Tutorial 7 Solution Emagnet

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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 -;


.+  m
b. 6 7 71 )7 8  9 

  1 -: )+  1 - ) +

,

:  ) 



1 -;

 .3  m
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

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