Sol-Ch31 - CHAPTER 31 E xercises 1. = B A(cos120 1 ) = 2.52...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1. Δ Ф = BA(cos120 ° – 1) = –2.52 nWb 2. (a) Ф = B(πr 2 )cos 60 ° = 1.41 × 10 –3 Wb (b) Δ Ф = –2 Ф = –2.82 × 10 –3 Wb 3. Ф = BAcos θ = µ o n I Acos θ . ξ   = N Δ Ф /Δt = µ o NnI Acos θ ΔI/Δt = 40.2 µV 4. (a) ξ   = BLv, thus v = I R/BL = 32 m/s (b) F = I LB = 2.5 × 10 –2 N 5. (a) Ф = BA = (3.2t – 2.4t 2 ) × 10 –4 Wb; (b) I = –N(d Ф /dt)/R = 1.33 mA 6. Ф = BAcos50 ° ; and ξ   = N d Ф /dt = NAcos50 ° (ΔB/Δt) = 18.2 mV. 7. ξ = – N d Ф /dt = –NA co il dB/dt = –µ o NnA coil d I /dt. Find   ξ = –30.9 cos(60πt) mV 8. ξ = – N d Ф /dt = –N(πr 2 ) dB/dt = – 0.682 cos (2πft) 9. (a) Δ Ф = (πd 2 /4)ΔB = –1.96 mWb; (b) ξ   = AdB/dt = 98.2 mV; (c) Counterclockwise 10. (a)   ξ   = BLv = 0.9 V; (b) F = I LB = ( ξ /R)LB = 3.38 × 10 –2 N (c) I 2 R = 0.675 W; (d) Fv = 0.675 W 11. (a) ξ   = BLv = 3.24 V; so I = ξ /R = 1.2 A (b) I 2 R = I LB = 0.13 N; (c) Fv = 3.9 W; (d) I 2 R = 3.9 W 12. ξ   = N(πr 2 )dB/dt = 2.5 1dB/dt. Need ξ 2 /R = 2 W, so dB/dt = +3.56 T/s 13. R = ρ L/A = ρ (20 × × 0.05)/(πd
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

Page1 / 4

Sol-Ch31 - CHAPTER 31 E xercises 1. = B A(cos120 1 ) = 2.52...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online