1. 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

2. 8.1 FORCE ON A MOVING POINT CHARGE
Force in electric field:
Fe QE
Force in magnetic field:
Fm QU B
Total force:
F Fe Fm or F QE U B
Also known as Lorentz force equation.

3. Force on charge in the influence of fields:
Charge Combination
Condition E Field BField E and B
Stationary QE - QE
Moving
QE QU B QE U B

4. 8.2 FORCE ON A FILAMENTARY CURRENT
__
The force on a differential current element , I dl
due to the uniform magnetic field, :
B
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.

5. 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
ˆ
field, B yB0 . 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.
__
r
F I dl B
x
F1 ˆ ˆ
x(2 Ir ) yB0 ˆ
z 2 IrB 0 (N) I

6. __
(b) For curve part, dl B will be in the –
ve z direction and the magnitude is y
proportional to sin B
r
F2 I dl B
0 x
I
ˆ
zI rB0 sin d ˆ
z 2 IrB0 (N)
0
Hence, it is observed that F2 F1 and it is shown that
the net force on a close loop is zero.

7. 8.3 FORCE BETWEEN TWO FILAMENTARY
CURRENT
z
Loop l2
Loop l1
__
R12 I 2 dl 2
__
ˆ
a R12
I1 dl1
I2
I1 P1(x1,y1,z1)
P2(x2,y2,z2)
y
x

8. We have : z
dF Id l x B (N) Loop l2
Loop l1
__
The magnetic field at point P2 due to the ˆ
a R12
R12 I 2 dl 2
__
filamentary current I1dl1 : I1 dl1
I2
ˆ
I1dl1 x aR12 I1 P1(x1,y1,z1)
dH 2 (A/m) P2(x2,y2,z2)
2
4 R12 y
x
o 1
ˆ
I dl1 x aR12
d dF2 I 2 dl2 x 2
4 R12
ˆ
I dl1 x aR12
o 1
dF2 I 2 dl2 x 2
I 2 dl2 x B2
l1 4 R12
where d F2 is the force due to I2dl2 and due to the magnetic field of loop l1

9. Integrate:
ˆ
I dl1 x aR12
o 1
F2 I 2 dl2 x 2
l2 l1 4 R12
II
o 1 2
ˆ
aR12 x dl1
F2 2
x dl2
4 l2 l1 R12
For surface current :
F2 J s 2 B2 ds
s
For volume current :
F2 J 2 B2 dv
v

10. 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
I1 F21
ˆI ˆ
x 0 I1 I2
1
B2 0H2 0 y
2 rc 2 d d
I1 = I2 = I
x
Hence:
1
ˆ
x 0 I1
1
ˆ
x 0 I1
F2 I 2 d 2 ˆ
I 2 ( zdz )
0
2 d 0
2 d
I1 I 2 I2
ˆ
y 0
ˆ
y 0
(N/m)
2 d 2 d

11. 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 force on the square loop.
y
Solution:
Field created in the square loop due to
filamentary current : (1,2,0) (3,2,0)
I 15 ˆ ˆ ˆ
al a R 15 A
H ˆ
z ˆ
z A/m 2 mA
2 x 2 x ˆ ˆ ˆ
y x z
3 10 -6 x
∴B 0 H 4 -7
10 H ˆ
z T z
(1,0,0) (3,0,0)
x

12. Hence: y
__ __
F I dl B I B dl
(1,2,0) (3,2,0)
3
ˆ
z
2
ˆ
z 15 A
F 2 10 3
3 10 6
ˆ
dxx ˆ
dyy 2 mA
x 1
x y 0
3 x
z
(1,0,0) (3,0,0)
1
ˆ
z
0
ˆ
z
ˆ
dxx ˆ
dyy
x 3
x y 2
1
3 1 2 1 0
F 6 10 9
ˆ
ln x 1 y y0 ˆ
x ˆ
ln x 3 y y2 ˆ
x
3
2 1
6 10 9
ˆ
ln 3 y ˆ
x ˆ ˆ
ln y 2 x
3 3
ˆ
8 x nN

13. 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 will produce bound current or
magnetization current.

