文档介绍：Chapter 11
Radiation
Problem
From Eq. , A = - ILo4POUJ7r r~sin[UJ(t - rlc)](cos(Jr - sin B8), so
ILoPoUJ I a 21 . 1 a . 2 1 .
= -- -- r -sm[UJ(t-r Ic)]coBB +-;--- -SIn B-smUJ(t-rlc)][
47r { r2ar [ r ] rsmB aB [ r ] }
= --ILoPoUJ I . UJr 2 sin B cos B .
47r { r2" (sm[UJ(t- riG)]- -c cos[UJ(t- riG)])coBB- r 2'sm () sm[UJ(t- riG)]}
PoUJ 1 . UJ
= ILOfO- 2"sm[UJ(t-rlc)]+-cos[UJ(t-rlc)] cosB .
{ 47rfO(r rc ) }
Meanwhile, from Eq. ,
av PoCOgB UJ2 UJ .
- = -- cos[UJ(t- riG)]- - sm[UJ(t- riG)]
at 47rfor { c r }
PoUJ I. UJ av
= -- 2" sm[UJ(t riG)]+ cos[UJ(t riG)] cosB. So V. A -ILOfO- . Qed
~ { r - -n - } = at
Problem
: V(r,t) = --
47r~C-sm[UJ(t-rlc)].r : A(r,t) = ---sm[UJ(t-rlc)].~ r
I UJ PO' f . I I ILoUJPo .
NowPo x f =posinBcP and f x (Po x f) =posinB(f x cP)= -posinB8, so
ILOUJ2f x (Po x f) ILOUJ2(Po x f)
Eq. : cos[UJ(t . cos[UJ(t riG)].
IE(r, t) = _47r r - riG)]. : B(r, t) = -- 47rC r -
ILOUJ4(Po x f)2 A
Eq. : (8)
I = _327r2c r 2 r.
Problem
P = [2 R = q5UJ2sin2(UJt)R (Eq. ) =?(P)= ~q5UJ2R. Equate this to Eq. :
1 2 2 ILOQ5fflUJ4 ILofflUJ2 . 27rc
= 127rc =? R = 67rc j or, SInceUJ= A'
2 2
ffl47r2C2 2 d 2 2 d d
R = ~:c ~ = 37rILoC:x( ) = 37r(47rx 10-7)(3X1O8):x( ) = 807r2(:x) n = I (dl A)
195
196 CHAPTER 11. RADIATION
For the wires in an ordinary radio, with d = 5 X 10-2 m and (say) >.= 103m, R = 790(5 X 10-5)2 = 2 X10-60,
which is pared to the Ohmic resistance.
Problem
By the superposition principle, we can add the potentials of the two dipoles. Let's first express V (Eq. )
POW
in Cartesian coordinates: V(x, y, z, t) = - . sin[w(t-r Ic)]. That's for an oscillatingdipole
41fEOC(x 2 + yz2 + z 2 )
along the z axis. For one along x or y, we just change z to x or y. In the present case,
P = Po[cos(wt) x+ cos(wt - 1f12) y], so the one along y is delayed by a phase angle 1f12:
sin[w(t - TIc)] -+ s