EKT 241: ELECTROMAGNETIC THEORY
SEM 2 2010/2011
TUTORIAL 3
1. In a cylindrical coordinate system, a 2-m-long straight wire carrying a current of 5 A in

the positive z-direction is located at r  4 cm ,   and - 1 m  z  1 m . If
2
B  rˆ 0.2 cos  (T) , what is the magnetic force acting on the wire?
Answer:
2. Two infinitely long, parallel wires carry 6A current in opposite directions. Calculate
the magnetic flux density at point P in Figure 1.
Figure 1
Answer:
3. Explain the meaning of self-inductance and mutual inductance by including diagrams
and equations in your explanation.
Answer:
Self-inductance, L of any conducting structure is the ratio of the magnetic flux linkage,
Λ to the current I flowing through the structure.
L
Where;

I
H 
 N2 
  N   
IS  (Wb)
Hence,
 l

L
N2
S solenoid
l

Mutual inductance – produced by magnetic coupling between two different conducting
structures.
Magnetic field B1 generated by current I1 results in a flux Φ12 through loop 2:
12   B1  dS
S2
If loop 2 consists of N2 turns all coupled by B1 in exactly the same way, the total
magnetic flux linkage through loop 2 due to B1 is:
12  N 212  N 2  B1  dS
S2
Hence, the mutual inductance:
L12 
12 N 2

B1  ds H 
I1
I1 s2
 1.5  10 6
cos a r A/m exist in free space. Find the
r
magnetic flux crossing the surface defined by 0     / 3 and 0  z  3 m .
4. A radial magnetic field, H 
Answer:
   B  ds
since B   0 H
   0  H  ds
 1.5  10 6
  0   
z 0  0
r

3
 0 
3
z 0
 /3

 cos a r  rddza r

 1.5  10 cos ddz
 /3
6
0
  0  1.5  10 6 sin  0
z 30
 4  10 7  1.5  10 6  0.866  3
 /3
 4.9 Webers
5. A long cylindrical conductor whose axis is coincident with the z-axis has a radius a
and carries a current characterized by a current density J  zˆ J 0 / r , where J 0 is a
constant and r is the radial distance from the cylinder’s axis. Derive an expression for
the magnetic field H for the regions 0  r  a and r  a .
6. The rectangular loop shown in Figure 2 is coplanar with the inifinitely long, straight
wire carrying the current I = 20 A. Determine the magnetic flux through the loop.
Figure 2
Answer:
The magnetic field due to the wire:
I
H
a
2r
B  0H
 I 
 4  10 7 
a 
 2r 
 20 
 4  10 7 
a 
 2r 

4  10 6
a
r
The magnetic flux through the loop:
   Β  dS
S
The differential area of the loop:
dS   drdza 
   Β  dS 
S
4  10 6

a   drdza 
z  0.05 r  0.05
r
 0.2 
 4  10 6 0.3 ln 

 0.05 
0.35

0.2

 1.66  10 6 Wb
7. A coaxial cable has inner conductor of radius a = 5 mm and outer conductor of radius
b = 25 mm. The center of the coaxial cable lies on the z axis and a dc current I flow in the
center conductor in the +az direction. The region between the conductors contains a
magnetic material with  r  2.5 located at a  r  3a and air located at 3a  r  5a . Find
60
A / m at r  10 mm .
B everywhere between the conductors if H  

Answer:
First, we know that H  
I
2r
from which we construct:
I
60

2
2 (10 ) 
Hence,
I  1.2 A
Since the interface between the two media lies in the aφ direction and H is in aφ
direction, H is continuous across the two media.
(Note: the unit of r is in milimeter)
In the region
5  r  25
1.2
a
2r
0.6

a kA / m
r 
H
In the magnetic material,
5  r  15
B  H
( 2.5)( 4x10 7 )(1.2)
a
2r
0 .6

a mT
r

In the region of air, 15  r  25
B  0 H
(4x10 7 )(1.2)

a
2r
0.24

a mT
r
(Note: the unit of r is in milimeter)