Electromagnetics Wave and Fields
Lecture 7
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Dr. M. Tanseer Ali
EMWF Lec 7 /1
Engineering Electromagnetics
Chapter 8
The Steady Magnetic Field
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Dr. M. Tanseer Ali
EMWF Lec 7 /2
Chapter 8
The Steady Magnetic Field
Steady Magnetic Field
 This chapter discusses the magnetic field with definition of
the magnetic field itself and show how it arises from a
current distribution.
 The source of magnetic field may be a permanent magnet,
an electric field changing linearly with time or a direct
current.
 Biot-Savart’s Law describes the relation between the
magnetic field produced by the differential DC element in
free space.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /3
Chapter 8
The Steady Magnetic Field
Biot-Savart Law
When current I is flowing through the differential length dL of
cylindrical conductor of circular cross section as the radius
approaches to zero; then in vector notation the Biot-Savart
Law can be given as:
dH 
IdL  aR IdL  R

2
4R
4R 3
Where H is the magnetic field
intensity, R is the distance at which the
magnetic field is measured and aR is
the unit vector along radial distance of
the point from the filament.
dH 
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I dL sin aR
4R 2
Dr. M. Tanseer Ali
θ
dH 2 
I1dL1  aR12
4R122
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Chapter 8
The Steady Magnetic Field
Biot-Savart Law
Integral form of Biot-Savart Law
H
IdL  aR
4R 2
Current can be expressed with surface density (K) or volume density (J)
I dL  K dS  J dV
K  aR dS
J  aR dV

vol 4R 2
S
4R 2
H
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Dr. M. Tanseer Ali
EMWF Lec 7 /5
Chapter 8
The Steady Magnetic Field
Biot-Savart Law
 Let us find the magnetic field intensity produced by an
infinite long filament carrying a current I. The filament lies on
the z axis in free space, flowing to az direction.
R   a   z ' a z and

4  2  z '2

I

4
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
3/ 2

4  2  z '2
z  

 2  z '2
I1dz ' a z   a   z ' a z 

Considering cylindrical coordinate
system, no variation with z or Φ
can exist.
a R12 
I1dz ' a z   a   z ' a z 
dH 2 
H
 a   z' a z

 
z  
I a 
3/ 2
dz ' a 
2
 z'

2 3/ 2

z'
2  2  2  z '2
Dr. M. Tanseer Ali


z  
I
2
a
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Chapter 8
The Steady Magnetic Field
Biot-Savart Law
 The direction of the magnetic-field-intensity vector is
circumferential. The streamlines are therfore circles about
the filament.
H
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Dr. M. Tanseer Ali
I
2
a
EMWF Lec 7 /7
Chapter 8
The Steady Magnetic Field
Biot-Savart Law
For a finite length of wire, H at any point can be expressed in
terms of α1 and α2
H
I
4
sin 2  sin1  a 
If one or both ends are below point
2, then α1 , or both α1 and α2 are
negative.
This equation can be used to find
the magnetic field intensity caused
by current filaments arranged as
sequence of straight line
segments.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /8
Chapter 8
The Steady Magnetic Field
Example
 Determine H at P2(0.4,0.3,0) in the field of an 8-A filament
current directed inwards from infinity to the origin on the positive
x axis and outward to infinity along y axis.
 x  0.3
8
sin 53.1  1a
H 2 x  
4 0.3
12
 a A/m

The unit vector aϕ on z = 0 plane is
same downward direction along az
with respect to x axis.
1x  90
2x
0.4
 tan
 53.1
0.3
-1
 H 2 x   
12

a z A/m
aΦ or -az
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Dr. M. Tanseer Ali
EMWF Lec 7 /9
Chapter 8
The Steady Magnetic Field
Example
 Determine H at P2(0.4,0.3,0) in the field of an 8-A filament
current directed inwards from infinity to the origin on the positive
x axis and outward to infinity along y axis.
 y  0.4
8
1  sin 36.9 a z 
H 2 y  
4 0.4
8
  a z A/m

