Introduction to Electrodynamics - ch11 - Chapter 11...

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Unformatted text preview: 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 Now Po 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) = ~q5UJ2R. 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.1 195 196 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 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 = ---- --...
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This note was uploaded on 02/17/2010 for the course PHYSICS mae142 taught by Professor Kim during the Spring '10 term at American College of Computer & Information Sciences.

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Introduction to Electrodynamics - ch11 - Chapter 11...

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