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Unformatted text preview: Physics 217: Problem Set 8 Solutions by Yao Liu Due Monday, April 13th, 2009 1 Purcell 8.7 To find the inductance of a toroid with N turn, see Problem 4 in Hw 6, B (2 πr ) = 4 πNI c = ⇒ B = 2 NI cr Φ = N Z b a Bhdr = 2 N 2 Ih c ln b a Definition of L is E = L dI dt = 1 c d Φ dt so we read it off L = 2 N 2 h c 2 ln b a For the resonant cavity of this problem, N = 1. The capacitance of a parallelplate capacitor is C = A 4 πs = πa 2 4 πs = a 2 4 s where s is the gap. Therefore, the frequency of this LC circuit is ω = 1 √ LC = s 4 c 2 s 2 ha 2 ln( b/a ) = c a s 2 s h ln( b/a ) You should imagine E and B fields alternate. E field is in between the gap, and B is circumferential inside the toroid. That of course is not a complete picture; changing E field produces B field in the gap, and changing B field also produces E field in the toroid. The study of resonant cavities is complicated. Here you just get a flavor of it. 2 Purcell 8.8 I will explain as much as I can the relevant things about damped harmonic oscillator. This would be more appropriate in a course on analytical mechanics, where the typical context would be mass on a spring. The similarity between RLC circuit and damped massonaspring is manifest in the form of differential equations governing the two systems: L d 2 Q dt 2 + R dQ dt + Q C = 0 m d 2 x dt 2 + β dx dt + kx = 0 As all ordinary differential equations go, we are solving it (the one on the left) subject to certain initial conditions. Since it’s secondorder, we need two conditions. Typically we have Q (0) = Q , and Q (0) = 0, i.e., the capacitor is fully charged to Q , and there is no current going through the circuit. 1 2.1 Solution First step in solving a homogeneous linear o.d.e. with constant coefficients is to consider the characteristic equation (or secular equation) Lλ 2 + Rλ + 1 C = 0 which has two roots λ = R ± p R 2 4 L/C 2 L = R 2 L ± r R 2 4 L 2 1 LC = γ ± q γ 2 ω 2 where γ = R/ 2 L is the damping factor, and ω = 1 / √ LC is the natural frequency of LC circuit. For purposes of this problem, we’d like to think L and C as fixed, and we can change R at our will. In other words, ω is fixed, and we are tweaking damping γ . If the two roots are real and distinct ( γ > ω ), we say it is overdamped. The general solution is Q ( t ) = Ae λ 1 t + Be λ 2 t where A and B are determined from the two initial conditions. Note that both λ 1 and λ 2 are negative, which means Q ( t ) is exponential decay. I’ll show you another way to write the solution, using hyperbolic functions defined by cosh x = e x + e x 2 sinh x = e x e x 2 just like ordinary trigonometric functions can be defined as cos x = e ix + e ix 2 sin x = e ix e ix 2 i Now the general solution can be written Q ( t ) = Ae γt cosh ωt + Be γt sinh ωt where ω = p γ 2 ω 2 (Naming it ω would make sense in the underdamped case). Imposing initial condition Q (0) = Q gives A = Q , and Q (0) = 0 gives A ( γ ) + Bω = 0, so...
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 Spring '06
 LECLAIR, A
 Inductance, R1 + R2

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