 111 SBΦ
0dd 222 SBΦ
0dc os SB
S
1dS
1? 1B?
2dS
2?
2B
磁场高斯定理物理意义,通过任意闭合曲面的磁通量必等于零
（ 故磁场是 无源的,）
Chapter 26 Sources of Magnetic Field
1d
2d
lI
xo
x
IB
π2
0 SB
//
xl
x
ISBΦ d
π2
dd 0
2
1
d
π2
d 0 ddS
x
xIlSBΦ
1
20 ln
π2 d
dIlΦ
例 如图载流长直导线的电流为,试求通过矩形面积的磁通量,
I
解 先求,对变磁场给出 后积分求,Φd Φ
B?
B?
Chapter 26 Sources of Magnetic Field
1,Classification of Magnetic Materials,P620
0B
B
r
When the interior of a solenoid is filled with some
material of relative permeability?r?相对磁导率 ),
26-10,Interaction between Magnetic Field and
Dielectrics Magnetic Materials:
'
0 BBB
介质磁化后的附加磁感强度真空中的磁感强度磁介质中的总磁感强度
Chapter 26 Sources of Magnetic Field
顺磁质
0BB
0BB
抗 磁质
0BB
铁 磁质 （铁、钴、镍等）
（铝、氧、锰等）
（铜、铋、氢等）
弱磁质
1r?
Ferromagnetism (铁磁质 )
Paramagnetism (顺磁质 )
1?r?
Diamagnetism (抗磁质 )
1?r?
Chapter 26 Sources of Magnetic Field
2,Magnetization(磁化 )of Magnetic Materials:
When the interior of a solenoid is fully filled with
some isotropic material?r,it will be magnetized (长直螺线管内部充满均匀的各向同性介质，将被均匀磁化 ),
磁介质是由大量分子或原子组成 分子内电子绕核旋转分子电流 i
i
分子磁矩?
Chapter 26 Sources of Magnetic Field
0?
无外场 Bo时，分子的磁矩排列杂乱无章，
有外场 Bo 时，分子磁矩沿外场转向，
分子磁矩的矢量和
0?
Bo
介质内分子磁矩的矢量和
Chapter 26 Sources of Magnetic Field
Cross area of
solenoid
Taking the paramagnetic material as an example,
Each molecular in it is
equivalent to a loop current,
I?
Chapter 26 Sources of Magnetic Field
'0 BBB
无外磁场顺磁质的磁化分子圆电流和磁矩
m?
I
0B
有外磁场
sI
Chapter 26 Sources of Magnetic Field
To express the extent to which a given medium is
magnetized,a vector quantity,magnetization
磁化强度,is introduced:
M?
The unit of is A/m.M?
V
M i
i
v?
0
lim
It can be proved,
L
ldMI
The net bound current enclosed in a closed loop
L equals to the circulation of magnetization along
the closed loop.
BM
r
r
0
1 and
意义 磁介质中单位体积内分子的合磁矩,
Chapter 26 Sources of Magnetic Field
3,Circulation Theorem with Magnetic Materials
L L ldMIldB
e n c0
0?
L IldMB e n c0
0
)(
MBH
0?
Let,
)( e n ce n c00 IIldBL
磁场强度
MBH
0?
where? is the permeability of magnetic materials.
Chapter 26 Sources of Magnetic Field
then,
L IldH e n c0
The circulation of along any closed loop equals
to the algebraic sum of free current encircled by
the closed loop L —— Theorem of circulation,
H?
H?
磁介质 中的 安培环路 定理
IlHl d
各向同性磁介质 For isotropic medium,
HHB r0
Chapter 26 Sources of Magnetic Field
Explanation:
(1) If the magnetic material were not present,
then the above equation reduce to the Ampere-
loop law in vacuum.
0?M?
ED o(2) One may compare with,which is another supplementary physical quantity.H?
(3) If with symmetry in distribution of free current,
then
BHI0
Chapter 26 Sources of Magnetic Field
I
r?
r
例 有两个半径分别为 和 的“无限长”同轴圆筒形导体，在它们之间充以相对磁导率为 的磁介质,当两圆筒通有相反方向的电流 时，试 求
（ 1）磁介质中任意点 P 的磁感应强度的大小 ;（ 2）圆柱体外面一点 Q 的磁感强度,
r?
rR
I
解 对称性分析
Rdr IlH
l
d
IdH?π2
d
IH
π2
d
IHB
π2
r0
d
I
R
Chapter 26 Sources of Magnetic Field
0,0π2 HdH
0 HB?
0, Brd
同理可求
Rdr
d
IB
π2
r0
Rd? 0 II
lHld
I
r?
r
I
R
d
Chapter 26 Sources of Magnetic Field
R1R2
Solution:
Example,A toroid is fully filled with magnetic
medium,Find,interior the medium,BH,
L NIrHldH?2
nIrNIH2
As same as the one of long solenoid.
nIHB
Chapter 26 Sources of Magnetic Field
26-9,Magnetic Field in Magnetic Materials;
Hysteresis p618-620
当外磁场由 逐渐减小时，磁感强度 B并不沿起始曲线 0P 减小,而是沿 PQ比较缓慢的减小，
这种 B的变化落后于 H的变化的现象，叫做 磁滞现象,简称 磁滞,
mH?
mB
mH?
P
rB
cH
mH?
'P mB?
H
B
O
磁滞回线
Q
1,Magnetization Curve
(磁化曲线 ) (P618-620),
Chapter 26 Sources of Magnetic Field
2,Hysteresis Loop (磁滞回线 )(P619):
矫顽力
cH
由于磁滞,当磁场强度减小到零（即 ）
时,磁感强度,而是仍有一定的数值,
叫做剩余磁感强度 (剩磁 ).r
B
0?H
0?B
rB
当外磁场由 逐渐减小时，磁感强度 B并不沿起始曲线 0P 减小,而是沿 PQ比较缓慢的减小，
这种 B的变化落后于 H的变化的现象，叫做 磁滞现象,简称 磁滞,
mH?
mB
mH?
P
rB
cH
mH?
'P mB?
H
B
O
磁滞回线
Q
Chapter 26 Sources of Magnetic Field
Heating a magnetic too can cause a loss of
magnetism,Above a certain temperature known
as the Curie temperature (1043 K for iron),a
magnet cannot be made at all.
3,铁磁性材料
H
B
O
软 磁材料
H
B
O
硬 磁材料
H
B
O
矩 磁铁氧体材料实验表明，不同铁磁性物质的磁滞回线形状相差很大,
Chapter 26 Sources of Magnetic Field
Homework,p622—628:
3,14,15,16,27,28,31,32,33,50,61
