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CHAPTER 7
MAGNETOSTATIC FIELD
(STEADY MAGNETIC)
7.1 INTRODUCTION - SOURCE OF MAGNETOSTATIC FIELD
7.2 ELECTRIC CURRENT CONFIGURATIONS
7.3 BIOT SAVART LAW
7.4 AMPERE’S CIRCUITAL LAW
7.5 CURL (IKAL)
7.6 STOKE’S THEOREM
7.7 MAGNETIC FLUX DENSITY
7.8 MAXWELL’S EQUATIONS
7.9 VECTOR MAGNETIC POTENTIAL
1
7.1 INTRODUCTION - SOURCE OF
MAGNETOSTATIC FIELD
Originate from:
• constant current
• permanent magnet
• electric field changing linearly with time
Analogous between electrostatic and magnetostatic fields
Attribute
Electrostatic
Magnetostatic
Source
Static charge
Steady current
Field
E and D
H and B
Factor


Related Maxwell
equations
∇ D   v
∇ B  0
∇ E  0
∇ H  J
Scalar V with
Vector A with
Potential
E  V
Energy density
1
we  E 2
2
B   A
1
wm  H 2
2
Two important laws – for solving magnetostatic field
• Biot Savart Law – general case
• Ampere’s Circuital Law – cases of symmetrical current
distributions
7.2 ELECTRIC CURRENT CONFIGURATIONS
Three basic current configurations or distributions:
• Filamentary/Line current, I dl
• Surface current, J s ds
• Volume current, J dv
Can be summarized:
I dl  J s ds  Jdv (A.m)
7.3 BIOT SAVART LAW
Consider the diagram
as shown:
A differential magnetic field strength, dH results from a
differential current element, I dl . The field varies inversely with
the distance squared. The direction is given by cross product of
I dl and aˆ R
dH 2 
I 1 dl1 x aˆ R1 2
4R122
(Am-1 )
Total magnetic field can be obtained by integrating:
H2 

I dl x aˆ R
-1
(
Am
)
2
4R
l
Similarly for surface current and volume current elements the
magnetic field intensities can be written as:
dH 2 
J s1 x aˆ R12 ds1
 H2 
4R122

s
J x aˆ R ds
4R 2
-1
(Am )
dH 2 
J 1 x aˆ R12 dv1
(Am )  H 2 
-1
4R122

v
J x aˆ R dv
4R 2
(Am-1 )
(Am-1 )
Ex. 7.1: For a filamentary current distribution of finite length and along
the z axis, find (a) H and (b) H when the current extends from - to
+.
Solution:
z
(0,0,z’)
b
} dl  z$ dz '
z’- z
R
z’
rc
z
I
a
x
f
rc
a2
a1
dH
(r c , f ,z)
y
b
H ( rc , f , z )  
a
@
dH =
I dl ' x aˆ R ( rc ' , f ' , z ' , rc , f , z )
4R 2 (rc ' , f ' , z ' , rc , f , z )
z
dl ' (r’, f’, z’)
I ( zˆdz ' ) x [ rˆc rc  zˆ ( z ' z )]
2
4 [ rc  ( z ' z ) 2 ]3 / 2
dH (r, f, z)
=> zˆ x rˆc  fˆ ; zˆ x zˆ  0
f
Ifˆrc dz '
dH =
2
4 [ rc  ( z ' z ) 2 ]3 / 2
fˆIrc b
dz '
H =
4 a [ rc 2  ( z ' z ) 2 ]3 / 2
Hence:
fˆI
H 
4rc
b

