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CHAPTER 8
MAGNETOSTATIC FIELD
(MAGNETIC FORCE, MAGNETIC MATERIAL
AND INDUCTANCE)
8.1 FORCE ON A MOVING POINT CHARGE
8.2 FORCE ON A FILAMENTARY CURRENT
8.3 FORCE BETWEEN TWO FILAMENTARY CURRENT
8.4 MAGNETIC MATERIAL
8.5 MAGNETIC BOUNDARY CONDITIONS
8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE
8.7 MAGNETIC ENERGY DENSITY
1
8.1 FORCE ON A MOVING POINT CHARGE
Force in electric field:
Force in magnetic field:
Total force:
Fe  QE
Fm  QU  B
F  Fe  Fm
or
F  QE  U  B 
Also known as Lorentz force equation.
Force on charge in the influence of fields:
Charge
Condition
E
Field
Stationary
QE
Moving
QE
B Field
-
QU  B
Combination
E and B
QE
QE  U  B 
8.2 FORCE ON A FILAMENTARY CURRENT
__
The force on a differential current element ,
field, B :
I dl due to the uniform magnetic
dF  Idl  B
__
__
F   I dl  B   I  B  dl
F   IB   dl  0
It is shown that the net force for any close current loop in the uniform
magnetic field is zero.
Ex. 8.1: A semi-circle conductor carrying current I, is located in plane xy
as shown in Fig. 8.1. The conductor is under the influence of uniform
magnetic field, B  yˆ B0 . Find:
(a) Force on a straight part of the conductor.
(b) Force on a curve part of the conductor.
Solution:
y
B
(a) The straight part length = 2r.
Current flows in the x direction.
__
F   I dl  B
F1  xˆ (2 Ir )  yˆB0  zˆ 2 IrB0 (N)
r

I
x
__
(b) For curve part, dl B will be in the
–ve z direction and the magnitude is
proportional to sin 
y
B

F2  I
r
dl  B



0

  zˆI
rB


0
I
sin d   zˆ 2 IrB0 (N)
0
Hence, it is observed that
F2   F1
the net force on a close loop is zero.
and it is shown that
x
8.3 FORCE BETWEEN TWO FILAMENTARY
CURRENT
z
Loop l2
Loop l1
__
__
aˆ R1 2
R12
I 2 dl 2
I1 dl1
I2
I1
P1(x1,y1,z1)
P2(x2,y2,z2)
y
x
We have :
z
dF  Id l x B
(N)
Loop l2
Loop l1
The magnetic field at point P2 due to the
filamentary current I1dl1 :
dH 2 
aˆ R1 2
__
I 2 dl 2
I1 dl1
I2
I1dl1 x aˆ R1 2
4R12
__
R12
I1
(A/m)
2
P1(x1,y1,z1)
P2(x2,y2,z2)
y
x
d dF2   I 2 dl2 x
dF   I dl x 
2
2
2
l1
o I1dl1 x aˆ R
12
4R12
o I1dl1 x aˆ R
12
4R12
2
2
 I 2 dl2 x B2
where dF2 is the force due to I2dl2 and due to the magnetic field of loop l1
Integrate:
  o I1dl1 x aˆ R1 2 
F2   I 2 dl2 x  

2

l2
l1 
 4R12
o I1I 2
F2 
4

 aˆ R x dl1
12

l l R122
2  1
For surface current :
F2   J s 2  B2 ds
s
For volume current :
F2   J 2  B2 dv
v

 x dl2

Ex. 8.2: Find force per meter between two parallel infinite conductor carrying
current, I Ampere in opposite direction and separated at a distance d meter.
z
Solution:
B2
B 2 at position conductor 2
 xˆ 0 I1
ˆI1
B2   0 H 2   0

2rc
2d
I2
d
x
Hence:
  xˆ0 I1 
  xˆ0 I1 
ˆ
F2   I 2 d 2  

I
(

z
dz
)