 H 2  H 2 x   H 2 y 
 2 y  90
0.3
1y  tan
 36.9
0.4
-1
aΦ or -az
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Dr. M. Tanseer Ali

12

20


az 
8

a z A/m
a z   6.37a z A/m
EMWF Lec 7 /10
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 In solving electrostatic problems, whenever a high degree of
symmetry is present, we found that they could be solved much
more easily by using Gauss’s law compared to Coulomb’s law.
 Again, an analogous procedure exists in magnetic field.
 Here, the law that helps solving problems more easily is known
as Ampere’s circuital law.
 The derivation of this law will waits until several subsection
ahead. For the present we accept Ampere’s circuital law as
another law capable of experimental proof.
 Ampere’s circuital law states that the line integral of magnetic
field intensity H about any closed path is exactly equal to the
direct current enclosed by that path,
 H  dL  I
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /11
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
• The line integral of H about the closed
path a and b is equal to I
• The integral around path c is less than I.
 The application of Ampere’s circuital law involves finding the
total current enclosed by a closed path.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /12
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 Let us again find the magnetic field
intensity produced by an infinite long
filament carrying a current I. The
filament lies on the z axis in free
space, flowing to az direction.
 We choose a convenient path to any
section of which H is either
perpendicular or tangential and
along which the magnitude H is
constant.
 The path must be a circle of radius ρ, and Ampere’s circuital
law can be written as
2
 H  dL  
0
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2
H  d  H   d  H 2  I
0
Dr. M. Tanseer Ali
 H 
I
2
EMWF Lec 7 /13
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 As a second example, consider an infinitely long coaxial
transmission line, carrying a uniformly distributed total current I
in the center conductor and –I in the outer conductor.
 A circular path of radius ρ, where ρ is larger than the radius of
the inner conductor a but less than the inner radius of the outer
conductor b, leads immediately to
I
H 
( a    b)
2
 If ρ < a, the current enclosed is
I encl  2 H  I
2
a2
 Resulting

H  I
(   a)
2
2 a
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /14
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 If the radius ρ is larger than the outer radius of the outer
conductor, no current is enclosed and
H  0
(   c)
 Finally, if the path lies within the outer conductor, we have
  2  b2 
2 H  I  I  2
2 
 c b 
I c2   2
H 
2 c 2  b2
(b    c)
• ρ components cancel,
z component is zero.
• Only φ component of H
does exist.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /15
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 The magnetic-field-strength variation with radius is shown
below for a coaxial cable in which b = 3a, c = 4a.
 It should be noted that the magnetic field intensity H is
continuous at all the conductor boundaries  The value of Hφ
does not show sudden jumps.
 Outside the coaxial cable, a complete cancellation of magnetic
field occurs. Such coaxial cable would not produce any
noticeable effect to the surroundings (“shielding”)
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /16
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 As final example, consider a sheet of current flowing in the
positive y direction and located in the z = 0 plane, with uniform
surface current density K = Ky ay.
 Due to symmetry, H cannot vary with x and y.
 If the sheet is subdivided into a number of filaments, it is
evident that no filament can produce an Hy component.
 Moreover, the Biot-Savart law shows that the contributions to
Hz produced by a symmetrically located pair of filaments cancel
each other.  Hz is zero also.
 Thus, only Hx component is present.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /17
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 We therefore choose the path 1-1’-2’-2-1 composed of straightline segments which are either parallel or perpendicular to Hx
and enclose the current sheet.
 Ampere's circuital law gives
H x1L  H x 2 ( L)  K y L  H x1  H x 2  K y
 If we choose a new path 3-3’-2’-2-3, the same current is
enclosed, giving
H x3  H x 2  K y
and therefore
H x3  H x1
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /18
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 Because of the symmetry, then, the magnetic field intensity on
one side of the current sheet is the negative of that on the other
side.
 Above the sheet
H x  12 K y ( z  0)
while below it
H x   12 K y ( z  0)
 Letting aN be a unit vector normal (outward) to the current
sheet, this result may be written in a form correct for all z as
H  12 K  a N
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /19
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 If a second sheet of current flowing in the opposite direction,
K = –Ky ay, is placed at z = h, then the field in the region
between the current sheets is
H  K  aN
(0  z  h)
and is zero elsewhere
H0
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( z  0, z  h)
Dr. M. Tanseer Ali
EMWF Lec 7 /20
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 The difficult part of the application of Ampere’s circuital law is
the determination of the components of the field which are
present.
 The surest method is the logical application of the Biot-Savart
law and a knowledge of the magnetic fields of simple form (line,
sheet of current, “volume of current”).
• Solenoid
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• Toroid
Dr. M. Tanseer Ali
EMWF Lec 7 /21
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 For an infinitely long
solenoid of radius a and
uniform current density
Ka aφ, the result is
H  Kaa z
(   a)
H0
(   a)
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 If the solenoid has a finite
length d and consists of N
closely wound turns of a
filament that carries a
current I, then
NI
well within
H
a z (the
solenoid)
d
Dr. M. Tanseer Ali
EMWF Lec 7 /22
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
 For a toroid with ideal case
0  a
inside
H  Ka
a (toroid
)