fˆI
z ' z


4rc [ rc 2  ( z ' z ) 2 ]1/ 2  a
Using Table : 
dx
x

[c 2  x 2 ]3 / 2 c 2 c 2  x 2 1/ 2


bz
az
 2
 2
2 1/ 2
2 1/ 2 
[
r

(
b

z
)
]
[
r

(
b

z
)
] 
c
 c
In terms of a1 and a2 :
fˆI
sin a 2  sin a1  A/m
H 
4rc
(b) When a = -  and b = +, we see that a1 = /2, and a2 = /2
fˆ I
H
(Am-1 )
2rc
The flux of H in the fˆ
direction and its density
decrease with rc as
shown in the diagram.
z
I
Filamentary
current
FluxH
Unit vecto r : fˆ  aˆl  aˆ R
H
Ex. 7.2: Find the expression for the H field along the axis of the
circular current loop carrying a current I.
Solution:
Using Biot Savart Law
dH
(0,0,z)
dH =
z
f
R
df
I
Current
loop, I
a$ R
dl  fˆrd f
(rc,f,z)
I (fˆrc df ) x ( zˆz  rˆ r )
4 [ r  z 2 ]3 / 2
2
and
fˆ x zˆ  rˆ  xˆ cos f  yˆ sin f
fˆ x (rˆ)  zˆ
Ir 2df
 dH = ẑ
4 (r 2  z 2 )3 / 2
where the r̂ component was
omitted due to symmetry
Hence:
2
zˆIr 2
H   dH 
2
2 3/ 2
4

(
r

z
)
0
zˆIr 2

2( r 2  z 2 ) 3 / 2
2
 df
; r  a
zˆIa 2
-1

(Am
)
2
2 3/ 2
2( a  z )
0
Ex. 7.3: Find the H field along the axis of a s solenoid closely wound
with a filamentary current carrying conductor as shown in Fig. 7.3.
flux
surface
current
Fig. 7.3: (a) Closely wound solenoid (b) Cross section (c) surface current, NI (A).
Solution:
• Total surface current = NI Ampere
• Surface current density, Js = NI / l
Am-1
• View the dz length as a thin current loop that
carries a current of Jsdz = (NI / l )dz
Solution from Ex. 7.2:
surface
current
zˆIa 2
-1
H
(Am
)
2
2 3/ 2
2(a  z )
Therefore dH at the center of the solenoid:
 NI

dz a 2



ˆ  2
dH  z
2( a  z 2 ) 3 / 2
Hence:
NIa 2
H  zˆ
2
/2
dz
NI
 zˆ
2
2 3/ 2
2
2 1/ 2
∫
(
a

z
)
(
4
a


)
- / 2
If
 >> a :
NI
H  zˆ
 zˆJ s

(Am-1)
at the center of
the solenoid
a
Field at one end of the solenoid is
obtained by integrating from 0 to  :
H  zˆ
If
NI
2(a 2   2 )1/ 2
 >> a :
Js
NI
H  zˆ
 zˆ
2
2
which is one half the value at the center.
Ex. 7.4: Find H at point (-3,4,0) due to the filamentary current as shown in
the Fig. below.
z
to ∞
3A
(-3,4,0)
3A
x
to ∞
Solution:
Total magnetic field intensity
is given by :
H  Hx  Hz
y
Hz :
To find
to ∞
fˆI
sin a 2  sin a1 
Hz 
4rc
Unit vector:
z
3A
fˆ  aˆl  aˆ R
-3
âR
4  4
3
 3
 - zˆ - xˆ  yˆ   xˆ  yˆ
5  5
5
 5
Hence:
(-3,4,0)
a1 = /2 ; a2 = 0
4
x
fˆI
sin a 2  sin a1 
Hz 
4rc
3  3
4
0  1
  xˆ  yˆ 
5  4 5
5
 38.2 xˆ  28.65 yˆ
aˆ R 
y
R  3 xˆ  4 yˆ  3 xˆ  4 yˆ


R
5
9  16
To find
Hx :
z
fˆI
sin a 2  sin a1 
Hx 
4rc
-3
Unit vector:
â R
fˆ  aˆl  aˆ R
3A
aˆ R 
Hence:
x
to ∞
3

