chap10 - .1 In Fig 10.4 let B = 0.2 cos 120t T and assume...

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CHAPTER 10 10.1. In Fig. 10.4, let B = 0 . 2 cos 120 πt T, and assume that the conductor joining the two ends of the resistor is perfect. It may be assumed that the magnetic field produced by I(t) is negligible. Find: a) V ab (t) : Since B is constant over the loop area, the flux is 8 = π( 0 . 15 ) 2 B = 1 . 41 × 10 2 cos 120 πt Wb. Now, emf = V ba (t) =− d8/dt = ( 120 π)( 1 . 41 × 10 2 ) sin 120 πt . Then V ab (t) = V ba (t) =− 5 . 33 sin 120 πt V . b) I(t) = V ba (t)/R = 5 . 33 sin ( 120 πt)/ 250 = 21 . 3 sin ( 120 πt) mA 10.2. Given the time-varying magnetic field, B = ( 0 . 5 a x + 0 . 6 a y 0 . 3 a z ) cos 5000 t T, and a square fila- mentary loop with its corners at (2,3,0), (2,-3,0), (-2,3,0), and (-2,-3,0), find the time-varying current flowing in the general a φ direction if the total loop resistance is 400 k  : We write emf = I E · d L =− d8 dt =− d dt ZZ loop area B · a z da = d dt ( 0 . 3 )( 4 )( 6 ) cos 5000 t where the loop normal is chosen as positive a z , so that the path integral for E is taken around the positive a φ direction. Taking the derivative, we find emf =− 7 . 2 ( 5000 ) sin 5000 t so that I = emf R = 36000 sin 5000 t 400 × 10 3 =− 90 sin 5000 t mA 10.3. Given H = 300 a z cos ( 3 × 10 8 t y) A/m in free space, find the emf developed in the general a φ direction about the closed path having corners at a) (0,0,0), (1,0,0), (1,1,0), and (0,1,0): The magnetic flux will be: 8 = Z 1 0 Z 1 0 300 µ 0 cos ( 3 × 10 8 t y)dx dy = 300 µ 0 sin ( 3 × 10 8 t y) | 1 0 = 300 µ 0 h sin ( 3 × 10 8 t 1 ) sin ( 3 × 10 8 t) i Wb Then emf =− d8 dt =− 300 ( 3 × 10 8 )( 4 π × 10 7 ) h cos ( 3 × 10 8 t 1 ) cos ( 3 × 10 8 t) i =− 1 . 13 × 10 5 h cos ( 3 × 10 8 t 1 ) cos ( 3 × 10 8 t) i V b) corners at (0,0,0), (2 π ,0,0), (2 π ,2 π ,0), (0,2 π ,0): In this case, the flux is 8 = 2 π × 300 µ 0 sin ( 3 × 10 8 t y) | 2 π 0 = 0 The emf is therefore 0 . 167
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10.4. Conductor surfaces are located at ρ = 1cm and ρ = 2cm in free space. The volume 1 cm <ρ< 2cm contains the fields H φ = ( 2 /ρ) cos ( 6 × 10 8 πt 2 πz) A/m and E ρ = ( 240 π/ρ) cos ( 6 × 10 8 πt 2 πz) V/m. a) Show that these two fields satisfy Eq. (6), Sec. 10.1: Have ∇× E = ∂E ρ ∂z a φ = 2 π( 240 π) ρ sin ( 6 × 10 8 πt 2 πz) a φ = 480 π 2 ρ sin ( 6 × 10 8 πt 2 πz) a φ Then B ∂t = 2 µ 0 ( 6 × 10 8 ρ sin ( 6 × 10 8 πt 2 πz) a φ = ( 8 π × 10 7 )( 6 × 10 8 ρ sin ( 6 × 10 8 πt 2 πz) = 480 π 2 ρ sin ( 6 × 10 8 πt 2 πz) a φ b) Evaluate both integrals in Eq. (4) for the planar surface defined by φ = 0, 1cm <ρ< 2cm, 0 <z< 0 . 1m ( note misprint in problem statement ), and its perimeter, and show that the same results are obtained: we take the normal to the surface as positive a φ , so the the loop surrounding the surface (by the right hand rule) is in the negative a ρ direction at z = 0, and is in the positive
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This note was uploaded on 08/12/2010 for the course ECE ECE 302 taught by Professor Ferguson during the Spring '10 term at Cal Poly Pomona.

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chap10 - .1 In Fig 10.4 let B = 0.2 cos 120t T and assume...

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