(outside
H0
toroid)
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 For the N-turn toroid, we have
the good approximations
Dr. M. Tanseer Ali
NI
H
a
2
H0
(inside
toroid)
(outside
toroid)
EMWF Lec 7 /23
Chapter 8
The Steady Magnetic Field
Curl
For an incremental closed path of sides ∆x and ∆y which
produces a reference value for H at the center.
H  L12  H y,12y
H 0  H x 0a x  H y 0 a y  H z 0 a z
1
 x
2
1
x
2
1
 y
2
1
y
2
H y  1 
H y ,12  H y 0 
 x 
x  2 
H


H  L12   H y 0  y  1 x  y
x  2 

H  L23   H x 0  H x  1 y  x

H  L34
y  2

H y  1 

  H y 0 
 x  y
x  2 

H  L41   H x 0  H x  1 y  x

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Dr. M. Tanseer Ali
y  2

EMWF Lec 7 /24
Chapter 8
The Steady Magnetic Field
Curl
 H y H x 
 H  dL   x  y  xy
 H  dL  I
Combining all the paths,
By amperes circuital law,
Here, I  J z xy
H 0  H x 0a x  H y 0 a y  H z 0 a z
∆I
 H y H x 
 H  dL   x  y  xy  J z xy
or,
 H  dL   H
xy
lim
xy  0
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Dr. M. Tanseer Ali

 x
y

H x 
  Jz
y 
 H  dL   H
xy

 x
y

H x 
  Jz
y 
EMWF Lec 7 /25
Chapter 8
The Steady Magnetic Field
Curl
The Curl of any vector is a vector, and any component of the
curl is given by the limit of the quotient of the closed line integral
of the vector, about a small path in a plane normal to that
component desired and the area enclosed.
By amperes circuital law,
H 0  H x 0a x  H y 0 a y  H z 0 a z
curl H N 
lim
S N  0
 H  dL    H
S N
∆I
H 
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Dr. M. Tanseer Ali
lim
S N  0
 H  dL  J
S N
z
EMWF Lec 7 /26
Chapter 8
The Steady Magnetic Field
Curl
Properties of Curl:
lim
xy  0
lim
yz  0
lim
zx  0
 H  dL   H
xy

 x
 H  dL   H
yz

H x 
  Jz
y 
H y 


  Jx

y

z


 H  dL   H
zx
y
 z

z
x

ax

H  J 
x
Hz
ay

y
Hz
az

z
Hz
H z 
 Jy

x 
 H z H y 
 H y H x 
 H x H z 
a x  
a z  J
  H  



a y  
z 
x 
y 
 z
 y
 x
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Dr. M. Tanseer Ali
EMWF Lec 7 /27
Chapter 8
The Steady Magnetic Field
Curl
Rectangular Coordinate:
ax