1 - 
4 4  
5
 23.88 zˆ
 zˆ
Hence:
sin a1 = -3/5
a2 = /2
y
 xˆ  yˆ  zˆ
fˆI
sin a 2  sin a1 
Hx 
4rc
(-3,4,0)
3
H  H x  H z  38.2 xˆ  28.65 yˆ  23.88 zˆ A/m
R
4 yˆ

 yˆ
R
16
7.4 AMPERE’S CIRCUITAL LAW
• Solving magnetostaic problems for cases of symmetrical
current distributions.
Definition:
The line integral of the tangential component of the magnetic
field strength around a closed path is equal to the current
enclosed by the path :
 H  dl  I
en
Graphical display for Ampere’s Circuital Law interpretation of Ien
 H  dl  I
en
l
I
(a)
I
(b)
I
(c)
I
(d)
0
Path (loop) (a) and (b)
enclose the total current I ,
path c encloses only part
of the current I and path d
encloses zero current.
Ex. 7.5: Using Ampere’s circuital law, find H field for the filamentary
current I of infinite length as shown in Fig. 7.6.
z
z
Solution:
Construct a closed
concentric loop as
shown in Fig. 7.6a.
to +
Filamentary current
of infinite length
}dl
I
r
y
f
x
H
I Amperian loop
to -
Fig. 7.6a
Fig. 7.6
l
y
dl  fˆrdf
x
 H  dl  I
df
2
enc
  fˆH f  fˆrdf  I  H f r  df  I  H f (2r )  I
l
I ˆ
H 
f (A/m)
2r
0
(similar to Ex. 7.1(b) using Biot Savart)
Ex. 7.5: Find H inside and outside an infinite length conductor of
infinite cross section that carries a current I A uniformly distributed
over its cross section and then plot its magnitude.
y
Solution:
For r  a (C2) :
conductor
C2
 H  dl  I enc
fˆ
C1
x
c2
H f ( 2r )  I
H 
I
2r
I
fˆ
a
r
A/ m
Amperian
path
H(r)
For r ≤ a (C1) :
 H  dl
H(a)
= I/2πa
 I enc
c1
 r 2
H f ( 2r )  I 
2

a

Ir ˆ
H 
f
2
2a



A/ m
H1
H2
r
a
Ex. 7.6: Find H field above and below a surface current distribution of
infinite extent with a surface current density J s  J y yˆ Am-1.
Solution:
Graphical display for finding H and using Ampere’s circuital law:
z
3
dH 2
1
dH r
z
3’
1’
dH 1
y
1
2
Filamentary current
x
x
2
2
1
J s  J y yˆ Am-1.
dH 1
dH r
dH 2
z
2’
x
Amperian path 1-1’-2’-2-1
1'
2'
y
2
1
 H  dl   H  dl  H  dl  H  dl  H  dl  I
l
1
1'
2'
Surface
current
2
en
 J yl
From the construction, we can see that H above and below the surface
current will be in the x̂ and  x̂ directions, respectively.
1'
H
2
x1
xˆ  xˆdx   H x 2 (  xˆ )  xˆdx  J y l
1
where
2'
2'
1
 and 
1'
Therefore:
 0
since H is perpendicular to dl
2
3
H x1l  H x 2l  J y l
Similarly if we takes on the path 3-3'-2'-2-3, the
equation becomes:
1
1’
2
H x 3l  H x 2l  J y l
Hence:
H x1  H x3  H x
z
3’
y
1
2
x
J s  J y yˆ Am-1.
2’
Amperian path 1-1’-2’-2-1
And we deduce that
equal, its becomes:
H
above and below the surface current are
H xl  H xl  J y l
1
Hx  Jy ;
2
1
Hx   Jy ;
2
In vector form:
1
H   J y xˆ
2
1
H   J y ( xˆ )
2
z0
z0
1
H  J  aˆ n
2
ˆn
z a
x
y
It can be shown for two parallel plate with separation h, carrying equal current
density flowing in opposite direction the H field is given by:
H  J  aˆ n ; ( 0  z  h )
0
;
( z  h and z  0 )
z
 0 ; ( z  h)
ˆn
z a
h
H  J  aˆ n
x
x
x
x
y
0
 0 ; (z  0 )
H  J  aˆ n ; ( 0  z  h )
0
;
( z  h and z  0 )
7.5 CURL (IKAL)
The curl of a vector field,
H is another vector field.
For example in Cartesian coordinate, combining the three components,
curl H can be written as:
 ∂
Hy 
Hz ∂
∇ H   xˆ 

