hw-6-fischler

# hw-6-fischler - 8 × 10-3 m 2(b Φ max = NBA = 60 × 4 ×...

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HW # 6 Solutions E4. We use F l = 2 k I 1 I 2 r for F/l = 1 . 6 × 10 - 5 N/m , k = 10 - 7 N/A 2 , and I 1 = I 2 = 2 A . Then, solving for r , we get r = 0 . 05 m . E12. Given N = 60, A = 0 . 02 m 2 , B = 1 . 5 T and t = 0 . 2 s , we have (a) Φ = NBA = 1 . 8 T m 2 ; (b) ϵ = Φ t = 9 V . E13(a) This is a step-up transformer. E13(b) Using the relation ( V 2 /V 1 ) = ( N 2 /N 1 ), we obtain V 2 = V 1 N 2 N 1 = (110) ( 60 15 ) = 440 volts E15. We have 120 60 = 300 N N = 15 E16. In this case, we have 1380 115 = 400 N N = 100 / 3 = 33 . 3 so either 33 or 34 is a good answer to get (roughly) the required voltage. SP1(a). We have F l = 2 k I 1 I 2 r = (2 × 10 - 7 × 5 × 10) 0 . 05 = 2 × 10 - 4 N/m (b) Repulsive. 1

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(c) Total force is F = ( F/l ) × l = 2 × 10 - 4 × . 3 = 6 × 10 - 5 N . (d) Using F = BIl , we obtain B = F Il = 6 × 10 - 5 (10) (0 . 3) = 2 × 10 - 5 T (e) Into the page. SP3(a). Area, A = 0 . 03 × 0 . 06 = 1 .
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Unformatted text preview: 8 × 10-3 m 2 . (b) Φ max = NBA = 60 × . 4 × . 0018 = 4 . 32 × 10-2 T m 2 . (c) Φ min = 0. (d) t = 1 / 4 s = 0 . 25 s . (e) ϵ avg. = Φ t = . 04232 . 25 = 0 . 173 volts. SP4(a) We have 22 110 = N 400 ⇒ N = 80 turns (b) We use the relation P 1 = P 2 ⇒ ∆ V 1 I 1 = ∆ V 2 I 2 , so I 2 = ∆ V 1 I 1 ∆ V 2 = (110) (5) (22) = 25 A (c) No. In general, ∆ V 2 I 2 ≤ ∆ V 1 I 1 where equality exists in the absence of all energy losses, in the form of heat or whatever else. 2...
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## This note was uploaded on 10/20/2011 for the course PHY 309L taught by Professor Sai during the Spring '08 term at University of Texas.

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hw-6-fischler - 8 × 10-3 m 2(b Φ max = NBA = 60 × 4 ×...

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