H  J 
x
Hx
ay

y
Hy
az

z
Hz
Cylindrical Coordinate:
 1 H z H 
 H
 1 H  1 H  
H z 
a     
a   
a z
  H  J  


z 
 
  
  
 z
  
Spherical Coordinate:
1  H sin H

H 

rsin  

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
1  1 H r rH  
1  rH  H r 
a r  
a  


a 
r  sin 
r 
r  r
 

Dr. M. Tanseer Ali
EMWF Lec 7 /28
Chapter 8
The Steady Magnetic Field
Curl
Example 5.2: Calculate  H  dL about a square path with side d
centered at (0, 0, z1) in the y = 0 plane where z1 > 2d.
2
H

0
.
2
z
a x for z > 0 and .H  0 elsewhere.
Where
2
2
1 
1 


H

dL

0
.
2
z

d
d

0

0
.
2
z

d d 0
 1

 1

2 
2 


 0.4 z1d 2
H 
lim
d 0
ax

H  J 
x
0.2 z 2
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Dr. M. Tanseer Ali
 H  dL 
d2
ay

y
0
lim 0.4 z1d 2
 0.4 z1
2
d 0 d
az


 0.2 z 2  a y  0.4 z a y
z z
0
EMWF Lec 7 /29
Chapter 8
The Steady Magnetic Field
Stokes Theorem
 From Curl of H
 H  dL
S
 H  dL
S
S
   H N
   H N  S
 H  dL S  S   H dS
Stokes’ Theorem
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Dr. M. Tanseer Ali
EMWF Lec 7 /30
Chapter 8
The Steady Magnetic Field
Stokes Theorem
 Example
Consider the portion of sphere shown in the figure. The surface is specified
by r  4, 0    0.1 , 0    0.3 and the closed path forming its parameter is
composed of three circular arcs. Given,H  6r sin  a r  18r sin  cos  a .
Evaluate each side of Stokes’ Theorem
dL  dr a r  rd a  r sin  d a
 H  dL
S
  H rd   H r sin d   H rd
1

0.3
0
2
3
184sin 0.1  cos  4sin 0.1 d
 288 sin 2 0.1 sin0.3   22.2 A
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Dr. M. Tanseer Ali
 H  0
EMWF Lec 7 /31
Chapter 8
The Steady Magnetic Field
Stokes Theorem
  H  
1
36r sin  cos  cos  a r  1  1 6rcos   36r sin  cos  a
r sin
r  sin 

Here, the surface dS  r 2sin d d a r
0.3
0.1
0
0
   H dS    36 cos  cos  16 sin  d d
S

0.3
0
0.1
1

576 sin 2  cos d
2
0
 288 sin 2 0.1 sin0.3   22.2 A
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /32
Chapter 8
The Steady Magnetic Field
Stokes Theorem
 Stokes theorem relates surface integral to a closed line integral.
 While divergence theorem relates a volume integral to a closed
surface integral
 Let consider A to be any vector, then
 A  T
Is T is vector or scalar?
vol
    A dv   Tdv
vol
S   A dS  volTdv
Divergence theorem,
SB  dS  vol  B dv
 dS approaches to zero (from curl conditions); hence closed
path also approaches to zero and stokes theorem gives
S   A dS  0  volTdv  0  T  0
 A  0
AIUB
Dr. M. Tanseer Ali
This is a useful identity in
vector calculus
EMWF Lec 7 /33
Chapter 8
The Steady Magnetic Field
Stokes Theorem
 In case of magnetic field,
 H  0
 J  0
 Ampere’s circuital law from stokes theorem
 H  J
    H  dS   J  dS
S
S
or,  H  dL   J  dS  I
S
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /34
Chapter 8
The Steady Magnetic Field
Magnetic Flux and Magnetic Flux Density
 In free space, the magnetic flux density
B  0 H
 Unit of magnetic flux density is webers per meter sq (Wb/m2) or
tesla (T)
 Magnetic flux,
   B  dS Wb
S
 Gauss’s law, states   D  dS  Q