y
∂
z 
  ∂
Hy ∂
∂
Hx ∂
H x 
Hz 
∂
yˆ 
 zˆ 


z
∂
x 
x
∂
y 
 ∂
 ∂
And can be simplified as:
xˆ
∂
∇ H 
∂
x
Hx
yˆ
∂
∂
y
Hy
zˆ
∂
∂
z
Hz
Expression for curl in cyclindrical and spherical coordinates:
 1  ∂H z ∂H f   ∂H r ∂H z  ˆ 1  ∂
∂H r  
rˆ  
 zˆ  cyclindrical
∇ H   
f   rH f  ∂f  
 r  ∂f ∂z   ∂z ∂r  r  ∂r
 ∂H f sin   ∂
H 

rˆ
∂

∂
f 

rH   - ∂H r
1 1 ∂
H r  ∂rH f   ˆ 1  ∂
  

  
r  sin   ∂
f 
∂
r 
r ∂
r
∂

1
∇ H 
r sin 
ˆ
f spherical

7.5.1 RELATIONSHIP OF H AND J
∇ H  J
Meaning that if
will produce
J
H is known throughout a region, then ∇ H  J
for that region.
Ex. 7.7: Find  x H for given H field as the following.
(a)
 I 
ˆ

H  f 
 2rc 
(b)
 Irc 
ˆ
H  f
2 
 2a 
(c)
(d)
for a filamentary current
in an infinite current carrying
conductor with radius a meter
J s for infinite sheet of uniformly
2 surface current Js
2
2


c

r
I
c
ˆ
 2 2  in outer conductor of coaxial cable
H f
2rc  c  b 
H  xˆ
Solution:
(a)
 I 
ˆ

H  f 
 2rc 
=>
 I 
ˆ
ˆ
, H rc  H z  0
H  fHf  f 
 2rc 
Cyclindrical coordinate
 1  ∂H z ∂H f   ∂H r ∂H z  ˆ 1  ∂
∂H r  
rˆ  
 zˆ 
∇ H   
f   rH f  ∂f  
 r  ∂f ∂z   ∂z ∂r  r  ∂r
=0
=0
=0
Hence:
zˆ
xH 
rc

 zˆ    I
rc Hf    

 rc
 rc  rc  2

  0

Solution:
(b)
 Irc 
ˆ
H  f
2 
 2a 
Cyclindrical coordinate
 1  ∂H z ∂H f   ∂H r ∂H z  ˆ 1  ∂
∂H r  
rˆ  
 zˆ 
∇ H   
f   rH f  ∂f  
 r  ∂f ∂z   ∂z ∂r  r  ∂r
=0
=0
Hence:
zˆ
xH 
rc
 ∂
Irc
 rc

2
∂
r
2

a

 c
I
 zˆ 2  zˆJ (Am -2 )
a
=0
 zˆ  2rc I 
  
2 
 rc  2a 
Solution:
(c)
Js
H  xˆ
2
Cartesian coordinate
 ∂
Hy 
Hz ∂
∇ H   xˆ 

y
∂
z 
  ∂
=0
Hy ∂
∂
Hx ∂
H x 
Hz 
∂
yˆ 
 zˆ 


z
∂
x 
x
∂
y 
 ∂
 ∂
=0
because Hx = constant and Hy = Hz = 0.
Hence:
xH 0
=0
Solution:
(d)
2
2


c

r
I
c
ˆ
 2 2 
H f
2rc  c  b 
Cyclindrical coordinate
 1  ∂H z ∂H f   ∂H r ∂H z  ˆ 1  ∂
∂H r  
rˆ  
 zˆ 
∇ H   
f   rH f  ∂f  
 r  ∂f ∂z   ∂z ∂r  r  ∂r
=0
Hence:
zˆ
xH 
rc
=0
 ∂
I
rc 

rc
2rc
∂
=0
 c 2  rc2 
 2

2 
 c  b 
 I 
2rc 
 2   c 2  b 2 



I
-2
  zˆ

J
(Am
)
2
2
 (c  b )
zˆ

rc
7.6 STOKE’S THEOREM
Stoke’s theorem states that the integral of the tangential component of a
vector field H around l is equal to the integral of the normal component of
curl H over S.
In other word Stokes’s theorem relates closed loop line integral


