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Introduction to Electrodynamics - ch11

# Introduction to Electrodynamics - ch11 - Chapter 11...

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Chapter 11 Radiation Problem 11.1 From Eq. 11.17, A = - ILo 4 POUJ ~ sin[UJ(t - rlc)](cos(Jr - sin B 8), so 7r r ILoPoUJ { I a [ 21 . I ] ] 1 a [ . 2 1 . [ ] } V.A = -- -- r -sm[UJ(t-r c) coBB +-;--- -SIn B-smUJ(t-rlc)] 47r r2 ar r rsmB aB r ILoPoUJ { I ( . UJr ) 2 sin B cos B . } = -- 4 2" sm[UJ(t- riG)] - - cos[UJ(t - riG)] coBB - 2' () sm[UJ(t- riG)] 7r r c r sm { PoUJ ( 1 . UJ ) } = ILOfO - 4 2"sm[UJ(t-rlc)]+-cos[UJ(t-rlc)] cosB . 7rfO r rc Meanwhile, from Eq. 11.12, av Po COg B { 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- a . Qed ~ r n t Problem 11.2 I UJ PO' f . I I ILoUJ Po . Eq.l1.14: V(r,t) = -- 4 -sm[UJ(t-rlc)]. Eq.l1.17: A(r,t) = ---sm[UJ(t-rlc)]. 7r~C r ~ r NowPo x f = posinBcP and f x (Po x f) = posinB(f x cP) = -posinB8, so ILOUJ2 f x (Po x f) ILOUJ2 (Po x f) Eq. 11.18: I E(r, t) = _ 4 cos[UJ(t - riG)]. Eq.11.19: B(r, t) = -- 4 . cos[UJ(t - riG)]. 7r r 7rC r ILOUJ4 (Po x f)2 A Eq. 11.21: I (8) = _ 32 2 2 r. 7r c r Problem 11.3 P = [2 R = q5UJ2 sin2(UJt)R (Eq. 11.15) =? (P) = ~q5UJ2 R. Equate this to Eq. 11.22: 1 2 2 ILOQ5fflUJ4 ILofflUJ2 . 27rc 2Qoc.v R = 127rc =? R = 67rc j or, SInceUJ= A' ffl47r2C2 2 ( d ) 2 2 ( d ) 2 ( d ) 2 R = ~:c ~ = 37rILoC:x = 37r(47rx 10-7)(3X1O8):x = 807r2 :x n = I 789.6(dl A)2 n.1 195

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196 CHAPTER 11. RADIATION For the wires in an ordinary radio, with d = 5 X10-2 m and (say) >.= 103m, R = 790(5 X10-5)2 = 2 X10-60, which is negligible compared to the Ohmic resistance. Problem 11.4 By the superposition principle, we can add the potentials of the two dipoles. Let's first express V (Eq. 11.14) in Cartesian coordinates: V(x, y, z, t) = - 4 POW ( 2 . z2 2 ) sin[w(t -r Ic)]. That's for an oscillatingdipole 1fEOC x + y + z 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)] -+ sin[w(t - TIc) -1f/2] = -cos[w(t - TIc)] (just let wt -+ wt -1f/2). Thus V - PoW { X. y } -- 4 2 2 2 sm[w(t - TIc)] - 2 2 2 cos[w(t - TIc)] 1fEOC x + y + z x + y + z - PoW sinO {coscPsin[w(t - TIc)] - sincPcos[w(t - rlc)]}. Similarly, 41fEOC r = A = -/-L 4 0POW {sin[w(t - TIc)] x - cos[w(t - TIc)] y}. 1fr We could get the fields by differentiating these potentials, but I prefer to work with Eqs. 11.18 and 11.19, using superposition. Since z = cosOf - sinOn, and cosO = ziT, Eq. 11.18 can be written 2 E = /-LO::: cos[w(t - TIc)] (z - ~ f). In the case of the rotating dipole, therefore, E = 2 . /-L~ {cos[w(t - TIc)] (x - ~f) + sin[w(t - TIc)] (y - ; f)} , 1~(fXE).1 B - s = ~(E x B) = ~ [E x (f x E)] = ~ [E2 f - (E .f)E] = E2 f (noticethat E. f = 0). Now /-Lo /-LoC /-Loc /-LoC E2 = ( /-LOPOw2 ) 2 {a2 COS2[W(t - TIc)] + b2 sin2[w(t - TIc)] + 2(a. b) sin[w(t- TIc)]cos[w(t - TIc)]} , 41fr where a ==x - (xlr)f and b == Y- (ylr)f. Noting that x. r = x and y. r = y, we have
197 2 X2 X2 X2 y2 Y X X Y xy xy a = 1 + - - 2- = 1 - -j b2 = 1 - -j a. b = --- - -- + - = --. r2 r2 r2 r2 r r r r r2 r2 E2 - (/Lo::;2) 2{(1- ::) COS2[W(t -TIc)] + (1- ~:) sin2[w(t -TIc)] - 2:; sin[w(t-rle)]cos[w(t-rle)]} (/Lo::;2) 2 { 1 - r12(x2COS2[w(t - TIc)] + 2xysin[w(t - TIc)]cos[w(t - TIc)] + y2sin2[w(t- TIc)]) } (/Lo::;2r {1- r12(xcos[w(t - TIc)] + ysin[w(t - rle)])2} But x = r sin B cos </Jand y = r sin B sin </J. (/Lo::;2) 2 {I - sin2 B (cos </J cos[w(t - rIe)] + sin </J sin[w(t - rIe)])2} (/Lo::;2) 2 {1- (sinBcos[w(t - TIc) - </J]}2}. z Intensity profile (1-~ sin26) y 8 = I~ (~r {1-(sinBcos[w(t-rle)-</J]}2}r. (8) = /Lo ( pOW2 ) 2 [ 1- ~sin2B ] r.

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