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MAGNETOSTATIC FIELD(STEADY MAGNETIC)
CHAPTER 7
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
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
Energy density
Scalar V withPotential
Related Maxwell equations
Factor
Field
Steady currentStatic chargeSource
MagnetostaticElectrostaticAttribute
D and E B and H
vD ∇
0∇ E JH
B
∇
0∇
VE 2
2
1Ewe 2
2
1Hwm
withVector A
AB
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,
• Surface current,
• Volume current,
dlI
dsJ s
dvJ
(A.m) dvJdsJdlI s
Can be summarized:
7.3 BIOT SAVART LAW
A differential magnetic field strength, results from a differential current element, . The field varies inversely with the distance squared. The direction is given by cross product of
dHdlI
RadlI ˆ and
)Am( 4
ˆ x 1-
212
12112
R
adlIdH
R
Consider the diagram as shown:
)Am( 4
ˆ x 1-
22
l
R
R
adlIH
Total magnetic field can be obtained by integrating:
Similarly for surface current and volume current elements the magnetic field intensities can be written as:
)(Am 4
ˆ x
)(Am 4
ˆ x
1-22
1-212
12
121
s
R
Rs
R
dsaJH
R
dsaJdH
)(Am 4
ˆ x
)(Am 4
ˆ x
1-22
1-212
112
12
v
R
R
R
dvaJH
R
dvaJdH
Ex. 7.1: For a filamentary current distribution of finite length and along the z axis, find (a) and (b) when the current extends from - to +.
H H
Solution:
} 'dl zdz
1
2
z’-z
x rc y
(r c, ,z)
dH R
rc
(0,0,z’)
I
a
z z’
b
z
2/322
2
])'([4
)]'(ˆˆ[ x )'ˆ(= @
),,,',','(4
),,,',','(ˆ x '),,(
zzr
zzzrrdzzIdH
zrzrR
zrzradlIzrH
c
cc
b
a cc
ccRc
b
acc
b
a c
c
c
c
zzr
zz
r
I
zzr
dzIrH
zzr
dzrIdH
2/122
2/322
2/322
])'([
'
4
ˆ
])'([
'
4
ˆ=
])'([4
'ˆ=
ˆ x ˆ crz 0ˆ x ˆ zz=> ;
2/12222/322 ][ :Table Using
xcc
x
xc
dx
2/1222/122 ])([])([4
ˆ
zbr
za
zbr
zb
r
IH
ccc
Hence:
In terms of 1 and 2 :
A/m sinsin4
ˆ12
cr
IH
(r’, ’, z’)
(r, , z)
Hd
'ld
z
(b) When a = - and b = +, we see that 1 = /2, and 2 = /2
)(Am 2
ˆ1-
cr
IH
I
z Filamentary current
Flux H
H
Rl aa ˆˆˆ :r Unit vecto
The flux of in the direction and its density decrease with rc as
shown in the diagram.
H
Ex. 7.2: Find the expression for the field along the axis of the circular current loop carrying a current I.
H
d H
d a R
rddl ˆ
( 0 , 0 , z )
z R
( r c , , z )
C u r r e n t l o o p , I
I
Solution:
2/322 ][4
)ˆˆ( x )ˆ(=
zr
rrzzdrIdH c
Using Biot Savart Law
and
zr ˆ)ˆ( x ˆ
sinˆcosˆˆ x ˆ yxrz rz ˆˆ x ˆ
2/322
2
)(4z=
zr
dIrdH
rwhere the component was omitted due to symmetry
)(Am )(2
ˆ
; )(2
ˆ
)(4
ˆ
1-2/322
2
2/322
2
2
02/322
22
0
za
Iaz
ar zr
Irz
dzr
IrzdHH
Hence:
Ex. 7.3: Find the field along the axis of a s solenoid closely wound with a filamentary current carrying conductor as shown in Fig. 7.3.
H
surface current
flux
Fig. 7.3: (a) Closely wound solenoid (b) Cross section (c) surface current, NI (A).
