Chap10 - .1 In Fig 10.4 let B = 2 cos 120 πt T and assume that the conductor joining the two ends of the resistor is perfect It may be assumed

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Unformatted text preview: CHAPTER 10 10.1. In Fig. 10.4, let B = . 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( . 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 = ( . 5 a x + . 6 a y − . 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 Z Z loop area B · a z da = d dt ( . 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 Z 1 300 µ cos ( 3 × 10 8 t − y) dx dy = 300 µ sin ( 3 × 10 8 t − y) | 1 = 300 µ 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 sin ( 3 × 10 8 t − y) | 2 π = The emf is therefore 0 . 167 10.4. Conductor surfaces are located at ρ = 1cm and ρ = 2cm in free space. The volume 1 cm < ρ < 2 cm 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 ( 6 × 10 8 )π ρ sin ( 6 × 10 8 πt − 2 πz) a φ = ( 8 π × 10 − 7 )( 6 × 10 8 )π 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, < z < . 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 a ρ direction at z = . 1. Taking the left hand side first, we find I E · d L = Z . 01 . 02 240 π ρ cos ( 6 × 10 8 πt) a ρ · a ρ dρ + Z . 02 . 01 240 π ρ cos ( 6 × 10 8 πt − 2 π( . 1 )) a ρ · a ρ dρ = 240 π cos ( 6 × 10 8 πt) ln ± 1 2 ² + 240 π cos ( 6 × 10 8 πt − . 2 π) ln ± 2 1 ² = 240 ( ln 2 ) h cos ( 6 × 10 8 πt − . 2 π) − cos ( 6 × 10 8 πt) i Now for the right hand side. First, Z B · d S = Z . 1 Z . 02 . 01 8 π × 10 − 7 ρ cos ( 6 × 10 8 πt − 2 πz) a φ · a φ dρ dz = Z . 1 ( 8 π × 10 − 7 ) ln 2 cos ( 6 × 10 8 πt − 2 πz) dz = −...
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This note was uploaded on 06/10/2008 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas at Austin.

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Chap10 - .1 In Fig 10.4 let B = 2 cos 120 πt T and assume that the conductor joining the two ends of the resistor is perfect It may be assumed

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