Page 206 - Maxwell House

186 Chapter 4

(, , ) where = sincos, = sinsin, = cos as shown in Figure 4.3.3b. Note
that the dipoles are shifted at the distance ±/√2 from origin along the y-axis. Therefore,

2
2
2
|| = � + +
⎫
⎪
2
2
⁄
| | = � + � − √2� + 2 (4.73)

⎬
2 ⎪
2
⁄
| | = � + � + √2� + 2
3 ⎭
Meanwhile, our goal to find the fields far away from the dipoles. It means that || ≫ √2 and
⁄
the expressions in (4.73) can be simplified as
2
| | = � − √2 + ⁄ ≅ �1 − ( √2) � = − ( √2) ⎫
2
2
⁄
⁄
⁄
2
⁄

(4.74)
⎬
2
2
2
2
⁄
⁄
⁄
⁄
| | = � + √2 + ⁄ ≅ �1 + ( √2) � = + ( √2)
3 ⎭
2
2
In both expressions (4.74) we omitted minor term ⁄ 2 ≪ and then apply the well-known
⁄
Taylor’s expansion (1 + ) 1/2 ≅ 1 + 2. Looking back at Figure 4.2.7b we see that =
sinsin. Thus,
⁄
| | ≅ − ( √2) sinsin � (4.75)
1
⁄
| | ≅ + ( √2) sinsin
3
Evidently, the dipoles 1 and 3 rotated counterclockwise ( ⟹ + 2) concur and align
⁄
with the dipoles 4 and 2 , respectively. Thus,
⁄
| | ≅ − ( √2) sincos � (4.76)
4
⁄
| | ≅ + ( √2) sincos
2
Taking into account the dipole orientation we obtain from (4.60) for free-of-loss medium
(−)
0 ( √2)sinsin
⁄
= 0 √2 ⎫
1
4
(−) ⎪
0 −( √2)sincos⎪
⁄
= 0 (− )√2
2
4 (4.77)
⁄
0 (−) −( √2)sinsin⎬
= 0 (− )√2
3
4
⎪
(−) ⎪
⁄
0 ( √2)sincos
= √2
4 0 ⎭
4
Combining the fields of the same direction dipoles we have
(−)
0
⁄
+ = √2 sin�( √2) sinsin�
1
3
0
2 � (4.78)
0 (−)
+ = − √2 sin�( √2) sincos�
⁄
4
2
0
2
In case of an electrically small loop (√2 ≪ 1)
sin�( √2) sinsin� ≅ ( √2) sinsin (4.79)
⁄
⁄