908_Physics ProblemsTechnical Physics

908_Physics ProblemsTechnical Physics - 248 Inductance a fa...

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248 Inductance P32.35 (a) UL I == 1 2 1 2 400 0500 2 2 .. H A af a f U = . J (b) When the current is 1.00 A, Kirchhoff’s loop rule reads +− = 22 0 1 00 5 00 0 ... V A a f Ω∆ V L . Then V L = 17 0 . V . The power being stored in the inductor is IV L ∆= = 100 170 . A V W . FIG. P32.35 (c) P 220 V a f = 11 0 . W P32.36 From Equation 32.7, I R e Rt L =− ε 1 ej . (a) The maximum current, after a long time t , is I R 200 A . At that time, the inductor is fully energized and = a f a f . V W . (b) lost W = IR 2 2 2 00 5 00 20 0 . (c) inductor drop 0 (d) U LI = 2 2 2 10 0 2 00 2 20 0 . A J P32.37 We have u E =∈ 0 2 2 and u B = 2 0 2 µ . Therefore ∈= 0 22 0 EB so BE 2 00 =∈ 2 =∈ = × × 5 8 3 680 10 300 10 227 10 . . . Vm ms T. P32.38 The total magnetic energy is the volume integral of the energy density, u B = 2 0 2 . Because B changes with position, u is not constant. For BB R r = F H G I K J 0 2 , u B R r = F H G I K J F H G I K J 0 2 0 4 2 . Next, we set up an expression for the magnetic energy in a spherical shell of radius
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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 .

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