• 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:
)(Am )(2
ˆ 1-2/322
2
za
IazH
surface current
Solution from Ex. 7.2:
2/322
2
)(2ˆ
za
adzNI
zHd
Therefore at the center of the solenoid:Hd
2/122
2/
2/-2/322
2
)4(ˆ
)(2ˆ ∫
a
NIz
za
dzNIazH
Hence:
If >> a :
a
at the center of the solenoidsJz
NIzH ˆˆ
(Am-1)
Field at one end of the solenoid is obtained by integrating from 0 to :
2/122 )(2ˆ
a
NIzH
If >> a :
2ˆ
2ˆ sJ
zNI
zH
which is one half the value at the center.
x
z
y
3A
3A
to ∞
(-3,4,0)
to ∞
Ex. 7.4: Find at point (-3,4,0) due to the filamentary current as shown in the Fig. below.
H
Solution:
zx HHH
Total magnetic field intensity is given by :
yxyxz ˆ5
3ˆ
5
4ˆ
5
4ˆ
5
3-ˆ-
12 sinsin4
ˆ
c
z r
IH
5
ˆ4ˆ3
169
ˆ4ˆ3ˆ
yxyx
R
RaR
Unit vector:
To find :zH
Rl aa ˆˆˆ
yx
yx
r
IH
cz
ˆ65.28ˆ2.38
1054
3ˆ
5
3ˆ
5
4
sinsin4
ˆ12
Hence:
1 = /2 ; 2 = 0
x
zto ∞
(-3,4,0)
y
3A
Ra
-3
4
To find :xH
Unit vector:
12 sinsin4
ˆ
c
x r
IH
Rl aa ˆˆˆ
zyx ˆˆˆ
x
z
to ∞
2 = /2
(-3,4,0)
y
Ra
3 A
-3
sin 1 = -3/5
yy
R
RaR ˆ
16
ˆ4ˆ
z
z
r
IH
cx
ˆ88.23
5
3-1
44
3ˆ
sinsin4
ˆ12
Hence:
A/m ˆ88.23ˆ65.28ˆ2.38 zyxHHH zx Hence:
7.4 AMPERE’S CIRCUITAL LAW
• Solving magnetostaic problems for cases of symmetrical current distributions.
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 :
Definition:
enIdlH
Graphical display for Ampere’s Circuital Law interpretation of Ien
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.
I
(a) (b) (c) (d)
en
l
IdlH
I I I 0
Ex. 7.5: Using Ampere’s circuital law, find field for the filamentary current I of infinite length as shown in Fig. 7.6.
H
I x
Amperian loop
rddl ˆ
H
d
r
Filamentary current of infinite length
z
dl} y
Solution:
Construct a closed concentric loop as shown in Fig. 7.6a.
Fig. 7.6a
z
y
x
to +
to -
Fig. 7.6
I
(A/m) ˆ2
)2(ˆˆ2
0
r
IH
IrHIdrHIrdHIldHll
enc
(similar to Ex. 7.1(b) using Biot Savart)
Ex. 7.5: Find 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.
H
C2
I a
xC1
r
yconductor
Amperian path
Solution:
For r a (C2) :
mAr
IH
IrH
IldH enc
c
/ ˆ2
)2(2
H(a)= I/2πa
a
H1
H2
H(r)
r
mAa
IrH
a
rIrH
IldH enc
c
/ ˆ2
)2(
2
2
2
1
For r ≤ a (C1) :
Ex. 7.6: Find field above and below a surface current distribution of infinite extent with a surface current density
H.Am ˆ -1yJJ ys
Solution:
lJIldHldHldHldHldH yen
l
1
2
2
'2
'2
'1
'1
1
x
y
z
.Am ˆ -1yJJ ys
1
1’
2
2’
3
3’
Amperian path 1-1’-2’-2-1
1
2
Graphical display for finding and using Ampere’s circuital law:H
x
12
z
2dH
1dH
1dH
2dH
rdH
rdH
Filamentary current
yx
z
Surface current
From the construction, we can see that above and below the surface current will be in the and directions, respectively.