  2


2

d
2

d

 0


0
1
0 I 2
I1 I 2
 yˆ 0
 yˆ
2d
2d
F21
I1
1
(N/m)
y
I1 = I2 = I
Ex. 8.3: A square conductor current loop is located in z = 0 plane with the edge
given by coordinate (1,0,0), (1,2,0), (3,0,0) and (3,2,0) carrying a current of 2 mA
in anti clockwise direction. A filamentary current carrying conductor of infinite
length along the y axis carrying a current of 15 A in the –y direction. Find the
y
force on the square loop.
Solution:
Field created in the square loop due
to filamentary current :
I
15
H
zˆ 
zˆ A/m
2x
2x
ˆ  aˆ l  aˆ R
(3,2,0)
15 A
  yˆ  xˆ  zˆ
3 10-6
∴B  0 H  4 10 H 
zˆ T
x
-7
(1,2,0)
2 mA
z
x
(1,0,0)
(3,0,0)
Hence:
y
__
__
F   I dl  B   I  B  dl
3
2

ˆ
z
zˆ
F  2 10 3  3 10 6    dxxˆ    dyyˆ
3
 x 1 x
y 0
1
0

zˆ
zˆ
   dxxˆ    dyyˆ 
x
1

x 3
y 2
(1,2,0)
15 A
z
2 mA
x
(1,0,0)
1 2
3
1
0


F  6  10 ln x 1 yˆ  y 0  xˆ   ln x 3 yˆ  y 2  xˆ 
3



2
 1
9 
 6  10 ln 3 yˆ  xˆ   ln  yˆ  2 xˆ 
3
 3


 8 xˆ nN
9
(3,2,0)
(3,0,0)
8.4 MAGNETIC MATERIAL
The prominent characteristic of magnetic material is
magnetic polarization – the alignment of its magnetic
dipoles when a magnetic field is applied.
Through the alignment, the magnetic fields of the
dipoles will combine with the applied magnetic field.
The resultant magnetic field will be increased.
8.4.1 MAGNETIC POLARIZATION (MAGNETIZATION)
Magnetic dipoles were the results of three sources of
magnetic
moments
that
produced
magnetic
dipole
moments : (i) the orbiting electron about the nucleus
(ii) the electron spin and (iii) the nucleus spin.
The effect of magnetic dipole moment
bound current or magnetization current.
will
produce
Magnetic dipole moment in microscopic
view is given by :
___
___
dm  I ds
Am2
___
where dm is magnetic dipole moment in discrete
and I is the bound current.
In macroscopic view, magnetic dipole moment
per unit volume can be written as:
 1 nv ___ 
M  lim   dmi 
 v 0 v
i 1


A/m
where M is a magnetization and n is the volume
dipole density when v -> 0.
If the dipole moments become totally
aligned :
___
___
M  n dm  nI ds Am-1
Magnetic dipole moments in a magnetic material
Ba  0
Ba  0
M 0
___
dmi
___
dmi
___
dmi
Macroscopic v
ds
Microscopic
base
___
dmi  0
M 0
___
dmi  0
dm' s tend to align
M 0
themselves
___
8.4.2 BOUND MAGNETIZATION CURRENT DENSITIES
J sm and J m
z
___
dm
y
Ba
x
M
Ba
M
I
M
___
dm
Ba
M
___
J sm
Alignment of dm' s within a magnetic material under uniform Ba
conditions to form a non zero J sm on the slab surfaces, and
a J m  0 within the material.
8.4.3 TO FIND J sm and J m
y
Bound magnetization current :
dI m  Indv
__
__
__
__
dI m  I (n ds dl )  (nI ds) (dl )
We have:
Hence:
___
___
M  n dm  nI ds Am-1
dI m  M  dl 
__
__
I m =  M  dl 
I m =  J m  ds
s
through the loop l’
on the surface bound
by the loop l’
Using Stoke’s Theorem:
I m =  J m  ds   M  dl      M  ds
s
l
s
Hence:
J m   M
(Am-2)
is the bound magnetization current
density within the magnetic material.
And to find Jsm :
From the diagram :
 M tan
M
dIm = Mtan dl'
dI m

 J sm
dl 
loop l’
On the slab
surface
J sm n̂

J sm  M  n (Am-1)
is the surface bound magnetization
current density
8.4.4 EFFECT OF MAGNETIZATION ON MAGNETIC FIELDS
Due to magnetization in a material, we have seen the formation of bound
magnetization and surface bound magnetization currents density.
Maxwell’s equation:
 H  J



B
o

 J  Jm
B
J
o

 
o

 J   M
B

    M   J
 o

;
B  o H
due to free charges and bound
magnetization currents
 M  J m
B
(free charge)
Define:

 B

H    M 
 o

  H  J
Hence:
B  o ( H  M )
Magnetization in isotropic
material:
M  m H
 m  magnetic susceptibi lity
B  o H (1   m )
  o r  permeability
Hence:
 r  (1   m )
 B  H
Ex. 8.4: A slab of magnetic material is found in the region given by 0 ≤ z ≤ 2 m
with r = 2.5. If B  10 yxˆ  5 xyˆ mWb/m 2 in the slab, determine:
(a) J (b) J m (c ) M (d ) J sm on z  0
Solution:
 dB y dBx 
1

 zˆ
(a) J    H   


7
0  r 4 10 2.5  dBx
dy 
106
 5  1010 3 zˆ  4.775 zˆ kA/m 2

B

(b) J m   m J   r  1J  1.5 4.775 zˆ  103  7.163zˆ kA/m 2
(c ) M   m H   m
B
0  r
ˆ  5 xy
ˆ  10 3
1.510 yx

4  10 7 2.5
ˆ  2.387 xy
ˆ kA/m
 4.775 yx
(c) M  4.775 yxˆ  2.387 xyˆ kA/m
(d)
J sm  M  nˆ
Because of z = 0 is under the slab region of 0 ≤ z ≤ 2 , therefore nˆ   zˆ
J sm   2.387 xxˆ  4.775 yyˆ kA/m
(4.775 yxˆ  2.387 xyˆ )   zˆ 
Ex. 8.5: A closely wound long solenoid has a concentric magnetic rod
inserted as shown in the diagram.In the center region, find: (a) H , B and M
in both air and magnetic rod, (b) the ratio of the B in the rod to the B in the
air, (c) J sm on the surface of the rod and J m within the rod. Assume the
permeability of the rod equals 5o .
0
magnetic
rod
P2
P3
b
r
P2’
P3’
P1
P4
a
z
0
Solution:
(a) Using Ampere’s circuital law to the closed path P1 - P2 - P3- P4 . If using
path P1 - P2’ - P3’ -P4 - P1, Hz in the rod will be the same as in the air since
Ampere’s circuital law does not include any Im in its Ien term.
Hence:
 H  d 

P3
 ( zˆH z )  ( zˆdz) H z d  I en 
P2
NI
Hz 
 Js

NI
d

0
magnetic
rod
M  zˆM z   m ( zˆH z ) in the rod
M  0 in air
(since m of air is zero)
B   0 ( zˆH z )
B   0  r ( zˆH z )
in air
in the rod
P2
P3
P2’
P3’
P1
P4
0
a
b
z
(b)
Brod 5o zˆH z 

5
Bair
o zˆH z 
(c) J sm  M  nˆ  ( zˆM z )  rˆc  ˆM z
J m    M    ( zˆM z )  0
0
Js
aa
flux Js
J sm
ẑ
b
M
nˆ  rˆ
Hz=NI/l
Hz
50Hz
Bz
Bz=0Hz
Mz
Js
flux Jsm
J sm
Mz=4Hz
Jsm
Jsm=4Hz
Js=Hz=NI/l
Plots of H, B and M, Js and Jsm
along the cross section of the
solenoid and the magnetic rod
8.4.5 MAGNETIC MATERIAL CLASSIFICATION
Magnetic material can be classified into two main groups:
Group A – has a zero dipole moment
dm  0
diamagnetic material eg. Bismuth
 m  1.66  10 5 ,  r  0.9999834
Group B – has a non zero dipole moment
(a) Paramagnetic material -
dm  0 ; M  0
When Ba is applied, there will be a slight alignment of the
atomic dipole moment to produce M  0
5
χ

