# Electromagnetics Wave and Fields ```Electromagnetics Wave and Fields
Lecture 7
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Dr. M. Tanseer Ali
EMWF Lec 7 /1
Engineering Electromagnetics
Chapter 8
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Dr. M. Tanseer Ali
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Chapter 8
 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.
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Dr. M. Tanseer Ali
EMWF Lec 7 /3
Chapter 8
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
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θ
dH 2 
I1dL1  aR12
4R122
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Chapter 8
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
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
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
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.
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Dr. M. Tanseer Ali
EMWF Lec 7 /8
Chapter 8
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
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
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
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Dr. M. Tanseer Ali
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Chapter 8
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.
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Dr. M. Tanseer Ali
EMWF Lec 7 /12
Chapter 8
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
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 H 
I
2
EMWF Lec 7 /13
Chapter 8
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
I
H 
( a    b)
2
 If ρ &lt; a, the current enclosed is
I encl  2 H  I
2
a2
 Resulting

H  I
(   a)
2
2 a
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Dr. M. Tanseer Ali
EMWF Lec 7 /14
Chapter 8
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.
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Dr. M. Tanseer Ali
EMWF Lec 7 /15
Chapter 8
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”)
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Dr. M. Tanseer Ali
EMWF Lec 7 /16
Chapter 8
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.
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Dr. M. Tanseer Ali
EMWF Lec 7 /17
Chapter 8
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
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Dr. M. Tanseer Ali
EMWF Lec 7 /18
Chapter 8
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
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Dr. M. Tanseer Ali
EMWF Lec 7 /19
Chapter 8
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
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
Ampere’s Circuital Law
 For an infinitely long
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
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
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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y  2

EMWF Lec 7 /24
Chapter 8
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
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
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
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
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 &gt; 2d.
2
H

0
.
2
z
a x for z &gt; 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
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
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
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
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
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
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 &amp; 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
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
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
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
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
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
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 
AIUB
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
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
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
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 &gt; 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 &gt; 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
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
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