838_Physics ProblemsTechnical Physics

838_Physics ProblemsTechnical Physics - 178 P29.56 Magnetic...

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Unformatted text preview: 178 P29.56 Magnetic Fields e je j If B = Bx i + By j + Bz k , FB = qv × B = e vi i × Bx i + By j + Bz k = 0 + evi B y k − evi Bz j . (a) Since the force actually experienced is FB = Fi j , observe that B x could have any value , B y = 0 , and B z = − Fi . evi FI kJ = e j FGH ev K F FI = qv × B = − ee v i j × G B i + 0 j − kJ = ev K H If v = − vi i , then (c) FB = qv × B = e − vi i × Bx i + 0 j − i If q = − e and v = vi i , then (b) FB i i x i i − Fi j . − Fi j . Reversing either the velocity or the sign of the charge reverses the force. P29.57 (a) The net force is the Lorentz force given by a f j e4i − 1 j − 2kj + e2 i + 3 j − 1kj × e2 i + 4 j + 1kj N F = qE + qv × B = q E + v × B e F = 3.20 × 10 −19 Carrying out the indicated operations, we find: e3.52i − 1.60 jj × 10 N . 3.52 FG F IJ = cos FG H FK GH a3.52f + a1.60f −18 F= θ = cos (b) P29.58 −1 −1 x 2 2 I JJ = K 24.4° A key to solving this problem is that reducing the normal force will reduce F the friction force: FB = BIL or B = B . IL When the wire is just able to move, ∑ Fy = n + FB cos θ − mg = 0 so n = mg − FB cos θ and f = µ mg − FB cos θ . Also, ∑ Fx = FB sin θ − f = 0 so FB sin θ = f : FB sin θ = µ mg − FB cos θ and FB = We minimize B by minimizing FB : dFB cos θ − µ sin θ = µmg = 0 ⇒ µ sin θ = cos θ . 2 dθ sin θ + µ cos θ b b b gb g FIG. P29.58 g g FG 1 IJ = tan a5.00f = 78.7° for the smallest field, and H µK F F µg I bm Lg =G J B= IL H I K sin θ + µ cos θ L a0.200fe9.80 m s j OP 0.100 kg m =M B MN 1.50 A PQ sin 78.7°+a0.200f cos 78.7° = 0.128 T Thus, θ = tan −1 −1 B 2 min Bmin = 0.128 T pointing north at an angle of 78.7° below the horizontal µmg . sin θ + µ cos θ ...
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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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