· E1= 171 turns×33.4142 V= 5713.83 VApply Ohm’s law, we get the current as fol-lowing:I=VR=ER=5713.83 V2.13 Ω= 2682.55 a008(part 1 of 2) 10.0 pointsA straight rod moves along parallel conduct-ing rails, as shown below. The rails are con-nected at the left side through a resistor sothat the rod and rails form a closed rectangu-lar loop. A uniform field perpendicular to themovement of the rod exists throughout theregion.Assume the rod remains in contact withthe rails as it moves. The rod experiences nofriction or air drag.The rails and rod havenegligible resistance.5 g8 Ω1.3 T1.3 T0.25 A0.7 mAt what speed should the rod be movingto produce the downward current in the resis-tor?Correct answer: 2.1978 m/s.
johnson (rj6247) – hw 11 – Opyrchal – (121014)5Explanation:Let :I= 0.25 A,ℓ= 0.7 m,m= 5 g,R= 8 Ω,andB= 1.3 T.mRBBIℓFrom Ohm’s Law, theemfinside the loop isE=I R .and the motionalemfisE=B ℓ v .Thus the velocity of the rod isv=EB ℓ=I RB ℓ=(0.25 A) (8 Ω)(1.3 T) (0.7 m)=2.1978 m/s.009(part 2 of 2) 10.0 pointsThe rod is1.moving either left or right, since bothdirections produce the same result.2.stationary.3.moving to the left.correct4.moving to the right.Explanation:The downward current through the resistorenhances the existing flux ΦBin the loop;i.e.the current produces magnetic flux inthe same direction as the existing flux, intothe paper.Such a response counteracts thereduction of enclosed flux within the rail androd loop that arises if the rod moves to theleft, as would follow from Lenz’ law.Therefore the rod must be moving to theleft.01010.0 pointsA single circular loop of wire in the planeof the page is perpendicular to a uniformmagnetic fieldvectorBdirected out of the page,as shown.BBIf the magnitude of the magnetic field isdecreasing, then the induced current in thewire loop is1.directed upward out of the paper.2.counterclockwise around the loop.cor-rect3.clockwise around the loop.4.zero. (No current is induced.)5.directed downward into the paper.Explanation:Using Lenz’s law, the induced current mustbe directed to counter the change of the mag-netic flux through the loop;i.e., the mag-netic field generated by the induced current ispointing upward out of the page. Thus, usingright hand rule, the induced current must becounterclockwisealong the loop.
johnson (rj6247) – hw 11 – Opyrchal – (121014)601110.0 pointsA copper bar has a constant velocity in theplane of the paper and perpendicular to amagnetic field pointed out of the plane of thepaper.BB++--If the top of the bar becomes positive rel-ative to the bottom of the bar, what is thedirection of the velocityvectorvof the bar?1.from left to right (⇒)2.from right to left (⇐)correct3.from bottom to top (⇑)4.from top to bottom (⇓)Explanation:Positive charges will move in the directionof the magnetic force, while negative chargesmove in the opposite direction.
- Spring '08
- Work, Magnetic Field