S
Divergence theorem,
SB  dS  vol  B dv
 The charge Q is the source of the lines of electric flux and these
lines begins & ends on positive and negative charge.
 While the magnetic flux lines are closed and do not terminate
on a “magnetic charge”.
 Hence Gauss’s law for the magnetic field is
 B  dS  0
S
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 B  0
Dr. M. Tanseer Ali
This is Maxwell’s
fourth Equations
EMWF Lec 7 /35
Chapter 8
The Steady Magnetic Field
Magnetic Flux and Magnetic Flux Density
 Maxwell’s Equations,
  D  v
E  0
H  J
B  0
 Electric Field Equations,
SD  dS  Q  vol  v dv
 E  dL  0
 Magnetic Field Equations,
SH  dS  I
 B  dS  0
AIUB
Dr. M. Tanseer Ali
  J  dv
vol
EMWF Lec 7 /36
Chapter 8
The Steady Magnetic Field
Magnetic Flux and Magnetic Flux Density
 Example: consider an infinitely long coaxial transmission line,
carrying a uniformly distributed total current I in the center
conductor and –I in the outer conductor.
 A circular path of radius ρ, where ρ is larger than the radius of
the inner conductor a but less than the inner radius of the outer
conductor b, leads immediately to
I
H 
( a    b)
2
0 I
 B  0 H 
a
2
d b 0 I
   B  dS   
a   d dz a 
S
0 a 2
 0 Id b

ln
2 a
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /37
Chapter 8
The Steady Magnetic Field
Scalar and Vector Magnetic Potentials
 Similar to electric potential, scalar magnetic potential
H  Vm
Taking Curl of both sides,
  H  J     Vm 
Curl of the gradient of any scalar is identically zero.
   Vm   0
Hence, H  Vm is true only if J  0
 Laplace’s Equation
  B   0  H  0
0   Vm   0
2Vm  0
AIUB
Dr. M. Tanseer Ali
J  0
EMWF Lec 7 /38
Chapter 8
The Steady Magnetic Field
Scalar and Vector Magnetic Potentials
 Example: consider an infinitely long coaxial transmission line,
carrying a uniformly distributed total current I in the center
conductor and –I in the outer conductor.
I
H
a  for a    b, J  0
2
I
1 Vm
 Vm   
2
 
Vm
I


2
I
Thus, Vm  

2
1

 2n  
4

n  0,1,2,
1

Vm   I  n  
8

n  0,1,2,
I
Vm  
2
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /39
Chapter 8
The Steady Magnetic Field
Scalar and Vector Magnetic Potentials
 Properties of Scalar Magnetic Potential
 Vm is not single-valued function of position
 Vm is not a conservative field
a
Vm,ab   H  dL
b
(specified path)
 Vector magnetic potential A
B   A
H
1
0
 A
H  J 
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1
0
Divergence of the Curl of
any vector field is zero
Curl of the curl of a
vector field is not zero
 A
Dr. M. Tanseer Ali
EMWF Lec 7 /40
Chapter 8
The Steady Magnetic Field
Scalar and Vector Magnetic Potentials
 Vector magnetic potential A defined by Biot-Savart Law
0 IdL
A
4R
0 IdL
dA 
4R
 For a filament at the origin in free space and allowing it to
extend in the positive z direction so that dL=dz az
dA 
dA z 
0 Idza z
dA   0
4  2  z 2
0 Idza z
4  2  z 2
dA  0
dA   0
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /41
Chapter 8
The Steady Magnetic Field
Scalar and Vector Magnetic Potentials
 Vector magnetic potential A defined by Biot-Savart Law
1  dAz 
 
