Q v b sin θ 1 60218 10 19 c3 77 10 6 m s 0 308 t sin

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=q v Bsinθ= (1.60218×1019C)(3.77×106m/s)×(0.308 T) sin 19.8=6.30183×1014N.006(part2of2)10.0pointsThe mass of proton is 1.67262×1027kgWhat is the proton’s acceleration?Correct answer: 3.76763×1013m/s2.Explanation:Let :m= 1.67262×1027kg,F=m a .a=Fm=6.30183×1014N1.67262×1027kg=3.76763×1013m/s2.00710.0pointsAindicates the magnetic fieldBpoint-ing up from the surface of the paper andaindicates the magnetic fieldBpointingdown into the surface of the paper.A copper bar has a constant velocityvinthe plane of the paper and is perpendicular toa magnetic fieldBas shown in the figures.How are charges distributed on the bar?1.vBB++--
gutierrez (bg25688) – Hw9 – keto – (55315)3++Using the right-hand rulevectorF=qvectorv×vectorB, thetop will be positive and the bottom will benegative.The other three choices are incorrect sincethey do not abide by the right-hand rule.2.vBB--cor-rect00810.0pointsA thin, horizontal copper rod is 0.876585 mlong and has a mass of 79.4284 g.3.vBB--++4.vBB++--Explanation:Positive charges will move in the directionof the magnetic force, while negative chargesmove in the opposite direction.Study the case where the magnetic fieldBis pointed out of the plane of the paper andthe bar is moving from right to left ().To produce the indicated charge separa-tion, the positive charges in the conductorexperience upward magnetic forces while thenegative charges in the conductor experiencedownward magnetic forces leaving the chargeseparation shown in the figure.FFvBB++--
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