2

10
, μr  1.00002
Eg. Aluminum - m
(b) Ferromagnetic material : has strong magnetic moment
the absence of an applied Ba field.
Eg: metals such as nickel, cobalt and iron.
dm
in
8.5 MAGNETIC BOUNDARY CONDITIONS
To find the relationship between B , H and M
Region 1:
n̂21
1 ,  m1
B/H
s
Boundary
Region 2:
l
a
h / 2
h / 2
b
d
c
 2 ,  m2
h / 2
h / 2
Js
To find normal component of
Consider a small cylinder as
__
 B  ds  B
1n
B and H
at the boundary
h  0 and use
s  B2 n s  0
 B1n  B2 n
 1 H1n  2 H 2 n
__
 B  ds  0
To find tangential component of
B and HRegion
at the1:boundary
1 ,  m1
Consider a closed
abcd as h  0
__
and use
H  dl  I
Boundary
n̂21 = aˆ n 21
B/H
s
enc
l
h / 2
h / 2
H1t l  H 2t l  I enc
Region 2:

l
a
b
d
h / 2
c
 2 ,  m2
Js
H1t  H 2t  J s
z
x
where
J s is perpendicular to the directions of H1t
In vector form :
and
H 2t
aˆ n 21  H1  H 2   J s
aˆ n 21 is a normal unit vector from region 2 to region 1
h / 2
x
y
zˆ  xˆ   yˆ
We have:
H1t  H 2t  J s
Hence:
B1t
1

B2t
2
We have:
M  mH
 Js
Hence:
M 1t
 m1
We have:
  M  Jm
and
 m2
 Js
We have:
   M dV   J mdV  I m
v

M 2t
v
__
 M  dl  I
l
Hence:
M1t  M 2t  J sm
m
1H1n  2 H 2 n
Hence:
1
M 1n
 m1
 2
M 2n
m2
If Js = 0 :
H1t  H 2t or
B1t
1

B2t
2
If the fields were defined by an angle  normal to the interface
B1 cos1  B1n  B2n  B2 cos 2
B1
1
sin 1  H1t  H 2t 
B2
2
(1)
sin  2
(2)
y
Divide (2) to (1):
tan 1 1  r1


tan  2  2  r 2
Region 1:
1 ,  m1
Boundary at
y = 0 plane
Region 2:
 2 ,  m2
B2 or H 2
2
B1 or H1
1
B1n
B1
B1t
x
Ex. 8.6: Region 1 defined by z > 0 has 1 = 4 H/m and 2 = 7 H/m in
region 2 defined by z < 0. J s  80 xˆ A/m on the surface at z = 0. Given
B1  2 xˆ  3 yˆ  zˆ mT,
find
B2
B1n
Solution:
Normal component
B1
#1
B1n  ( B1  nˆ12 )nˆ12
 2 xˆ  3 yˆ  zˆ    zˆ  zˆ 
#2
B1t  B1  B1n  2 xˆ  3 yˆ
1

J s  80 xˆ
H 1t
B1t
B2 n B2t
B2
H 2t
z
n̂12   ẑ
 B2 n  B1n  zˆ
H 1t 
n̂21  ẑ
Z=0
 zˆ
B1t
B1
2 xˆ  3 yˆ 10 3
4  10  6
 500 xˆ  750 yˆ A/m
y
x
ˆ12  J s
H 2t  H 1t  n
ˆ   zˆ   80 xˆ
 500 xˆ  750 y
ˆ  80 y
ˆ
 500 xˆ  750 y
ˆ A/m
 500 xˆ  670 y
B2t  2 H 2t  7 106 500 xˆ  670 yˆ   3.5xˆ  4.69 yˆ
 B2  B2t  B2n  3.5xˆ  4.69 yˆ  zˆ mT
Ex. 8.7: Region 1, where  r1 = 4 is the side of the plane y + z < 1 . In
region 2 , y + z > 1 has  r2 = 6 . If B1  2 x
ˆ  yˆ find B2 and H 2
Solution:
z
The unit normal:
nˆ  ( yˆ  zˆ ) / 2
y+z=1
B1n  ( B1  nˆ12 )nˆ12
B1n

