EKT 241 Electromagnetic Theory
TUTORIAL 4
1.
A conductive loop on the x-y plane is bounded by ρ = 3.0 cm, ρ = 5.0 cm, φ =
0 and φ = 90. 1.0 A of current flows in the loop, going in the aφ direction on
the ρ = 3.0 cm arm. Determine H at the origin.
2.
A current filament of 3 ax A lies along the x - axis. Find H in cartesian
components at P (−1, 3, 2).
3.
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):
4.
Given the field H = 3y2 ax, find the current passing through a square in the x-y
plane that has one corner at the origin and the opposite corner at (2, 2, 0).
5.
Given a 3.0 mm radius solid wire centered on the z-axis with an evenly
distributed 2.0 amps of current in the +az direction, state the magnetic field
intensity, H everywhere.
6.
An infinite length coaxial cable exists along the z-axis, with an inner shell of
radius a carrying current I in the +az direction and outer shell of radius b
carrying the return current. Find the magnetic flux passing through an area of
length h along the z-axis bounded by radius between a and b.
7.
The magnetic flux density in a region of free space is given by B = −3x ax +
5y ay − 2z az T. Find the total force on the rectangular loop shown in figure
below if it lies in the plane z = 0 and is bounded by x = 1, x = 3, y = 2, and y =
5, all dimensions in cm.
8.
A coaxial transmission line has a = 5 mm and b = 20 mm. Let its center lie on
the z axis and let a dc current I flow in the +az direction in the center
conductor. The volume between the conductors contains a magnetic material
for which µR = 2.5, as well as air. Find H and B everywhere between
conductors if Hφ = 600/π A/m at ρ = 10 mm, φ = π/2, and the magnetic
material is located at a < ρ < 3a.
Solution:
A conductive loop on the x-y plane is bounded by ρ = 3.0 cm, ρ = 5.0 cm, φ = 0 and
φ = 90. 1.0 A of current flows in the loop, going in the aφ direction on the ρ = 3.0
cm arm. Determine H at the origin.
1.
It can be figured as below:
Using Biot-Savart’s law :
H 
Idl  a R
4R 2
ab
dl  dya y
ba
dl  dxax
a R  a y
a R  a x
dl  a R  dya y  a y  0
dl  a R  dya x  a x  0
aa
dl  da
bb
  a ; dl  ada
  b ; dl  bda
a R  a 
a R  a 
a  a
dl  a R  0 ad
1 0
 2
H

0
dl   da
az
0  ada z
0
Iad
1
a  az
2 z
8a
4a
a
dl  a R  0
1
 2
H

0
a
 bd
0
az
0  bda z
0
I (bd )
1
az   az
2
8b
4b
For a  3cm, b  5cm


1
1
H 

 a z  1.667a z A m
2
2
8 5  10 
 8 3  10

 

2. A current filament of 3 ax A lies along the x - axis. Find H in cartesian
components at P (−1, 3, 2).
We use the Biot-Savart law,
H
IdL x a R
4R 2
Where I dL = 3dx ax , aR = [−(1 + x)ax + 3ay + 2az]/R, and R  x 2  2 x  14 .
Thus,
HP 





3dxax   (1  x)a x  3a y  2a z

4 ( x 2  2 x  14) 2


3
(9a z  6a y )dx
3

(9a z  6a y )( x  1)

2
 2 x  14) 2 4 (13) x  2 x  14  
2(9a z  6a y )

 0.110a z  0.073a y A / m
4 (13)
  4 ( x
2
3. 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):
The Biot - Savart law is applied to the two wire segments using the following setup:
H

IdL x a R
4R 2
  Idza  (  za  a )  Idxa  (  xa  a )
z
z
y
x
x
y

3
3
0
0
4 ( z 2  1) 2
4 ( x 2  1) 2





Idza x

3

0 4 ( z 2  1) 2
Idxa z

3
0 4 ( x 2  1) 2



xa z
I  za x




2
4  z 2  1
x 1 0 
0


I
a x  a z   0.40a x  a z mA / m

4
4.
It can be figured as below:
We evaluate the circulation of H around the square path.
b
c
d
a
a
b
c
d
 H  dL  I enc        
b
 3(0)
2
a x  dxa x  0
a
c
 3y
b
2
a x  dya y  0
d
 3(2)
0
2
a x  dxa x  12 dx  24
c
2
a

0
d
So we have Ienc = 24 A. The negative sign indicates current is going in the
-az direction.
5.
It can be figured as below:
Figure shows the situation along with the Amperian Paths. We have:
 H  dL  I enc where H  H a and dL  da , therefore 2H 
This will be true for each Amperian path.
Amperian Path 1:
 I enc
I enc   J  dS , J 
So: H 
I
a 2
a z ; I enc 
I
a 2

2
0
0
 d  d 
I 2
a2
I
 a for   a
2 a 2
Amperian Path 2: I enc  I
So: H 
I
2
a for   a
6.
From formula,
H
 I
a , B  0 a
2
2
I
For a < ρ < b,
   B  dS 
 0 I a
 I
b
ddza  0 ln(  ) a h

2 
2
 I b
 0 ln  Wb
2  a 
7.
First, note that in the plane z = 0, the z component of the given field is zero, so
will not contribute to the force. We use:
F
loop
IdL x B
Which in our case becomes with, I  30 A
 30dxax   3xax  5 y y  0.02 a y   30dya y   3x x  0.03 a x  5 ya y 
0.03
0.05
0.01
0.01
0.02
0.02
F

 30dxax   3xax  5 y y  0.05 a y   30dya y   3x x  0.01 a x  5 ya y 
0.03
0.05
Simplifying this becomes,
0.03
0.05
0.01
0.01
0.02
0.02
 30(5)(0.02)a z dx    30(3)(0.03) a z dy
F

 30(5)(0.05)a z dx    30(3)(0.01) a z dy
0.03
0.05
 0.06  0.081  0.150  0.027  a z N  36a z mN
8.
First, we know that H  
I
2 (10
2

)
600

I
2
from which we construct:
 I  12 A
Since the interface between the two media lies in the aφ direction, we use the
boundary condition of continuity of tangential H and write:
H 5    20 
12
6
a 
a A / m
2

In the magnetic material, we find:
B 5    15  H 
(2.5)( 4x10 7 )(12)
6
a  a T
2