Hx x
lJdxxxHdxxxH yxx ˆ)ˆ(ˆˆ2
'2
2
'1
1
1
x
y
z
.Am ˆ -1yJJ ys
1
1’
2
2’
3
3’
Amperian path 1-1’-2’-2-1
1
2
1
2
'2
'1
0 and
where
since is perpendicular to dlH
lJlHlH yxx 21
Hence:
Similarly if we takes on the path 3-3'-2'-2-3, the equation becomes:
lJlHlH yxx 23
xxx HHH 31
Therefore:
HAnd we deduce that above and below the surface current are equal, its becomes:
lJlHlH yxx
0z ; 2
1
0z ; 2
1
yx
yx
JH
JH
)ˆ( 2
1
ˆ 2
1
xJH
xJH
y
y
In vector form:
naJH ˆ2
1
yx
naz ˆ
It can be shown for two parallel plate with separation h, carrying equal current density flowing in opposite direction the field is given by: H
) 0 z andh z ( ; 0
)h z0 ( ; ˆ
naJH
xxx
h)z ( ; 0 z
0
h
yx
naz ˆ
ˆnaJH
) 0 z andh z ( ; 0
)h z0 ( ; ˆ
naJH
) 0 (z ; 0
7.5 CURL (IKAL)
The curl of a vector field, is another vector field.H
y
H
x
Hz
x
H
z
Hy
z
H
y
HxH xyzxyz
∂
∂-
∂
∂ˆ
∂
∂-
∂
∂ˆ
∂
∂-
∂
∂ˆ∇
For example in Cartesian coordinate, combining the three components, curl can be written as:H
zyx HHHzyx
zyx
H∂
∂
∂
∂
∂
∂ˆˆˆ
∇
And can be simplified as:
z
HrH
rrr
H
z
Hr
z
HH
rH rzrz ˆ
∂
∂-
∂
∂1ˆ∂
∂-
∂
∂ˆ
∂
∂-
∂
∂1∇
ˆ∂
∂-
∂
∂1ˆ∂
∂-
∂
∂
sin
11
ˆ∂
∂-
∂
sin∂
sin
1∇
rr H
r
rH
rr
rHH
r
rHH
rH
Expression for curl in cyclindrical and spherical coordinates:
cyclindrical
spherical
7.5.1 RELATIONSHIP OF ANDH J
JH ∇
Meaning that if is known throughout a region, then H
will produce for that region.J
JH ∇
Ex. 7.7: Find for given field as the following.H x H
cr
IH
2ˆ(a) for a filamentary current
22ˆ
a
IrH c
(b) in an infinite current carrying
conductor with radius a meter
2ˆ sJxH (c) for infinite sheet of uniformly
surface current Js
(d) in outer conductor of coaxial cable
22
22
2ˆ
bc
rc
r
IH c
c
Solution:
02
ˆˆ x
I
rr
zHr
rr
zH
ccc
cc
cr
IH
2ˆ(a) 0 ,
2ˆˆ
zr
c
HHr
IHH
c =>
Cyclindrical coordinate
z
HrH
rrr
H
z
Hr
z
HH
rH rzrz ˆ
∂
∂-
∂
∂1ˆ∂
∂-
∂
∂ˆ
∂
∂-
∂
∂1∇
= 0 = 0 = 0
Hence:
22ˆ
a
IrH c
Solution:
(b)
)(Am ˆˆ
2
2ˆ
2∂
∂ˆ x
2-2
22
Jza
Iz
a
Ir
r
z
a
Irr
rr
zH c
c
cc
cc
z
HrH
rrr
H
z
Hr
z
HH
rH rzrz ˆ
∂
∂-
∂
∂1ˆ∂
∂-
∂
∂ˆ
∂
∂-
∂
∂1∇
= 0 = 0 = 0
Cyclindrical coordinate
Hence:
Solution:
(c)
Cartesian coordinate
Hence:
2ˆ sJxH
0 x H
y
H
x
Hz
x
H
z
Hy
z
H
y
HxH xyzxyz
∂
∂-
∂
∂ˆ
∂
∂-
∂
∂ˆ
∂
∂-
∂
∂ˆ∇
= 0= 0 = 0
because Hx = constant and Hy = Hz = 0.
