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.Using the right-hand rule withvectorF=qvectorv×vectorBto produce this force on positive charges, thevelocityvectorvmust be directed from right to left(⇐).FFvBB++--01210.0 pointsA copper bar has a constant velocity in theplane of the paper and perpendicular to amagnetic field pointed into the plane of thepaper.vBBIf the bar is moving from left to right (⇒),how are charges distributed on the bar?1.Both the top and bottom of the bar willbe negative.2.The top will be positive and the bottomwill be negative.correct3.Both the top and bottom of the bar willbe positive.4.The top will be negative and the bottomwill be positive.Explanation:Positive charges will move in the directionof the magnetic force, while negative chargesmove in the opposite direction.The right-hand rule withvectorF=qvectorv×vectorBpro-duces a force on positive charges such that thepositive charges in the conductor experienceupward magnetic forces while the negativecharges in the conductor experience down-ward magnetic forces leaving the charge sep-aration shown in the figure below.FFvBB++--The top will be positive and the bottomwill be negative.
johnson (rj6247) – hw 11 – Opyrchal – (121014)7013(part 1 of 3) 10.0 pointsIn an AC electric generator, a rigid loop ofwire rotates in an external magnetic field.Say the loop is positioned as shown at timet= 0.ASNwaterfallsliding contactsWhich graph best represents the inducedcurrenti(t) at later times?Takei >0 for current flowing in directionshown by arrows.1.i0vectort2.i0vectortcorrect3.i0vectortExplanation:The current is proportional to the timederivative of magnetic flux through the loop.The magnetic field is constant in this case butthe loop area perpendicular to the magneticfield is varying because the loop is rotating.Therefore, the magnetic flux is proportionalto the loop area which has the time depen-dence of cos(ωt) in this case. In other words,the current should have the same time depen-dent form as the time derivative of cosωt,i.e.,i(t)∝sinωt .014(part 2 of 3) 10.0 pointsThe AC generator consists ofN= 4 turns ofwire each of areaA= 0.116 m2and total resis-tance 13.1 Ω. The loop rotates in a magneticfieldB= 0.704 T at a constant frequency of72.3 Hz.Find the maximum induced emf.Correct answer: 148.391 V.Explanation:Let :N= 4 turns,A= 0.116 m2,B= 0.704 T,R= 13.1 Ω,andf= 72.3 Hz.Faraday’s Law for solenoid:E=-N·dΦBdt.Ohm’s Law:I=VRFirst note thatω= 2π f= 2π(72.3 Hz )= 454.274 rad/s.
- Spring '08
- Work, Magnetic Field