895_Physics ProblemsTechnical Physics

895_Physics ProblemsTechnical Physics - 235 Chapter 31...

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Chapter 31 235 P31.64 The magnetic field at a distance x from a long wire is B I x = µ π 0 2 . Find an expression for the flux through the loop. d I x dx B Φ= 0 2 A af so Φ B r rw I dx x I w r == + F H G I K J + z 00 22 1 AA ln Therefore, ε =− = + d dt Iv r w B Φ 0 2 A and I R Rr w + 0 2 A . P31.65 We are given Φ B tt 600 180 32 .. ej Tm 2 and + d dt B Φ 18 0 36 0 2 . Maximum E occurs when d dt t + = 36 0 36 0 0 which gives t = 100 . s . Therefore, the maximum current (at t = s ) is I R −+ = 18 0 36 0 . V 3.00 A . P31.66 For the suspended mass, M : FM gTM a = . For the sliding bar, m : FTIBm a =− = A , where I R Bv R A Mg R mM a −= + A a f or a dv dt Mg RM m + + A dv v dt vt αβ = zz bg where α = + Mg Mm and β = + B A . Therefore, the velocity varies with time as ve MgR B e t Bt R = 11 A A . P31.67 (a) εµ N d dt NA dB dt NA d dt nI B Φ 0 where A = area of coil N = number of turns in coil and n = number of turns per unit length in solenoid. Therefore, NA n d dt tN A n t 4 120 480 120 sin cos
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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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