Solution:
(d)
Cyclindrical coordinate
Hence:
22
22
2ˆ
bc
rc
r
IH c
c
z
HrH
rrr
H
z
Hr
z
HH
rH rzrz ˆ
∂
∂-
∂
∂1ˆ∂
∂-
∂
∂ˆ
∂
∂-
∂
∂1∇
= 0 = 0 = 0
)(Am )(
ˆ
2
2
ˆ
2∂
∂ˆ x
2-22
22
22
22
Jbc
Iz
bc
rI
r
z
bc
rc
r
Ir
rr
zH
c
c
c
cc
cc
7.6 STOKE’S THEOREM
Stoke’s theorem states that the integral of the tangential component of a vector field around l is equal to the integral of the normal component of curl over S.H
H
sdHldHsl
In other word Stokes’s theorem relates closed loop line integral
to the surface integral
ldH sdH
surface S
Hna
s
unit vector normal to s
Path l
Consider an open surface S whose boundary is a closed surface l
It can be shown as follow:
naHs
dlHˆ∇∫
dls
Hna
∫ dlH naH ˆ ∇ s sH ∇
≈∫∑lk
k
m
k
dlH 1
k
m
k
sH
∇∑1
dsHdlHSl
∫ ∇
From the diagram it can be seen that the total integral of the surface 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 therefore
s
0Δ →ks ∞=m
Hence:
surface S
Hna
Path l
s k
dlk
where loop l is the path that enclosed surface S and this equation is called Stoke’s Theorem.
Ex. 7.8: 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.
)/(2
ˆ2
mAa
IrH
II
rdrdza
Izd
I
rdrdza
Irad
a
Ir
sdHldH
a
arl
sl
ˆˆ2
ˆ2
ˆˆ2
ˆ
)(
0
2
02
2
0
22
Solution:
222ˆ
2
2ˆ
2∂
∂ˆ x
a
Iz
a
Ir
r
z
a
Irr
rr
zH c
c
cc
cc
7.7 MAGNETIC FLUX DENSITY
Magnetic field intensity : Teslas HB o )/( 2mWb
H/mo 104 7- where permeability of free space
∫s
m dsB Magnetic flux : that passes through the surface S.
ds
nadsds ˆ
s
B
S
mdΨ B= ds cos
dsB =
s
m sdBΨHence:
)( 0 WbsdBΨs
m
I
H
In magnetics, magnet poles have not been isolated:
0
0
B
dvBsdBvs
4th. Maxwell’s equation for static fields.
rH 310Ex. 7.9: For (Am-1), find the m that passes through a plane
surface by, ( = /2), (2 r 4), and (0 z 2).
Wb10 x 8.150
121010
ˆ10ˆ
4-
32
0
4
2
3
2
0
4
2
3
oo
o
s
m
rdrdz
drdzrdsBΨ
Solution:
7.8 MAXWELL’S EQUATIONS
POINT FORM
INTEGRAL FORM
vD encQv
dvvs
sdDv
dvD
0 x E 0 x l
ldEs
sdE
JH x encIs
sdJl
ldHs
sdH x
0 B 0 s
sdBv
dvB
ED
HB
Electrostatic fields :
Magnetostatic fields:
7.9 VECTOR MAGNETIC POTENTIAL
0s
sdB magnet poles have not been isolated
To define vector magnetic potential, we start with:
0∇∫ v
dvBdsB
Using divergence theorem:
0∇∇ A
From vector identity:
0∇ B
Therefore from Maxwell and identity vector, we can defined if is a vector magnetic potential, hence:
<=>
AB ∇
A
where is any vector.A
=>
SUMMARY
en
s sl
IsdHsdJldH Stoke’s theorem
Ampere’s circuital law
Maxwell’s equations
s v
m dvBsdB 00
Divergence theorem
Gauss’s law
Magnetic flux lines close on themselves (Magnet poles cannot be isolated)
Maxwell’s equations