Ampere Law - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Ampere Law - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! PC3231 Electricity and Magnetism 2 Typed by Lim Yen Kheng Suggested solutions for AY2007/08 Question 1 Part (A) If we consider regions with no current, Ampere’s law says: ∇× ~B = 0. This allows us to write ~B as a gradient of some scalar function, ~B = ∇V (~r, t). For the magnetic vector potential, from the fact that there are no monopoles, ∇ · ~B = 0 ⇒ ~B = ∇× ~A. Choose ∇ · ~A = 0, we have ∇2 ~A = µ0 ~J , which are three Poisson’s equations, one for each vector component. For line currents, the solutin is analogous to the case of electrostatics: (note: ~r − ~r′ = r) ~A = µ0 4π ∫ ~I ~r dl′ In this case, the scalar potential does not exist. Part (B) Gauss’s Law in media: Given auxiliary fields: ~D = 0 ~E + ~P ∇ · ~D = ρf → ∮ ~D · d~a = Qf,encl ~D = Q 4πr2 r̂ Linear media, ~D =  ~E. And inside the metal sphere, ~E = ~P = ~P = 0 (a conductor) ~E = { Q 4πr2 r̂ for a < r < b Q 4π0r2 r̂ for r > b V = − ∫ 0 ∞ ~E · d~l = ∫ b ∞ Q 4π0r2 dr − ∫ b a Q 4πr2 dr + 0 = Q 4π ( 1 0b + 1 a − 1 b ) , V (~r →∞) = 0 1 At the inner surface, r = a,~n directed inside, so σb = ~P · ~n = − 0χeQ 4πa2 Outer surface, r = b, ~n directed outside, σb = ~P · ~n = + 0χeQ 4πb2 Total energy: U = 1 2 0 ∫ E2dτ E2 =  ( Q 4π )2 1 r4 , a < r < b( Q 4π0 )2 1 r4 , r > b U1 = 1 2 0 ∫ ( Q 4π )2 1 r4 r2 sin θdrdφdθ = 1 2 0 Q2 4π22 ∫ 2π 0 dφ︸ ︷︷ ︸ s ∫ π 0 sin θ︸ ︷︷ ︸ 2 ∫ b a 1 r2 dr︸ ︷︷ ︸ a−3−b−3 = 1 2 0 Q2 π2 ( a−3 − b−3 ) Similarly, integrating the electric field for the region outside, U2 = 1 2 0 Q2 π20 1 b So the total energy: U = 1 2 0 [ Q2 π20b + Q2 π2 ( 1 a3 − 1 b3 )] Question 2 Part (A) TE00 mode, Ez = 0, from Gauss’ Law: ∂Ex ∂x + ∂Ey ∂y = 0. 2