886_Physics ProblemsTechnical Physics

886_Physics ProblemsTechnical Physics - 226 P31.36 Faradays...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
226 Faraday’s Law P31.36 For the alternator, ω π = F H G I K J F H G I K J = 3000 21 314 rev min rad 1 rev min 60 s rad s bg ε =− × =+ × −− N d dt d dt tt B Φ 250 2 50 10 314 250 2 50 10 314 314 44 .c o s . s i n Tm s s 22 ej a f (a) = 19 6 314 .s i n V af a f t (b) max . = 19 6 V P31.37 Bn I == × = × µπ 0 71 3 4 10 200 15 0 3 77 10 TmA m A T e j .. For the small coil, Φ B NN B A t N B r t =⋅ = = BA cos cos ωπω 2 . Thus, επ = d dt NB r t B Φ 2 sin = 30 0 3 77 10 0 080 0 4 00 4 00 28 6 4 00 3 2 1 . . s i n . . s i n . T m s m V . P31.38 As the magnet rotates, the flux through the coil varies sinusoidally in time with Φ B = 0 at t = 0 . Choosing the flux as positive when the field passes from left to right through the area of the coil, the flux at any time may be written as ΦΦ B t max sin so the induced emf is given by εω = d dt t B Φ Φ max cos . –1 –0.5 0 0.5 1 0 0.5 1 1.5 2 t/T = ( t/2 ) I/I max FIG. P31.38 The current in the coil is then I RR tI t = ωω Φ max max cos cos . *P31.39 M 850 mA 120 V To analyze the actual circuit, we model it as
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online