to the surface integral   H  ds
 H  dl     H  ds
l
s
H  dl
It can be shown as follow:
Consider an open surface S whose boundary is a closed surface l
H
â n unit vector normal to s
surface S
s
Path l
H â n
s
dl
H  dl
∫
 ∇ H  aˆ
s
∫H  dl  ∇ H   aˆn s  ∇  H   s
n
m
∑ ∫H  dlk
k 1
lk
≈ ∑ ∇  H   sk
m
k 1
From the diagram it can be seen that the
total integral of the surface s enclosed
by the loop inside the open surface S is
zero since the adjacent loop is in the
opposite direction. Therefore the total
integral on the left side equation is the
perimeter of the open surface S.
If
Δsk → 0
therefore
H
â n
m = ∞
surface S
Path l
dlk
sk
Hence:
 H  dl  ∫∇  H   ds
l
S
where loop l is the path that enclosed surface S and this equation is
called Stoke’s Theorem.
Ex. 7.8:
 Ir 
ˆ
H f
( A / m)
2
 2a 
was found in an infinite conductor of
radius a meter. Evaluate both side of Stoke’s theorem to find the current
flow in the conductor.
Solution:
 H  dl   (  H )  ds
l
s


Ir
 ˆ Ir 

ˆ
ˆ
f

f
ad
f



f
2
l  2a 2 
 
2

a
r a
2
a 2
I
0 2 df  0
I

ˆ
z
0  a 2

  zˆrdrd f


  zˆrdrd f

I I
xH 
zˆ
rc
 ∂
Irc  zˆ  2rc I 
I
ˆ
r


z




c


rc  2a 2  rc  2a 2 
a 2
∂
7.7 MAGNETIC FLUX DENSITY
Magnetic field intensity :
where
B  o H
o  4 10-7 H/m
Magnetic flux :
m  ∫
B  ds
Teslas (Wb / m 2 )
permeability of free space
that passes through the surface S.
s
B
dΨ m = ds B cos a
=
B  ds
a
Hence:
Ψ m   B  ds
s
s
S
ds
ˆn
ds  ds a
In magnetics, magnet poles have
not been isolated:
Ψ m   B  ds  0 (Wb)
s
H
 B  ds     B dv  0
s
v
 B  0
4th. Maxwell’s equation for static fields.
I
Ex. 7.9: For H  fˆ10 r (Am-1), find the m that passes through a plane
surface by, (f = /2), (2  r  4), and (0  z  2).
3
Solution:
2 4


Ψ m   B  ds     ofˆ103 r  fˆdrdz 
s
0 2
2 4
  o 103   rdrdz   o 103 12 
0 2
 150.8 x 10 -4 Wb
7.8 MAXWELL’S EQUATIONS
POINT
FORM
INTEGRAL FORM
  D  v
   D dv   D  ds    v dv  Qenc
v
s
v
xE  0
xH  J
B  0


  x E  ds   E  dl  0
s
l


  x H  ds   H  dl   J  ds  I enc
s
l
s
   B dv   B  ds  0
v
s
Electrostatic fields :
Magnetostatic fields:
D  E
B  H
7.9 VECTOR MAGNETIC POTENTIAL
To define vector magnetic potential, we start with:
 B  ds  0
s
=> magnet poles have not
been isolated
Using divergence theorem:
B  ds  ∇ B dv  0
∫
<=>
∇ B  0
v
From vector identity:
∇ ∇ A   0
where
A is any vector.
Therefore from Maxwell and identity
vector, we can defined if A is a
vector magnetic potential, hence:
B ∇ A
SUMMARY
Maxwell’s equations
 H  dl   J  ds     H  ds  I
l
s
en
s
Stoke’s theorem
Ampere’s circuital law
Maxwell’s equations
 m   B  ds  0     B dv  0
s
v
Divergence theorem
Gauss’s law
Magnetic flux lines close on themselves
(Magnet poles cannot be isolated)