14. 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 A/m
v 0 vi1
where M is a magnetization and n is the volume
dipole density when v -> 0.

15. 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 ___
dmi 0 dm' stend to align
Microscopic ___
base dmi 0 M 0 themselves
M 0

16. 8.4.2 BOUND MAGNETIZATION CURRENT DENSITIES
J sm and J m
z
___
y dm
x Ba
M
Ba
M
I M
___
dm Ba
M
J sm
___
Alignment of dm' s within a magnetic material under uniform B a
conditions to form a non zero J sm on the slab surfaces, and
a J m 0 within the material.

17. 8.4.3 TO FIND J sm and J m
y

18. Bound magnetization current :
dI m Indv
__ __ __ __
dI m I (n ds dl ) (nI ds ) (dl )
___ ___
We have:
M n dm nI ds Am-1
Hence:
dI m M dl
__ __
I m = M dl through the loop l’
I m = J m ds on the surface bound by
s the loop l’

19. Using Stoke’s Theorem:
I m = J m ds M dl M ds
s l s
Hence:
is the bound magnetization current density
Jm M (Am-2)
within the magnetic material.
And to find Jsm :
M
From the diagram : dIm = Mtan dl'
loop l’
dI m On the slab ˆ
J sm n
M tan J sm surface
dl
J sm M n (Am-1) is the surface bound magnetization
current density

20. 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 (free charge)
B
J ; B o H
o
B due to free charges and bound
J Jm magnetization currents
o
M Jm Define:
B B
J M H M
o o
B
M J
o
H J

21. Hence:
B o (H M )
Magnetization in isotropic material:
M H m magnetic susceptibility
m
Hence:
B o H (1 m ) permeability
o r
r (1 m )
B H

22. Ex. 8.4: A slab of magnetic material is found in the region given by 0 ≤ z ≤ 2 m with r = 2.5.
If ˆ 5x ˆ
B 10 yxin the y mWb/m 2
slab, determine:
( a ) J (b) J m (c ) M (d ) J sm on z 0
Solution:
B 1 dBy dBx
(a) J H ˆ
z
0 r 4 10 7 2.5 dBx dy
106
ˆ
5 10 10 3 z ˆ
4.775z kA/m 2
(b) J m mJ r 1J 1.5 ˆ
4.775 z 10 3 ˆ
7.163 z kA/m 2
B
(c ) M mH m
0 r
ˆ ˆ
1.5 10 yx 5 xy 10 3
4 10 7 2.5
ˆ ˆ
4.775 yx 2.387 xy kA/m

23. (c ) M ˆ
4.775 yx ˆ
2.387 xy kA/m
(d) J sm M n
ˆ
Because of z = 0 is under the slab region of 0 ≤ z ≤ 2 , therefore ˆ
n ˆ
z
J sm ˆ
(4.775 yx ˆ
2.387 xy ) ˆ
z
ˆ ˆ
2.387 xx 4.775 yy kA/m

24. 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 mon the surface of the rod and J sm
within the rod. Assume the permeability of the rod equals
5 o.
P2 P3
0
b
magnetic
a
rod r P2’ P3’ z
0
P1 P4
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.

25. Hence:
P3
NI
H d ˆ ˆ
( zH z ) ( zdz) H z d I en d
 P2

NI
Hz Js P3
 0 P2
b
magnetic a
P2’ P3’
rod z
0
M ˆ
zM z m ˆ
( zH z ) in the rod
P1 P4
M 0 in air (since m of air is zero)
B 0 ˆ
( zH z ) in air
B 0 r ˆ
( zH z ) in the rod

