Cylindrical Gaussian Surface - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Cylindrical Gaussian Surface - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! PC 3231 - Electricity and Magnetism 2 AY05/06 SEM 1 Suggested Solutions Q1 (i) Within the coaxial cable, drawing an Amperian loop of radius a < s < b coaxial with the cable, ∫ ~B · d~l = µ0 ∫ ~J · d~a ⇒ 2πsB = µ0I ⇒ ~B = µ0I 2πs φ̂ Drawing a cylindrical gaussian surface of radius a < s < b with its axis coinciding with that of the coaxial cable, ∫ ~E · d~a = 1 0 ∫ ρdτ ⇒ (2πslE) = λl 0 ⇒ ~E = λ 2πs0 ŝ Then, ~S = 1 µ0 ~E × ~B = Iλ 4π2s20 ẑ (ii) P = ∫ ~S · d~a = Iλ 4π0 ∫ b a 1 s2 (2πs) ds = Iλ 20 ln b a 1 (iii) ~pem = µ00 ∫ ~Sdτ = µ00IλL 4π ∫ b a 1 s20 (2πs) dsẑ = µ0λIL 2π ln ( b a ) ẑ (iii) Drawing a retangular Amperian loop with its normal in the φ̂ direction, one edge of length l at the middle of the coaxial cable and the opposite edge at a < s < b,∫ ~E · d~l = − ∂ ∂t ∫ ~B · d~a ⇒ El = −µ0 2π dI dt ∫ s a 1 s (lds) ⇒ ~E = −µ0 2π dI dt ln s a ẑ (iv) For the −λ at s = b, ~F = q ~E = (−λL) ( −µ0 2π dI dt ln b a ) ẑ = λµ0L 2π dI dt ln b a ẑ For the λ at s = a, ~E = 0 and ~F = 0. (v) ~P = ∫ ~F dt = [ λLµ0 2π ln b a ∫ dI dt dt ] ẑ = λLµ0 2π ln b a (I − 0)ẑ which is precisely the momentum that was originally stored in the fields. 2 , so r(1− r̂ · ~v/c) = c(t− tr) [ 1− ~v c · (~r − ~vtr) c(t− tr) ] = c(t− tr)− ~v · ~r c − v 2 c tr = 1 c [(c2t− ~r · ~v)− (c2 − v2)tr] = 1 c √ (c2t− ~r · ~v) + (c2 − v2)(r2 − c2t2) Therefore, V (~r, t) = 1 4π0 qc√ (c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2) and ~A(~r, t) = µ0 4π0 qc~v√ (c2t− ~r · ~v)2 + (c2 − v2)(r2 − c2t2) . 5 Q3 (i) ∇V = ∂V ∂r r̂ + 1 r ∂V ∂θ θ̂ = − p0ω 4π0c { cos θ[− 1 r2 sin ω(t− r/c)− ω rc cos ω(t− r/c)]− sin θ r2 sin ω(t− r/c)θ̂ } ≈ p0ω 2 4π0c2 ( cos θ r ) cos ω(t− r/c)r̂, ∂ ∂t ~A = −µ0p0ω 2 4πr cos[ω(t− r/c)](cos θr̂ − sin θθ̂), so ~E = −∇V − ∂ ~A ∂t = −µ0p0ω 2 4π ( sin θ r ) cos ω(t− r/c)θ̂. Meanwhile, ~B = ∇× ~A = 1 r [ ∂ ∂r (rAθ)− ∂Ar ∂θ ] φ̂ = −µ0q0ω 4πr { ω c sin θ cos ω(t− r/c) + sin θ r sin ω(t− r/c)] } φ̂ ≈ −µ0p0ω 2 4πc ( sin θ r ) cos[ω(t− r c ]φ̂. (ii) < ~S > = 1 µ0 (< ~E × ~B >) = µ0 c { p0ω 2 4π ( sin θ r ) < cos ω(t− r/c) > }2 r̂ = ( µ0p 4 0ω 4 32π2c ) sin2 θ r2 r̂. 6 (iv) < P > = ∫ < ~S > ·d~a = µ0p 2 0ω 4 32π2c ∫ sin2 θ r2 r2 sin θdθdφ = µ0p 2 0ω 4 12πc (v) P = I2R = q20ω 2 sin2 ωtR Average power, < P >= 1 2 q20ω 2R Equating this to the power of a dipole, < P >= µ0q 2 0ω 4d2 12πc , R = µ0d 2 6πc ω2 = µ0d 2 6πc 4π2c λ2 = 2 3 πµ0c ( d λ )2 7