891_Physics ProblemsTechnical Physics

891_Physics ProblemsTechnical Physics - Chapter 31 P31.52...

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Chapter 31 231 P31.52 I R = + εε induced and ε induced =− d dt BA af Fm dv dt IBd dv dt IBd m Bd mR dv dt Bd mR Bvd == + induced bg To solve the differential equation, let uB v d du dt Bd dv dt Bd du dt Bd mR u −= 1 so du u Bd mR dt u ut 0 2 0 zz . FIG. P31.52 Integrating from t = 0 to tt = , ln u u Bd mR t 0 2 or u u e BdtmR 0 22 = . Since v = 0 when t = 0 , u 0 = and v d Bvd e . Therefore, v Bd e 1 ej . *P31.53 The enclosed flux is Φ B BA B r π 2 . The particle moves according to Fa = m : qvB mv r sin90 2 °= r mv qB = . Then Φ B Bmv qB = . (a) v m B ×⋅ × × −− Φ 2 2 69 2 16 2 5 15 10 30 10 0 6 21 0 254 10 Tm C T kg ms 2 e j . . (b) Energy for the particle-electric field system is conserved in the firing process:
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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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