26. Brod 5 o ˆ
zH z (c) J sm ˆ ˆ ˆ
M n ( zM z ) rc ˆM
(b) 5 z
Bair o ˆ
zH z
Jm M ˆ
( zM z ) 0
0
flux Js
Js aa
J sm ˆ
z
b
M
ˆ
n ˆ
r
Hz=NI/l
Hz
Bz 5 0Hz
flux Jsm
B z = 0 Hz
Mz
J sm
Mz=4Hz
Jsm Plots of H, B and M, Js and Jsm along
Js
the cross section of the solenoid
Js=Hz=NI/l
Jsm=4Hz and the magnetic rod

27. 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 Bais applied, there will be a slight alignment of the atomic dipole
moment to produce M 0
Eg. Aluminum - χm 2 10 5 , μ r 1.00002
(b) Ferromagnetic material : has strong magnetic moment in the dm
absence
of an applied field. Ba
Eg: metals such as nickel, cobalt and iron.

28. 8.5 MAGNETIC BOUNDARY CONDITIONS
To find the relationship between B , H and M
ˆ
n21
Region 1: 1 , m1 B/H
s l
a
h/2 b
Boundary h/2
h/2 d
c h/2
Region 2: 2 , m2 Js
To find normal component of B andthe boundary
at H
__
Consider a small cylinder as h 0 and use B ds 0
__
B ds B1n s B2 n s 0
B1n B2 n
1 H1n 2 H 2n

29. To find tangential component of ˆ ˆ
n21 = a n 21
B and H at the boundary1
Region 1: 1, m B/H
Consider a closed abcd as h 0
__ s l
and use H dl I enc a
h/2 b
lBoundary h/2
h/2 d
h/2
H1t l H 2t l I enc c
Region 2: 2 , m2 Js z
H1t H 2t Js
x x
where J sis perpendicular to the directions of andH1t H 2t y
In vector form : ˆ
z ˆ
x ˆ
y
ˆ
an 21 H1 H 2 Js
ˆ
a n 21 is a normal unit vector from region 2 to region 1

30. We have:
We have:
H1t H 2t Js
Hence: M m H
B1t B2t Hence:
Js
1 2 M 1t M 2t
Js
We have: m1 m2
M Jm We have:
M dV J m dV Im 1 H1n 2 H 2n
v v
__ Hence:
and M dl Im
l M 1n M 2n
Hence: 1 2
m1 m2
M 1t M 2t J sm
If Js = 0 :
B1t B2t
H 1t H 2t or
1 2

31. If the fields were defined by an angle normal to the interface
B1 cos 1 B1n B2 n B2 cos 2 (1)
B1 B2
sin 1 H1t H 2t sin 2 (2)
1 2
y
Divide (2) to (1): B1 or H 1
tan Region 1: 1 , m1 B1n
1 1 r1 1 B1
tan 2 2 r2 Boundary at y
= 0 plane
B1t
x
Region 2: 2 , m2
2
B2 or H 2

32. 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 x A/m on the surface at z = 0. Given
ˆ
B1 ˆ ˆ
2x 3y ˆ
z mT , find B 2 B1n
B1 ˆ
n21 ˆ
z
Solution: Js ˆ
80 x
Normal component B1 #1 B1t
H 1t
B1n ˆ ˆ
( B1 n12 )n12 Z=0
#2
ˆ ˆ ˆ
2x 3 y z ˆ
z ˆ
z B2 n B2t
B2 H 2t z
ˆ
z
ˆ
n12 ˆ
z
B2 n B1n ˆ
z y
x
B1t B1 B1n ˆ
2x ˆ
3y ˆ
H 2t H 1t n12 Js
B1t ˆ ˆ
2 x 3 y 10 3
ˆ
500 x 750 yˆ ˆ
z ˆ
80 x
H 1t
1 4 10 6 ˆ
500 x ˆ
750 y ˆ
80 y
ˆ
500 x ˆ
750 y A/m ˆ
500 x ˆ
670 y A/m
B2t 2 H 2t
ˆ ˆ
7 10 6 500x 670 y ˆ ˆ
3.5 x 4.69 y
B2 B2t B2 n ˆ ˆ ˆ
3.5 x 4.69 y z mT