2 xˆ  yˆ    yˆ  zˆ 


2
1
2
1
nˆ12  0.5 yˆ  0.5 zˆ  B2 n
2
B1t  B1  B1n  2 xˆ  0.5 yˆ  0.5 zˆ
B1n 
H1t 
1
0
â n  n̂12
#2
0.5 xˆ  0.125 yˆ  0.125 zˆ   H 2t
B2t   0  r 2 H 2t  3 xˆ  0.75 yˆ  0.75 zˆ
#1
 r2 = 6
y
O
 r1 = 4
 B2  B2t  B2 n
B2  3xˆ  1.25 yˆ  0.25 zˆ
H2 
1
0
(T)
0.5 xˆ  0.21yˆ  0.04 zˆ  (A/m)
8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE
Simple electric circuit that shows the effect of energy stored in a
magnetic field of an inductor :
Magnetic
Coil
flux
_
I
VL
+
From circuit theory the induced potential across a wire wound coil
such as solenoid or a toroid :
dI
VL  L
dt
where L is the inductance of the coil, I is the time varying current
flowing through the coil – inductor.
In a capacitor, the energy is stored in the electric field :
In an inductor, the energy is stored in
the electric field, as suggested in the
diagram :
t t 0
1
WE  CV 2
2
Magnetic
flux
Coil
t t 0
 dI 
Wm   VL Idt    L  Idt
dt 
t 0
t 0 
_
I
VL
t t 0
1
  LI dI  LI 2
2
t 0
where

switch
( Joule)
Define the inductance of an inductor :
+

L
Henry
I
(lambda) is the total flux linkage of the inductor

L
H ( Henry )
I
  m N
I2
Weber turns
Circuit 2
N2 turns
Hence :
L
 mN
I
H
I1
Circuit 1
N1 turns
Two circuits coupled by a common
magnetic flux that leads to mutual
inductance.
Mutual inductance :
M 12

 12
I1
 12 is the linkage of circuit 2
produced by I1 in circuit 1
For linear magnetic medium
M12 = M21
Ex. 8.8: Obtain the self inductance of the long solenoid shown in the diagram.
Solution:
Assume all the flux ψ m links all N turns and
that B does not vary over the cross section
area of the solenoid.
flux
 
   m N  B a 2 N
We have B  H
N turns
 NI  2
  H  a N  
 a N
 l 
 
 
2
 N 2 I
 
 l
 N 2a 2
L  
I
l
 2
 a

 
Ex. 8.9: Obtain the self inductance of the toroid shown in the diagram.
Solution:
  c  a 
B
4
  mN

L 

I
I
I
NI

SN
 2b
I
2
Mean path


N


Cross sectional
m
 0 area S
b
c
a
I
N 2 S
L 
2b
where b - mean radius
S - toroidal cross sectional area
__
ds aˆl Bave
N turns
Feromagnetic
core
Ex. 8.10: Obtain the expression for self inductance per meter of the coaxial
cable when the current flow is restricted to the surface of the inner conductor
and the inner surface of the outer conductor as shown in the diagram.
Solution:
The ψ m will exist only between a and b
and will link all the current I
 m
L


I
I
1 b

H drc dz 
0 a
I drc dz 
 
2rc
I
0 a

b

ln
2
a
1 b
I
b
ψm
a
Ex. 8.11: Find the expression for the mutual inductance between
circuit 1 and circuit 2 as shown in the diagram.
I2
Solution:
Let us assume the mean path :
Circuit 2
N2 turns
2b >> (c-a)
M 12
12  m (12) N 2


I1
I1
  c  a 
B12 
4


I1
2

c


N2


N1 I1
SN 2
N1 N 2 S
2b

I1
2b
b
a
I1
Circuit 1
N1 turns
Two circuits coupled by a common
magnetic flux that leads to mutual
inductance.
8.7 MAGNETIC ENERGY DENSITY
We have :
L

I
Henry

m
b
c
a
1
1 2 1
Wm  LI 2 
I  I
2
2 I
2
I
S
Joule
Consider a toroidal ring : The energy in the magnetic field :
1
1
Wm   m NI  BSNI
2
2
N turns
Multiplying the numerator and denominator by 2b :
1 NI
S 2b 
Wm  B
2 2b
where
NI
 H and ( S 2b) is the volume V of the toroid
2b
Hence :
1
Wm  BHV
2
wm 
Wm 1
1
 BH  H 2
V
2
2
Jm 3
In vector form :
1
wm  B  H
2
Hence the inductance :
2
2 1
L  2 Wm  2  B  H dv
I
I v2
Ex. 8.12: Derive the expression for stored magnetic energy density in a
coaxial cable with the length l and the radius of the inner conductor a and
the inner radius of the outer conductor is b. The permeability of the
dielectric is  .
Solution:
H 
I
2r
1
I 2
2
Wm    H dv 
2 v
8 2
I 2 b 1
Wm 
 2rl dr
2 
2
8 a r
I 2l  b 

ln  
4
a
(J)
1
v r 2 dv
b
ψm
a