 a 
dH 
  dA 
0
0   
1
Idz

dH 
4  2  z 2

IdL  KdS
0KdS
A
S
4R
AIUB

3/ 2
a
IdL  Jdv
0 Jdv
A
vol 4R
Dr. M. Tanseer Ali
EMWF Lec 7 /42
Chapter 8
The Steady Magnetic Field
Assignment 4
Problem 1. Two filamentary current of 8 A each are directed inward from infinity to the origin on the
positive x axis, and outward to infinity along the y axis. Determine H at P2(0.4, 0.3, 0).
Problem 2. A current filament carrying 15 A in the az direction lies along the entire z axis. Find H in
rectangular coordinates at: (a) PA(√20, 0, 4); (b) PB(2,−4, 4).
Problem 3. A coaxial cable is excited from a current source of 20 A. The radius of inner conductor is 4 m.
Radius of inner surface of Outer conductor is 4.25 m. The Outer conductor having thickness of 0.75 m. Find
the magnitude of magnetic field intensity at the region: (i) 2.5 m, (ii) 4.8 m, (iii) 4.15 m, (iv) 5.75 m.
Problem 4. Suppose that H = 0.2 z2 ax for z > 0, and H = 0 elsewhere. Calculate about a square path with
side d, centered at (0, 0, z1) in the y = 0 plane where z1 > d/2.
Problem 5. Calculate the value of the vector current density: (a) in rectangular coordinates at PA(2, 3, 4) if
H = x2z ay − y2x az; (b) in cylindrical coordinates at PB(1.5, 90◦, 0.5) if H  2 cos0.2 a  ; (c) in spherical

coordinates at PC(2, 30◦, 20◦) if H  1 a
sin 
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /43
Chapter 8
The Steady Magnetic Field
Assignment 4
Problem 6. Consider the portion of a sphere where the surface is specified by r = 4, 0 ≤ θ ≤ 0.1π, 0 ≤ φ ≤
0.3π, and the closed path forming its perimeter is composed of three circular arcs. Evaluate each side of Stokes’
theorem if H = 6r sin φ ar +18r sin θ cos φ aφ.
Problem 7. Evaluate both sides of Stokes’ theorem for the field H = 6xy ax − 3y2 ay A/m and the rectangular
path around the region, 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, z = 0. Let the positive direction of dS be az.
(a) Find H in cartesian components at P(2, 3, 4) if there is a current filament on the z axis carrying 8 mA in the
az direction. (b) Repeat if the filament is located at x = −1, y = 2. (c) Find H if both filaments are present.
Problem 8. Let a filamentary current of 5 mA be directed from infinity to the origin on the positive z axis and
then back out to infinity on the positive x axis. Find H at P(0, 1, 0).
Problem 9. When x, y, and z are positive and less than 5, a certain magnetic field intensity may be expressed
as H = [x2yz/(y + 1)] ax + 3x2z2 ay − [xyz2/(y + 1)] az. Find the total current in the ax direction that crosses
the strip x = 2, 1 ≤ y ≤ 4, 3 ≤ z ≤ 4, by a method utilizing: (a) a surface integral; (b) a closed line integral.
Problem 10. Evaluate both sides of Stokes’ theorem for the field H = 6xy ax − 3y2 ay A/m and the rectangular
path around the region, 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, z = 0. Let the positive direction of dS be az.
Problem 11. Given H = (3r 2/ sin θ)aθ + 54r cos θ aφ A/m in free space: (a) Find the total current in the aθ
direction through the conical surface θ = 20◦, 0 ≤ φ ≤ 2π, 0 ≤ r ≤ 5, by whatever side of Stokes’ theorem you
like the best. (b) Check the result by using the other side of Stokes’ theorem.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 7 /44