33. 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 B ˆ ˆ
2 x y find B2 and H 2
1
Solution: z
The unit normal: ˆ
n ˆ
(y ˆ
z) / 2
y+z=1
#2 ˆ
an ˆ
n12
B1n ˆ ˆ
( B1 n12 ) n12
#1
ˆ
2x ˆ
y ˆ
y ˆ
z 1 r2 =6
B1n
2 2 O y
1 =4
B1n ˆ ˆ ˆ
n12 0.5 y 0.5 z B2 n r1
2 B2 B2t B2 n
B1t ˆ ˆ
B1 B1n 2 x 0.5 y 0.5 zˆ
B2 ˆ ˆ ˆ
3x 1.25 y 0.25z (T)
1
H1t ˆ ˆ ˆ
0.5 x 0.125 y 0.125z H 2t 1
0 H2 ˆ ˆ ˆ
0.5 x 0.21y 0.04 z (A/m)
0
B2t 0 r2 H 2t ˆ ˆ ˆ
3 x 0.75 y 0.75 z

34. 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
flux Coil
_
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.

35. In a capacitor, the energy is stored in the electric field : 1
WE CV 2
2
In an inductor, the energy is stored in the
Magnetic
electric field, as suggested in the diagram : flux Coil
t t0 t t0
dI
Wm VL Idt L Idt I
_
t 0 t 0
dt VL +
t t0
1 2 switch
LI dI LI ( Joule)
t 0
2
Define the inductance of an inductor :
L Henry
I
where (lambda) is the total flux linkage of the inductor

36. L H ( Henry) I2
I
m N Weber turns Circuit 2
N2 turns
Hence :
m N
L H I1 Circuit 1
I N1 turns
Two circuits coupled by a common
magnetic flux that leads to mutual
inductance.
Mutual inductance :
12 is the linkage of circuit 2
12
M 12 produced by I1 in circuit 1
I1
For linear magnetic medium M12
= M21

37. 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 thatB
does not vary over the cross section area of the flux
solenoid.
mN B a2 N
N turns
We have B H
2 NI
H a N a2 N
l
N 2I
a2
l
N 2 a2
L
I l

38. Ex. 8.9: Obtain the self inductance of the toroid shown in the diagram.
Solution:
Mean path
2
c a
B N Cross sectional
mN
4 0 area S
L b
c
I I I m
a
NI I
SN Feromagnetic core
2 b __
ˆ
ds al Bave
I
N 2S
L
2 b N turns
whereb - mean radius
S - toroidal crosssectional area

39. 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 mwill exist only between a and b and will
link all the current I
1 b
m H drc dz
L b
I I 0 a
I
1 b ψm
I drc dz a
0 a
2 rc I
b
ln
2 a

40. 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
2 b >> (c-a) N2 turns
c
12 m (12 ) N2 b
M 12 a
I1 I1
I1 Circuit 1
2
c a N1 turns
B12 N2
4 Two circuits coupled by a common
magnetic flux that leads to mutual
I1 inductance.
N1 I 1
SN 2
2 b N1 N 2 S
I1 2 b

41. 8.7 MAGNETIC ENERGY DENSITY
We have :
L Henry c
b
I m
a
I
1 2 1 1
Wm LI I2 I Joule S
2 2 I 2
Consider a toroidal ring : The energy in the magnetic field :
N turns
1 1
Wm m NI BSNI
2 2
Multiplying the numerator and denominator by 2 b :
1 NI
Wm B S2 b
2 2 b
NI
where H and ( S 2 b) is the volumeV of the toroid
2 b

42. Hence :
1
Wm BHV
2
Wm 1 1
wm BH H2 Jm 3
V 2 2
In vector form :
1
wm B H
2
Hence the inductance :
2 2 1
L 2
Wm B H dv
I I2 v
2

43. 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:
I
H
2 r
1 I2 1
Wm H 2 dv dv
2 v 8 2 vr2
b
b
I2 1
Wm 2 rl dr
8 2
r 2 ψm a
a
I 2l b
ln (J)
4 a