867_Physics ProblemsTechnical Physics

# 867_Physics ProblemsTechnical Physics - Chapter 30 e ja g...

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Chapter 30 207 P30.61 (a) B I r == ×⋅ µ π 0 7 4 2 41 0 2 4 0 20 0 1 7 5 274 10 TmA A m T ej af bg . . . (b) At point C , conductor AB produces a field 1 2 4 . ± ×− T e j j , conductor DE produces a field of 1 2 4 . ± T e j j , BD produces no field, and AE produces negligible field. The total field at C is 4 . ± T j . (c) FB k j i B I =×= × × −= × −− A 24 0 0 035 0 5 2 74 10 115 10 43 .. ± . ± . ± A m T N e j e j e j (d) a F i i × × = m 30 10 0384 3 3 . ± . . ± N kg ms 2 (e) The bar is already so far from AE that it moves through nearly constant magnetic field. The force acting on the bar is constant, and therefore the bar’s acceleration is constant . (f) vv a x fi 22 2 0 2 0 384 1 30 =+ = + m 2 a f , so vi f = 0999 . ± *P30.62 Each turn creates field at the center 0 2 I R . Together they create field µµ 0 12 5 0 7 2 0 2 11 1 4 1 0 2 1 505 1 515 1 995 1 10 50 6 93 I RR R I II ++ + F H G I K J = × + F H G I K J …… Tm A m m3 4 7 m 0 . . B I FIG. P30.62 P30.63 At equilibrium, FI I a mg BA B AA 0 2 or I am g I B A = 2 0 A I B = = 2 0 025 0 0 010 0 9 80 0 1 5 0 81 7 7 . . m k g m m s A A 2 b g P30.64 (a) The magnetic field due to an infinite sheet of current (or the
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