TUTORIAL 7
(SOLUTION)
EEE 3133 ELECTROMAGNETIC FIELDS AND WAVES DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING
1.
The conducting triangular loop in Figure 1 carries a current of 10 A. Find H at (0,0,5) due to: a. Side 1 of the loop b. Side 3 of the loop
Figure 1 Ans: H1 = -59.1 ay mA/m, H2 = -30.63ax + 30.63 ay mA/m Solution:
a.
mA/m
)
!
")
) ) - .1
b. )
+
, +
*
)
!
")
-3 . 3 , 3 . 3 mA/m 2.
) ) *
-./ 0.1
)
A thin ring of radius 5cm is placed on plane z = 1cm so that its center is at (0,0,1 cm). If the ring carries 50 mA along aφ, find H at: a. (0,0,-1cm) b. (0,0,10 cm) Ans: (a)400 az mA/m, (b)57.3 az mA/m Solution:
a.
.4
5
.4 0 4 )4
6 7 71)7
8 +9
1 -: )+ 1 - ) +
,
: + )
1 -;
9 =
.+ m9
b. 6 7 71 )7 8 9
1 -: )+ 1 - ) +
,
: )
1 -;
9 >.3 m 9
3.
Plane y=1 carries current K = 50 az mA/m. Find H at: a. (0,0,0) b. (1,5,-3) Ans: (a)25 ax mA/m, (b)-25 ax mA/m Solution:
? @
a. 7 7 ) ) + mA/m
b. 177 3) ) + mA/m
4.
An infinitely long transmission line whose cross section in Figure 2 consisting of two concentric cylinders having their axes along the z-axis. The inner conductor has radius, a and carries current I while the outer conductor has inner radius, b and thickness, t carries return current, -I. Assuming the current is uniformly distributed in both conductors, determine the expression of H by applying Ampere’s Law along the Amperian path of the following regions: i. ii. iii. iv.
0≤ρ≤a a≤ρ≤b b ≤ p ≤ b+t ρ ≥ b+t
Figure 2
Solution: a. Region 0 ≤ ρ ≤ a
A B CD EF@G J K B CL
HI
K
E CL CC M
EF@G
E E E J K B CL J J NCC M M M
.
P! OP!
E J CD +M HI
E +M b. Region a ≤ ρ ≤ b
A B CD EF@G E +M E
H4
E +M
c. Region b ≤ p ≤ b+t
A B CD B +M EF@G
H5
EF@G E , J K B CL K
EF@G
E MQR , 6) R S
O
E R E, J J NCC E U1 V MQR , 6) R S 6 , +R6
E R U1 V +M 6 , +R6
P! OPT
d. Region ρ ≥ b+t
A B CD E , E)
H5
5.
A current distribution gives rise to the vector magnetic potential A = x2y ax + y2x ay – 4xyz az Wb/m. Calculate the following: a. B at (-1,2,5) b. The flux through the surface defined by z = 1, 0 ≤ x≤ 1, -1 ≤ y ≤ 4 Ans: (a)20 ax + 40 ay + 3 az Wb/m2, (b)20 Wb Solution:
a. W X < =Y8 ) , , =Z8) ,Z Y ) W17+7) + , = ,3 Wb/m+
b. ψ \ W B CL \P- \P!Z Y )CYCZ
J Z CZ ) J Y CY P-
P!
1 3= , 1) ) + Wb 3 3