Monoopole Term - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Monoopole Term - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! PC 3231 - Electricity and Magnetism 2 AY04/05 SEM 1 Suggested Solutions Q1 A ρ(r′, θ′) = kR r′2 (R− 2r′) sin θ′ Vmono = 1 4π0 ( 1 r ∫ ρ(τ ′) dτ ′) = 2π 4π0 1 r ∫ R 0 kR r′2 (R− 2r′)r′2 dr′ ∫ π 0 sin θ sin θ dθ Consider the dr′ integration :∫ R 0 (R− 2r′) dr′ = Rr′ − r′2 |R0 = 0 so the monoopole term is zero. Vdipole = 1 4π0 1 r2 ∫ r′ cos θ′ρ dτ ′ = kR 20r2 ∫ R 0 r′(R− 2r′) ∫ π 0 sin2 θ′ cos θ′ dθ′ Consider the dθ′ integration:∫ π 0 sin2 θ′ cos θ′dθ′ = 1 3 sin3 θ|π0 = 0 so the dipole term is zero. 1 Vquad = 1 4π0 1 r3 ∫ r′2( 3 2 cos θ − 1 2 )ρ dτ ′ = kR 4π0r3 ∫ R 0 (R− 2r′)r′2 dr′ ∫ π 0 (3 cos2 θ − 1) sin θ′ dθ′ = kR 40r3 ( Rr′3 3 − 1 2 r′4)|R0 (cos θ − cos3 θ)|π0 = kR 40r3 (−R 4 6 )(−2− 2) = kR5 60r3 (B) (i) ~Ein = −∇Vin = − ρ 30 ( ∂ ∂r (r cos θ)r̂ + 1 r ∂ ∂θ (r cos θ)θ̂ = − ρ 30 (cos θr̂ − sin θθ̂) = − ρ 30 ẑ (ii) ~Eout = −∇Vout = −ρR 3 30 ( ∂ ∂r cos θ r2 r̂ + 1 r ∂ ∂θ cos θ r2 θ̂) = −ρR 3 30 (−2 cos θ r3 r̂ − sin θ r3 θ̂) = ρR3 30r3 (2 cos θr̂ + sin θθ̂) 2 Q3 (A) As ~Aabove = ~Abelow at every point on the surface, ∂ ~A ∂x and ∂ ~A y are continuous, and the discontinuity is confined to ∂ ~A ∂z . ~B = ∇× ~A ~Babove − ~Bbelow = x̂ ( − ∂Ay(above) ∂z − ∂Ay(below) ∂z ) + ŷ ( ∂Ax(above) ∂z − ∂Ax(below) z ) = µ0( ~K × n̂) = −µ0Kŷ Equating x̂ and ŷ components, y component : ( ∂Ay ∂z ) (above) = ( ∂Ay ∂z ) below x component : ( ∂Ax ∂z ) (above) − ( ∂Ax ∂z ) (below) = −µ0k The normal derivative of the component of ~A parallel to ~k suffers a discontinuity ∂ ~Aabove ∂n − ~Abelow ∂n = −µ0~k . (B) For the static case the magnetic field due to a sheet current is ~E = µ0K 2 ŷ. It is independent of the distance x from the plane. The magnetic vector potential is then given by ~A = µ0K 2 xẑ. Thus the retarded potential is expected to be in the z direction. For above the plane, in the region ±x, we expect therefore ~A = µ0 4π ∫ ~K r da = µ0ẑ 4π ∫ K(tr)√ r2 + x2 2πr dr = µ0ẑ 2 ∫ K(t− √ r2 + x2c√ r2 + x2 r dr 5 The maximum r is given by t − √ r2 + x2/c = 0, or rmax = √ c2t2 − x2. (since K(t) = 0 for t < 0). ~A = µ0K0ẑ 2 ∫ ∞ 0 r√ r2 + x2 dr = µ0K0ẑ 2 √ r2 + x2|r∞0 = µ0K0(ct− x) 2 ẑ ~E(x, t) = ∂ ~A ∂t = −µ0K0c 2 ẑ for ct > x, and 0, for ct < x. ~B(x, t) = ∇× ~A = µ0K0ŷ 2 for ct > x, and 0, for ct < x. 6 Q4 (A) (i) The plane wave solution for ~E is ~E(z, t) = ~E0e i(kz−ωt) Substituting inside ∇2 ~E = µ∂ 2 ~E ∂t2 + µσ ∂ ~E ∂t , gives k2 = −µω2 − iµσω . Write k = k+ + ik−. The above becomes k2+ − k2− + 2ik−k+ = µω2+iµσω . Comparing real and imaginery parts gives k2+ − k2− = µω2 2k−k+ = µσω k− = µσω 2k+ k2+ − ( µσω 2k+ )2 = µω2 k4+ − k2+µω2 − (µσω 2 )2 = 0 k2+ = µω2 ± √ (µω2)2 + 4 ( µσω 2 )2 2 = µω2 2 ± µω 2 2 √ 1 + (µσω)2 (µω2)2 = µω2 2 [ 1± √ 1 + ( σ ω )2] k+ = ω √ µ 2 [ 1 + √ 1 + ( σ ω )2] 12 A similar derivation for k− results in k± = ω √ µ 2 [√ 1 + ( σ ω )2 ± 1 ] 1 2 7