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3209-2009-Solutions

3209-2009-Solutions - ECE 3209 — Electromagnetic Fields...

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Unformatted text preview: ECE 3209 — Electromagnetic Fields University of Virginia Fall 2009 Homework # 2 Solutions 1. To verify Gauss’s Theorem, let’s first consider the surface integral, which we can break into two parts: I S ~ F · d~a = Z side ~ F · d~a side + Z top ~ F · ~a top where d~a side = ˆ θr sin θ drdφ, and d~a top = ˆ rr 2 sin θ dθdφ Thus, I S ~ F · d~a = 4 Z r = R r =0 Z φ =2 π φ =0 r 3 cos θ sin θ fl fl fl θ =30 ◦ drdφ + Z θ =30 ◦ θ =0 Z φ =2 π φ =0 r 4 sin 2 θ fl fl fl r = R dθdφ = 8 π cos θ sin θ fl fl fl θ =30 ◦ Z r = R r =0 r 3 dr + 2 πR 4 Z θ =30 ◦ θ =0 sin 2 θ fl fl fl r = R dθ = 2 √ 3 π h r 4 4 i R + πR 4 h θ- sin 2 θ 2 i π/ 6 = √ 3 πR 4 2 + πR 4 ‡ π 6- √ 3 4 · I S ~ F · d~a = πR 4 ‡ √ 3 4 + π 6 · Now, let’s take the divergence of ~ F : ~ ∇ · ~ F = 1 r 2 ∂ ∂r ‡ r 2 F r · + 1 r sin θ ∂ ∂θ ‡ F θ sin θ · + 1 r sin θ ∂F φ ∂φ = 1 r 2 ∂ ∂r ‡ r 4 sin θ · + 1 r sin θ ∂ ∂θ ‡ 4 r 2 cos θ sin θ · + 1 r sin θ ∂ ( r 2 tan θ ) ∂φ ~ ∇ · ~ F = 4 r sin θ + 4 r ‡ cos 2 θ- sin 2 θ sin θ · = 4 r cos 2 θ sin θ Integrating this over the volume of the “ice cream cone,” we have Z V ~ ∇ · ~ F dv = 4 Z R r 3 dr Z θ = π/ 6 θ =0 cos 2 θdθ Z φ =2 π φ =0 dφ = πR 4 ‡ θ + sin 2 θ 2 · π/ 6 = πR 4 ‡ π 6 + √ 3 4 · Which is the same result we obtained for the surface integral. This shows that Gauss’s Theorem is satisfied. 2. Cheng, P. 2-32. Consider the vector field, ~ D = ˆ r cos 2 φ r 3 between two spherical shells of radius r = 1 and r = 2 . (a) Evaluating the surface integral, we have I S ~ D · d~a = Z π sin θdθ ‡ 1 r Z 2 π cos 2 φdφ fl fl fl r =2- 1 r Z 2 π cos 2 φdφ fl fl fl r =1 · = 1 4 h cos θ i π ‡ φ + sin2 φ 2 · 2 π =- π (b) Evaluating the volume integral of the divergence, we have ~ ∇ · ~ D = 1 r 2 ∂ ∂r ‡ r 2 cos 2 φ r 3 · =- cos 2 φ r 4 Z V ~ ∇ · ~ D dv =- Z R =2 R =1 dr r 2 Z θ = π θ =0 sin θdθ Z φ =2 π φ =0 cos 2 φdφ = 1 r fl fl fl R =2 R =1 h- cos θ i π h φ 2 + sin2 φ 4 i 2 π =- 1 2 (2)( π ) =- π which is the same as part (a), as expected. 3. Cheng, P. 2-36. Given the vector function, ~ A = ˆ φ sin( φ/ 2) we have (for the only non-zero components of the curl), ~ ∇ × ~ A = ˆ r 1 r sin θ ∂ ∂θ (sin( φ/ 2)sin θ )- ˆ θ 1 r ∂ ∂r ( r sin( φ/ 2)) = sin( φ/ 2) r ‡ ˆ r cot θ- ˆ θ · Integrating this over the hemispherical surface, we get Z S ~ ∇ × ~ A · d~a = r Z θ = π/ 2 θ =0 cos θdθ Z φ =2 π φ =0 sin( φ/ 2) dφ...
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3209-2009-Solutions - ECE 3209 — Electromagnetic Fields...

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